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It is barely imaginable that materials based on the physical properties of solid inorganic compounds have played such decisive roles in present-day technology and industry. One may think that this field belongs more to material physics. However, apart from the theories of physical properties, the contribution of chemistry and chemists to the preparation of materials and their structural analysis has been greater than that of other branches of science. Material science is the application of the fundamental physical properties of materials such that basic theories and their applications converge. Therefore, by surveying the applications in such fields, the outlines of research themes and their purposes can be understood.
Important inorganic materials are surveyed from the chemical point of view by focusing on the relationship between preparation and isolation, and structure and physical properties.
(a) Electric properties
A semiconductor is an electrical conductor with electrical resistance in the range of about 104 to 108 ohms. A typical semiconductor is a super-high grade silicon that is manufactured on a large scale and is widely used for information processing devices such as computers and energy conversion devices such as solar cells. VLSI (very large-scale integrated circuits) are printed on wafers made from almost defectless silicon single crystals with diameters of no less than 20 cm, prepared from polycrystalline silicon by the Czochralski method. Memory chips with a very high degree of integration as well as highly efficient computer chips have recently been realized.
In a short periodic table, silicon is a group IV element and has four valence-electrons. Although silicon semiconductors currently represent 90% or more of all semiconductors, isoelectronic 1:1 compounds of II-VI or III-V groups form compound semiconductors and are also used for optical or ultra high-speed electronic devices. For example, ZnS, CdS, GaAs, InP, etc. are typical compound semiconductors and the development of technologies to grow single crystals of these materials is remarkable. Light emitting diodes (LED) or semiconductor lasers are important applications of compound semiconductors.
As thin films of compound semiconductors are made by MBE (molecular beam epitaxy) or MOCVD (metallorganic chemical vapor deposition), special organometallic compounds, such as trimethyl gallium Ga(CH3)3 and trimethylarsenic As(CH3)3, whichpreviously found little application, are now used industrially.
Exercise $3$
What compounds other than the examples given are suitable as semiconductors?
Answer
ZnO, CdS, PbS, GaP, and InSb
Superconductivity is a phenomenon of zero electrical resistance below a critical temperature, Tc, and was discovered in 1911 by Kamerlingh Onnes (1913 Nobel Prize for physics), who succeeded in liquefying helium during his experiments to measure the electrical resistance of mercury at ultra low temperatures. About 1/4 of the elements, such as Nb (Tc = 9.25 K), In, Sn, and Pb behave as superconductors and more than 1000 alloys and intermetallic compounds are also superconductors, but only Nb-Ti alloy (Tc = 9.5 K) and Nb3Sn (Tc = 18 K) find application. Nb3Sn, Nb3Ge, V3Ga, etc., are cubic A-15 type compounds, in which transition metal atoms are aligned in chains, and interatomic distances are shorter than those in the crystalline bulk metal, raising the density of states of the conduction band and the critical temperature, Tc, of the compound.
Among inorganic compound superconductors, chalcogenide compounds MxMo6X8 (X = S, Se, Te, and M = Pb, Sn, etc.) of molybdenum called Chevrel phases and high-temperature superconductors of copper oxide derivatives, which J. G. Bednortz and K. A. Müller discovered in 1986 (1987 Nobel Prize for physics), have attracted attention. Chevrel phases have structures (refer to Section 4.4) in which hexanuclear cluster units of molybdenum are joined and the highest Tc is only 15 K of PbMo6S8, but the superconductive state is not broken even in strong magnetic fields. In the copper oxide system, more than 100 similar compounds have been prepared since the first discovery and the highest Tc so far discovered is 134 K. A typical compound, YBa2Cu3O7-x, has a structure (Figure $4$) in which CuO5 square pyramids and CuO4 planes are connected by corner-sharing, Ba and Y are inserted between them, and the oxygen content is non-stoichiometric.
On the other hand, molecular superconductors have also been studied. Representative donor-acceptor complexes are composed of TTF and BEDT-TTF (Figure $5$) as electron donors, and ClO4- or [Ni(dmit)2]2- as electron acceptors. The first example of this kind of superconductor was discovered in 1980, and of the about 50 complexes known at present, the highest Tc is 13K. Recently (1991) fullerene C60 doped with alkali metals showed a Tc of about 30K.
Although thousands of superconductors are known, only a few of them find ap pound superconductors are very brittle; either it is difficult to ake plication. Because com m them into wires or only small single crystals are obtained. It will take considerable time before some of them find practical use. Therefore, mainly Nb-Ti wires are used as the superconducting magnets of analytical NMR, medical MRI (magnetic resonance imaging instrument) or maglev trains, etc. Efforts are concentrated on discovering materials that have suitable mechanical and other properties by cooperation between inorganic chemistry and solid-state physics.
Various metal oxides are used as thermistors (temperature sensitive resistance device), varistors (nonlinear resistance device), capacitors, etc. For example, BaTiO3, with a perovskite structure, and SrTiO3, etc. can be used for any of the above-mentioned purposes. Ionic conduction materials are also called solid electrolytes and $\alpha$-AgI, $\beta$-Al2O3, stabilized zirconia (a part of Zr in ZrO2 is replaced by Ca or Y), etc. are used in solid state batteries or fuel cells.
(b) Magnetism
Magnetic materials are divided into hard (permanent magnets) and soft magnetic materials. Permanent magnets are indispensable to machines using motors and MRI, which requires a high magnetic field. Japan has a strong tradition in the development of magnets, and has made many epoch-making magnetic materials for practical use. Alnico magnets with Fe, Ni, and Al as their main constituents, ferrite magnets composed of solid solutions of CoFe2O4 and Fe3O4, cobalt-rare earth magnets such as SmCo5, and Nb-Fe-B magnets were especially significant achievements. Since soft magnetic materials are strongly magnetized in weak magnetic fields, they are most suitable for use as core materials in transformers. Hard magnetic properties are necessary for the stable maintenance of information whereas soft magnetic properties are required for recording and over-writing information in magnetic recording materials such as magnetic tapes, floppy disks, and hard disks. Although $\gamma$-Fe2O3 is a typical magnetic powder used for these purposes, Co+ or crystalline CrO2 is added to it to improve its magnetic properties. Recording materials as well as semiconductor devices are indispensable to our modern information-oriented society, and the role played by inorganic chemistry in the improvement of the performance of these materials is significant. Recently, ferromagnetism of organic compounds or metal complexes has been discovered, in which unpaired spins are aligned parallel in a molecule and coupled ferromagnetically. The study of molecular magnets has the subject of intensive investigation. Molecular design to couple paramagnetic metal complexes and to make spins parallel is an interesting subject in coordination chemistry.
(c) Optical properties
Mainly inorganic substances are used as materials for optical applications. The optical fiber in particular has been used for optical communications on a large scale, and has had a major social influence in information communication. A necessary property of good optical glass materials is the transmission of information to distant places with little optical loss. Silica fibers are manufactured by lengthening silica glass rods produced from silica grains. The silica is made from ultra pure SiCl4, which is oxidized in the vapor phase by an oxyhydrogen flame. As the optical loss along fibers obtained by this method has already reached its theoretical limit, fluoride glasses are being used in the search for materials with lower levels of loss.
Compound semiconductors such as GaP are widely used as laser light emitting diodes for optical communications, CD players, laser printers, etc. A high output YAG laser is made from neodymium-doped yttrium aluminum garnet, Y3Al5O12, which is a ouble oxide of Y2O3 and Al2O3. Single crystals, such as lithium niobate, LiNbO3, are used for changing wavelength of light by means of SHG (second harmonic generation) of nonlinear optics phenomena.
problems
8.1
Write a catalytic reaction cycle of the hydroformylation reaction which uses [RhH(CO)(PPh3)3] as a catalyst.
8.2
Describe differences between ammonia synthesis by the Harber-Bosch process and biological nitrogen fixation reactions.
8.3
A-15 type intermetallic compounds such as Nb3Sn are cubic crystals with the A3B composition. Consider how to locate each atom in such a unit cell.
Structure-function correlation
Since all the naturally occurring elements have been discovered, various bonding modes are established and the structures of compounds can be readily determined, studies of the chemical properties of inorganic compounds will give way to studies of reactions and physical properties. The synthesis of new compounds and the elucidation of structure-function correlations will be the foundations of these studies, although the is distant.
It is considerably difficult quantitatively to explain the thermal stability of a known inorganic compound using our present knowledge of theoretical chemistry and it is almost impossible fully to design compounds by a rational method. Although the selectivity of a catalytic reaction can be explained to some extent, the theoretical calculation of a reaction rate remains difficult. The relation between superconductivity and structure is not understood well, and critical temperatures cannot be predicted. Many of the structures and functions of the metalloenzymes that are the basis of biological activities are unknown. The research problems confronting the next generation of inorganic chemists are extensive, and novel solutions can be anticipated. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(Saito)/08%3A_Reaction_and_Physical_Properties/8.03%3A_Physical_properties.txt |
1.1
• \(\;_{1}^{1}H\)(1,0,1)
• \(\;_{1}^{2}H\)(1,1,1)
• \(\;_{6}^{12}C\)(6,6,6)
• \(\;_{6}^{13}C\)(6,7,6)
• \(\;_{6}^{14}C\)(6,8,6)
• \(\;_{8}^{16}O\)(8,8,8)
• \(\;_{8}^{17}O\)(6,9,8)
• \(\;_{6}^{18}O\)(8,10,8)
9.02: Chapter 2
2.1
$\begin{split} \Delta &= 428 - \sqrt{432 \times 239} \ &= 106.6\; kJ\; mol^{-1} \ &= 25.48\; kcal\; mol^{-1} \end{split}$
$\chi_{Cl} - \chi_{H} = 0.208 \times \sqrt{25.48},$
$\chi_{Cl} = 2.1 + 1.05 = 3.15$
2.2
In a $\sigma$ bond, a bonding orbital is a centrosymmetric g orbital, whereas an antibonding orbital is a non-centrosymmetric u orbital.
9.03: Chapter 3
3.1
• O2 (0)
• H2O2(-1)
• H2O (-2)
$E_{0} = \frac{0.70 + 1.76}{2} = 1.23\; V$
Since the reduction potential is positive, the reaction is spontaneous.
3.2
As the pKa of the conjugate acids NH4 is 9.25, and C5H5NH+ 5.25, ammonia is more protophilic than pyridine and is a stronger base.
3.3
Electronegativities of halogens are in the order F > Cl > Br. A boron trihalide bonded to more electronegative halogens attracts more electrons and the Lewis acidity should become larger. However, opposite tendency is observed and this is considered to be due to $\pi$ bonding between boron and halogen.
9.04: Chapter 4
4.1
$3 LiAlH_{4} + 4 BF_{3} \cdot O(C_{2}H_{5})_{2} \rightarrow 2 B_{2}H_{6} + 3 AlF_{3} + 3 LiF + 4 (C_{2}H_{5})_{2}O$
4.2
$PCl_{3} + 3 C_{2}H_{5}MgBr \rightarrow P(C_{2}H_{5})_{3} + 3 MgBrCl$
4.3
$Os + 2 O_{2} \rightarrow OsO_{4}$
4.4
$3 NH_{4}+ + 12 MoO_{4}^{2-} + H_{2}PO_{4} + 22 H^{+} \rightarrow (NH4)_{3}[PMo_{12}O_{40}] + 12 H_{2}O$
4.5
$PdCl_{2} + 2 HCl \rightarrow H_{2}PdCl_{4}$
4.6
$CoCl_{2} + 6 H_{2}O \rightarrow [CoCl_{2}(H_{2}O)_{4}] \cdot 2H_{2}O$
9.05: Chapter 5
5.1
$2 Li + C_{4}H{9}Br \rightarrow LiC_{4}H_{9} + LiBr$
5.2
Because six ether oxygen atoms of a benzene-soluble crown ether (e.g. 18-dibenzo-crown-6) in a cyclic arrangement coordinate to potassium cation K+.
5.3
Two methyl groups in the dimeric Al2(CH3)6 bridge two aluminum atoms to form formally 8 covalent bonds requiring 16 electrons. The compound is called electron-deficient because only 12 six electrons are supplied from two aluminum atoms (6) and six methyl groups (6).
9.06: Chapter 6
6.1
Oxide ions are weak-field ligands and transition metal ions assume high-spin states. Fe3+ ions have d5 electron configurations and the LFSE is zero either in the octahedral or in tetrahedral coordination. On the other hand, F2+ ions tend to enter octahedral holes, because the LFSE for the octahedral coordination by six oxide ligands is larger than the one for tetrahedral coordination by four oxide ligands. This is one of the reasons why magnetite Fe3O4 has an inverse spinel structure B3+[A2+B3+]O4.
6.2
$trans-[PtCl_{2}(PEt_{3})_{2}] + EtMgBr \rightarrow trans-[PtCl(Et)(PEt_{3})_{2}] + MgBrCl$
• CpV(CO)4
• [CpFe(CO)2]2
6.4
The trans effect of Cl- is larger than that of NH3. Therefore, it is possible to synthesize geometrical isomers selectively by choosing starting compounds.
$[Pt(NH_{3})_{4}]^{2+} + 2 Cl^{-} \rightarrow trans-[PtCl_{2}(NH_{3})_{2}]$
$[PtCl_{4}]^{2-} + 2 NH_{3} \rightarrow cis-[PtCl_{2}(PEt_{3})_{2}]$
6.5
Unless [Cr36Cl(NH3)5]2+ forms by the addition of an isotope ion 36Cl to the aqueous solution of the reaction $[CoCl(NH_{3})_{5}]^{2+} + [Cr(OH_{2})_{6}]^{2+} \rightarrow [Co(OH_{2})(NH_{3})_{5}]^{+} + [CrCl(OH_{2})_{5}]^{2+},$it is concluded that the chloride ion coordinated to cobalt transfers to chromium by the inner-sphere mechanism via a bridged structure [(NH3)5-Co-Cl-Cr(OH2)5]4+.
9.07: Chapter 7
7.1
Most of lanthanide ions are stable in the 3+ oxidation states but Ce4+ and Eu2+ are stable ions and solubilities and adsorption ability in the solvent extraction are significantly different from those of other lanthanides which makes the separation easier.
7.2
$\frac{N}{N_{0}} = e^{- \lambda \times \frac{\ln 2}{\lambda} \times 10} = e^{- \ln 10} = 2^{-10} = 9.77 \times 10^{-4}$
9.08: Chapter 8
8.1
Figure A.1 shows the catalytic cycle. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(Saito)/09%3A_Solution_of_problems/9.01%3A_Chapter_1.txt |
Dr. Kai Landskron (Lehigh University). If you like this textbook, please consider to make a donation to support the author's research at Lehigh University: Click Here to Donate.
01: Atomic Structure
Development of Atomic Theory
Since the ancient times, humans thought of explaining the material world in all of its complexity. The basic idea behind all element theories is that the matter that surrounds us is composed of more simple matter. This matter may be composed of even simpler matter. The most simple matter would be called an element.
The first element theories were one element theories: The greek philosopher Thales thought water was the only element, and everything was a form of water. Anaximenes thought air was the only element, and Heraclitus believed fire was the only element (Figure 1.1.1).
However, with only one element assumed it was difficult to explain the material world in all its complexity satisfactorily. This could be done more satisfactorily by a multi-element theory. The first philosopher who introduced a multi-element theory was Empedocles. He suggested a four-element theory with fire, air, water, and soil as the elements (Figure 1.1.2).
First atomistic ideas
The element theories of Thales, Anaximenes, Heraklitus, and Empedocles were all non-atomistic. This means that they did not include the idea that elements were made of small particles that were indivisible. The first greek philosophers that introduced atomistic element theory were Leukippes, and Democrites (Figure 1.1.3). They assumed that particles cannot be divided into smaller particles infinitely often. Ultimately, after many divisions, one would arrive at particles that could not be further divided, and these particles would be called atoms. Atomistic element theory allows for many different elements which helps to explain the complexity of the material world satisfactorily. However, Leukippes and Democrites did not know how many different elements there were, and how different atoms of different elements were actually distinguished. This question would not be answered until about 2000 years later.
Modern Atomic Theories
Although the atomistic idea was already known in the antique, it became forgotten for a long time and was only reintroduced about 2000 years later with Dalton’s atom hypothesis. What lead to Dalton’s atom hypothesis? The first discovery that was important to the development of modern atomic theory was the law of the conservation of mass by Antoine Lavoisier (Fig. 1.1.4). Historically, the conservation of mass and weight was kept obscure for millennia by the buoyant effect of the Earth's atmosphere on the weight of gases, an effect not understood until the vacuum pump first allowed the effective weighing of gases using scales. Until then, in many instances mass seemed to appear or disappear. For example, the mass of wood seems to disappear when it is burned. However, the mass of the wood actually does not disappear it is just converted into the mass of gases, mainly carbon dioxide. When scientists realized that mass never disappeared they could for the first time embark on quantitative studies of the transformations of substances. This in turn led to the idea of chemical elements, as well as the idea that all chemical processes and transformations are simple reactions between these elements. The law of conservation of mass states that the mass of a closed system of substances will remain constant, regardless of the processes acting inside the system. An equivalent statement is that matter cannot be created or destroyed, although it may change form. This implies that for any chemical process in a closed system, the mass of the reactants must equal the mass of the products.
Law of Constant Composition
The discovery of the law of the conservation of mass led to the discovery of the law of the constant composition (also called law of definitive proportions) by Joseph Proust (Fig. 1.1.5). This law was the result of chemical analysis that determined the mass ratio of elements in pure substances. It was found that a pure substance always contains exactly the same proportion of elements by mass. For example the element analysis of pure substances containing the elements carbon and oxygen would be either 42.9% carbon and 57.1% oxygen or 27.3% carbon and 72.7% oxygen.
The law was questioned by Proust's fellow Frenchman Claude Louis Berthollet, who argued that the elements could combine in any proportion. The very existence of this debate was because at the time the distinction between pure chemical compounds and mixtures had not yet been fully developed. When two pure compounds of two different elements are mixed, then the mixture can have a continuous mass ratio between the two elements because compounds can be mixed at any ratio. So, for the example in the two compounds between carbon and oxygen, in mixtures of the two there could be carbon to oxygen ratios varying continuously between 27.7 % to 42.9% carbon, and 42.9% to 72.7% oxygen. It was an accomplishment of Proust that he was able to correctly distinguish between pure compounds and mixtures.
The Law of Multiple Proportions
John Dalton (Fig. 1.1.6) discovered the law of multiple proportions by close analysis of Proust’s law of constant composition. He noticed that if two elements form more than one compound between them, then the ratios of the masses of the second element which combines with a fixed mass of the first element will be the ratios of small whole numbers, also called integer numbers. Let us see what is the meaning of the law of the multiple proportions by an example:
Example $1$
The first compound contains 42.9% by mass carbon and 57.1% by mass oxygen.
The second compound contains 27.3% by mass carbon and 72.7% by mass oxygen.
Show that the data are consistent with the Law of Multiple Proportions!
Solution
In 100 g of the first compound (100 is chosen to make calculations easier) there are 57.1 g O and 42.9 g C.
The mass of O per gram C is:
57.1 g O / 42.9 g C = 1.33 g O per g C
In 100 g of the second compound, there are 72.7 g O and 27.3 g C.
The mass of oxygen per gram of carbon is:
72.7 g O / 27.3 g C = 2.66 g O per g C
Dividing the mass O per g C of the second (larger value) compound by the mass O per g C of the first compound gives:
2.66 : 1.33 = 2 : 1
2 and 1 are both integer numbers.
Dalton's Atom Hypothesis
Dalton argued that atoms could explain the law of the multiple proportions. If one assumed that elements were made of the same type of indivisible particles, that are identical in mass and all properties then, because these particles can only come in integer numbers, different atoms can be combined to form compounds also only in integer numbers. Thus, in a pure compound atoms of unlike elements would be combined in small whole number ratios. Consequently a given compound always has the same relative number and types of atoms. Chemical reactions would involve reorganization of atoms, but the atoms would retain their identity. Atoms could be rearranged in a chemical reaction but not created or destroyed.
Figure 1.1.7 is a simple illustration how atoms can explain the law of multiple proportions. The blue and red balls symbolize different atoms of different elements. It is possible to combine a red atom with one, two, three, or four blue atoms to make four different compounds. It is for example also possible to combine two red atoms with one, two or three blue atoms. In all of these compounds as well as in all other thinkable ones, the blue and red atoms can only be combined in integer numbers.
Dalton's Atom Symbols
Dalton also thought of symbols for the atoms of the different elements. The symbols are different from the element symbols that are used today, but the concept is the same. You can see some of them in Figure 1.1.8. For example oxygen is represented by a white ball, while carbon is symbolized by a black ball. You can also see that Dalton already combined atom symbols of different elements to illustrate compounds. For example he combined one white ball with one black ball to indicate carbon monoxide, while he combined two white balls with one black ball to indicate carbon dioxide. In this case he correctly identified the ratios of atoms in the two compounds. However, you can see that this is not always the case. For example, he combined one hydrogen atom (white ball with a dot in the middle) with one oxygen atom to indicate the composition of water. However, as we know today, a water molecule has two hydrogen atoms and one oxygen atom. Why has he been right with the carbon-oxygen compounds, but wrong about the water? The answer is that at the time he could determine the atom ratios only from the law of the multiple proportions. At the time there was only one compound known that was made of hydrogen and oxygen, water, so he assumed the simplest atom ratio of 1:1. If he had known hydrogen peroxide, which has the composition H2O2, and thus half as much hydrogen per oxygen, he would have probably correctly assigned the correct hydrogen to oxygen ratio to water. In the case of the carbon oxides, he assigned the compositions correctly because both carbon monoxide and carbon dioxide were known at the time.
Subatomic Particles
Dalton’s atomic theory stated that different elements were made of different atoms, but did not explain what made the atoms different. The answer is that atoms are composed of subatomic particles, protons, neutron, and electrons, and the number of the protons and electrons in an atom defines the element it represents. So after all, atoms are not so indivisible. However, an atom can still be seen as the smallest particle that represents the full properties of an element. If an atoms gets divided further into subatomic particles then these properties are being lost. To understand atoms further, we must therefore understand the subatomic particles they are made of. The first subatomic particle that was discovered was the electron. It was found through the investigation of so-called cathode rays which were a mysterious phenomenon at the time. Cathode rays occur when a high voltage of about 10-20 kV is applied to an evacuated tube (0.0001 mm Hg). You can see such a tube with a cathode ray going through in Figure 1.1.9. The big question was: What causes these rays?
The answer was found by John Joseph Thomson. He applied electric and magnetic fields to these rays and found them deflected. When an electric field was applied the beam was deflected toward the positive pole. When an equally strong electric and magnetic field was applied, and the magnetic field was perpendicular to the electric field then these fields canceled out, and the rays were no longer deflected. Thomson concluded that these rays must consist of negatively charged particles emitted by the negatively charged electrode because they were attracted by the positive pole of the electric field: Electrons. Knowing the forces associated with the electrical and magnetic fields, he was able to calculate a mass to charge ratio of the electrons which is 5.6857 x 10E-9 g/C. After Thomson, Robert Millikan performed another experiment, the so-called oil drop experiment, that allowed to determine the charge and the mass of the electron. The charge of an electron is -1.6 x 10E-19 C, and its mass is 9.1 x 10E-28 g.
Thomson’s plum pudding atomic model
The discovery of the electron led Thomson to the development of a first atomic model that would include a subatomic particle. It is named plum pudding model (Fig. 1.1.10). In the Thomson atom model electrons are embedded as little particles in a positively charged mass like raisins are embedded in a cake. This atom model is still far from the atom model that we accept today, but it represented an important step forward, because it introduced for the first time the idea that atoms are not indivisible, but contain subatomic particles.
Rutherford's Atom Model
Ernest Rutherford (Fig. 1.1.11) wanted to test Thomson’s plum pudding atom model by experiment. To do so, he placed a photographic film around a gold foil (Figure 1.1.12). In the following he bombarded the gold foil with alpha radiation. Alpha radiation is a form of radioactivity and consists of helium nuclei, also called alpha-particles. It was known at this time the alpha-particles were positively charged and had a large mass compared to an electron. Rutherford’s hypothesis was that if Thomson’s atom model was true, then the alpha particles should all go straight through the gold foil like a bullet going through a plum pudding. Therefore, a blackening of the photographic film, which served as the alpha radiation detector, should only occur directly behind the gold film.
However, what Rutherford observed was that while most of the alpha-particles went straight through the gold foil, some of them were deflected, and a few of them were even reflected. Rutherford concluded that Thomson’s plum pudding model must be wrong. A bullet shot at a plum pudding would just never be reflected by the plum pudding and come back at you.
He suggested that the atom would be made of a positively charged nucleus where almost the entire mass of the atom would be concentrated. In the event of a collision between the alpha particle and that nucleus, the alpha particle would be reflected. In the event of an alpha particle passing by the nucleus close, the alpha particle would be deflected. In the event of an alpha particle passing the nucleus in larger distance, the alpha particle would pass the atom practically non-deflected. The observation that most of the alpha particles were not deflected led Rutherford to the conclusion that the overall atom must be much larger than the nucleus, in fact, the data allowed him to calculate that the radius of the atom would be about 10,000 times larger than the radius of the nucleus. This implies that the atoms would be mostly empty space. To explain the empty space, Rutherford assumed that the electrons would move in orbits around the nucleus like planets around the sun. This would be called the planetary model of the atom (Fig. 1.1.13).
Bohr's Atom Model
Rutherford’s atom model was another big step forward in the development of atomic theory, however there were inherent problems with it as it violated fundamental principles of physics. An electron in an orbit is a self-accelerating electrically charged particle, and according to the laws of physics such particles must emit electromagnetic radiation. However, under normal circumstances atoms do not emit electromagnetic radiation. Secondly, even if the electron emitted electromagnetic radiation, then this would mean that the electron would lose energy because electromagnetic radiation is a form of energy. However, an electron constantly losing energy would make it unstable in its orbit around the nucleus, and it should spiral downward closer to the nucleus until the atom was eventually collapsed. It is however, experimentally not observed that atoms collapse, they are quite stable species. These difficulties of the Rutherford atom model meant that it could not be the final answer to atomic structure. Niels Bohr (Fig. 1.1.14) was aware of the problems of the Rutherford model, and two new developments in physics, namely the concept of the quantization of energy and atomic spectra helped him to develop an improved atom model, known as Bohr model. To understand this model let us look first at the quantization of energy and atomic spectra.
Blackbody Radiation
The quantization of energy was discovered in the context of the physical phenomenon called “blackbody radiation”. Blackbody radiation is the electromagnetic radiation any object sends out due to its temperature (Fig. 1.1.15).
It’s distribution of wavelengths follows curves that depend on the temperature and are shown in Figure 1.1.16. You can see that for each temperature the intensity first increases with increasing wavelength, then goes through a maximum, and finally decreases again. You can also see that the overall intensity of the blackbody radiation increases with temperature and that the maximum of the curve shifts to smaller wavelengths with increasing temperature. Objects at room temperature do not emit much blackbody radiation and the wavelengths are far longer than visible for the human eye. However, when an object is heated high enough, intensity increases and wavelength decreases, and the object starts to glow. For instance, lava (Figure 1.1.15) has a temperature high enough so that its blackbody radiation is visible for the human eye. With increasing temperature objects first glow red, then orange, then yellow, and eventually white. This is because red is the color associated with the largest wavelength, and as temperature increases other colors mix into red, until the object glows white. At extremely high temperatures the blackbody radiation also has a significant intensity in the UV region. Such temperatures occur in welding processes, for example, and for this reason welders need to wear glasses that block the UV radiation. The wavelength and intensity of blackbody radiation can be easily measured, but at the beginning of the 20th century there was no good explanation for its behavior. Classical theory predicted that intensity would continuously increase with decreasing wavelength at any temperature which was not in accordance with experimental observation.
To bring experiment and theory in accordance, Max Planck made the radical assumption that the energy associated with radiation of a given wavelength or a given frequency was quantized. It would be an integer multiple n of that frequency $\nu$, multiplied with proportionality constant h, know today as the Planck constant.
Equation 1.1.1 The Planck-Einstein equation.
Using this assumption he was able to derive the Planck equation (Equation 1.1.2) which correctly describes the intensity and wavelength distribution of the blackbody radiation for any temperature. The correct description of blackbody radiation strongly supported Planck’s assumption that energy was quantized, but it did not prove it or explain it. The proof and the explanation was found only later. Initially Planck assumed it merely to fit the data.
Equation 1.1.2 Planck's equation
Bohr's Atom Model
The second development that contributed to Bohr’s atom model was the absorption and emission spectra of atoms. It was experimentally observed that under certain circumstances atoms would send out or absorb electromagnetic radiation of discreet wavelengths that were characteristic for an atom. For example, H atoms would absorb or emit at four discreet wavelengths in the visible region of the electromagnetic spectrum (Fig. 1.1.17).
This is known as the Balmer series named after its discoverer Johann Balmer. Friedrich Paschen and Chester Lyman later found that H also absorbs and emits at discreet wavelengths in the IR and UV region respectively. These wavelengths are known as the Paschen and Lyman series, respectively (Fig. 1.1.18).
For the absorption and emission spectra of H a simple mathematical relationship between the energies associated with the wavelengths can be found. The energy is proportional to a constant, today known as the Rydberg constant, times one over an integer number square minus one over another integer number square, whereby the second integer number would be larger than the first one (Eq. 1.1.3). The observation of integer numbers showed that also the energy and the wavelengths of the atomic absorption and emission spectra are quantized. The question was what the quantization of the absorption spectra meant for atomic structure.
Equation 1.1.3 Rydberg formula for the emission of light with particular energies.
Bohr answered the question the following way. He argued, that like in the Rutherford model the electrons would move in orbits around the nucleus. A balance of opposite centrifugal forces and Coulomb attractions would hold the electron stable in the orbit. However, because energy is quantized, also the angular momentum of the electron would be quantized, and thus, only discreet radii would be allowed. The most inner orbit would have the quantum number n=1, the next higher orbit the quantum number n=2, and so forth (Fig. 1.1.19).
The quantization of the electron energies and the radii would be the explanation why electrons, despite self-accelerating, do not continuously emit electromagnetic radiation. Electromagnetic radiation is only emitted when an electron jumps from an outer orbit of higher energy to an inner orbit of lower energy. This radiation must have a discreet wavelength because the energy difference between two orbits is discreet. Vice versa, an atom can adsorb electromagnetic radiation of specific wavelength and energy that is suitable to make the electron jump from an inner to an outer orbit. In sum, the quantization of the energy and the radii would explain the quantization of the absorption and emission spectra. The question is: Can the radii be calculated, and what are the associated energies of the electrons in the orbits?
Definition: Bohr's Postulates (1913)
1. An electron in an atom moves in a circular orbit about the nucleus under the influence of the Coulomb attraction between the electron and the nucleus, obeying the laws of classical mechanics.
2. Instead of the infinity of orbits which would be possible in classical mechanics, it is only possible for an electron to move in an orbit for which its orbital angular momentum L is an integral multiple of h/2π.
3. Despite the fact that it is constantly accelerating, an electron moving in such an allowed orbit does not radiate electromagnetic energy. Thus, its total energy E remains constant.
4. Electromagnetic radiation is emitted if an electron, initially moving in an orbit of total energy, discontinuously changes its motion so that it moves in an other orbit of total energy. The frequency of the emitted radiation is equal to the quantity divided by h.
To calculate the radii and energies of the electron in the H atom Bohr made two assumptions: Firstly, the centrifugal force associated with the electron moving in the orbit would be assumed equal to the Coulomb force between the electron and the proton so that the electron would be stable in the orbit. Secondly, the angular momentum of the electron would be quantized and an integer multiple of the Planck constant h. The factor 2$\pi$ is because E = h/2π x angular frequency. The angular frequency (also called angular speed) is the angular displacement (in degree or rad) per time unit. Angular frequency is frequency x 2π. We can rearrange this equation by solving it for v and then insert the equation into equation I giving the result as shown in Fig. 1.1.20.
We can then multiply this equation by 4πε0r2, divide by e2, and multiply it by r. This gives us the term r=(n2h2ε0)/(πme2). Analyzing the term shows us that r is only a function of the quantum number n, specifically the radius is proportional to n2 (Fig. 1.1.21). All other terms are constants, namely the Planck constant h, the dielectric constant of the vacuum, π, the mass of the electron, and the elementary charge. By inserting the quantum numbers into the equation we can calculate the actual values for the radii. For example, when we insert 1 for the quantum number n, we get the radius for the most inner orbit of the electron in the H atom. It is 5.29 x 10-11 m. If we inserted 2 for n, we would get the second radius, if we inserted 3, we would get the third radius and so on.
Now let us calculate the energies of the electron on these radii. Generally, the overall energy of the electron is the sum of the kinetic and potential energies. The kinetic energy of a moving object is given by Ekin=1⁄2 mv2. We know that m must be the mass of the electron, but what is the velocity of the electron? We can derive it from equation I we previously used by solving it for v2 (Fig. 1.1.20). We can then insert the term for v2 into the equation Ekin=1⁄2 mv2 , which gives Ekin=e2/(8πε0r), Fig. 1.1.23.
Now let us consider the potential energy. The potential energy is the Coulomb energy between the proton and the electron in the H atom. The formula for the Coulomb energy of two particles having two opposite elementary charges is Epot=-e2/(4πε0r). Note that this energy has a negative algebraic sign because the forces between the proton and the electron are attractive (Fig. 1.1.23).
We can now add up the kinetic and the potential energy to give the overall energy which is E = Ekin+Epot= [e2/(8πε0r)] - [e2/(4πε0r)] = e2/(πε0r) (1/8-1/4) = -e2/(8πε0r). We can then use the term we previously calculated for r (Fig. 1.1.21) and insert it into the term for the overall energy. As a result the overall energy becomes E=-(e4m)/(8ε02n2h2), Fig. 1.1.24. As we can see, the energies for the electrons in the different orbits are also only a function of the quantum number n, specifically, they are a function of 1/n2. Note also that the overall energy is negative. This is because the energy is a binding energy. Because of the negative algebraic sign, a higher quantum number n means a higher, because less negative energy. At very high quantum numbers n the value for E would approach zero, meaning that the binding energy for the electron would approach zero. The higher the orbit of the electron the more energy it has and the less strongly it is bound to the atom.
Finally let us calculate energy differences between electrons in different orbits. Subtraction of the terms for the energies of two electrons in two different orbits gives ∆E=(e4m)/(8ε02h2) (1/(nlow2 ) - 1/(nhigh2 )), Fig. 1.1.25. The calculated and empirically found ΔE match excellently, the empirically found Rydberg constant matches the theoretically derived constant (e4m)/(8ε02h2). Thus experiment and theory are in accordance. Bohr’s theory is able to explain the H spectra very well, and can predict both radii of electron orbits and energies. The Bohr model for the first time introduced the quantization of electron states in atoms, and in this regard in was a big step forward. However, there were still problems with Bohr’s theory. It could only explain the H spectra well, but failed to explain the spectra of all other atoms. Secondly, Bohr’s postulates seemed ad hoc and lacked an explanation. There was no good explanation why an electron in a quantized orbit would not emit electromagnetic radiation continuously. Thus, the Bohr model could still not be the final answer to atomic theory. In fact, it lacks to take an important property of the electron into account: The wave-particle dualism of the electron.
Definition: Problems with Bohr's Theory
1. It can explain only the H spectrum and fails to explain all the spectra of all other atoms.
2. Bohr’s postulates are ad hoc. They lack an explanation.
Dr. Kai Landskron (Lehigh University). If you like this textbook, please consider to make a donation to support the author's research at Lehigh University: Click Here to Donate. | textbooks/chem/Inorganic_Chemistry/Inorganic_Coordination_Chemistry_(Landskron)/01%3A_Atomic_Structure/1.01%3A_Historical_Development_of_Atomic_Theory.txt |
Wave Particle Dualism
The phenomenon of the wave-particle dualism was first discovered for electromagnetic radiation, and the extended to all other particles including the electron. It began with the investigation of the photoelectric effect by Albert Einstein (Fig. 1.2.1 and Fig. 1.2.2).
The photoelectric effect occur when a metal surface is irradiated by light. Above a certain frequency, or below a certain wavelength, light is able to eject electrons from the metal surface. The threshold frequency depends on the metal. Below the threshold frequency no electrons get ejected. Einstein investigated the maximum kinetic energy of the ejected electrons as a function of the frequency of the light. He found that there was a linear relationship. He analyzed the slope of this line and found that the slope was the Planck constant h. This would mean that electrons had an energy E=h$\nu$ minus an energy EB that would be needed to overcome the binding energy, also called the work function, of the electron in the metal (Figure 1.2.3, left).
The equation E=h$\nu$ was previously derived by Planck (Fig. 1.2.4) based on the assumption that energy was quantized, and now Einstein had experimentally found it again in the quest to explain the photoelectric effect. This would mean that light was quantized. The quantization would be explained by the fact that light would not only have wave but also particle properties, and these particles would be called photons. Assuming photons the photoelectric effect could be easily explained (Figure 1.2.3, right). When light hits the metal surface the photon collides with the electron. Only when the photon had an energy larger than the work function of the metal, the electron would be ejected and would have a kinetic energy equal to the difference between the energy of the photon and the binding energy. The wave-particle dualism of light, and electromagnetic radiation in general can also be mathematically derived. Because mass can be converted into electromagnetic radiation according to the equation E = mc2, and the energy of electromagnetic radiation is E=h$\nu$, mc2=h$\nu$. We can solve the equation for $\nu$, and then it is $\nu$=mc2/h. With $\nu$=c/λ, and solved for λ, the equation becomes λ=h/mc. This equation shows the wave particle-dualism of electromagnetic radiation because it relates a wavelength to a mass. In fact, the mass of the particle associated with electromagnetic radiation, the photon, is inverse proportional to the wavelength of the electromagnetic radiation. The discovery of the wave-particle dualism of electromagnetic radiation was a radically new concept that is difficult to grasp intellectually up to this date because the human mind tends to see waves and particles to be mutually exclusive. However, it is one of the most fundamental principles of nature. As we will see later, not only electromagnetic radiation shows the wave particle dualism, but all particles including electrons.
Wave Particle Dualism of Massive Particles
The wave-particle dualism was originally thought to be valid for the photon only. A young French physics PhD student, Louis De Broglie had the radical idea that not only the photon, but all particles would exhibit the wave particle dualism, including the electron. Einstein’s formula λ = h/mc would just need to be slightly rearranged into λ = h/mv, whereby m would be the mass of the particle, and v would be the velocity of the particle. This idea was much disputed at the time, and Louis De Broglie’s PhD thesis was almost not accepted. However, eventually the wave-particle dualism of the electron was proven by electron diffraction experiments, and Louis De Broglie was awarded the Nobel prize in 1929. Today, we believe that all particles show the wave-particle dualism, and no experiment up to this date indicates an exception.
Standing Waves
With the discovery of the wave-particle dualism of the electron, and the observation of the quantization of electronic states in atoms led physicists focus on a field in physics in which waves are quantized. The field of standing waves (Fig. 1.2.6).
Standing waves are a quite common phenomenon. You can for example trigger a standing waves by plucking a guitar string. This causes the guitar string to vibrate in standing waves with discreet, quantized wavelengths. The vibration with the longest wavelength is the so-called ground vibration. Its wavelength is two times the length of the guitar string. In addition, so-called higher harmonics are possible. The first harmonic has a wavelength equal to the length of the guitar string, the second one, has a wavelength equal to two-thirds of the length of the string, the third one has a wavelength equal to half of the length of the string, the fourth one has a wavelength equal to one third of the length of the guitar string, and so fourth (Fig. 1.2.7).
It can be easily seen that the possible wavelengths at which the string can vibrate follow the equation λ=2L/n, whereby n is an integer, or a quantum number, and L is the length of the guitar string. Thus, we can say that the waves associated with the guitar string vibrations are quantized. Why are these waves called standing waves? This is because the positions of the crests and the troughs and the nodes do not move. They remain at the same position on the guitar string at any point in time. It should be said here that the standing nature of the waves is actually an illusion. There are actually two waves traveling in opposite direction on the guitar string, and these waves interfere with each other so that a standing wave is produced (this is illustrated in Fig. 1.2.6). When the guitar string is plucked two waves are sent into opposite direction on the guitar string toward the two opposite ends on the guitar string. Once they have reached these ends they get reflected and sent into the opposite direction until they again reach the ends of the guitar string where they get reflected again. During this process, which happens over and over again, the two waves interfere and produce the standing wave. In sum, the fact that the wave is confined within the guitar string, leads to the quantized standing waves.
Electron in a One-Dimensional Box
Let us know go from a vibrating guitar string to an electron in a one-dimensional box of length a having infinitely high walls. Inside the box the potential energy of the electron is zero, in the walls the potential energy is infinite (Figure 1.2.8).
Due to its kinetic energy the electron can travel on a line within the box until it hits the wall (Fig. 1.2.9, A). At the wall it is getting reflected and forced to travel into the opposite direction until it again hits the wall where it reverses direction again, and so forth. Consider now that the electron does not only have particle, but also wave properties. Because of that also a wave travels along the line, gets reflected at the wall, travels into opposite direction, gets reflected again and so on. These waves can interfere with it other just like the waves traveling on the guitar string to produce a standing wave. Thus, the electron in the one-dimensional box should behave like a standing wave, and this wave should be quantized (Fig. 1.2.9, B-F).
How can we mathematically describe the electron in the box as a standing wave? Generally, you can describe a standing wave by a wave function. A wave function tells the amplitude of the standing wave at a particular position in the one-dimensional box. How can we find the wave function? We can start out from a differential equation that is generally valid for standing waves (Eq. 1.2.1).
Equation 1.2.1 Standing wave differential equation.
It says that the second derivative of the amplitude of the wave function at the position x is equal to –(2π/λ)2 multiplied with the amplitude of the wave function at the position x. Let us now consider that the kinetic energy of the electron is E = 1/2mv2 and expand the equation by m.
Equation 1.2.2 Solving for λ from kinetic energy equation and De Broglie equation.
Let us then solve the equation for (mv) . Now let us solve the De Broglie equation for mv and insert mv by h/λ in the previous equation. Finally, let us solve the equation by λ and we will get λ = h/[2mE]1/2, Eq. 1.2.2. We can now substitute λ by h/[2mE]1/2 in the differential equation. Slightly rearranged this equation becomes the Schrödinger equation for the electron in the one-dimensional box (Equation 1.2.3).
Equation 1.2.3 Schrödinger Equation for the electron in a 1-D box.
The Schrödinger equation is a differential equation. To get the wave function that describes the electron in the box we need to solve the differential equation. One possibility solve a differential equation is to guess its solution, and after that show that the solution is right. This is the approach we want to pursue here. A very general wave function is one that is a sum of a sinus term and a cosinus term of the coordinate x whereby we shall assume two general coefficients r and s in front of x, and two other general coefficients A and B in front of the sinus and the cosinus term, respectively.
Equation 1.2.4 Generic solution to the differential equation.
We can now think of so-called boundary conditions for the wave function which will make the wave function more specific. A boundary condition is a property the wave function must have to be a sensible solution to the differential equation. We can assume a first boundary condition which assumes that at the position x=0 the amplitude of the wave function must be 0. This can be assumed because at these positions the "electron wave" gets reflected at the wall. That means that the wave function cannot have a cosinus term and thus B must be 0. If the wave function had a cosinus term it would not be 0 at x = 0 because the cosinus of 0 is not 0.
Equation 1.2.5 Boundary condition 1 for the wave function
The second boundary condition is that the amplitude of the wave function is zero at x = a. Again, this is because the electron hits the wall at x = a and reverses its direction. The sinus function is only zero at x = a when ra is an integer number n times π: ra=nπ. This means that r must be nπ/a.
Equation 1.2.6 Boundary condition 2 for the wave function
Thus, the wave function must be A=sin(nπx/a). We can see that the wave function that describes the electron as a standing wave in the one-dimensional box is quantized because the quantum number n appears.
Equation 1.2.7 The wave function considering boundary conditions 1 and 2.
Inserting the quantum numbers n into the wave functions produces all the standing waves the electron can adopt (Fig. 1.2.10). You can see that the number of nodes and the wavelengths of the waves depend on the quantum number n. For n = 0 there is no node and the wavelength is twice the length of the box, for n = 2 there is one node and the wavelength is equal to the length of the box, for n =3 there are three nodes and the wavelength of the wave is 2/3 of the lengths of the box and so forth. We can illustrate that the wave function describes the waves depicted in Figure 1.2.10 by an example. The amplitude for the wave for n=2 is zero in the middle of the box where x = a/2. If we insert a/2 into the equation for Ψ then Ψ = A sin(nπ2a/a)= A sin(nπ) = 0 because it is the property of a sinus function to be 0 at an integer number multiple of π. We could also insert other values for x and n into the wave function, and would get the expected amplitude. This shows that the wave function correctly represents the standing waves the electron can adopt.
We are still not quite finished with the wave function because we have not determined the parameter A in front of the sinus term. To obtain it we need to consider a third boundary condition (Eq. 1.2.8).
Equation 1.2.8 Boundary condition 3.
It says that the integral of the square of the wave function over the length of the box must be equal to one. We can understand this boundary condition when we consider that the square of the wave function represents the probability to find the electron at a particular position in the box. The value the square of Ψ adopts at a position x within the box represents the probability to find the electron at this position when we consider the electron as a particle. This is called the Born interpretation of the wave function, named after the German physicist Max Born.
Because the probability to find the electron anywhere in the box must be 100%, the integral of the square of the wave function over the entire box must be 100% or 1. One can show, and we omit the necessary mathematical steps for clarity here, that the boundary condition is only fulfilled when A is equal to the square root of 2/a. The final wave function is then psi is equal to 2/a sin(nπx/a). The factor square root of 2/a is called the normalization constant of the wave function because it adjusts the amplitude of the wave function so that the probability to find the electron anywhere in the box is 100%.
The Schrödinger Equation for the H Atom
Let us now go from an electron in a 1-dimensional box to the electron in the hydrogen atom. What is similar and what is different between these two cases. A similarity is that the electron is confined in an atom just like the electron is confined in the 1D box. Thus, like in the 1-dimensional box the electron should behave like a standing wave. While in the one-dimensional box there is only one coordinate to consider, there are three coordinates to consider for the electron in the atom. This is because an atom is spherical and thus three coordinates x, y, and z are necessary to describe a position within the atom. A second major difference is that the potential energy of the electron is zero at any position within the box, while it is not zero in the hydrogen atom. This is because in an atom there are attractive Coulomb forces between the nucleus and the electron. The further away the electron is from the nucleus the higher its potential energy. This is because it takes energy to pull the electron away from the nucleus.
We therefore need to modify the Schrödinger equation that we used previously for the one-dimensional box the following way. Firstly, we need to expand the operator for the kinetic energy from one to three dimensions and introduce the coordinates y and z in addition to x. Secondly, we have to add an operator for the potential energy to the equation. The potential energy is the Coulomb energy between the proton and the electron. It is similar to the term we previously used for the calculation of the potential energy of the electron in the Bohr model. Instead of the radius r we use now the square root of the sum of the square of the three coordinates x, y, and z, to indicate the Coulomb energy of the electron at any position within the atom. Our wave function will now be a function of three coordinates x, y, and z. This means our wave function will now be three-dimensional and represent three-dimensional standing waves. Three-dimensional waves are harder to imagine compared to 1-dimensional ones, but have the same properties, which are that the position of the crests, troughs, and nodes does not move.
Equation 1.2.9 Schrödinger equation for the H atom.
In order to get the wave functions for the electron we need to solve the Schrödinger equation for the hydrogen atom. To do so, we need to think about the boundary conditions for the wave function. One condition is that the square of the wave function approaches zero when we go very far from the nucleus, and r, the distance from the nucleus approaches infinite. Secondly, like in the one-dimensional box, the integral over the square of the wave function must be one. This is because the probability to find the electron somewhere in the atom must be 100%. Thirdly, it would be sensible to assume that the wave function must be continuously differentiable and single valued (Fig. 1.2.13).
The Spherical Polar Coordinate System
The mathematical process to solve the Schrödinger equation is beyond the scope of this course and you are referred to Physical Chemistry classes and textbooks for the details. We shall only provide an brief outline of the process here. It is mathematically simpler to solve the Schrödinger equation in spherical polar coordinates instead of cartesian coordinates. Therefore, we obtain the solutions of the Schrödinger equation, the wavefunctions, in polar coordinates. The position of a point is specified by three numbers: the radial distance of that point from a fixed origin, its polar angle measured from a fixed zenith direction, and the azimuthal angle of its orthogonal projection on a reference plane that passes through the origin and is orthogonal to the zenith, measured from a fixed reference direction on that plane (Fig. 1.2.14).
Solutions of the Schrödinger Equation
The wavefunction is then a function of r, θ and φ, in particular it is a product of a radial wave function which is function of r, a colatitude wave function, which is a function of θ, and an azimuthal wave function which is a function of φ (Eq. 1.2.10). You can see the explicit forms of the radial, colatitude, and azimuthal functions in Fig. 1.2.15.
Equation 1.2.10 The wave function for the electron in the H atom as a function of r, θ and φ
We do not need to understand all details here, but need to realize that these wavefunctions are functions of quantum numbers. In contrast to the electron in the one-dimensional box not only one, but three quantum numbers need to be considered. Beyond the quantum number n, which is the called the principal quantum number, there is also a so-called orbital quantum number l, and a magnetic quantum number m. The quantum number n only occurs in the radial wave function, the quantum number l only occurs in the radial and colatitude function, and the quantum number m only occurs in the colatitude and the azimuthal part of the wave function. The values the orbital quantum number l can adopt depends on n, it can be between 0 and n-1 for a given quantum number n. The magnetic quantum number m depends on the quantum number l, and can run from –l to +l. So, for example when n is equal to 2, l can vary between 0 and 1, and for l equal to 2, m can adopt any value between -2 and 2, namely -2,-1, 0, +1, and +2 (Fig. 1.2.16).
These wave functions have a particular name in chemistry. They are called orbitals. One can understand an orbital as the three-dimensional wave function that describes the electron in an atom as a standing wave. Thus, an orbital is a state the electron can adopt, and when we say that an electron is in a particular orbital we mean that the electron is in a particular state.
The spin quantum number s
Within an orbital an electron can adopt two different spins described by the spin quantum number s. This quantum number is not a result of the Schrödinger equation, but was found experimentally. The spin quantum number s can adopt two values: +1/2 and -1/2. We say an electron is spin up when s = +1/2, and spin down when s = -1/2.
The spin quantum number is understood most easily when you view the electron as a particle rotating around its own axis. A counter-clockwise rotation would be associated with s=+1/2, a clockwise rotation would be associated with s=-1/2. The rotation produces a magnetic field with the direction of the field lines depending on the direction of rotation.
The 1s Orbital
Let us now insert the quantum numbers into the general form of the wave function for the electron in the hydrogen atom, and calculate their explicit mathematical forms. Let us start with the smallest possible numbers. When we insert n=1, l=0, and m=0 we get the wave function for the 1s orbital (Eq. 1.2.11).
Equation 1.2.11 Wave function for the 1s orbital
You can see that the colatitude and azimuthal parts of the wave functions, together also called the angular part of the wave function, are simple numbers, namely one over the square root of two pi, and one over the square root of two respectively. The angles theta and phi do not appear and that means that the wave function is angle-independent. This implies that the orbital has a spherical shape.
The radial part of the wave function has an exponential term e to the power of –r over a0. The number e is the Euler number, that is the base of the natural logarithm. It is approximately equal to 2.71828. a0 is the Bohr radius, which is the distance of the electron from the proton for the electron on its first orbit according to the Bohr atom model. We calculated it previously, and it was 5.29 x 10-11m. r is the distance from the nucleus. This exponential term has a constant in front of it which is 1 over the Bohr radius to the power of three over two. Because the radial part of the wave function is e to the power of –r, the amplitude of the radial part of the wave function declines exponentially with increasing distance r from the nucleus. This is true for the entire wave function because the angular parts of the wave function are just simple numbers. You can see the exponential decline of the amplitude of the wave function as a function of the radius r in units of the Bohr radius a0 in the graph below (Fig. 1.2.19).
This means that the 1s orbital must be a spherical orbital with its amplitude exponentially declining with increasing distance from the nucleus. This can be graphically depicted via a density of points around the nucleus (Fig. 1.2.18). A higher density of points means higher amplitude and a lower density of points means smaller amplitude. The fact that the amplitude is the highest closest to the nucleus means that the probability to find the electron in a particular point in space is the highest closest to the nucleus. This is because the square of the wave function for the 1s orbital represents the probability to find the electron at a particular point in space. However, the probability to find the electron at a certain distance from the nucleus is not the highest closest to the nucleus. This is because the probability to find the electron at a certain distance from the nucleus is the probability to find the electron at a particular point in space times the number of points in space that have that particular distance. The number of points in space that have a particular distance r from the nucleus is related to the surface of a sphere with the a radius r. The surface of a sphere A is given by the formula A=4πr2. Thus, the probability to find the electron at a particular distance r from the nucleus is equal to the surface of a sphere with the radius r times the radial wave function square. This defines the radial probability function RP=4πr2R2.
The graph for the radial probability function is depicted in Figure 1.2.19. You can see that it is zero when r=0 which is due to the r2 term which becomes zero at r = 0. At small r values the 4πr2 term dominates the overall function and the probability to find the electron increases with r. However, because the exponential term of the radial function becomes more and more significant at larger r values, the radial probability function goes through a maximum and then declines. At distances r approaching infinite the function approaches zero. The maximum of the curve represents the distance at which it is most likely to find the electron. Interestingly, it is equal to the Bohr radius r. That means that Bohr was not so wrong after all. However, he was wrong in the sense that he viewed the electron as a classical particle being only at that radius. Instead, the electron, due to its wave properties, it is not only at the Bohr radius r0, but most likely we can find it there.
The result that the probability to find the electron at a specific point in space is the greatest at the nucleus does away with the question of why the H atom does not collapse. Because the amplitude of the wave function is the greatest closest to the nucleus we can actually say in a way that the atom is collapsed. Nonetheless, the atom is mostly empty space because of the delocalization of the wave function that describes the state of the electron. The delocalization of the electron due to its wave properties also makes the question of the size of an orbital non-trivial. The wave function approaches zero only for r approaching infinite, but never becomes zero. This means that to account for all electron density the orbital would be infinitely large. This however would not be a sensible definition. The commonly accepted definition is that the size of an orbital is defined by the space that contains 90% of its electron density. We can depict orbitals according to this definition. For the case of the 1s orbital the radius of the sphere that represents the 1s orbital is chosen so that the probability to find the electron within a sphere of that radius is 90%.
The 2s Orbital
Inserting n=2, l=0, and m=0 into the general wave function gives us the wave function for the 2s orbital (Eq. 1.2.12).
Equation 1.2.12 Wave function for the 2s orbital
Like in the case of the 1s orbital, the angular parts of the wave function are only simple numbers. Thus, the 2s orbital is also a spherical orbital. Like in the 1s orbital, the radial part of the wave function has an exponential term of the type e-r and a simple coefficient. However, there is an additional term 2-r/a0. Due to the exponential term, the amplitude of the wave function declines exponentially with r. However, because of the term 2-r/a0 the wave function becomes negative, and then approaches 0. The radius at which the wave function changes its algebraic sign is called a spherical node. The node is spherical because it describes the surface is a sphere. On the surface of this sphere the amplitude of the wave function is zero. The radius at which the wave function changes its algebraic sign is 2a0. This becomes clear when we consider that the term 2-r/a0 becomes zero when r=2a0 because 2-2a0/a0=0. When this term is zero then the entire wave function becomes zero. We can illustrate the change of the algebraic sign in the 2s orbital using colors as shown below (Fig. 1.2.20).
The orange sphere indicates the space in which the amplitude of the 2s orbital is positive, and the blue area indicates where it is negative. The interface area where the color changes represents the node. By multiplying the square of the radial wave function R with 4πr2 we obtain the radial probability function for the 2s orbital (Fig. 1.2.21).
You can see that similar to the 1s orbital, the probability to find the electron on a radius closest to the nucleus is 0. However, unlike the 1s orbital there are now two maxima instead of only one. This second maximum is imposed by the spherical node of the 2s orbital. The second maximum is larger than the first one. Generally, the probability to find the electron further away from the nucleus is greater compared to the 1s orbital. That means that the 2s orbital is larger than the 1s orbital. This reflects the general trend that orbitals with a higher quantum number n tend to be larger and the probability to find the electron further away from the nucleus is greater.
The 3s Orbital
Let us look at one more s orbital, the 3s orbital (Eq. 1.2.13).
Equation 1.2.13 Wave function for the 3s orbital.
The angular parts of the wave function for the 3s orbital are again simple numbers, which implies that the orbital is again spherical. Generally all s orbitals are spherical.
The radial part of the wave function can be divided into the parts. The first part is a simple number, then there is the term 27-18r/a0+ 2r2/a02, and the exponential term e-r/3a0. Because of the exponential part the amplitude of the wave function declines exponentially with increasing r. However, then the wave function becomes negative, goes through a minimum, then becomes positive again, goes through a small maximum, and finally approaches zero. The fact that the wave function changes its algebraic sign two times means that the 3s orbital has two spherical nodes. These spherical nodes are due to the second term which is a square function of the type ax2 + bx + c, whereby in this case a = 2/a02, b =-18/a0 and c=27. When this term becomes zero the entire wave function becomes zero. Square functions have two solutions according the formula x=(-b±√(b2-4ac))/2a. The solutions give the radii for the two spherical nodes of the 3s orbital. We omit the exact calculation of the radii for briefness and clarity reasons here. You may calculate them as a homework.
Because of the two spherical nodes, the radial probability function Rp has three maxima. A very small one close to the nucleus, are larger one further away from the nucleus, and the largest one at even greater distance from the nucleus. Overall, the 3s orbital is larger than the 2s orbital confirming the general trend that s orbitals increase in size with increasing quantum number n.
The 2pz Orbital
Now let us look at orbitals in which the orbital quantum number l is larger than 0. When l is equal to 1, then n must be at least 2. An orbital with n=2 and l=1 is a 2p orbital. Overall, three 2p orbitals must exist because for l=1, the magnetic quantum number m can adopt three values: -1, 0, and +1. The orbital with m=0 is called the 2pz orbital. Let us look at the mathematical form of this wave function (Eq. 1.2.14).
Equation 1.2.14 Wave function for the 2pz orbital
Analyzing the angular parts of the wave function shows that while the azimuthal part is still a simple number, the colatitude part is not, it is a function of θ. This means that the amplitude of the wave function is now angle dependent, which implies that the orbital cannot be spherical. Instead the 2p orbital has the shape of a dumbbell which is oriented along the z-axis. One lobe is above the xy plane, and the other lobe is below the xy plane. These two different lobes have a different algebraic sign, indicated by different colors (Fig. 1.2.24).
Within the xy plane the amplitude of the wave function is zero. The xy plane represents a so-called planar node. A planar node is a type of an angular node. The name angular node is because it is due to the fact that the angular part of the wave function is not a simple number, but a function of an angle, in this case θ. We can show that the angular part of the wave function produces the planar node in the xy plane when we convert it from spherical into Cartesian coordinates (Eq. 1.2.15).
Equation 1.2.15 Angular part of the wave function of 2pz in cartesian coordinates. The function becomes zero when z=0.
We can see that the angular part of the wave function is only a function of z. We can also see, that the angular part of the wave function becomes zero when z=0. When the angular part of the wave function becomes zero, then the wave function of the entire orbital becomes zero. Z is zero only in the xy plane, and this explains why the xy plane is a planar node.
Now let us look at the radial part of the wave function. It is made of three terms: a simple constant, the term r/a0, and the exponential term e-r/2a0. Because of the term r/a0 the radial part of the wave function is zero at r=0. This means that it is very unlikely to find the electron at a point in space directly at the nucleus. This behavior is opposite to s orbitals. Because the r/a0 term dominates the behavior of the wave function at small r value, the amplitude first increases. However, at larger r values the exponential term becomes more an more important which forces the amplitude of the radial function through a maximum, after which it declines and approaches zero. The radial wave function never changes algebraic sign. Therefore, the 2pz orbital does not have a spherical node. The radial probability function has a similar shape compared to the radial function. It is zero at r=0 and goes through a maximum before it declines again and approaches zero (Figure 1.2.25).
The 2px and 2py Orbital
The 2px and the 2py orbitals are obtained when the quantum numbers m are 1, and -1, respectively.
Equation 1.2.16 Wave function for the 2px and 2py orbital
They have the same shape as the 2pz orbital, but their orientation is different. The dumbbell of the 2px orbital is oriented along the x-axis, and that of the 2py orbital is oriented along the y-axis.
This is because the 2px orbital has a planar node in the yz plane, and the 2py orbital has a planar node in the xz plane. The planar nodes are due to the angular part of the wave functions of these orbitals.
Equation 1.2.17 Angular part of the wave function of 2px and 2py, respectively. The wave functions become 0 for x=0 and y=0, respectively.
As one can see, the angular parts of the wave function contain both the angles θ and Φ. When converted into Cartesian coordinates (Eq. 1.2.17) one can see that the angular function of the 2px orbital is only a function of x, and the angular function of the 2py orbital is only a function of y. The functions become zero for x=0 and y=0, respectively. X is zero only in the yz plane, and y is zero only in the xz plane, hence these nodes are planar nodes. The radial wave function is exactly the same as for the 2pz orbital.
The 3p Orbitals
Now let us look at the 3p orbitals. In a 3p orbital the quantum number n=2 and l=1. Like for the case of the 2p orbitals there are three p orbitals because the magnetic quantum number m can adopt the values -1, 0, and +1. The wave function of the 3pz orbital is shown below (Eq. 1.2.18).
Equation 1.2.18 Wave function for the 3pz orbital
One can see that the angular part of the wave function is exactly the same as for the 2pz orbital. This means that like in the 2pz orbital there must be a planar node in the xy plane. When you look at the radial part of the wave function you can see that it is composed of four terms, a simple constant, a term 6-r/a0, the term r/a0, and the exponential term e-r/3a0. Because of the term r/a0 the wave function becomes zero at r=0. Because of the term 6-r/a0 the wave function has a spherical node. This is because when 6-r/a0=0 the entire wave function becomes 0. We can determine the radius at which the wave function becomes zero by solving the equation 6-r/a0=0 for r, which gives r=6a0. Overall, the radial function is zero at r=0, goes through a maximum, changes its algebraic sign at the spherical node at r=6a0, has a minimum, and then approaches zero for very large distances r (Fig. 1.2.28).
The radial probability function has two maxima, a small one close to the nucleus, and a larger one further away from the nucleus. Overall the radial probability function is further away from the nucleus compared to the 2p orbitals, which implies that the 3p orbital is larger.
The overall shape of the 3pz orbital is determined by the both the spherical and the planar node (Fig.1.2.29). Due to the planar node in the xy plane it has a dumbbell shape and is oriented in z direction. However, because of the spherical node the algebraic sign of the wave function changes within the two lobes of the dumbbell. Finally let us consider the 3px and the 3py orbitals. They would have the same shape as the 3pz orbital, but they would be oriented in the x and the y direction respectively. We could also think about what would happen to the p orbitals if we increased the quantum number n further. In this case the dumbbell shape would remain, but additional spherical nodes would be introduced. A 4p orbital would have two spherical nodes, a 5p orbital three spherical nodes, and so on. The size of these p orbitals would also increase with the quantum number n.
The 3dz2 Orbital
Now let us consider the orbitals with the orbital quantum number l=2. These orbitals are so-called d-orbitals. Because the principal quantum number n always has to be at least one integer number larger than the quantum l, the d orbitals with the smallest quantum number n are the 3d orbitals. For each quantum number n there must be five d orbitals because for l=2 the magnetic quantum number m can adopt the values -2, -1, 0, -1, +1, and +2. Let us first look at the 3d orbital with m=0. This orbital is called the 3dz2 orbital. Its wave function is shown below (Eq. 1.2.19).
Equation 1.2.19 Wave function for the 3dz2 orbital
You can see that the radial function is a product of three terms: a constant, a term r2/a02, and the exponential term e-r/3a0. The exponential term is the same as for the 3p orbitals. Because of the term r2/a02 the amplitude of the wave function is 0 at r=0. The amplitude then increases with r, goes through a maximum, and then approaches zero. This means that the wave function never changes its algebraic sign, and therefore it does not have spherical nodes. The radial probability function has a similar shape compared to the radial function (Figure 1.2.30). The electron probability is somewhat further away from the nucleus compared to the 3p orbitals, meaning that the 3d orbital is somewhat larger than the 3p orbital.
You can see that the angular part of the wave function is a function of θ which means that the orbital will be non-spherical and there will be at least one angular node. In this case the angular nodes are two conical nodes. These nodes describe two cones that give the orbital its characteristic shape. It is a doughnut ring in the xy plane around a dumbbell pointing into z direction. The dumbbell and the doughnut have different algebraic signs indicated by different colors (Figure 1.2.31). Note that in contrast to the p orbitals the two lobes of the dumbbell have the same algebraic sign.
We can understand that the 3dz2 orbital has two conical nodes when we convert the angular part of the wave function into Cartesian coordinates (Eq. 1.2.20).
Equation 1.2.20 Angular part of the wave function of the 3dz2 orbital in cartesian coordinates.
You can see that it is a function of 2z2-x2-y2. This is the mathematical form of cones. The angular part of the wave function becomes 0 when 2z2=x2-y2 which is true on the surface of two cones, one above the xy plane and one below the xy plane. The name of the 3dz2 orbital is because it is a function of z2.
The 3dx2-y2 Orbital
Let us next look at the 3dx2-y2 orbital (Eq. 1.2.21).
Equation 1.2.21 Wave function for the 3dx2-y2 orbital
The radial part of the wave function is the same as the one for the 3dz2 orbital. The angular part of the wave function is both a function of θ and Φ and thus there are angular nodes to expect. We can find them again when we convert the angular part of the wave function from spherical to Cartesian coordinates.
Equation 1.2.22 Angular part of the wave function of the 3dx2-y2 orbital in cartesian coordinates.
You can see that the angular function is a function of x2-y2. The function becomes zero when x2-y2=0. This is the case when x=y or x=-y. X is equal to y on two lines that bisect the 90 degree angle between the x and the y axis. The coordinate z is completely variable. Thus, overall, the wave function becomes zero on two planes that stand perpendicular to the xy plane and bisect the 90 degree angle between the x and the y axis. These two planes are therefore the two planar nodes of the orbital.
Because of the two planar nodes the orbital has four lobes lying on the x and the y axis, respectively. Lobes lying on the x-axis have the opposite algebraic sign compared to those lying on the y-axis. The name of the dx2-y2 orbital is because it is a function of x2-y2.
The 3dxy, 3dyz and 3dxz orbitals
The remaining d orbitals are the 3dyz, the 3dxz, and the 3dxy. The radial function of these orbitals is just like the ones we previously discussed. The angular part of the wave functions is shown here directly in Cartesian coordinates. The dyz orbital is a function of yz. The function becomes zero when either y=0 or z=0. y is zero in the xz plane, and z is zero in the xy plane. Therefore these planes are planar nodes of the orbital. Because of these nodes, the orbital has four lobes which lie in the yz plane whereby the lobes are in between the y and the z axes. Note that adjacent lobes have opposite algebraic sign, and opposite lobes have the same algebraic sign.
The 3dxz orbital is a function of xz, and thus the function becomes zero with x=0 or z=0. This defines that yz and the xy planes as the two planar nodes. The orbital has the same shape as the 3dyz, except that the four lobes lie within the xz plane instead the yz plane.
The 3dxy-orbital is a function of xy, and thus its wave function becomes zero at x=0 or y=0. Therefore, the xz and the yz planes are the planar nodes and the orbital has four lobes in the xy plane in between the x and the y axis. Note that the 3dxy and the 3dx2-y2 look very similar but are different. The difference is that the 3dx2-y2 orbital has its lobes on the x and the y axis, while the 3dxy orbital has the lobes in between the x and the y axis. One can also say that the 3dx2-y2 orbital is rotated by 45 degree around z with respect to the 3dxy orbital.
Rules for Angular and Spherical Nodes
Of course, we could now discuss the size and shape of many other orbitals such a the 4d orbitals or the 4f orbitals, but this will be beyond scope. Instead, let us think about if there are general rules that allow the prediction of the number of spherical and angular nodes in an orbital. The number of radial nodes is always equal to n-l-1, the number of angular nodes is equal to l, and the total number of nodes is always n-1. This means that the number of nodes increases with the quantum number n. If we increase the quantum number l for a given quantum number n, then we replace spherical nodes by angular nodes. This is summarized in Figure 1.2.36.
The Orbital Energies for the H Atom
The Schrödinger equation does not only allow to calculate the orbitals of the hydrogen atom, but also their energies. The energies of Bohr and Schrödinger model match: E = constant/n2 : This means that orbitals with the same quantum number n have the same energies. The energies are not a function of the quantum numbers l and m. The energies are negative because they are binding energies. In other words: Adding an electron to a proton is an exothermic process. The binding energy for the energy increases as the orbital energy decreases (Fig. 1.2.37).
For the 1s orbital of H the energy is -13.6 eV, the energy of the 2s and the 2p orbitals are ¼ of that, the energy of the 3s, 3p, and 3d orbitals are 1/9 of the energy of the 1s orbital and so forth. Note at ¼ and 1/9th of -13.6 eV is more than -13.6 eV due to the negative algebraic sign. The electron volt is a unit of energy. It is the amount of kinetic energy gained by a single unbound electron when it passes through an electrostatic potential difference of one volt, in vacuum.
Equation 1.2.23 The electron volt to joule unit conversion.
In other words, it is equal to one volt (1 volt = 1 joule per coulomb) times the charge of a single electron (in coulombs). It is a very small unit of energy which is practical for orbital energy calculations because the orbital energies are very small (Eq. 1.2.23).
Multi-electron Atoms
Thus far we have only considered the orbitals for the hydrogen atom which contains only one electron. Can we also solve the Schrödinger equation for atoms that have more than one electron and get the exact energies of the orbitals? The answer is no, this mathematically not possible. The process is just already too complex even for only two electrons. The Schrödinger equation can only be solved for one electron systems. Therefore, the description of the atomic structure of all other atoms must work with approximations. Let us first consider the He atom. It has only one more electron than hydrogen. It is a useful approach to approximate multi-electron atoms as one-electron systems first, and then approximate the electron-electron interactions. Electron energies in an atom with more than one proton should follow the equation En=-Z2x13.6 eV/n2, whereby Z is the number of protons.
Equation 1.2.24 The orbital energies for atoms with more than one proton
The binding energy of the electron increases proportionally to the square of the number of protons because the attractive Coulomb forces that act on the negatively charged electrons increases with the number of positively charged protons in the nucleus. One can experimentally measure the orbital energies via ionization energies. The energy required to remove an electron in a particular orbital from the atom is equal to the binding energy for the electron in that orbital. Therefore the ionization energy IE = -En (En = orbital energy). According to the Schrödinger model the orbital energy for a helium electron in a 1s orbital should be E1s=-(22×13.6 eV)/11 =-54.4 eV (Eq. 1.2.25).
Equation 1.2.25 Energy for a helium electron in a 1s orbital.
However, the experimentally measured ionization energy of the electron is +24.6 eV, which means that the real orbital energy is -24.6 eV, and not -54.4 eV. On the other hand, the ionization of a He+ ion is +54.4 eV which is exactly what we would expect. We can explain this phenomenon by the fact that the Schrödinger model works for a single electron only and must neglect electron interactions. In a He+ ion there is only one electron, therefore the Schrödinger model correctly predicts the energy of the electron. However, in a helium atom there are two electrons, and the Schrödinger model cannot account for the electron-electron interactions. Therefore, it does not give the correct energy for the electrons in a Helium atom. The electron-electron interactions can be viewed as shielding effects. This means that the first electron shields part of the nuclear charge from the second electron. Therefore, the second electron experiences a reduced Coulomb force from the nucleus. Because of the reduced Coulomb force, the binding energy is smaller. It is reduced from -54.4 eV to -24.6 eV. It should be pointed out that the two electrons in the 1s orbital of the He atom are indistinguishable, that means they both have the reduced binding energy. The binding energy only increases to -54.4 eV after one of the two electrons has been removed.
The net positive charge from the nucleus after accounting for shielding effects is called the effective nuclear charge. For the helium atom the effective nuclear charge is 1.34. We can calculate the effective nuclear charge from the experimentally measured first ionization energy. Solving the equation for Zeff gives Zeff=1.34 (Eq. 1.2.26).
Equation 1.2.26 Calculation of Zeff in the Helium atom from first ionization energies.
The Lithium Atom
Now let us go to the atom with three electrons, the lithium atom. Following the so-called Aufbau principle we would fill the first two electrons into the 1s orbital, because the 1s orbital is the orbital with the the lowest energy. However, the third electron does not fit into the 1s orbital because of the Pauli principle.
The Pauli principle states that no two electrons in an atom can have the same four quantum numbers. For that reason an orbital cannot accommodate more than two electrons. Within an orbital two electrons must have different spin quantum numbers s. We can indicate this by writing the electrons as arrows pointing up and down into a square box which represents the orbital. The third electron of the lithium would need to go to the orbital with the next higher energy.
According to the Schrödinger model, the energy of an orbital is only a function of the quantum number n, and we would have two choices: The 2s and the 2p orbitals. Both of them would have the same energy. However, in the lithium atom the 2s orbital has a somewhat lower energy than the 2p orbital. This can again be explained by shielding effects. The 2s orbital penetrates the 1s orbital slightly better than the 2p orbitals. Because of that, the 1s orbital shields the nuclear charge from the nucleus less for the 2s orbital compared to the 2p orbitals. The ground state electron configuration, which is the atom electron configuration having the lowest energy, must therefore be 1s2 2s1. We can see that the 2s orbital penetrates the 1s orbital slightly better than the 2p orbital from the graph for the radial probability functions of the orbitals shown in Figure 1.2.40.
The second large maximum of the function of the 2s orbital is actually further away from the nucleus compared to the maximum associated with the 2p orbital. However, because the 2s orbital has a small second maximum very close to the nucleus, the 2s orbital overall penetrates the 1s orbital better, and therefore the shielding effect of the 1s orbital on the 2s orbital is smaller compared to the 2p orbitals.
Generally we can say, the smaller the quantum number l for a given quantum number n, the better the penetration ability of this orbital. Due to the better penetration ability, the shielding is less and thus the effective nuclear charge that acts on the electron in the orbital is higher. The higher effective nuclear charge leads to a lower energy of this orbital. For that reason the energy sequence of orbitals with the same quantum number n increases from s to p to d to f.
For example, a 3s orbital has a lower energy then a 3p orbital which has a lower energy than a 3d orbital (Fig. 1.2.41).
Slater's Rules
Because an orbital energy can be calculated from the effective nuclear charge, it would be useful if the effective nuclear charge could be somehow estimated by simple approximations.
The so-called Slater rules, named after their developer John C. Slater (Fig. 1.2.42), are a simple tool that give a good estimate for the effective nuclear charge of orbitals in multi-electron atoms. The Slater rules estimate a shielding constant σ that allows to calculate the effective nuclear charge Zeff from the nuclear charge according to Zeff=Z-σ.
The rules for the estimation of σ are the following. Firstly, we need to recognize that the rules for s and p electrons are slightly different than those for d and f electrons. This is because s and p electrons of the same quantum number n have somewhat higher penetration abilities compared to d and f electrons. For both groups the first step is the same. We write out the electron configuration of the atom according to their quantum numbers n and l as shown below (Fig. 1.2.43).
Note that this order does not exactly reflect the order of energy. Electrons that have similar shielding effects are grouped together in parentheses. You can see that s and p orbitals of the same quantum number n build a group, all other orbitals are their own group.
To estimate the shielding constant for an s or a p orbital we can now apply the following rules:
a) Electrons to the right of the (ns, np) group contribute nothing to σ. This is because these electrons are further away from the nucleus than the electron of consideration.
b) Each of the other electrons in the same (ns, np) group contribute 0.35 to σ. These electrons are in the same shell as the electron of consideration, and thus have similar distance to the nucleus. For this reason their shielding is modest only and can be estimated to be about 35% of a full elementary charge, or 0.35.
c) Each electron in the n-1 shell contributes 0.85 to σ. Those electrons are significantly closer to the nucleus compared to the electron of consideration. For that reason they shield significantly better, approximately 85% of an elementary charge, or 0.85.
d) Each electron in the n-2 shell or below contributes 1.00 to σ. Those electrons are much closer to the nucleus than the electron of consideration and can fully shield an elementary charge. Its contribution to the shielding constant is therefore approximated to be 1.
Now let us look at the Slater rules for d and f electrons.
a) Electrons to the right of the (nd) or (nf) group contribute nothing to σ. This is again because the electron is further away from the nucleus than the electron of consideration, and thus cannot contribute to shielding.
b) Each of the other electrons in the same (nd) or (nf) group contribute 0.35 to σ. This is again because these electrons have similar distance compared to the electron for which we calculate the shielding constant, and thus the shielding is modest.
c) Each electron to the left of the (nd) or (nf) group contributes 1 to σ. Those electrons are considered much closer to the nucleus, and thus they shield a full elementary charge.
Note that the values 0.35, 0.85, and 1 have been chosen so that the results from the Slater rules are in the best possible accordance with experimental measurements of orbital energies. Remember, when we discussed the helium atom we said that the measurement of ionization energies can provide experimental orbital energies.
Slater's Rules Applied to Multi-Electron Atoms
Let us practice the Slater rules by a few examples. We can, for instance, calculate the effective nuclear charge that acts on a 2p electron in an oxygen atom. To answer this question we need to first write out the electron configuration according to the quantum numbers n and l and group the orbitals correctly (Fig. 1.2.44).
Next, we need to consider that the effective nuclear charge Zeff is the nuclear charge Z minus the shielding constant σ. Then, because oxygen has eight protons, the nuclear charge is 8. From that we need to subtract the shielding constant. Firstly, we need to realize that there are 5 electrons in the same group as the 2p electron of consideration. These are the two 2s electrons and the three other 2p electrons. These five electrons shield with a factor of 0.35 because they are in the same group as the 2p electron for which we want to calculate the shielding constant. Note that this fourth 2p electron does not get a factor because it is the electron for which we calculate the shielding constant. In addition, we need to consider the two 1s electrons. They are in an n-1 shell, therefore they contribute with a factor of 0.85. Overall, the effective nuclear charge on the 2p electron is 8-(2 x 0.85)-(5 x 0.35)= 4.55 (Fig. 1.2.45).
We can also ask what is the effective nuclear charge on a 1s electron in the oxygen atom. It is 8-0.35=7.65 (Fig. 1.2.46).
This is because we only need to consider the second electron in the 1s orbital to calculate the shielding constant σ. It contributes with a factor of 0.35 because it is in the same group as the 1s electron of consideration. The 2s and the 2p electrons are in a group right to the 1s orbital, therefore they do not contribute to the shielding.
Slater's Rules and the Aufbau Principle
The Slater rules are of great help to understand the so-called Aufbau principle, which says that for an atom in the ground state the electrons are in the orbitals of the lowest energy. To some extent the orbitals energies follow the quantum number n, but this is not always the case. For example, the element potassium has the electron configuration 1s2 2s2 2p6 3s2 3p6 3d0 4s1 and not 1s22s22p6 3s23p6 3d1. This means that the energy of the 3d orbital must be higher than the energy of the 4s orbital. Can the Slater rules predict that? Let us therefore calculate the energy of the 3d and 4s orbital using the Slater rules, and see if the energy of the 3d orbital comes out higher than the electron of the 4s orbital.
For the 3d orbital, the shielding constant is 18x1=18 because there are 18 electrons besides the 3d electron of consideration, and all the 18 electrons are in a lower shell. Therefore, they all contribute with a factor of 1. The nuclear charge Z of the potassium is 19. Therefore, the effective nuclear charge Zeff= 19-18=1 (Fig. 1.2.47).
Now let us do the analogous calculation for the 4s electron. For the 4s electron the Slater rules for s electrons apply. According to that, the two 3s and the six 3p electrons contribute 0.85 to the shielding constant because they are in an n-1 shell. The remaining two 1s, two 2s, and six 2p electrons are in an n-2 or lower shell and therefore contribute with the factor of 1. Therefore, the overall shielding constant is 8 x 0.85 + 10 x 1 = 16.8. The effective nuclear charge Zeff is therefore 19-16.8=2.2 (Fig. 1.2.48).
We can see that a higher effective nuclear charge acts on a 4s electron compared to a 3d electron, which indicates that the orbital energy for the 4s electron will be lower compared to the 3d electron. We can calculate the orbital energies by using the formula E=(Zeff2)/n2 ×13.6 eV (Eq. 1.2.27).
Equation 1.2.27 Equation for orbital energies
For the 3d electron we insert Zeff=1 and n=3 which gives -1.51 eV, Eq. 1.2.28.
Equation 1.2.28 Equation for orbital energy of the 3d electron of potassium
For the 4s electron insert Zeff=2.2 and n=4 which gives -4.14 eV.
Equation 1.2.29 Equation for orbital energy of the 4s electron of potassium
As expected, the energy of the 4s electron is lower than the energy of the 3d electron. This explains why the electron configuration with the 4s electron as the valence electron is preferred over that with the 3d electron as the valence electron. Overall, we can see that the Slater rules can correctly predict electron configurations of atoms and the Aufbau principle.
Let us do another, more complex example: There are two conceivable electron configurations for Fe. One in which the 4s subshell is filled before the 3d subshell, and one in which the 3d subshell is filled before the 4s subshell. For the first case the electron configuration is 1s22s22p63s2 3p6 4s2 3d6. For the second case, the second electron configuration is 1s2 2s2 2p6 3s2 3p6 3d8 4s0. We can see that the two electron configurations are the same for the 1s, 2s, 2p, 3s, and 3p electrons. We therefore focus on the remaining eight electrons, denoted in red, and calculate their energies.
In this case, we need to compare the sum of the electron energies for both electron configurations to be able to decide which electron configuration is favored. We calculate the sum of the energies of the eight electrons for the first electron configuration the following way: In the first step we need to rewrite the electron configuration in Slater form: (1s2)(2s2 2p6)(3s2 3p6)(3d6)(4s2).
Next, we can calculate the shielding constant for a 3d electron. We can ignore the two 4s electrons because they are to the right of the d electron of consideration, and do not contribute to the shielding. However, we do need to consider the five other 3d electrons that are in the same group as the 3d electron of consideration. They contribute with the factor of 0.35. The other 18 electrons contribute 1.00 because they are in a lower shell. Because the effective nuclear charge of Fe is 26, the effective nuclear charge Zeff is Zeff=26-19.75=6.25 (Eq. 1.2.30).
Equation 1.2.30 Equation for the shielding constant and effective nuclear charge of the 3d electron of an iron atom
For a 4s electron, there is one other 4s electron in the same group which contributes with a factor of 0.35. In addition, there are 14 electrons in an n-1 shell, namely the 3s, the 3p, and the 3d electrons. They contribute 0.85 to the shielding constant. The remaining 10 electrons are in an n-2 shell or lower, therefore they contribute with the factor 1.00. This gives an overall shielding constant of 22.25. From that we can calculate the effective nuclear charge for a 4s electron which is 26-22.25=3.75.
Equation 1.2.31 Equation for shielding constant and effective nuclear charge of the 4s electron of an Fe atom
We can now calculate the electron energies. It is the sum of the energies for the 3d and 4s electrons. The electron energy for a single electron is E=-(Zeff2)/n2 ×13.6 eV= -(6.252)/32×13.6 eV=-59.03 eV. Because we have six 3d electrons, the overall energy for the six electrons is 6 x -59.03 eV = -354.17 eV (Eq. 1.2.32).
Equation 1.2.32 Equation for the electron energies of the six 3d electrons in an iron atom
The energy of a single 4s electron is E=-(3.752)/42 ×13.6 eV=-11.95 eV. For two 4s electrons the energy is 2 x -11.95 eV= -23.90 eV.
Equation 1.2.33 Equation for the electron energies of the 4s electron in iron
The sum of all electrons is then -354.17 eV + (-23.90 eV) = -378.1 eV (Eq. 1.2.34).
Equation 1.2.34 Equation for the sum of all electrons
Now let us calculate the energy for the second electron configuration. First we need this electron configuration in Slater configuration. Next, we can calculate the shielding constant for the d electrons. It is σ3d = 7(0.35) + 18(1.00) = 20.45 because there are seven other 3d electrons in the same group and 18 electrons in lower shells. The effective nuclear charge is then Zeff (3d) = 26- 20.45 = 5.55.
Equation 1.2.35 Equation for the shielding constant and effective nuclear charge of the 3d electrons of iron (second configuration)
The sum of the electron energies for the eight d electrons is then E(Σ 3d)=-8(5.552/32 ) eV=-372.3 eV, Eq. 1.2.36.
Equation 1.2.36 Equation for the sum of the electron energies for the eight d electrons of iron
We can can see the energy of the second electron configuration is somewhat higher, and thus less preferred. This is in accordance with the experimentally observed ground state electron configuration of the Fe atom. Again, we see that the Slater rules can correctly predict the ground state electron configuration of atoms.
The Aufbau Principle and the Spin Pairing Energy
The Slater rules are suitable to account for shielding effects in multi-electron atoms which are of electrostatic nature. However, not only electrostatic effects influence the orbital energies, there are also magnetic effects. This is because each electron has a spin with a magnetic field associated with it. The electrons behave like little magnets that can interact with each other. These interactions can be either attractive or repulsive, which influences the electron energies. The electron energies are usually minimized when the number of electrons with the same spin is maximized. This is known as Hund’s (Fig. 1.2.49) rule of maximum spin multiplicity.
In the previously example we have seen that the electron configuration 4s2 3d6 is favored in Fe because of electrostatic shielding effects. In addition, it also favored by magnetic effects. In this electron configuration there are four unpaired electrons, while in the electron configuration 3d8 there are only two unpaired electrons.
We can now understand the Aufbau principle for multi-electron atoms (Figure 1.2.51).
For H and He the 1s orbital gets filled. From lithium to beryllium the 2s subshell gets filled, and from boron to neon the 2p subshell gets filled under consideration of Hund's rule. You can see that for carbon the two 2p electrons are both spin up in different 2p orbitals. For nitrogen there are three unpaired electrons in the 2p orbitals. From the element oxygen on, we have to start pairing spins until all spins are paired in neon. From Na to Ar the 3s and the 3p orbitals get filled according to the same principles.
After that, the 4s orbitals of K and Ca get filled, and then the 3d orbitals of the elements from Sc to Zn (Fig. 1.2.52). We have seen previously that the Slater rules nicely show why. You can see that there are two anomalies though. For the chromium, the electron configuration is 4s1 3d5 and not 4s2 3d4. This is because there are more unpaired spins in this electron configuration, and a half-filled 3d subshell with unpaired spins only represents a particularly stable electron configuration. The second anomaly occurs for the element Cu which has an electron configuration 4s13d10 and not 4s23d9. This is because a d10 electron configuration is a particular stable electron configuration. After the 3d subshell is filled, the 4p subshell is filled for the elements from Ga to Kr. Again, the electrons are filled into the orbitals according to Hund’s rule. We could extend our consideration to even heavier elements, but will be beyond scope.
Dr. Kai Landskron (Lehigh University). If you like this textbook, please consider to make a donation to support the author's research at Lehigh University: Click Here to Donate. | textbooks/chem/Inorganic_Chemistry/Inorganic_Coordination_Chemistry_(Landskron)/01%3A_Atomic_Structure/1.02%3A_The_Schrodinger_Equation.txt |
The Periodic Table of Elements
Having understood the Aufbau principle we can now understand the periodic table of the elements. The periodic table was actually invented a long time before the quantum-mechanical atom model was developed. It was first introduced by Mendeleev and Meyer who ordered the elements according to their masses in a row. Whenever an element had a property similar to an element that already was previously considered a new row was begun. This gave a table with rows and columns. Within a row the mass of the elements would increase, within a column one would find the elements that had similar properties. The rows would be called periods, and the columns would be called groups, hence the name periodic table of the elements. At the time Mendeleev and Meyer invented the periodic table, there were still many elements undiscovered visible as “holes” within the periodic table. However, over time the periodic table became complete (Fig. 1.3.1).
Today we know 118 elements completing seven periods and 18 groups. Today, the elements are not ordered by mass, but by the atomic number (number of protons), however this did not change the original periodic table much because except for a very few elements the mass of the elements follows the atomic number. Before the quantum-mechanical model was developed, it was not understood why elements within the same group had similar properties. When you analyze the electron configuration of the atoms you can see that all atoms in the same group have the same type of outmost electrons, also called valence electrons. These electrons are the electrons that have the greatest distance from the nucleus and have the highest energies. Because they have the highest energies, they are also the electrons that are most reactive, and determine the chemical and physical properties of the element. For example, in the first group all elements have one s electron as the valence electron. Because these electrons have the highest energy and are most reactive, they determine the properties of the atom, and thus the elements in the first group have similar properties.
When you look into group 2, all the electrons have two s electrons as the valence electrons. An exception is the element helium which is placed into group 18 which is called the group of the noble gases. The other group 18 elements have two s and six p electrons of the same quantum number n as valence electrons. The behavior of He is much more similar to that of the other group 18 elements, namely it is a gas with extremely low reactivity. This property is due to the fact that like the other noble gases He has a full shell. Elements with a full shell are particularly unreactive because full shells represent particularly stable electron configurations, also called noble gas configurations.
Group 1, group 2 and He are called the s-block of the periodic table because they are the elements that only have s electrons as valence electrons. From group 3 to group 12 there are elements with s and d electrons as valence electrons. These elements are the d-block of the periodic table. The d-block begins with group 3 where there is only one d electron together with two s electrons, and ends with group 12 where there are ten d and two s electrons. In group 12 the d subshell is completely filled with electrons. From group 13 to group 18 we find electrons that have p electrons as valence electrons. There are six groups because p orbitals can accommodate up to six p electrons. In group 13 there is only one p valence electron, in group 14 there are two, in group 15 there are three, in group 16 four, in group 17 five, and in group 18 there a six except the Helium which we previously discussed already. Group 13 to 18 is called the p block of the periodic table. The s-block and the p block elements together are also called the main group elements, the elements of the d-block are called the transition metals. The elements of the main groups have specific names. Group 1 elements except hydrogen are called the alkali metals, group 2 elements the earth alkaline metals, group 13 elements the triels, group 14 elements the tetrels, group 15 elements the pnictides, group 16 elements are the chalcogens, group 17 elements the halogens, and group 18 elements the noble gases.
Finally, there is the f block of the periodic table. The f-block contains the elements with f-valence electrons. Because there are seven f-orbitals for a given quantum number n, the f orbitals can accommodate up to 14 electrons, and thus there are 14 groups. The f block is actually located in between the s block and the d block of the periodic table, but is typically written underneath the rest of the periodic table. This is done because the periodic table would become too wide if it was placed between the s and the d block. Note that the f-block elements do not have a group number. Note also that the elements on the left end of the f-block, lanthanum and actinium, actually do not have f-valence electrons. They are actually group 3 elements located underneath the elements scandium and yttrium. The first row of the f-block is are called the lanthanides, which is the greek meaning the elements following the lanthanum. Only the elements following the lanthanum have f-valence electrons. Analogously, the actinium is not an actual f-block element, only the elements that follow the actinium. They are called the actinides.
The quantum mechanical model of the atoms also explains why the periodic table has periods. Each period is associated with a quantum number n, the first period with n=1, second period with n=2 and so forth. Within a period the available s and p valence orbitals of a specific quantum number n are getting filled. Subshells of a lower quantum number may also be filled within the period. For example, the 3d subshell gets filled within the fourth period, and the 4f subshell gets filled in the 6th period.
Elements with electron configurations that deviate from the expected electron configuration are highlighted in bold (Figure 1.3.1). There are only exceptions for the d-block and f-block elements. We discussed some of them already, for example the Cr and the Cu. In addition, for instance, Pd has the electron configuration 4d10 instead of 5s24d8, and Pt has the electron configuration 6s15d9 instead of 6s25d8. We can explain these exceptions when considering the complex electrostatic and magnetic interactions of electrons in multi-electron atoms. Not everything can be explained by the simple concepts of Slater rules and Hund’s rule.
As chemists we should always know our periodic table. Learning is easier with music, and there is a fabulous song by Tom Lehrer that can help you to learn the periodic table. The link to a respective YouTube video is provided below.
https://www.youtube.com/watch?v=DYW50F42ss8
Atomic Radii
Let us now look at what is called the periodic properties of atoms. These are properties which change due to the position of the element within the periodic table. Let us first consider how the atomic radii of the atoms depend on the position of the atom in the periodic table, Fig. 1.3.2.
When you look at how the atomic radii change within a period you can easily see that they decrease from the left to the right in the periodic table. The decrease is more significant for the s and p block elements compared to the d block elements. We can explain this effect using the concept of effective nuclear charge. As we go from the left to the right in the periodic table we add electrons to the same shell. Because also a proton is added to the nucleus with each electron added, the electron experiences a greater attractive force from the nucleus. This effect tends to increase the effective nuclear charge. However, there is also an opposite effect. Because there is one more electron, there are more electron-electron repulsion and shielding effects, and this tends to decrease the effective nuclear charge. Because the new electron is added to the same shell, the shielding effects do not fully compensate for the additional Coulomb force coming from the additional proton in the nucleus, and thus the effective nuclear charge increases. This higher effective nuclear charge pulls the electrons closer to the nucleus, and thus the atom size decreases. For the d-block elements new d electrons get added to a more inner subshell, for example for the d-block elements of the 4th period the d electrons get added to the 3d subshell. Because of that, these 3d electrons can more effectively shield nuclear charge from the outmost two 4s electrons, and thus the decrease of the atomic radius is less pronounced. In summary:
Trend 1: Atomic radius decreases continuously within a period.
A second trend is that each time a new period is begun the atomic radius jumps dramatically. For example it jumps from 31 pm to 167 pm as we go from the helium to the lithium, and from 38 pm to 190 pm as you go from the neon to the sodium. This trend is because the electrons get filled into a new, more outer shell when a new period is begun. This outer shell is located much farther away from the nucleus, and thus the radius of the atom increases. In addition, the electrons in the inner shells can very effectively shield the nuclear charge from the electron added to the new shell which tends to increase atom radius. In summary:
Trend 2: Atomic radius jumps when a new period is begun.
As a third trend you can see that the atomic radii increase as you go down a group. This is because within a group there are the same number and type of valence electrons, but the valence electrons are in a more outer shell associated with a higher quantum number n. In a somewhat different view we can also argue that the first two trends add up to give the third trend. The effect of jump of atomic radius when a new period is begun is greater than the effect of decline of the radius within a period. As a result, the radius tends to increase as you go down a group. In summary:
Trend 3: Atomic radius increases continuously within a group.
The determination of an atomic radius is not as trivial at it may seem. You can obtain it either from quantum mechanical calculations, or determine it experimentally. If you calculate it, then the radius is defined as the radius that defines the volume in which the electron can be observed with 90% probability. The radii that you see in Figure 1.3.2 are actually calculated radii. If you determine the radius experimentally, then there are again different subtypes of radii. The so-called metallic, or crystal radii, the vanderWaals radii, and the covalent radii. The covalent radius is the half of the distance between two, same atoms in a molecule held together in a single covalent bond.
This is unambiguous for molecules such as Cl2, the other halogens, and for other cases such as hydrogen, carbon (as diamond), sulfur, and a few other cases. However for oxygen, O2, the situation is less clear as the order of the oxygen-oxygen bond is 2. In this case, it is necessary to infer the covalent radius from molecules containing O-O single bonds or from molecules containing an E-X single bond (E=element) in which the covalent radius of X is known. The van der Waals radius is half the distance between two same atoms in a crystal when there is no actual bonding between the atom except weak van der Waals forces. For example, in solid argon, there are only van der Waals forces between the atoms, and half the distance between the atoms in solid argon is the Van der Waals radius. For chlorine, for example, the van der Waals radius is half the distance between two chlorine atoms of two different Cl2 molecules in solid Cl2. Note that the van der Waals radius of chlorine is different than the covalent radius of chlorine. This again shows that there is a certain ambiguity when determining an atomic radius, and therefore it is always important to say what definition of atomic radius you use. Lastly, there is also the metallic radius that applies to solid metals in which metal atoms are held together by metallic bonding. The metallic radius is the defined as half the distance between the centers of two metal atoms within a metal.
First Ionization Energy
The first ionization energy is another important periodic property of the elements. It is defined as the energy required to remove an electron from a neutral atom in the gas phase. It is important to understand that we refer to an atom in the gas phase here, because only in the gas phase atoms do not significantly interact with each other. Therefore, only when looking at an atom in the gas phase we can truly determine the properties of single, isolated atoms. Let us look at the two images (Figure 1.3.4) to determine periodic trends.
First of all we need to realize that all 1st ionization energies are positive. This means that it always requires energy to remove an electron from a neutral atom. You can understand this when you consider that all electrons in an atom are bound to the nucleus via a binding energy. If an ionization energy was negative then this would mean that the atom would spontaneously lose an electron, and spontaneously ionize. This would be unreasonable.
Now let us see how the first ionization energy changes within a period. We can see that it increases as we go from the left to the right within a period. This trend can be explained by the fact that the effective nuclear charge increases from the left to the right within a period. As the effective nuclear charge increases, the pull of the nucleus on the outmost electron increases, and thus the harder it is to remove this electron from the atom. You can see, however, that within the main group elements the trend is two times broken. The group 13 elements have a lower ionization energy than the group 2 elements, and the group 16 elements have a lower ionization energy than the group 15 elements. This phenomenon can be explained by the fact that the filled and half-filled subshells in group 2 and group 15 represent particularly stable electron configurations, and thus the electrons within these subshells have unusually low energy. For that reason these electrons are harder to remove from the atoms. The group 2 elements have filled s subshells, and the group 15 elements have half-filled p subshells. You can also notice that the increase in ionization energy is less pronounced within the d-block, and even less pronounced with in the f-block. This is because the effective nuclear charge on the outmost electrons does not increase as much because inner d and f orbitals are getting filled. You can also see that the group 13 elements have a lower ionization energy than the group 12 elements. This is because in group 13 a new p subshell is begun and the new electrons get added to the outmost shell where shielding effects are smaller.
As a second trend you can see that whenever a new period is begun, the ionization energy drastically drops. This is because the new electron is added to a new, more outer shell where the effective nuclear charge acting on the new electron is much smaller. The third trend is that the ionization energy becomes smaller as we go down a group. This is because the effective nuclear charge on the valence electrons decreases because the quantum number n of the valence electrons increases. There is one peculiarity that needs additional explanation. The drop of ionization energy from hydrogen to lithium in the first group is unusually large. This is because hydrogen is the only element for which there are no shielding effects, simply because the hydrogen has only one electron. For that reason the valence 1s electron of the hydrogen is far harder to remove than the valence s electrons of the other group 1 elements.
Electron Affinity
Now let us consider the first electron affinity (Figure 1.3.5). It is defined as the energy required or released when you add an electron to a neutral atom in the gas phase. Again, we look at an atom in the gas phase, because we want to consider an isolated atom that does not make significant interactions with other atoms.
We can see from the graph that most elements have negative electron affinities, meaning that the addition of a free electron to a neutral atom is exothermic, and releases energy, however, this is not always the case. Some elements have negative electron affinities, in particular the noble gases, but also Be, N, Mg, Zn, Cd, and Hg. Zn, Cd, and Hg, are group 12 elements with a full d subshell which is particularly stable. An additional electron would need to be added to a p orbital of a higher shell which is energetically quite unfavorable. Similarly, Be and Mg have filled s subshells which are also fairly stable. Adding an additional electron would need to start an new subshell, which is energetically not favorable. You can see that Ca and Sr have electron affinities of about 0, and that of Ba is only slightly negative showing that higher periods make the addition of an electron to a new p shell slightly more favorable. Nitrogen has a half-filled p subshell, which is also a quite stable electron configuration, and therefore adding an electron is not favorable. The addition of an electron is slightly more favorable for the other group 15 elements, however, these also tend to have rather low electron affinities.
Generally, for the main group elements the electron affinity tends to increase from the group 1 to group 17. However, the trend is broken for group 2 and group 15 elements because these elements have full and half-filled subshells, respectively. In addition, it is noteworthy that group 1 elements have higher electron affinities than group 3 elements. This is because adding an electron to a group 1 element produces a full s subshell which is fairly stable. For the d-block elements, the electron affinity tends to increase from group 3 to group 11, but the trend is broken multiple times showing that each element would need to be investigated individually which is beyond the scope here. There is a big drop in electron affinity from group 17 to group 18 which is easily explained by the fact that the addition of an electron to a group 17 element produces a filled shell, while a group 18 element already has a full shell, and the addition of an electron would start a new shell. Similarly, there is a sharp drop in electron affinity from group 11 to group 12, as an addition of an electron to a group 11 atom produces a full subshell, while the addition of an electron to a group 12 element would require the start of new, more outer subshell. For the p block elements the electron affinity tends to first increase from period 2 to 3, and then decrease. For the s-block there is a steady decrease down a group. For d block elements there is no clear trend for the electron affinity.
Electronegativity
Lastly, let us study the electronegativity as a periodic property. The electronegativity is the measure of the ability of an atom to attract electrons within a chemical bond. It is related to the electron affinity in the sense that both period properties measure the force by which an electron is attracted to a neutral atom. However, the electron affinity is associated with isolated atoms, that do not make bonds to neighbored atoms, while the electronegativity refers to atoms that are bonded to neighbored atoms. The importance of the electronegativity stems largely from its ability to predict and understand the nature of the bonding between atoms, in particular, covalent, ionic, and metallic bonding.
Electronegativity Scales
Electronegativity was a concept first developed by Linus Pauling to describe the relative polarity of bonds and molecules. He argued that the electronegativity of an atom could be derived from bond energy differences between homoleptic and heteroleptic bonds (Eq. 1.3.1).
Equation 1.3.1 Electronegativity according to Pauling, eV = electron volts.
For example, the energy required to break the bond in H2 is 432 kJ mol-1, and for F2 it is 159 kJ mol-1. However the energy required to break a bond in HF is 565 kJ mol-1, which is much higher than expected just by averaging the energy of the two homonuclear bonds (296 kJ mol-1). Pauling argued that the difference could be assigned to electrostatic attractions between the F and H “ends” of the molecule whereby the F end has more electron density and the H end has less electron density. The greater the difference between the average energy of the homoleptic bonds and the heteroleptic bonds, the greater the electronegativity difference between the atoms, and the greater the polarity of the heteroleptic bond.
First, Pauling used the arithmetic means of the heteroleptic bond energies, later he used the geometric means, because he found empirically that it worked better (Equation 1.3.1). The geometric mean of the homoleptic bond energies is the square root of the product of the homoleptic bond energies. This method only provided electronegativity differences. He therefore needed to define a reference atom with an arbitrarily defined electronegativity value, and then determine the electronegativity values of all other atoms relative to that value. He chose the most electronegative atom, the fluorine atom, as the standard and assigned a value of 4.0. The values of all other atoms vary between 0.7 (Fr) and 3.44 (O), and are shown in the periodic table on the below (Fig. 1.3.6).
What are the periodic trends? One can see that within a period the electronegativity values of the main group elements strictly increase with the group number. Note though that the noble gases do not have a Pauling electronegativity because, with few exceptions, noble gases do not make compounds and thus no bond energies are available for them. We can also see that for main group elements the electronegativity strictly decreases down a group. This makes the fluorine the most electronegative atom, and the cesium the least electronegative atom. Cesium has an electronegativity value of 0.7. We ignore the radioactive francium here. For the d-block, the trends are less strict, but there is a tendency of electronegativity increase from group 3 to group 11. All group 12 elements have smaller electronegativities compared to their neighbored group 11 elements. What about trends within a group? In group 3 to group 5 there is a small decrease in electronegativity down a group. For group 6 and 9, the period 5 elements have higher values than their neighbored elements in period 4 and 6. For group 10 to 12 the electronegativity increases down a group. The d-block element with the highest electronegativity is the gold, it has a value of 2.4. Note that this is similar to the electronegativity of non-metals such as I, S, and P. For these reasons, sometimes gold can behave like a non-metal in compounds with very low electronegativity. For example, there is the compound cesium auride (CsAu) which is a transparent, ionic crystalline compound due to the high electronegativity difference between cesium and gold. The f-block elements generally have low electronegativity.
The Pauling electronegativity scale is derived from empirical data on bond energies. It works very well in practice and it is up to date the most used electronegativity scale. However, it does not to relate electronegativity to other periodic properties, and the quantum-mechanical model of the atoms. One electronegativity scale that addresses this short-coming the Allred-Rochow scale. It relates the electronegativity to the Coulomb force that acts on an electron on the surface of an atom. The Coulomb force is proportional to the effective nuclear charge Z* and inverse proportional to the atomic radius square. Z*/r2 is multiplied with a factor of 3590 and a value of 0.744 is added to this term. These numbers are empirically chosen so that the values of the Allred-Rochow scale become comparable to the Pauling scale.
Equation 1.3.2 Allred-Rochow definition of electronegativity
Another frequently used scale is the Mulliken electronegativity scale. The Mulliken scale relates the electronegativity to the sum of the first ionization energy IE and the first electron affinity EA. We would intuitively agree that the ability of an atom to attract electrons within a chemical bond is the higher, the harder it is to remove an electron from an isolated atom, and the easier it is to add an electron to an isolated atom. The numbers 0.118 and -0.207 are empirically chosen to make comparisons to the Pauling scale possible. The Mulliken scale is related to the Allred-Rochow scale in the sense that you can explain ionization energies and electron affinities of atoms with the concept of the effective nuclear charge.
Equation 1.3.3 Mulliken definition of electronegativity
Another electronegativity scale has been developed by Leland C. Allen. It designed for main group elements in particular. It argues that the electronegativity is proportional to the average energy of the s and p valence electrons in the atom. This electronegativity concept also has a relationship to that of Allred-Rochow, because orbital energies can be calculated from the effective nuclear charge.
Equation 1.3.4 Leland C. Allen definition of electronegativity
Overall, you can see that the electronegativity scales are inter-related, they say essentially the same, but present electronegativity from a somewhat different perspective.
Electronegativity of the Elements and Chemical Bonding
One of the most powerful attributes of electronegativity is that it can predict what type of chemical bond is to expect in elements and compounds. Let us first look at the bonding between elements.
In the depicted periodic table (Figure 1.3.7) you can see the type of chemical bonding between atoms indicated by different colors. There is metallic bonding in metals, indicated by the color blue. In metallic bonds the electrons are shared between atoms but are delocalized over many atoms in the metal. The green color indicates covalent bonding seen in non-metals. In covalent bonding the electrons are shared, but localized in between typically two atoms. Metalloids have chemical bonding with a mix of covalent and metallic character shown in orange. There is electron-sharing with a moderate degree of delocalization. The Pauling electronegativity values of the elements are written underneath the element symbols.
Can we relate the electronegativity values to the chemical bonding in the elements? We can clearly see that elements with high electronegativity values significantly above 2.0, the non-metals, tend to make covalent bonds in between them. All metalloids with a hybrid covalent-metallic character, have intermediate electronegativity around 2.0. Most metals have low electronegativity values below 2.0 except noble metals like Pt and Au. Overall, we can say that there is a clear relationship between the bonding type in an element and its electronegativity. Now let us see if electronegativity can also predict chemical bonding in compounds.
Ketelaar's Triangle
The ability of electronegativity to predict the bonding type in compounds can be understood by Ketelaar’s triangle, named after J.A.A. Ketelaar.
You can see that Ketelaar’s triangle has three corners (Figure 1.3.8). Two of the corners are occupied by the elements F and Cs, and the third corner is occupied by CsF. All elements are located on the horizontal edge of the triangle, and there many compound indicated by dots either on the two other edges or within the triangle. There are no compounds or elements outside of the triangle. How can we understand this triangle and what does it say about the relationship between bonding character and electronegativity? To understand this, look at the two axes. The horizontal x-axis represents the average electronegativity of the atoms in an element or a compound. The vertical y-axis represents the electronegativity difference between the atoms in the elements or the compound. For elements the electronegativity difference between the atoms is zero, because in an element all atoms are of the same type. Therefore, all the elements lie on the horizontally oriented edge of the triangle. Because Cs is the element with the lowest electronegativity, it lies furthest to the left on this edge. Fluorine is the element with the highest electronegativity, and thus lies the furthest on the right side. The other elements are in located in between the cesium and the fluorine on the edge. The higher the electronegativity of the elements, the further right their position on this edge of the triangle.
For compounds, the electronegativity difference between atoms is never exactly zero, therefore all compounds are located above the horizontal edge of the triangle. The compound with the highest electronegativity difference is the CsF. Its average electronegativity is the sum the electronegativity of Cs and F divided by 2: Therefore, CsF defines the third corner of the triangle. All other compounds must lie on the edges or within the triangle. All cesium compounds are located on the edge between the Cs and CsF, and all the fluorine compounds are located on the edge between CsF and F2. The compounds of all other elements are inside the triangle. The position of the compound on or in the triangle defines its bonding character. The closer the position of the compound toward the Cs corner, the more metallic, the closer the position is to the F2 corner, the more covalent, and the closer it is to the CsF corner, the more ionic. For example Li2O is located close to CsF, and the bonding would be predominantly ionic, although there is also a small degree of covalent and metallic bonding. In contrast to that, SrMg has mostly metallic bonding, a little bit of covalent bonding, and even less ionic bonding. The bonding situation in LiH is in between metallic and ionic, with a little bit of covalent bonding character mixed in. Note that a 100% ionic bond is not possible because the electronegativity difference is finite, and thus there must always be a certain degree of electron sharing. In contrast to that, a 100% covalent bond is possible because electrons can be equally shared between two atoms when the electronegativity difference is zero.
The take home message is that in compounds and elements there is usually a mix of bonding types, and not a single bonding type, even through one bonding type may strongly dominate. The concept of electronegativity, and Ketelaar’s triangle in particular, is extremely helpful to predict to which degree the three bonding types are present in a substance.
Dr. Kai Landskron (Lehigh University). If you like this textbook, please consider to make a donation to support the author's research at Lehigh University: Click Here to Donate. | textbooks/chem/Inorganic_Chemistry/Inorganic_Coordination_Chemistry_(Landskron)/01%3A_Atomic_Structure/1.03%3A_Periodic_Properties_of_Atoms.txt |
Concept Review Questions
Section 1
1. Which were the elements in the antique non-atomistic element theories?
2. Explain the definition of the term “atom” as introduced by Leukippes and Demokrites.
3. What does the law of the conservation of mass state?
4. Who discovered the law of the conservation of mass?
5. What obscured the discovery of the law of the conservation of mass for a long time?
6. Explain why the discovery of the law of the conservation of mass was important for the development of atomic theory.
7. What does the law of the constant compositions state?
8. Who discovered the law of the constant compositions?
9. What does the law of the multiple proportions state?
10. Who discovered the law of the multiple proportions?
11. Explain why the law of the multiple proportions can be explained by Dalton’s hypothesis.
12. Explain what cathode rays are and how they are being produced.
13. Explain how Thomson identified electrons in cathode rays.
14. Explain why Rutherford’s gold foil experiment disproves Thomson’s atom model.
15. Describe Rutherford’s atom model and explain why it is in agreement with his gold foil experiment.
16. Which two opposing forces keep electrons stable in the orbit around the nucleus according to Rutherford’s atom model?
17. What are the two basic problems with Rutherford’s atom model?
18. What is blackbody radiation?
19. What assumption did Max Planck make to explain the spectra of blackbody radiation?
20. Objects first start to glow red, then, orange, and finally white with increasing temperature. Explain why.
21. What is meant by the “Balmer series”, the “Paschen series” and the “Lyman series”?
22. Explain why the atomic absorption and emission spectra inspired the Bohr atom model?
23. What is the meaning of an angular momentum in physics?
24. Name the four postulates of Niels Bohr that lead to his atom model.
25. Mathematically derive the terms for the allowed radii and the allowed energies for the electron in an H atom according to Bohr.
26. Explain why the results of the above calculations explain the atomic spectra of the H atom.
27. What are the two major problems of the Bohr atom model?
Section 2
1. What is the photoelectric effect?
2. Explain how Einstein verified Planck’s law by investigating the photoelectric effect.
3. Explain why the wave-particle dualism can explain the photoelectric effect.
4. What is the name of the particle associated with electromagnetic radiation?
5. Derive the formula for the wave-particle dualism of electromagnetic radiation.
6. Give a definition for standing matter waves.
7. Explain why standing matter waves are quantized.
8. Derive the Schroedinger equation for the electron in the one-dimensional box.
9. What are the boundary conditions for the wavefunctions describing the electron in a one-dimensional box.
10. Derive the wavefunction for the electron in the one-dimensional box.
11. Explain qualitatively what are the major similarities and differences for an electron in a one-dimensional box compared to an electron in a hydrogen atom.
12. What are the quantum numbers in the solution of the Schroedinger equation for the hydrogen atom?
13. The radial probability function is the product of the formula for the surface of a sphere multiplied with the square of the radial function. Explain why this product gives the probability to find the electron at a certain distance (radius) from the nucleus.
14. At which radius r has the radial probability function of the 1s orbital its maximum?
15. What is a spherical node in an orbital?
16. What is an angular node in an orbital?
17. What types of angular nodes do we know?
18. Can an s orbitals have angular nodes?
19. Can p and d orbitals have spherical nodes?
20. Name an orbital that has a conical node.
21. What are the formulas for the number of radial, angular, and total number of nodes in an atom?
22. On which quantum number do the orbital energies in the H atom depend on according to the wavemechanical model of the H atom?
23. The energy of an electron in the ground state of the hydrogen is -13.6 eV. Explain why this energy is negative.
24. What are the energies of the electron in the hydrogen atom when the electron is in a 2p and 4f orbital respectively?
25. Explain the concept of shielding in multi-electron atoms.
26. Explain why the energy of a 2s orbital in Li is smaller than that of a 2p orbital.
27. What are the Slater rules for s,p,d, and f electrons respectively?
28. What is special about the electron configurations of Cr and Cu?
29. Explain the concept of the spin-pairing energy.
30. What does Hund’s rule state?
Section 3
1. Write down the periodic table of the elements for the elements with the atomic numbers 1 to 86. Include element symbols only.
2. Listen to Tom Lehrer’s “The Elements” song. Relax.
3. Which are the three periodic trends for the atomic radii of the elements of the periodic table? Explain these trends.
4. What is the definition of the ionization energy, the electron affinity, and the electronegativity?
5. Which are the three periodic trends for the first ionization energies of the elements of the periodic table? Explain these trends INCLUDING their exceptions.
6. The electron affinities of the group 2, 12, and 18 elements as well as the electron affinity of N are negative. Explain.
7. Alkali metals have a relatively high electron affinity. Explain.
8. Explain how Pauling derived his electronegativity scale.
9. What the periodic trends for the electronegativity of the elements of the periodic table?
10. Explain why CsAu is an ionic crystal.
11. Explain why electronegativity differences are related to the dissociation energy differences of homonuclear and heteronuclear bonds.
12. Explain the concepts of the Allred-Rochow, Mulliken, and Allen electronegativity scale.
13. Which of the four electronegativity scales we talked about in class cannot make statements about the electronegativity of noble gases like Ne or He. Why?
14. We combine two elements of low electronegativity to form a compound. Which bonding type would you expect for this compound?
15. If two elements with high electronegativity are combined, which bonding results then?
16. If two elements of very different electronegativities are combined, what is the bonding in the compound?
17. Explain the concept of Ketalaar’s triangle.
18. Can there be any compounds outside Ketelaar’s triangle?
Dr. Kai Landskron (Lehigh University). If you like this textbook, please consider to make a donation to support the author's research at Lehigh University: Click Here to Donate. | textbooks/chem/Inorganic_Chemistry/Inorganic_Coordination_Chemistry_(Landskron)/01%3A_Atomic_Structure/Concept_Review_Questions_Chapter_1.txt |
Homework Problems
Section 1
Exercise 1
Cobalt forms two sulfides, CoS and Co2 S3. Predict the ratio of the masses of sulfur that combine with a fixed mass of cobalt to form CoS and Co2S3
Answer
In Co2S3 there are 1.5 S atoms per Co atom
In CoS there is 1 S atom per Co atom
Mass ratio of S in the two compounds is 1:1.5 = 2:3
Exercise 2
What is true about the Rutherford atom model?
1. It explains why atoms do not send out electromagnetic radiation permanently
2. It explains why alpha particles get scattered by atoms
3. It explains the atomic spectrum of the hydrogen atom
4. It explains the wave-particle dualism
Answer
b) It explains why alpha particles get scattered by atoms
Exercise 3
The Bohr radius (the radius of the electron orbit for the H atom in its ground state) is 5.29 x 10-11 m. Calculate the radius of an electron in the third shell of the H atom according to the Bohr atom model.
Answer
r3 = 32 x 5.29 x 10-11 m = 47.61 x 10-11 m.
Section 2
Exercise 1
The electron in an H atom undergoes an electronic transition from the 3rd to the 2nd shell. What frequency does the light that is emitted have? The energy of the electron in the first shell is -2.18 x 10-18 J.
Answer
Energy of electron in the 3rd shell: E3 = -2.18 x 10-18 J / 32 = -0.24 x 10-18 J
Energy of electron in the 2nd shell: E2 = -2.18 x 10-18 J / 22 = -0.545 x 10-18 J
Energy difference between the two electrons: E3-E2 = 0.305 x 10-18 J
Frequency of emitted light: n=(E3-E2)/h = 0.305 x 10-18 J/ 6.63 x 10-34 Js = 4.60 x 1014 s-
Exercise 2
What is the mass of a photon with a wavelength of 400 nm that travels though space?
Answer
λ = h/mc --> m = h/λc = 6.63 x 10-34Js / (400 x 10-9 m x 3.00 x 108 m/s) = 5.525 x 10-36 kg.
Exercise 3
Two objects are moving at the same speed. Which (if any) of the following statements are true?
1. The DeBroglie wavelength of the heavier object is longer than that of the lighter one.
2. If one object has twice as much mass as the other, its wavelength is one-half of the other
3. Doubling the speed of one of the objects will have the same effect on its wavelength as doubling its mass.
Answer
a) If one object has twice as much mass as the other, its wavelength is one-half of the other
b) Doubling the speed of one of the objects will have the same effect on its wavelength as doubling its mass
Exercise 4
The power of a red laser with a wavelength of 630 nm) is 1.00 Watt (1.00 Js). How many photons per second does the laser emit?
Answer
Exercise 5
Which of the following waves would you consider to be standing matter wave?
1. The vibration of a drum.
2. Sound traveling through open air.
3. A tsunami.
4. None of the above.
Answer
a) The vibration of a drum.
Exercise 6
Assume an electron travels in a one-dimensional box (as discussed in class) of the length of 1 m. Look up relevant constants in the internet or a suitable textbook.
1. What is the wavelength of the associated standing matter wave in the ground state?
2. What is the velocity of the electron in the ground state?
3. What is the energy of the electron in the ground state?
Answer
a)
b)
c)
Exercise 7
An electron in the first excited state travels with a velocity of 15 m/s within a one-dimensional box. What is the length of the box?
Answer
In the first excited state n =2. For n = the length of the box is equal to the wavelength. Therefore the length of the box is 4.9 x 10-5 m.
Exercise 8
What are orbitals (more than one answer can be correct)?
1. Wave functions that describe the electrons as three-dimensional standing matter waves in an atom.
2. Spaces inside an atom in which the electron travels as a classical particle.
3. Solutions of the Schrödinger equation for the hydrogen atom.
Answer
a) Wave functions that describe the electrons as three-dimensional standing matter waves in an atom.
c) Solutions of the Schrödinger equation for the hydrogen atom.
Exercise 9
Which quantum numbers l are allowed when the quantum number n is 4?
Answer
l can be 3,2,1,0
Exercise 10
Which quantum numbers m are allowed when the quantum number l is 3?
Answer
-3,-2,-1,0,+1,+2,+3
Exercise 11
What is true about the following wave function?
$\Psi=\frac{1}{\sqrt{2 \pi}} \quad \frac{\sqrt{6}}{2} \cos \theta \quad \frac{4}{81 \sqrt{6} a_0 3 / 2}\left[6-\frac{r}{a_0}\right] \frac{r}{a_0} e^{-r / 3 a_0} \nonumber$
1. Its does not have angular nodes
2. Its does not have spherical nodes
3. Its amplitude is 0 at the nucleus
4. The wave function represents an s orbital
Answer
c) Its amplitude is 0 at the nucleus
Exercise 12
The angular part of the wavefunction of an orbital has the following form:
$\Theta \Phi(x, y, z)=\frac{1}{4} \sqrt{\frac{15}{\pi}} \frac{\left(x^2-y^2\right)}{r^2} \nonumber$
Which planes are the planar nodes in this orbital?
Answer
The wavefunction is 0 for x=y and x=-y. So it is two planes that bisect the x and the y axis.
Exercise 13
Order the following orbitals with respect to their penetration abilities:
4s, 4p, 4d, 4f
Answer
4s > 4p > 4d > 4f
Exercise 14
What is the effective nuclear charge on an electron in an He+ ion?
Answer
Exercise 15
Calculate the orbital energy of a 3p electron in a sulfur atom using the Slater rules.
Answer
(1s2) (2s2 2p6) (3s2 3p4)
Exercise 16
Calculate the orbital energy of a 3d electron in a palladium atom using the Slater rules.
Answer
(1s2) (2s2 2p6) (3s2 3p6)(3d10)(4s24p6)(4d8)(5s2)
Exercise 17
Calculate how much higher the first ionization energy of an oxygen atom is compared to a fluorine atom. Use the Slater rules to answer the question.
Answer
O: (1s2 ) (2s2 2p4)
Dr. Kai Landskron (Lehigh University). If you like this textbook, please consider to make a donation to support the author's research at Lehigh University: Click Here to Donate. | textbooks/chem/Inorganic_Chemistry/Inorganic_Coordination_Chemistry_(Landskron)/01%3A_Atomic_Structure/Homework_Problems_Chapter_1.txt |
Dr. Kai Landskron (Lehigh University). If you like this textbook, please consider to make a donation to support the author's research at Lehigh University: Click Here to Donate.
02: Symmetry and Group Theory
Symmetry in Chemistry
Symmetry is actually a concept of mathematics and not of chemistry. However, symmetry, and the underlying mathematical theory for symmetry, group theory, are of tremendous importance in chemistry because they can be applied to many chemistry problems. For example it helps us to classify the structures of molecules and crystals, understand chemical bonding, predict vibrational spectra, and determine the optical activity of compounds. We will therefore first discuss the general foundations of symmetry and group theory, and then apply them to chemical problems, in particular chemical bonding.
Let us first find a definition for symmetry. Symmetry is very familiar to us as we associate symmetry with beauty, but very familiar things are not necessarily easy to define scientifically. One common definition is that symmetry is the self-similarity of an object. The more similar parts it has the more symmetric it appears. For example, we would argue that the two wings of the butterfly depicted look similar. If the left wing was very different from the right wing the butterfly would look less symmetric.
How can we measure the self-similarity, or symmetry of an object quantitatively? We can do this using the concept of the symmetry operation. It is defined as a movement of an object into an equivalent indistinguishable orientation. The number and kind of symmetry operations that can be carried out defines the symmetry of the object.
Definition: Symmetry Operation
Movement of an object into an equivalent indistinguishable orientation
Symmetry operations are carried out around so-called symmetry elements. A symmetry element is a point, line, or plane about which a symmetry operation is carried out. Let us understand know what symmetry elements and operations exist.
Definition: Symmetry Element
A point, line or plane about which a symmetry operation is carried out
The Identity Operation (E)
The most simple operation is the identity operation. It can be denoted by the Schoenflies symbol E. Schoenflies symbols are the most common symbols to denote a symmetry operation. The identity operation says that each object is self-similar to itself when you do not move it in any way. This is a trivial statement, but as we will see later, the identity operation is necessary to make the mathematical framework of symmetry, group theory, complete. The identity operation is present in any object. In the example of the depicted snail shell it is the only operation (Fig. 2.1.2).
The Proper Rotation Operation
The proper rotation operation is a counter-clockwise rotation about a proper rotational axis by an angle of 360° over an integer number n. After that rotation the object must be indistinguishable from its original form. That means that the object after the rotation must superimpose the original object before the rotation. For example, when n=4, then we rotate around 90°, and after that the object must superimpose its original form. The proper rotational axis is the symmetry element associated with the proper rotation operation. It’s Schoenflies symbol is Cn, whereby n is called the order of the rotational axis. A proper rotational operation has the symbol Cnm, whereby m counts the number of times the operation is carried out. Overall we rotate by and angle of 360°× m/n when we carry out an operation Cnm. This means that when m=n, then we have rotated around 360° (Fig. 2.1.3). Then all points in the object are at their original position. It is as though we had done nothing with the object. We can also say we have reached the identity E. In mathematical form we can say that Cnn=E. If we rotated one more time, Cn(n+1), then this would be equal to rotating only one time, and thus Cn(n+1)=Cn1. For example if n=4, then rotating four times around 90° will produce the identity. We have rotated around 4 x 90°=360° which is the same a though we had not rotated at all, because all points in the object are in their original position after a 360° rotation. If we rotated 5 times around 90°, it would the same as rotating only one time.
If an object has several axes with different order n, then the one with the highest order is called the principal axis. If there is more than one axis of the same order, then they get distinguished by primes if they are not conjugate. We will learn about the exact definition of conjugation somewhat later, we can however often see by inspection if two axes are conjugate. This is usually the case when they pass through the object in a equivalent way, and rotate the object in an equivalent way. Axes that pass through less many bonds, get less many primes. This is just a convention, but you have to follow it. An additional rule is that an axis which is in the same position as the principal axis gets the least number of primes. For example, a square planar molecule such as PtCl42- has a C4 principal axis standing perpendicular to the square plane of the molecule. There is an additional C2 axis where the C4 axis runs. It is present because one can also rotate around 180°, and not only around 90°. This C2 axis does not get any prime. You an see that there are four additional C2 axes, two of them are denoted C2’, and two others are denoted C2’’. You can see that the two axes which only have one prime pass through two Pt-Cl bonds, while the ones that have two primes, do not pass through any bonds. The two C2’ axes are conjugate, meaning that they transform the object in an equivalent way. The two C2’’ axes are also conjugate.
We can now think about how we can write out the axes and their associated proper rotational operations in a systematic way. Let us first look at the symmetry elements, the proper axes: We conventionally write the principal axis first, and then all other axes from their highest to their lowest order. When there are axes of the same order, those with the least number of primes get denoted first, and those with the highest number of primes last. If there are conjugate axes then their number is placed in front of their Schoenflies symbol. For the proper axes of the PtCl42- the notation would therefore be: C4, C2, 2C2', 2C2’’. Now let us see how to denote the rotation operations that are associated with these symmetry elements. The notation follows the same rules as for the symmetry elements. In addition, we have to consider that we must not count identical operations twice, we also do not denote the operations that are the same as the identity. For the C4 axis there are four operations until we reach the identity. C41, C42, C43, and C44. For each of the C2, C2’, and C2’’ elements there are two operations until the identity is reached. C21, C22, C21’, C22’, C21’’, C22’’. The C44, the C22, the C22’, and the C22’’are the same as the identity and therefore we do not consider them. In addition we can see that the C42 is the same as the C21. This is because the C4 and the C2 axes are in the same location, and rotating two times around 90° is the same as rotating one time around 180°. By convention we eliminate the operation associated with the higher order, thus the C42. The overall notation would then be: C41, C43, C21, 2 C21’ 2 C21’’ (Fig. 2.1.5).
The Reflection Operations (σ)
Let us look at reflection operations which are carried out around reflection planes, or mirror planes. Mirror planes have the Schoenflies symbol σ. When we carry out a reflection operation, then we move any point of the object to the other side of the mirror plane. There are two types of mirror planes, so-called horizontal mirror planes and vertical mirror planes. A horizontal mirror plane always stands perpendicular to the principal axis. For example in the depicted BH3 molecule there is a horizontal mirror plane that stands perpendicular to a C3 principal axis (Fig. 2.1.6). A horizontal mirror plane has the Schoenflies symbol σh.
A vertical mirror plane has the property that it contains the principal axis, this means that it is part of the principal axis. It is denoted σv. The BH3 molecule has three vertical axes that pass through the three B-H bonds. You can see that each of them contain the principal C3 axis (Fig. 2.1.7). The three vertical mirror planes are conjugate, and therefore they are not distinguished by primes. We can write a coefficient 3 in front of the symbol σv to indicate that there are three conjugate vertical mirror planes.
Now let us look at how many symmetry operations are associated with a particular mirror plane. Fortunately, things are simple here: There is only one reflection operation associated with one mirror plane. This because reflecting two times at a mirror plane produces the identity E: σv2=E. More generally, when we reflect n times, and n is an even number then this is the same as the identity or, σvn=E (n is even). When n is odd the reflecting n times is the same a reflecting only one time or, σvnv1 (n=odd).
The horizontal mirror plane of the BH3 molecule deserves an additional comment. Carrying out the horizontal reflection does not change the position of any atom. It is important to understand why the operation does exist despite the fact it does not change the position of any atom. The criterion is not whether the position of an atom is changed, but whether the position of the points in the object changes. In the case of the BH3 molecule the part of the molecule that is located above the mirror plane will be located below the mirror plane after the reflection operation has been carried out. Vice versa any part of the molecule that was formerly below the mirror plane will be located above the mirror plane after the reflection operation has been carried out. For example the lower half of the B atom with be above the mirror plane, and the half above the plane will be below the plane after the execution of the reflection operation.
Like non-conjugate proper rotations are distinguished by primes, also non-conjugate vertical mirror planes must be distinguished by primes. The smaller the number of bonds the vertical mirror plane contains, the larger the number of primes.
For example in the water molecule (Fig. 2.1.8) there are two non-conjugate vertical mirror planes. One contains the two O-H bonds, the other stands perpendicular to the first mirror plane. It is easy to see that these two mirror planes will not move the points in the atom in an equivalent way, and therefore they are not conjugate. The first mirror plane does not change the position of any atom, while the second one swaps the positions of the two hydrogen atoms. Therefore, the second mirror plane not containing any O-H bonds gets one prime the one that contains the O-H bonds doe not get a prime.
Dihedral Reflection Planes (σd)
A special case of a vertical mirror plane is a dihedral mirror plane, denoted σd. A dihedral mirror plane bisects the angle between two conjugate C2 axes.
For example, in the PtCl42- anion (FIg. 2.1.9) there are two vertical mirror planes that bisect the angle between the two conjugate C2 axes that pass through the Pt-Cl bonds. Therefore, these mirror planes are vertical mirror planes σd.
The Inversion Operation (i)
Let us look next at the inversion operation which is symbolized by a Schoenflies symbol i. The symmetry element associated with an inversion, is the inversion center, also called center of symmetry. It is a single point. When an inversion operation is performed, then each point of the object is moved through the inversion center to the other side. Each coordinate in the object (x,y,z) is inverted into the coordinates (-x,-y,-z).
For example, the octahedral molecule SF6 has an inversion center in the center of the molecule (Fig. 2.1.10). When the inversion operation is carried out, then each fluorine atom is moved through the inversion center to the other side. This means that the fluorine atoms 1 and 2 swap up their positions, the fluorine atoms 3 and 5 swap up their positions, and so do the fluorine atoms 4 and 6. The sulfur atom does not change its position. There is only one inversion operation associated with each inversion center. Inverting two times, or more generally, an integer number of two times produces the identity. Inverting an odd number of times is the same as inverting one time.
The Rotation-Reflection Operation (Sn)
The rotation-reflection operation Sn is the most complex symmetry operation. It is carried out in two steps. First, a rotation around an improper axis is carried out. The angle is determined by the order n of the improper axis, and is 360°/n. This axis is called improper, because the object does not need to superimpose the original object after the rotation. Achieving superposition requires the second step which is the reflection at a mirror plane that stands perpendicular to the improper axis. Only after the second step the operation is complete. The presence of the rotation-reflection does not require a proper rotational axis or a regular mirror plane σ to exist, however it also do not preclude their existence.
An example of a molecule with an improper axis is the methane molecule (Fig. 2.1.11). It has an S4 improper axis. The axis bisects the H-C-H tetrahedral bond angle. The order of the axis is four which requires that we rotate by 90° around this axis. You can see that after we carry out the rotation, the molecule does not superimpose the original molecule. Only after we reflect the rotated molecule at a mirror plane standing perpendicular to the improper axis the molecule superimposes the original molecule.
Properties of the Rotation-Reflection Operation (Sn)
The rotation-reflection has a number of interesting properties. One of the them is that an S1 operation is the same as a reflection. This is because the order 1 implies a rotation around 360° which produces the identity, and all points within the object are in their original position. This is the same as though we had not rotated at all. This means actually we only did the second step, the reflection, and therefore the S1 is identical to a “regular” reflection. The second property is that an S2 operation is the same as an inversion. When you rotate around 180° and then reflect perpendicular to the improper axis of rotation, the positions of the points in the object change exactly the same way as they do when you invert through an inversion center.
Look for example at the SF6 molecule again (Fig. 2.1.12) which has an inversion center in the center of the atom. We previously saw that when we carry out the inversion, the atoms 1 and 2 swap their position, and so do the atoms 3 and 5, as well as the atoms 4 and 6. Let us carry out the rotation-reflection, and see if the atoms change the same way. Firstly we rotate 180° around an axis that goes through the atoms 1 and 2. This leaves the positions of the atoms 1 and 2 unchanged, but swaps up the positions of the atoms 3 and 5, as well as the atoms 4 and 6. Next we must do a reflection at a plane that stands perpendicular to the improper axis. It is the plane defined by the atoms 3, 4, 5, and 6. Reflection at this plane does not change the positions of the atoms 3, 4, 5, and 6, but it swaps up the positions of the atoms 1 and 2, which lie above and below the plane, respectively. We can see that the positions of the atoms are the same as after the inversion.
The fact that we can express a reflection by an S1 rotation-reflection, and an inversion by a S2 rotation-reflection means that the reflection and the inversion are not independent symmetry operations, and we would not need them. The symmetry of an object could be fully described by the identity, proper rotations, and rotation-reflections. However, by convention, we use reflections and inversions instead of S1 and S2, simply because it is easier for the human mind to perform 1-step operations, rather than 2-step operations.
Improper rotations have different properties depending on whether the order of the axis is even or odd. For even orders, the presence of an Sn improper rotational axis implies that there must also be a proper rotational axis with an order n/2. For improper axes of even orders n the identity is produced after n rotation-reflections (Fig. 2.1.13).
For improper axes of odd order n, carrying out the rotation-reflection n times is the same as carrying out a reflection at a horizontal mirror plane. We need to do the rotation-reflection 2n times to reach the identity (Fig. 2.1.4).
Rotation-Reflections of PF5 (an example with odd order)
Let us show these properties using two examples. Let us first look at an example with an improper rotation of odd order. The PF5 molecule has trigonal pyramidal shape with three F atoms in the equatorial plane, and two additional F atoms above and below the plane. The F atoms have been numbered from 1 to 5 as indicated by the subscripts that follow the element symbol (Fig. 2.1.15). The PF5 has an improper rotational axis S3 that stands perpendicular to the equatorial plane going through the P-F1 and P-F2 bonds. Let us carry out the S3 symmetry operations step by step and see how the atoms move.
We would expect that after we carry out the operation six times the identity will be produced. After three rotation-reflections we would expect that the object has been moved the same way a horizontal mirror plane would do. After the first rotation-reflection the atoms F1 and F2 have swapped their positions, and the three F atoms in the equatorial plane have been rotated by 120 degrees counter-clockwise. The F5 atom now occupies the position of the F4, the F4 has been rotated into the position of the F3, and the F3 has been moved into the position of the F5. The second S3 operation again swaps up the F2 and F1 atoms, and rotates the remaining F atoms by 120°. Now the F5 atoms points toward us, the F4 atom points away from us, and the F3 atoms is in the paper plane. Carrying out the operation a third time again swaps the F1 and F2 and rotates F3, F4, and F5 counter-clockwise by 120°. Now let us compare the position of the atoms with the atom positions of the molecule we started with. We see that the position of the equatorial F atoms are same as in the beginning, but the position of the axial F1 and F2 atoms have been swapped up. This is equivalent to a reflection at a horizontal mirror plane located within the equatorial plane. This horizontal mirror plane would only swap up the axial F atoms, but would not move the equatorial ones. The S34 operation again swaps up the position of the F1 and the F2 atoms, and rotates the equatorial F atoms by 120° counter-clockwise. Now, F4 points toward us, F3 points away, and F5 is in the paper plane. After the fifth rotation-reflection F1 and F2 are again swapped so that F1 is below and F2 is above the equatorial plane. The atoms F3, F4 and F5 are rotated so that F3 is in the paper plane, F5 points toward us, and F4 points away from us. The sixth rotation-reflection again swaps up F1 and F2, and rotates the other F atoms around 120° counter-clockwise. We can now see that the produced molecule is the same as the one we started with, and the identity has been produced.
It is also noteworthy that the S32 and the S34 operation can be expressed by simpler operations, namely C32 and C31 proper rotation operations. We can understand this considering that an S32 operation requires that we reflect two times, and reflecting two times is just like not reflecting at all. So effectively, we only have two rotations around 120° which is equivalent to C32. Similarly, the S34 operation requires to reflect four times, which is the same as not reflecting at all. Thus, effectively, we only rotate four times around 120°. This is the same as rotating only one time around this angle, which is equivalent to a C31 operation.
Using similar considerations, we can also understand why the S33 operation is equivalent to a σh. In this case, we rotate 3 x 120° = 360°, and rotating around 360° is the same as not rotating at all. In addition, we carry out a reflection three times. Reflecting an odd number of times is the same as reflecting only one time. Thus overall, we effectively carry out a single reflection only.
Overall, only the S31 and the S35 operations are unique, all others can be expressed by other, simpler operations.
Rotation-Reflections of CH4(an example with even order)
Let us illustrate the properties of an improper rotational axis with even order using the example methane (Fig. 2.1.16).
We previously saw that the methane molecule has an S4 improper axis. Executing the symmetry operation once moves the H atoms 1 and 2 to the right, whereby atom 1 points to the back and atom 2 points to the front. After the second rotation-reflection the atoms 1 and 2 are reflected back to the left side, but because we have rotated two times by 90° atom 2 points up, and atom 1 points down. Similarly, atoms 3 and 4 are reflected back to the right side, but atom four points toward us and atom 3 points to the back. After three rotation-reflections, the atoms 1 and 2 are again on the right side with atom 2 pointing to the back and atom 1 pointing to the front. Atoms 3 and 4 are on the left with atom 4 pointing up, and atom 3 pointing downward. After the operation is carried out four times, all atoms are back in their original position, meaning the identity has been produced. This is what we expected for an improper rotation of even order. We can see in addition that the S42 operation is the same as an C2 operation. Rotating two times about 90° is the same as rotating about 180°. Reflecting two times is the same as not reflecting at all. Therefore, effectively, we only rotated by 180° which is the same as what a C21 operation does. Because an S42 can be expressed by a C21, and an S44 is the same as the identity, only the S41 and the S43 are unique symmetry operations.
The Symmetry of a Molecule
We have now discussed all types of symmetry elements and operations that can exist in an object. Next, let us think about how we can define the overall symmetry of an object. The overall symmetry of an object is defined as the sum of all the symmetry operations for this object. If two objects have exactly the same symmetry elements and operations then their symmetry is the same. Using the mathematical language of group theory, the mathematical theory for symmetry, we can say they belong to the same point group. The name point group comes from the fact, that it has at least one invariant point. The number of symmetry operations belonging to a point group is called the order of the point group. Let us see next what different point groups we know and how we can systematically classify them.
Dr. Kai Landskron (Lehigh University). If you like this textbook, please consider to make a donation to support the author's research at Lehigh University: Click Here to Donate. | textbooks/chem/Inorganic_Chemistry/Inorganic_Coordination_Chemistry_(Landskron)/02%3A_Symmetry_and_Group_Theory/2.01%3A_Symmetry_Elements_and_Operations.txt |
The Low Symmetry Point Groups
C1 Point Group
Overall, we divide point groups into three major categories: High symmetry point groups, low symmetry point groups, dihedral point groups, and rotational point groups. Let us begin with the low symmetry point groups. As the name says, these point groups only have few symmetry elements and operations. The point group C1 is the point group with the lowest symmetry. Molecules that belong to this point group only have the identity as symmetry element.
An example is the bromochlorofluromethane molecule (Fig. 2.2.1). It has no symmetry element, but the identity. The name C1 comes from the symmetry element C1. A C1 operation is the same as the identity.
Cs Point Group
The point group Cs has a mirror plane in a addition to the identity. An example is the 1,2-bromochloroethene molecule (Fig. 2.2.2).
This is a planar molecule and the mirror plane is within the plane of the molecule. This mirror plane does not move any atoms when the reflection operation is carried out, nonetheless it exists because any point of the molecule above the mirror plane will be found below the mirror plane after the execution of the operation. Vice versa, any point below the mirror plane will be above the mirror plane. This mirror plane does not have a vertical or horizontal mirror plane designation because no proper rotational axes exist.
Ci Point Group
The point group Ci has the inversion as the only symmetry element besides the identity. The point group Ci is sometimes also called S2 because an S2 improper rotation-reflection is the same as an inversion. An example is the 1,2-dibromo 1,2-dichloro ethane (Fig. 2.2.3).
This molecule looks quite symmetric, but it has inversion center in the middle of the carbon-carbon bond as the only symmetry element. Upon execution of the inversion operation, the two carbons swap up their positions, and so do the two bromine, the two chorine, and the two hydrogen atoms.
The High Symmetry Point Groups
The symmetry elements of the high symmetry point groups can be more easily understood when the properties of platonic solids are understood first. Platonic solids are polyhedra made of regular polygons. In a platonic solid all faces, edges, and vertices (corners) are symmetry-equivalent. We will see that this is a property that can be used to understand the symmetry elements in high symmetry point groups. There are only five possibilities to make platonic solids from regular polygons (Fig. 2.2.4).
The first possibility is to construct a tetrahedron from four regular triangles. The second platonic solid is the octahedron made of eight regular triangles. The third possibility is the icosahedron made of twenty triangles. In addition, six squares can be connected to form a cube, and twelve pentagons can be connected to form a dodecahedron. There are no possibilities to connect other regular polygons like hexagons to make a platonic solid.
The icosahedron is the most complex of all platonic solids. If you would like to see and study an icosahedron from the outside and inside, there is one for study on the playground of the Allentown Cedar Beach Park, in Allentown, Pennsylvania.
The Td Point Group
The tetrahedron, as well as tetrahedral molecules and anions such as CH4 and BF4- belong to the high symmetry point group Td. Let us find the symmetry elements and symmetry operations that belong to the point group Td. First, we should not forget the identity operation, E. Next, it is useful to look for the principal axes.
The tetrahedron has four principal C3 axes (Fig. 2.2.6). It is a property of the high-symmetry point groups that they have more than one principal axis. The C3 axes go through the vertices of the tetrahedron. Because each C3 axis goes through one vertex, there are four vertices, and we know that in a platonic solid all vertices are symmetry-equivalent, we can understand that there are four C3 axes. How many unique C3 operations are associated with these axes? After three rotations around 120° we reach the identity. Therefore C33=E, and we only need to consider the C31 and the C32 rotation about 120 and 240° respectively. Because there are four C3 axes, there are four C31 and four C32 operations and eight C3 operations overall. We can express this by writing the respective numbers as coefficients in front of the Schoenflies symbol for the operations (Fig. 2.2.7).
In addition to the C3 axes there are C2 axes (Fig. 2.2.8).
You can see that a C2 axis goes through two opposite edges in the tetrahedron. Because a tetrahedron has six edges, and each C2 axis go through two edges there are 6/2=3 C2 axes. There is only one C2 symmetry operation per C2 axis because we produce the identity already after two rotations. Therefore there are three C21 operations overall (Fig. 2.2.9).
In addition, the Td point group has S4 improper rotation reflections. Like the C2 axes, they pass through the middle of two opposite edges. This also means that they are superimposing the C2 axes. Because there are six edges, and two S4 axes per edge there are 6/2=3 S4 axes (Fig. 2.2.10).
How many operations are associated with these S4 axes? The order of the axes are even, and therefore we need four S4 operations to produce the identity. The S42 operation is the same as a C21 operation because reflecting two times is equivalent to not reflecting at all, and rotating two times by 90° is the same as rotating about 180°. Therefore overall, only S41 and S43 operations are unique operations. S42 and S44 can be expressed by the simpler operations C21 and E respectively. Because there are 3 S4 axes, there are three S41 and three S43 operations. Overall there are six S4 operations (Fig. 2.2.11).
There are also mirror planes (Fig. 2.2.11). The planes contain a single edge of the tetrahedron, thereby bisecting the tetrahedron. There a six edges in a tetrahedron, and therefore there are 6/1=6 mirror planes.
These planes are dihedral planes because each plane contains a C3 principal axis and is bisects the angle between two C2 axes. Overall, there are three C2 axes and three C2 operations. There is one reflection operation per mirror plane because reflecting two times produces the identity. Therefore, there are six σd reflection operations (Fig. 2.2.12, right and Fig. 2.2.13).
In sum (Fig.2.2.14 and 2.2.15) we can denote the overall symmetry of the Td point group the following way: E, 8C3, 3C2, 6S4, 6σd. In detail the unique symmetry operations are E, 4C31, 4C32, 3C21, 3S41, 3S43, 6σd.
The Rotational Subgroup T
The high symmetry point group T is the so-called rotational subgroup of the point group Td. A rotational subgroup is a point group in which all symmetry operations but the identity and the proper rotations have been removed from a high-symmetry point group. For the point group T this leaves the 4C3, the four C32 and the three C2 operations (Fig. 2.2.17). The S4 rotation-reflections and the mirror planes have been removed. The point group T is rare.
An example is the depicted Ga4L6 cage (Fig. 2.2.16). The Ga atoms occupy the vertices of a tetrahedron, but the point group is not Td but T because of the shape of the ligands that connect the four Ga atoms.
The Octahedral Point Group Oh
Another high symmetry point group is the point group Oh. Both the octahedron as well as the cube belong to this point group despite their very different shape (Fig. 2.2.18). Because they belong to the same point group they must have the same symmetry elements and operations. There are many octahedrally shaped molecules, such as the SF6.
Molecules with cubic shapes are far less common, because a cubic shape often leads to significant strain in the molecule. An example is cubane C8H8. Let us determine the symmetry elements and operations for the point group Oh using the example of the octahedron. If we used the cube, we would get exactly the same results.
There are three C4 principal axes in the octahedron. They go through two opposite vertices of the octahedron (Fig. 2.2.19). There are three C4 axes because an octahedron has six vertices which are all symmetry-equivalent because the octahedron is a platonic solid.
We can see that there are also C2 axes where the C4 axes run. This is because rotating two times around 90° is the same as rotating around 180°. What are the symmetry operations associated with these symmetry elements? Rotating four times around 90° using the C4 axes produces the identity. So we have to consider the operations C41, C42, C43 and C44. How many of these are unique? C44 is the same as the identity, so it is not unique, In addition a C42 is identical to a C21, and thus C42 is also not unique, and can be expressed by the simpler operation C21. That leaves the C41 and the C43 as the only unique symmetry operations. Because we have three C4 axes, there are 2x3=6 C4 operations, in detail there are 3C41 and three C43 operations. In addition, there are the three C21 operations belonging the the three C2 axes (Fig. 2.2.20).
In addition, there are four C3 axes (Fig. 2.2.21). They are going through the center of two opposite triangular faces of the octahedron.
You see above a single C3 axis, and on the right hand side all four of these axes. How can we understand that there are four axes? An octahedron has overall eight triangular faces, and each C3 axis goes through two opposite faces, so there are 8/2=4 C3 axes. Each C3 axis has the C31 and the C32 as unique symmetry operations. The C33 is the same as the identity. So overall we have 4x2=8 operations, four of them are C31, and four of them are C32 (Fig. 2.2.22).
In addition to the C2 axes that superimpose the C4 axes, there are C2’ axes which go though two opposite edges of the octahedron (Fig. 2.2.23). How many of them are there? An octahedron has twelve edges, and because each C2’ passes through two edges, there must be 12/2=6 C2’ axes. These axes have primes because they are not conjugate to the C2 axes that superimpose the C4 axes. For each C2’ axis there is only the C21 as the unique symmetry operation, and therefore there are overall 6 C21 symmetry operations (Fig. 2.2.24).
Let us look at the mirror planes next (Fig. 2.2.25). There are horizontal mirror planes that stand perpendicular to the C4 principle axes. You can see a single one of them below on the left.
Note that this mirror plane also contains two axes, in addition to the one to which it stands perpendicular. Because it contains two principal C4 axes, it has also properties of a vertical mirror plane. Nonetheless, we call it a horizontal mirror plane because it stands perpendicular to the third C4. The horizontal properties trump the vertical ones, so to say. You can see that a single mirror plane contains four edges of the octahedron. Because there are twelve edges, there are 12/4=3 horizontal mirror planes. There is one mirror plane per principal C4 axis. There are three horizontal reflection operations because there is always only one reflection operation per mirror plane (Fig. 2.2.26).
Next let us look for vertical mirror planes (Fig. 2.2.27). A vertical mirror plane is depicted below on the left.
You can see that - contrast to the horizontal mirror planes - it does not contain any edges. Rather, it cuts through two opposite edges. You can see that this plane contains a C4 axis, but it does not stand perpendicular to the other two C4 axes. Therefore it has only the properties of a vertical mirror plane. You can see however, that the mirror plane bisects the angle between two C2’ axes which also depicted. This makes the vertical mirror planes dihedral mirror planes, σd. How may of them do we have? As previously mentioned, each mirror plane cuts through two opposite edges. There are twelve edges in an octahedron, and thus there are 12/2=6 dihedral mirror planes. You can see all of them on the right side of Fig. 2.2.27. Each mirror plane is associated with one reflection operation, therefore there are six dihedral reflection operations (Fig. 2.2.28).
Next we can ask if the point group Oh has an inversion center? Yes, there is one in the center of the octahedron (Fig. 2.2.29)!
Each point in the octahedreon can be moved through the inversion center to the other side, and the produced octahedron will superimpose the original one. There is always one inversion operation associated with an inversion center (Fig. 2.2.30).
Next, let us look for rotation-reflections. You can see an S6 rotation-reflection operation below (Fig. 2.2.31, left).
The improper S6 axis passes though the centers of two opposite triangular faces. One can see that rotation about 60° alone does not make the octahedron superimpose. The reflection at a plane perpendicular to the improper axis is required to achieve superposition. Overall, the rotation-reflection swaps up the position of the two opposite triangular faces. How many S6 improper axes are there? Since each S6 passes through two faces, and an octahedron has 8 faces there must be 8/2=4 S4 axes. You can see all of them above (Fig. 2.2.31, right). Note that they are in the same position as the 4C3 axes we previously discussed. How many unique operations are associated with them? For an S6 axis we need to consider operations from S61 to S66. S66 is the same as the identity so it is not unique. The S62 is the same as a C31 because rotating two times round 60° is the same as rotating around 120°, and reflecting twice is the same as not reflecting at all. Similarly, an S64 is the same as an C32. Rotating four time by 60° is the same as rotating two times by 120° and reflecting four times is the same as not reflecting at all. Further, an S63 is the same as an inversion. After three 60° rotations we have rotated by 180°. If we reflect after that, then this is the same as an S21 operation which is the same as an inversion. Therefore, only the S61 and the S65 operations are unique, all other operations can be expressed by simpler operations (Fig. 2.2.32).
The octahedron also has S4 improper axes, and you can see one of them below (Fig. 2.2.33, right).
It goes through two opposite corners of the octahedron. The S4 improper axis seemingly does the same as the C4 axis that goes through the same two opposite vertices, but actually does not. While rotating around 90° already makes the octahedron superimpose with its original form, executing the reflection operation after the rotation swaps up the position of the two vertices, and generally all points of the octahedron above and below the plane, respectively. Overall the S4 moves the points within the object differently compared to the C4 which makes it an additional, unique symmetry element. There are overall three S4 improper axes because the octahedron has six vertices and one S4 passes through two vertices (Fig. 2.2.34).
Here is an overview over all the symmetry elements and operations (Fig. 2.2.35 and 2.2.36). Overall, there are 48 different unique operations that one can perform!
Like the point group Td , also the point group Oh has a rotational subgroup, named O. It has the identity and the same proper rotations as the point group Oh, but no other symmetry operations (Fig. 2.2.38). An example is the polyoxometalate cluster core shown below (Fig. 2.2.37). Polyoxometalates are cluster anions of the group 5 and 6 elements.
The point group O is generally rare.
Another high symmetry point group is the point group Th. It can also be derived from the point group Oh. In this case the S4, the C4, the C2’, and the σd operations are removed from the octahedral symmetry. An example is the hexapyridyl iron (2+) cation (Fig. 2.2.39).
You can see that the N-atoms of the pyridyl-ligands surround the Fe atoms octahedrally, but the symmetry is reduced from Oh to Th because of the planar shape of the pyridyl-ligands. In particular the C4 symmetry is reduced to C2. This reduction in symmetry leads to elimination of the S4, the C2’, and the σd symmetry elements (Fig. 2.2.40).
The Ih Point Group
The two remaining platonic solids, the icosahedron and the dodecahedron, belong both to the icosahedral point group Ih. This is despite they are made of different polygons (Fig. 2.2.41).
Because they belong to the same point group, they have exactly the same symmetry operations. An example for a molecule with icosahedral shape is the molecular anion B12H122-. An example for a molecule with dodecahedral shape is the dodecahedrane C20H20.
Let us determine the symmetry elements and symmetry operations for the example of the icosahedron. We could also use the dodecahedron, and the results would be the same. The principal axes of the icosahedron are the C5 axes. You can see one of them, going through the center of the pnetagon comprised of five triangular faces below (Fig. 2.2.43).
You can understand that there is a C5 when considering that there are five triangular faces making a pentagon. The C5 axis sits in the center of the pentagon. We can see that when we rotate around this C5 axis, then the produced icosahedron superimposes the original one. The C5 axis goes through two opposite vertices of the icosahedron. Because an isosahedron has 12 vertices, there must be six C5 axes overall. You can see all of them below (Fig. 2.2.44).
There are four unique symmetry operations associated with a single C5 axis, namely the C51, the C52, the C53, and the C54. The C55 is the same as the identity. Because there are six C5 axes, there are overall 6x4=24 C5 symmetry operations (Fig. 2.2.45).
In addition, there are C3 axes. One of them is shown below, and you can see that it passes through the centers of two opposite triangular faces (Fig. 2.2.46).
As one rotates by 120° the atoms on the triangular faces change their position, and the resulting icosahedron superimposes the original one. As the name icosahedron says, there are twenty faces overall.
Because one C3 passes through two opposite axes, there are 20/2=10 C3 axes overall (Fig. 2.2.47). Each C3 axis is associated with two symmetry operations, namely C31, and C32. Thus, there are overall 10x2=20 C3 symmetry operations.
There are also C2 axes (Fig. 2.2.49). They pass through the centers of two opposite edges of the icosahedron. Rotating around the C2 axis shown makes the icosahedron superimpose.
An isosahedron has overall 30 edges. Because one C2 axis passes through the centers of two opposite edges, we can understand that there are 30/2=15 C2 axes. There is one unique C2 operation per axis, and therefore there are 15 C2 operations (Fig. 2.2.50).
We have now found all proper rotations. Let us look for mirror planes, next. You can see a mirror plane below (Fig. 2.2.51).
It contains two opposite edges. It also bisects two other edges. An icosahedron has overall 30 edges, therefore there are 30/2=15 mirror planes. You can see all of them below (Fig. 2.2.52 and Fig. 2.2.53)).
The icosahedron also has an inversion center in the center of the icosahedron (Fig. 2.2.54 and Fig. 2.2.55)).
As we carry out the associated, one symmetry operation, all points in the isosahedron move through the inversion center to the other side.
Let us now look for improper rotations. The improper rotational axes with the highest order are S10 axes. They are located in the same position as the C5 axes, and go through two opposite corners (Fig. 2.2.56).
The S10 exists because in an icosahedron there are pairs of co-planar pentagons that are oriented staggered relative to each other. The rotation around 36° brings one pentagon in eclipsed position relative to the other, but superposition is only achieved after the reflection at the mirror plane perpendicular to the rotational axis. Because one S10 passes through two opposite vertices, and there are 12 vertices there are 6 S10 improper axes. For each axis there are four unique symmetry operations, the S101, the S103, the S107, and the S109. Therefore, there are overall 4x6=24 operations possible (Fig. 2.2.57).
Are the lower order improper rotational axes? Yes, there are S6 axes that pass through the centers of two opposite triangular faces (Fig. 2.2.58). This symmetry element exists because the two triangular faces are in staggered orientation to each other. Rotation alone brings one face in eclipsed orientation relative to the other, but reflection at a mirror plane perpendicular to the axis is required to achieve superposition. The S6 axes are in the same location as the C3 axes.
There are 10 S6 axes because there are twenty faces and one axis passes through two opposite faces. Only the S61 and the S65 operations are unique S6 operations, all others can be expressed by simpler operations. Therefore there are overall 10 S61+10 S65 = 20 S6 operations (Fig. 2.2.59).
We have now found all symmetry operations for the Ih symmetry. There are overall 120 operations making the point group Ih the point group with the highest symmetry (Fig. 2.2.60 and Fig. 2.2.61).
Also the point group Ih has a rotational subgroup.
It is called I. An example of an object with this symmetry is the snub-dodecahedron (Fig. 2.2.62). It has the identity, and all the proper rotation operations of the point group Oh, but the inversion, the rotation-reflections, and the mirror planes are eliminated (Fig. 2.2.63).
Cyclic Point Groups
After having discussed high and low symmetry point groups, let us next look at cyclic point groups. They have the property that they have only a single proper n-fold rotational axis, but no other proper axes. In the most simple case they do not have any additional symmetry element such as mirror planes or rotation-reflections. These point groups are denoted Cn whereby n is the order of the proper axis. An example is the hydrogen peroxide molecule H2O2 (Fig. 2.2.64).
It has a so-called roof-structure due to its non-planarity. One hydrogen atom points toward us, and the other points away from us. This structure is due to the two electron-lone pairs at each sp3-hybridized oxygen atom. These electron-lone pairs consume somewhat more space than the H atoms, and there is electrostatic repulsion between the electron lone pairs. Therefore, the electron lone pairs at the different oxygen atoms try to achieve the greatest distance from each other. This forces the H-atoms out of the plane, leading to the roof-structure of the hydrogen peroxide. Because the H2O2 molecule is not planar, it only has a single C2 axis, but no other symmetry element besides the identity. The C2 axis passes through the center of the O-O bond. Execution of the C2 operation swaps up both the O and the H atoms.
Definition: Cyclic Groups Cn
Cyclic groups have one rotational axiPyramidal Groups
Another class of groups are the pyramidal groups, denoted Cnv. They have n vertical mirror planes containing the principal axis Cn in addition to the principal axis Cn. Generally molecules belonging to pyramidal groups are derived from an n-gonal pyramid. An n-gonal pyramid has an n-gonal polygon as the basis which is capped (Fig. 2.2.66).
For example a trigonal pyramid has a triangular basis which is capped, a tetragonal pyramid has a square which is capped, and so on. The proper axis associated with a specific pyramid has the order n and goes through the tip of the pyramid and the center of the polygon. An example of a molecule with a trigonal pyramidal shape is the NH3 (Fig. 2.2.67).
The three H atoms form the triangular basis of the pyramid, which is capped by the N atom. The NH3 molecule belongs to the point group C3v. The C3 axis goes though the N atom which is the tip of the pyramid, and the center of the triangle defined by the H atoms. There are three vertical mirror planes that contain the C3 axis. Each of them goes through an N-H bond (Fig. 2.2.68).
Definition: Pyramidal Groups Cnv
Pyramidal groups have n vertical plane(s) in addition to the principal axis Cn.
The Linear Group C∞v
A special n-gonal polygon is the cone. A cone can be conceived as an n-gonal pyramid with an infinite number n of corners at the base (Fig. 2.2.69).
In this case the order of the rotational axis that passes through the tip of the cone and the center of the circular basis is infinite. This also means that there is an infinite number of vertical mirror planes that contain the C axis (Fig. 2.2.71). The point group describing the symmetry of a cone is called the linear point group C∞v. Polar, linear molecules such as CO, HF, N2O, and HCN belong to this point group. You can see the HCN molecule with its C axis and its infinite number of vertical mirror planes below (Fig. 2.2.70).
The infinite number of mirror planes, shown in blue are forming a cylinder that surround the molecule.
Definition: Linear group Coov
The linear group C∞v has an infinite number of vertical mirror planes containing a Caxis
Reflection Groups
If we add a horizontal mirror plane instead of n vertical mirror planes to a proper rotational axis Cn we arrive at a the reflection point group type Cnh. The presence of the horizontal mirror planes also generates an improper axis of the order n. This is because when one can rotate and reflect perpendicular to the rotational axes independently, then it must also be possible to do it in combination. An example of a molecule belong to a reflection group is the trans-difluorodiazene N2F2 (Fig. 2.2.72).
It is a planar molecule with a C2 axis going through the middle of the N-N double bond, and standing perpendicular to the plane of the molecule. The horizontal mirror plane stands perpendicular to the C2 axis, and is within the plane of the molecule. There is an additional inversion center because an S2 must exist which is the same as an inversion center. The inversion center is in the middle of the N-N bonds. Overall, the molecule has the symmetry C2h.
Definition: Reflection Group Cnh
A reflection group has a horizontal plane perpendicular to the principal axis Cn
Dihedral Groups
Dihedral groups are point groups that have n additional C2 axes that stand perpendicular to the principal axis of the order n. If there are no other symmetry elements, then the point group is of the type Dn.
For example in the point group D3 there is a C3 principal axis, and three additional C2 axes, but no other symmetry element (Fig. 2.2.75). The tris-oxolato ferrate (3-) ion belongs to this point group (Fig. 2.2.75). You can see that the C3 axis stands perpendicular to the paper plane, and there are three C2 axes in the paper plane.
Definition: Dihedral Groups Dn
In a point group of the type Dn there is a principal axis of order n, n C2 axes, but no other symmetry elements.
If a horizontal mirror plane is added to the Cn axis and the n C2 axes we arrive at the prismatic point groups Dnh (Fig. 2.2.76). The addition of the horizontal mirror plane generates further symmetry elements namely an Sn and n vertical mirror planes.
Generally, molecules belonging to this point group derive from n-gonal prisms. The order of the principal axis is the same as the number of corners of the polygons the prism are made of.
Definition: Prismatic Groups Dnh
In prismatic point groups there is a horizontal mirror plane perpendicular to the principal axis Cn. There are also n C2 axes.
An example for a molecule belonging to a prismatic point group is PF5 (Fig. 2.2.77).
It has a trigonal bipyramidal shape. The C3 axis goes through the axial F atoms of the molecule, and the three C2 axes go through the three equatorial F atom. The horizontal mirror plane stands perpendicular to the principal C3 axis and is located within the equatorial plane of the molecule. In addition, there are the vertical mirror planes that contain the C3 axis, and go through the three equatorial P-F bonds. There is also an S3 axis which superimposes the C3 axis. In sum:
A special case of a Dnh group is the linear group D∞h. An object that has this symmetry is a cylinder. A cylinder can be conceived as a prism with an infinite number of vertices. Thus, the principal axis that passes through a cylinder has infinite order. Because of the infinite order of the principal axis, there is an infinite number of C2 axes that stand perpendicular to the principal axis. You can see one such C2 going though the cylinder (Fig. 2.2.79).
There is now also an improper axis of infinite order, as well as an infinite number of vertical mirror planes. Non-polar linear molecules like H2, CO2, and acetylene C2H2 belong to the point group D∞h. You can see the C axis passing through a CO2 molecule below (Fig. 2.2.80).
You can see the infinite number of vertical mirror planes as a blue cylinder. The infinite number of C2 axes is shown a yellow lines going around the molecule. In sum:
Definition: Linear Group Dooh
In the point group D∞h there is an infinite number of n C2 axes in addition to the principal axis of infinite order, an infinite number of vertical mirror planes, and one horizontal mirror plane.
If we add n vertical mirror planes to the principal axis and the n C2 axes, we arrive at the point group Dnd. The vertical mirror planes are dihedral mirror planes because they bisect the angle between the C2 axes. An example is the ethane molecule in staggered conformation which has the symmetry D3d (Fig. 2.2.82).
The C3 axis goes along the C-C bond, and the 3C2 axes pass through the middle of the carbon-carbon bond, and bisect the angle between two hydrogens and one carbon atom. The three dihedral mirror planes pass through the C-H bonds. In addition, the ethane molecule has an S6 axis, and an inversion center. In sum:
Definition: Dnd
In this point group type there are n dihedral mirror planes that contain the Cn and bisect the angle between adjacent C2 axes
Improper Rotational Point Groups
The last class of point groups to be discussed are the improper rotation point groups. The only have one proper rotational axis, and an improper rotational axis that has twice the order of the proper rotational axis (Fig. 2.2.85). There may be an inversion center present depending on the order of the proper and improper axes. An example the tetramethylcycloocta-tetraene molecule (Fig. 2.2.84).
It has an S4 and an C2 axis as the only symmetry elements besides the identity. Rotating by 90° alone does not superimpose the molecule because two C-C double bonds lie above the plane and two below the plane. In addition, two opposite methyl groups lie above and below the plane respectively. Therefore it needs the additional reflection to achieve superposition. There is also a C2 axis which is in the same locations as the S4 axis.
Guide for the Determination of Point Groups
With the knowledge you have, you can unambiguously identify the point group of a molecule. The key to success is that you are able to see the symmetry elements in the molecule. This takes practice. With enough practice you can identify the point group of a molecule immediately. Until, you can use guides, that you can follow to identify a point group. Such a guide asks systematic questions about the presence or absence of a symmetry element. Depending on whether you answer the question with yes or no you can follow the guide in a particular direction. Eventually, after having answered enough questions the guide will lead you to the respective point group. You can see such a chart below (Fig. 2.2.86).
You can first ask if there is at least one Cn present. If not, then the molecule must be in a low symmetry point group. If there is an inversion center, the point group is Ci. If not we can next ask, if there is a mirror plane. If yes it is Cs, and if not the point group is C1. If a low symmetry point group can be ruled out, then we can ask next, if there is a high symmetry point group. This is the case when there are either 4C3, 3C4, or 6C5 rotational axes present, standing for tetrahedral, octahedral, and icosahedral symmetry, respectively. If there is an inversion center in case a C5 is present, the point group is Ih. If not, it is I. Similarly, if there are 3C4 axes, and an inversion center, the point group must be Oh. If there is no inversion center the point group is O. If there are 4C3 axes, the point group must be T type. If there is a horizontal mirror plane in addition, then the point group must be Th. If not, we can ask next is there are dihedral mirror planes. If yes, the point group is Td, otherwise it is T. Now we have checked for all high symmetry point groups.
If a high-symmetry point groups can be ruled out, we ask if there are n C2 axes in addition to the Cn principal axis. If this is so, then we must have a dihedral group of the D type. Next we ask, if there is a horizontal mirror plane. If yes, the point group must be Dnh. If not, we ask if there are dihedral mirror planes. If yes, the point group is Dnd. If there are no mirror planes at all, then the point group is Dn. If there are no n C2 axes in addition to the Cn, the the group must be either a rotational group or an improper rotation group. We next ask, if there is a horizontal mirror plane. If yes, then the point group is of the type Cnh. If not, we ask if there are vertical mirror planes. If yes, then the point group is Cnv. If that is not the case, we ask if there is an S2n in addition to the Cn. If yes, the point group is S2n. If not, it is Cn.
Example: Dibromonaphtalene
Let us practice the point group guide by one example. Let us look at the dibromonaphtalene molecule (Fig. 2.2.87).
The first question we would ask is: Can you see at least one proper rotational axis? The answer is yes. There is a C2 proper rotational axis that stands perpendicular to the plane of the molecule and goes through the center of the C-C bond that is shared by the two aromatic rings. Next we ask: Are the 6C5, 3C4 or 4C3 axes? That is clearly not the case, and thus we do not have a high symmetry point group. Next we can think about if the there are 2C2 axes in addition to the C2 axes we already found. The answer is no, so the point group cannot be a dihedral group. Next, we would ask: Is there a horizontal mirror plane. This is indeed the case. There is a horizontal mirror plane in the plane of the molecule. It does not move any atoms around, but as we discussed before, a mirror does not need to do this to exist. This identifies the point group as C2h.
Chiral Point Groups
A chiral point group is a point group that only has proper rotation operations in addition to the identity. This is equivalent to the statement that no improper rotations must exist in a chiral point group. This includes mirror planes and inversion centers because a mirror plane is the same as an S1, and an inversion center is the same as an S2. If a molecule belongs to a chiral point group, then it has a mirror image that cannot be superimposed with the original molecule. The two mirror images are called enantiomers. An example is the bromochlorofluoromethane. You can see two enantiomers separated by a dotted line (Fig. 2.2.88).
The dotted line represents a mirror plane. Note that this mirror plane is not a mirror plane in the meaning of a symmetry element. You can see that the two molecules to the left and the right of the mirror plane are mirror images respective to each other. The molecules cannot be superimposed meaning that they are enantiomers. Note that the bromochlorofluoromethane molecule on the far right is not an enantiomer to the other two. It the same molecule as the enantiomer on the far left. We only need to rotate clockwise around the C-Cl axis to make the two molecules superimpose meaning that they are the same.
It should be pointed out that the fact that a molecule has a carbon with four different substituents is not sufficient to make it a chiral molecule. The dibromodichloroethane molecule shown above (Fig. 2.2.89) has four different substituents around the carbon atom, but it is not chiral because there is an inversion center in the middle of the C-C bond.
Chiral High Symmetry Point Groups
Chiral groups not necessarily need to have low symmetry, in fact, the high symmetry rotational subgroups I, O, and T are chiral groups because they only have proper axes in addition to the identity (Fig. 2.2.90).
Dr. Kai Landskron (Lehigh University). If you like this textbook, please consider to make a donation to support the author's research at Lehigh University: Click Here to Donate. | textbooks/chem/Inorganic_Chemistry/Inorganic_Coordination_Chemistry_(Landskron)/02%3A_Symmetry_and_Group_Theory/2.02%3A_Point_Groups.txt |
Matrix Representations of Symmetry Operations
Thus, far we have looked at symmetry operations qualitatively, and determined their nature by inspection. Now let us have a look how we can describe symmetry operations from a more mathematical point of view. Why do we do this? We do this to be able to understand so-called character tables which we will later need to apply symmetry to molecular orbital theory. Generally, symmetry operations can be described by matrices. When a matrix representing a symmetry operation is multiplied with coordinates of an object, then this gives the new coordinates of the object after the symmetry operation was carried out. Therefore let us briefly review matrices, and matrix multiplications in particular. A matrix is nothing but an array of numbers arranged into rows and columns. When we multiply to matrices, we must multiply each row with each column of the matrix. When the first matrix Aik has i rows and k columns and the second matrix Bkj has k rows and j columns, then the product matrix Cij has i rows and j columns. A requirement for a matrix multiplication is that the number of columns of the first matrix is the same as the number of the rows of the second column. Otherwise, one can just not multiply each row with each column of the two matrices.
Let us do this by a few examples.
Example 1 of Matrix Multiplication
In the first example we multiply two 2x2 matrices, meaning that they both have two rows and two columns (Fig. 2.3.2). First, we need to check if the two matrices can be multiplied. We can see that the number columns of the first matrix is 2, and the number of rows of the second matrix is also 2. Therefore, we can multiply the two matrices.
The product matrix is expected to be also a 2x2 matrix. Now we need to multiply the first row of the first matrix with the first column of the second matrix, and that means that we to multiply 1x7 and 5x4. The sum of 1x7+5x4=27 gives us the first character of the first row of the product matrix. Next, we multiply the first row of the first matrix with the second column of the second matrix. So we have to multiply 1x3 and 5x8. The sum of the two products gives the second character of the first row of the product matrix. Because the second matrix has no additional column, there is no further column with which the first row of the first matrix could be multiplied. Therefore, we now go to the second row of the first matrix, and multiply it with the first column of the second matrix. This gives the products 2x7 and 6x4. When added together, then this gives 38, and this is now the first character of the second row of the product matrix. Lastly, we multiply the second row of the first matrix with the second row of the second matrix. This gives 2x3 and 6x8 which gives 54 when added together. This is the second character in the second row of the product matrix.
Example 2 of Matrix Multiplication
In the next example, we multiply a 1x3 matrix with a 3x3 matrix (Fig. 2.3.3). This is possible because the number of the columns of the second matrix is 3, and the number of rows of the second matrix is also 3.
The product matrix would be expected to have one row and three columns. The first matrix has only one row, so we multiply it with the three columns of the second matrix. Multiplying the first row of the first matrix with the first column of the second matrix gives (1)(1) + (2)(0) + (3)(0) = 1. This is the first character in the first, and only row in the product matrix. The multiplication of the first row of the first matrix with the second column of the second matrix is (1)(0) + (2)(-1) + (3)(0) = -2 which is the second character of the first row of the product matrix. The multiplication of the first row of the first matrix with the third column of the second matrix gives (1)(0) + (2)(0) + (3)(1) = 3. This is the third character of the first row of the product matrix. Overall, the product matrix is [1 -2 3].
Example 3 of Matrix Multiplication
In the last example let us multiply a 3x3 matrix with a 3x1 matrix (Fig. 2.3.4). Again, the number of the columns of the first matrix is the same as the number of the rows of the second matrix, they are both 3. Thus, we can multiply the two matrices.
Multiplying the first row of the first matrix with the first, and in this case only column of the second matrix gives 1x1 + 0x2 + 0x3 = 1. This is the first and only character of the first row of the product matrix. The second matrix has only one column, therefore we multiply the second row of the first matrix with the only column of the second matrix in the next step. This gives 0x1 + (-1)x2 + 0x3 = -2. This the character of the second row of the product matrix. Lastly, we multiply the third row of the first matrix with the column of the second matrix. This gives 0x1 + 0x2 + 1x3 = 3. This is the character of the third row of the product matrix.
Applying Matrix Multiplications to Symmetry Operations
So what do symmetry operations have to do with the multiplication of matrices? The answer is that a symmetry operation can be described as a matrix, and the multiplication of this matrix with the matrix that represents the coordinates of the position of the points in an object, will give the new coordinates of the object after the symmetry operation has been carried out (Fig. 2.3.5).
Example H2O
A matrix that represents a symmetry operation is a 3x3 matrix and the matrix that describes coordinates is a 3x1 matrix. When two are multiplied then this gives a 3x1 matrix that describes the new coordinates of the object. For example, let us look at the water molecule H2O.
We can define a coordinate system in the water molecule so that the molecule is within the xz plane, whereby the z-axis is chosen so that it bisects the H-O-H bond angle. The y-axis would stand perpendicular and point into the board plane (Fig. 2.3.6). Any point within the water molecule, would have coordinates x,y,z that are defined by vectors , , that point to these coordinates. For example, the center of a particular hydrogen atoms would have specific coordinates that would be described by three vectors that when added up, would point to the center of the hydrogen atom. Now let us think how the coordinates will change as we carry out a specific symmetry operations. For instance, take the C2 symmetry operation that is carried out around the z-axis. How will the rotation change, the x,y, and z coordinates, respectively?
The x-coordinate is defined by a vector pointing into x-direction. As we rotate this vector 180° around the z-axis, it will retain its length but will point into the opposite direction (Fig. 2.3.6). Hence, we can say that the coordinate x has changed its algebraic sign, and is now –x. Now what about the y-coordinate? A vector pointing into y-direction will also be rotated around 180°, and point into the opposite direction. That means it is –y after the execution of the symmetry operation. Finally, how will the z-coordinate change? Because we rotate around z, there will not be any change to the z-vector, and thus the new coordinate z will be identical to the old coordinate z. If we represent the three vectors of the old coordinates by a 3x1 matrix , , , the new coordinates of the matrix ', ', ' are represented by the matrix -,- , . The matrix that when multiplied with the matrix for the old coordinates gives the matrix for the new coordinates, would be the matrix that would represent the symmetry operation C2. In the case of the C2 symmetry operation the matrix has the form below (Fig. 2.3.7).
We can show that this matrix correctly represents the symmetry operation C2 by applying the multiplication rules for matrices. Multiplying the first row of the matrix for C2 with the only column of the matrix for the old coordinates would give (-1 x ) + (0 x ) + (0 x ) = - for the first character of the product matrix. Multiplying the second row of the matrix representation of C2 with the only column of the matrix for the old coordinates gives (0 x ) + (-1 x ) + (0 x ) = - for the second character of the product matrix. Multiplication of the third row of the matrix for C2 with the column of the matrix for the old coordinates gives (0 x ) +(0 x ) + (1 x ) = This proves that the matrix correctly represents the symmetry operation C2.
We can develop the matrix representation for the σ(xz) reflection operation along the same line we did for the C2 operation. Let us first think about how the vectors representing the coordinate changes as the reflection operation is carried out. The vectors and do not change because they are within the xz mirror plane. However, the vector changes its direction as it is reflected to the other side of the mirror plane. Thus, the y-coordinate changes its algebraic sign.
The matrix representing the new coordinates therefore has the form , - , . The matrix that when multiplied with the matrix of the old coordinates, gives the matrix of the new coordinates must be the matrix representing the symmetry operation σxz. It has the form below (Fig. 2.3.10).
We could again show using the multiplication rules for matrices that multiplication of this matrix with the matrix for the old coordinates gives the matrix for the new coordinates (Fig. 2.3.11).
We can apply the same process for the σ(yz) symmetry operation. The σ(yz) mirror plane contains the coordinates y and z, and thus their vectors do not change upon the application of the symmetry operation. However, the vector changes its direction upon reflection at the yz plane, and thus the algebraic sign of the x-coordinate changes (Fig. 2.3.12).
The matrix for the new coordinates is thus -, , . In this case the matrix that when multiplied with the matrix of the old coordinates , , gives the matrix with the new coordinates -, , has the form below (Fig. 2.3.13).
We could again use the multiplication rules for matrices to show that the above matrix is the correct matrix representation of the symmetry operation σ(yz), Fig. 2.3.14.
Lastly, we can also determine the matrix representation of the identity operation (Fig. 2.3.15). It has the form below (Fig. 2.3.16).
Like previously, we could use the multiplication rules to show that this matrix produces the correct matrix for the new coordinates. Since the identity does nothing to an object (Fig. 2.3.15), the new coordinates are the same as the old coordinates (Fig. 2.3.17).
Reducible and Irreducible Representations
Let us look closer at the four matrices that we just derived (Fig. 2.3.18).
It is noteworthy that all characters are zero, except those on a diagonal that goes from the top left corner in the matrix to the bottom right corner. This diagonal is called the trace of the matrix. With good justification we can say that the characters on the trace of the matrix tell us what the symmetry operation does with a coordinate. If it is a -1 the algebraic sign of the coordinate changes, if it is +1, it does not. If we write the characters of the trace of the matrices that belong to a specific coordinate on a line underneath the symmetry operations, we get what is called an irreducible representation for the specific coordinate (Fig. 2.3.19).
For example, for the x-coordinate the characters of the traces of the matrices for the symmetry operations E, C2, σyz, σxz, are 1, -1, -1, and 1, respectively. For the y-coordinate the characters would be 1, -1, 1, and -1, and for the z- coordinate they are 1, 1, 1, and 1. The use of an irreducible representation is that it tells us directly in a concise form what the symmetry operations do to a specific coordinate. The sum of two or more irreducible representations is a so-called reducible representation. If we sum up the three irreducible representations of Fig. 2.3.19, then this gives a reducible representation with the characters 1+1+1=3, (-1)+(-1)+1=-1, (-1)+1+1=1, and 1+(-1)+1=1. We will see about the use of reducible representations in a little bit.
Symmetry Types of Irreducible Representations
The specific characters in an irreducible representation determine the symmetry type of the irreducible representation. It is denoted by a capital letter with subscripts and/or superscripts. For example the irreducible representation for the x-coordinate is of the type B1 (Fig. 2.3.20)
B means that the symmetry is anti-symmetric with regard to the principal axis (Fig. 2.3.21).
Anti-symmetric means that the algebraic sign of the coordinate changes as we rotate. In this case the principal axis is the C2 axis. We can see that the algebraic sign changes because the character of the irreducible representation for the x-coordinate underneath C2 is -1. The subscript 1 means that the representation is symmetric to a C2 perpendicular to the principal axis, or if lacking to a vertical σv (Figure 2.3.22)
In this case, we do not have a C2 perpendicular to the C2 which is our principal axis, but we have two vertical mirror planes. Symmetric means that the coordinate does not change its algebraic sign as the symmetry operation is carried out. We can see that this is true for the reflection with the lower number of primes which comes first in our considerations. It is indicated by the character +1 underneath the symmetry operation.
The irreducible representation of the y-coordinate has the symmetry type B2 (Fig.2.3.20) The symmetry type is again B, because the character underneath C2 is -1. The subscript is 2 in this case, and this means that the representation is antisymmetric to a C2 perpendicular to the principal axis or, if lacking to a vertical plane σv (Fig. 2.3.23)
We can see that the σv(xz) is anti-symmetric as indicated by the character -1 in the irreducible representation. Lastly, let us look at the symmetry type of the irreducible representation for the z-coordinate which is A1 (Fig. 2.3.20) A means symmetric with regard to rotation around the principle axis (Fig. 2.3.24).
We can confirm the symmetric situation for the rotation around z by verifying that the character underneath the principal axis C2 is +1.
Symmetry Types of Orbitals
Irreducible representations are more powerful than only telling you the symmetry type of a specific coordinate in a point group. More generally, they can tell the symmetry type of a mathematical function in a specific point group. Remember, that orbitals are mathematically wave functions. Thus, it should be possible to assign orbitals to a symmetry type within a given point group. Let us look for example, at the 2s and the 2p orbitals of the oxygen atom in the water molecule. Let us consider the 2s orbital first, and determine what the symmetry operations do with it. We can see that no symmetry operation changes the 2s orbital in any way, thus all characters of the irreducible representation that belongs to the orbital should be +1. We can see that this is only the case for the irreducible representation with the symmetry type A1. Therefore we can say that the 2s orbital of the O-atom in the water molecule has the symmetry type A1 (Fig. 2.3.25).
Next, let us determine the symmetry type of the 2pz orbital. The 2pz orbital is oriented along the z-axis around which we rotate. When carrying a C2 operation we can see that this operation does not make any changes to the orbital. The same is true for the two reflection operations. Again, we can see that the symmetry operations do not change the 2pz orbital in any way, and therefore it also must belong to the symmetry type A1 (Fig. 2.3.26)
For the 2px orbital the situation is different though (Fig. 2.3.27). We can see that rotating around the C2 axis changes the algebraic sign of the orbital. This means that the 2px orbital should belong to a symmetry type which has a character of -1 for the C2 operation. We can see that this can be the B1 or the B2 symmetry type, but not A1. This rules out the A1 symmetry type. We still need to decide if the symmetry type is B1 or B2. For B1 reflection at the σv(xz) mirror plane would need to be symmetric, and for B2 it would need to be anti-symmetric. We can see that the orbital does not change when we carry out the σv (xz) reflection, and thus the symmetry type must be B1.
Lastly let us determine the symmetry type of the 2py orbital (Fig. 2.3.28). The 2py orbital is oriented perpendicular to the paper plane, the blue lobe points to the front, and the orange one, hardly visible points to the back. We can see that when we rotate around 180° the orange lobe points to the front, and the blue one points to the back. That means that the wave function of the orbital has changed its algebraic sign. Therefore, it must be of B symmetry type. Which one is it? B1 or B2? We can see that in this case the σv(xz) does change the algebraic sign of the wave function because the front lobe of the orbital gets reflected to the back, and the lobe in the back gets reflected to the front. Thus, it is antisymmetric with respect to σv(xz) and thus it must belong to the symmetry type B2.
We could have determined the symmetry types of the orbitals also from a more mathematical perspective. Remember when we discussed the atomic orbitals we saw that the wave function of a pz orbital is a linear function of the z-coordinate (Fig. 2.3.29). Therefore it has the same symmetry type as the z coordinate, namely A1. A py orbital is a linear function of y, therefore it has the same symmetry type as the coordinate y: B2. A px orbital is only a function of x, therefore its symmetry type is that of the coordinate x: B1. The 2s orbital is not a function of any coordinate. Because of that it must belong to the symmetry type in which all characters are +1 which is the A1 symmetry type in the point group C2v.
By the same means we can also determine the symmetry type of the d orbitals of the O atom in H2O. They are not occupied, and not involved in the bonding, but nonetheless they are possible states for the electrons. Let us determine them using the table below (Fig. 2.3.30)
Does the 3dz2 – orbital change when the symmetry operations are carried out (Fig. 2.3.31)? No, it does not! Therefore, it has the symmetry type A1.
What about the 3dxz? The 3dxz orbital changes its algebraic sign when the C2 operation is carried out, but not when the reflections are carried out. Therefore, it is of B1 symmetry type (Fig. 2.3.32).
The 3dx2-y2-orbital also does not change when the symmetry operations are executed, thus it is of symmetry type A1 (Fig. 2.3.33)
For the dyz orbital rotation and the σv(xz) reflection changes the algebraic sign, therefore it is B2 (Fig. 2.3.24).
For the 3dxy we can see that rotation does not change the orbital, but the σv(xz) reflection does. This means that the symmetry type must be A2 (Fig. 2.3.35).
This is a new symmetry type which also belongs to the point group C2v. No single coordinate is of this symmetry type, but mathematical functions that are the product of the coordinates x and y, such as the 3dxy orbital are. Similarly, because the 3dz2 orbital is a function of z2 and the 3dz2 has the symmetry type A1 any function of z2 has this symmetry type. Analogously, any function which is the product of the x and the z coordinate belongs to the symmetry type B1, any function which is a function of x2-y2 belongs to the symmetry type A1, and any function which is a product of y and z belongs to the symmetry type B2.
Character Tables
There is only a finite number of irreducible representations and symmetry types in a specific point group.
Generally, all the irreducible representations that are possible in a specific point group define the so-called character table of the point group Fig. 2.3.36).
Definition: Character Table
A complete set of irreducible representations in a point group defines its character table
Character tables describe what the symmetry operations of a point group do with a mathematical function. You can see the complete character table of the point group C2v below (Fig. 2.3.36).
It is made of several rows and columns. In the first column you can find the point group symbol to the very left, and then the symbols for the different symmetry types of the irreducible representations the point group has. Then, there are several columns with symmetry operations in the first row, and the characters for the irreducible representations below. Each of these columns is called a class. Symmetry operations that are conjugate are grouped into the same column, or the same class. For the point group C2v, each operation gets its own column and that means that they are all non-conjugate. But this is not always true, in many other point groups there are several symmetry operations that are conjugate, and listed in the same column. We have already discussed qualitatively that conjugate operations transform the coordinates of an object in a similar way. Now we have a more exact definition which is that whenever the characters for the operations in their irreducible representations are the same, they belong to the same class.
The sum of characters underneath the identity operation E defines the dimension of the point group. For C2v the dimension of the point group is 1+1+1+1=4. The order (h) of the point group is just the sum of all symmetry operations in the point group. For C2v the order is 4. In the two far right columns of the character table you can find the mathematical functions that belong to the different irreducible representations. In the second column from the right linear and rotational functions are listed, and in the far right column square functions are listed. For example, the letter z is in the row associated with the symmetry type A1. This means that any linear function of the z-coordinate has the symmetry type A1. The letters x and y are in rows for the symmetry type B1 and B2 respectively. This means that any linear function of x and y has the symmetry type B1 and B2 respectively.
You can also see the symbols Rz, Rx, and Ry in the same column. These symbols stand for rotational functions around the z, x, and y axis respectively. Rz is in the row of the symmetry type A2, which means that any rotational function of z has the symmetry type A2. How can we understand that? Consider a rotational vector around z that indicates clockwise rotation (Fig. 2.3.37). If we rotate this vector by 180° counter-clockwise it will not change its direction, it will still point clock-wise, and thus the character underneath C2 in the character table should be a 1. This rules out that the symmetry is B type, it must be either A1 or A2 type. We can further see that when we reflect this vector at the two vertical mirror planes, it reverses its direction, now pointing counter-clockwise. Therefore, the characters underneath the two mirror planes should be both -1. This means that the symmetry type must be A2. If we did the same exercise with the Rx and the Ry rotational vectors we could show that they belong to the symmetry types B2 and B1 respectively. In the last column the square functions are listed. The row in which the function is located gives you the information about the symmetry type of the square function. For example any function of x2, y2, or z2 has the symmetry type A1, a function that is a product of x and y has the symmetry type A2 and so forth.
Matrix Representations of Symmetry Operations in C3v
Now let us go from the point group C2v to the point group C3v and determine the matrix representations of the symmetry operations. We will see that for this, and more generally for any point group with rotational axes with an order higher than 2, a symmetry degeneracy is possible, which means that two or even all three coordinates belong to the same symmetry type. The coordinates do not change independently upon the execution of a symmetry operation, they are linked to each other. Let us look at the ammonia molecule that belongs to the point group C3v.
We can define the coordinate system so that the x-axis points to the right, the z-axis point to the to top , and the y-axis stands perpendicular to the paper plane (Fig. 2.3.38). The ammonia molecule is oriented so that its pyramid points into z-direction, and one of the three N-H bonds is within the xz plane.
Now let us see how the coordinates change when we rotate the molecule by an arbitrary general angle θ around the z-axis. In the diagram shown below, we now look along the z-axis, and the xy plane is within the paper plane (Fig. 2.3.39).
We can see that the rotation changes both the x and and the y coordinate together, but leaves the z-coordinate unaffected. According to trigonometry the new x-coordinate x’ is x’ = x cos θ - y sin θ. The new coordinate y’ is y’ = x sin θ + y cos θ.
The matrix representation of a symmetry operation that rotates around an arbitrary angle must be able to convert the old coordinates x,y,z into the new coordinates below (Fig. 2.3.40).
The matrix which is able to do this has the form below (Fig. 2.3.41)
We could again use the multiplication rules for matrices to show that this matrix is the correct matrix.
The matrix that we just developed rotates around a general angle θ. What does this matrix look like in the point group C3v where we rotate around 120° and 240°? Well, all we need to do is to insert the values of 120° and 240° for θ into the matrix. The sinus of 120° is -√3/2, and the sinus of 240° is √3/2. The cosinus of 120° and 240° are both -1/2 (Fig. 2.3.42).
Note that these are the values for clockwise rotation. Because the C3 operations rotate anti-clockwise, we must use the values for the 240° rotation for the C31 and the values for the 120° rotation for the C32 operation. Considering this, the matrix for the C32 operation and the matrix for the C31 operation are the matrices shown below (Fig. 2.3.43).
We can see there is a significant difference to the all the matrices we saw before. There are non-zero characters not only on the trace of the matrix, but also on other positions within the matrix, in particular in the first two rows which represent the x and the y coordinates. This is a consequence of the fact that the x and the y coordinates are no longer independent in C3v, they change in a dependent way. We say that they are degenerate.
Now can we determine the remaining symmetry operations in C3v. There are three conjugate vertical mirror planes to consider. The first mirror plane stands within the xz plane and passes through the N-H bond that is within the xz plane. The matrix for it is simple. It has the form below (Fig. 2.3.44).
It is straightforward to understand why the matrix has this form. Because the mirror plane is in the xz plane the coordinates x and z are not affected by the reflection at this plane. Therefore there are two +1 characters on the trace of the matrix in the rows for the x and the z coordinates, respectively, and all other characters in these rows are 0. The character in the second row on the trace of the matrix is -1 because the algebraic sign of the y-coordinate changes as we reflect at the xz plane. However, the other two mirror planes are not co-planar with any two coordinates, and for these planes the x and the y coordinates are dependent again. The matrix for the second mirror plane and the matrix for the third one are shown below (Fig. 2.3.45).
We will not derive them in detail here. It is enough for us to understand here that these matrices have more complex forms with non-zero values outside the trace of the matrix because of the degeneracy of the x and y coordinates. Lastly, we should not forget the identity. Because the identity does nothing with an object, its matrix is always the matrix below in any point group (Fig. 2.3.46).
Because we know now the matrix representations of the symmetry operations we can understand the character table for the point group C3v (Fig. 2.3.47)
The character table has three classes for the three groups of non-conjugate symmetry operations. In the first class there is only the identity. You can see that the second class contains the C31 and the C32 operations, summarized as 2C3. The third class contains the 3 conjugate σv operations. In the point group C3v there are three irreducible representations. The first two have the symmetry types A1 and A2 respectively, the third of is of type E. The last one is new to us. It stands for “double degeneracy”, and this symmetry type exists due to the double-degeneracy of the x and the y coordinate. You can see that the character table explicitly states that x and y are double-degenerate when you look into the last column of the character table. The letters x and y are in parentheses separated by a comma. The parentheses indicate the degeneracy of these two coordinates. More generally, any two functions that are linear functions of x or y are degenerate in C3v. You can see that there are also other functions that are degenerate. For instance functions of x2-y2 and xy are degenerate. Also functions of xz and yz are degenerate.
You can see that the characters in the irreproducible representation of the E symmetry type are also unusual. Previously, we only encountered the characters 1 and -1. However, the character underneath the identity is 2, and the character in the class for the vertical mirror planes is 0. How can we understand these characters? The answer has to do with the double-degeneracy of the x and y coordinates. Because of this double-degeneracy, the characters in the irreducible representations are the sum of the characters on the trace of the matrices for the x and y coordinate. Let us check what they are. You can see the three matrices representing the three classes of the symmetry operations below (Fig. 2.3.48).
For the identity the sum of the two characters is 1+1=2 which explains the character 2 in the irreducible representation E. For the matrix representing the C3 operations the characters on the trace of the matrix for x and y are -1/2. When summed up -1/2 + (-1/2) gives -1 which explains the character -1 in the irreducible representation. Lastly for the matrix representing the vertical mirror planes the characters on the trace of the matrix for the x and y coordinates are 1 and -1, respectively. -1+1=0 which explains the character 0 in the irreducible representation of the symmetry type E.
The z coordinate has the symmetry type A1 which is straightforward to understand because the characters for z on the trace of the matrices for the three different classes of symmetry operations are all 1. You can also see from the character table that there is again an irreducible representation to which no coordinate belongs to. It is the one with the symmetry type A2. In C3v only rotational functions around z have that symmetry type.
Symmetry Types of Degenerate Irreducible Representations
There are not only double-degenerate irreducible representations, denoted by a symbol E, there can also be triply-degenerate ones, indicated by a symbol T. In this case all three coordinates x, y, and z are degenerate. Triple-degeneracy does not appear in the point group C3v, but in other point groups, in particular the high-symmetry point groups (Fig. 2.3.49).
Both the E and T symbols can carry subscripts and primes that have specific meanings. The prime (‘) and (“) double prime in the symmetry representation label indicates “symmetric” or “anti-symmetric” with respect to σh, e.g. E` or E``(Fig. 2.3.50).
The subscripts g and u stand for the German words “gerade” and “ungerade” and a g and a u indicates “symmetric” or “anti-symmetric” with respect to inversion, i, respectively (Fig. 2.3.51).
For example, there can be symmetry types like Tg or Eu. There are other subscripts possible, and we do not need to go through them exhaustively. It is enough to understand that they indicate a certain symmetry property in a symmetry type.
Character Tables and Degenerated Orbitals
One of the most useful properties of character tables are that one can identify degenerate orbitals very easily. For example, a quick look into the character table of the point group C3v reveals that the 2px and the 2py orbital in NH3 must be degenerate (Fig. 2.3.52).
This is because the 2px orbital is a linear function of x, and the 2py orbital is a linear function of y, and linear functions of x and y are listed in parentheses in the double-degenerate irreducible representation of the type E. This double-degeneracy could also be derived by inspecting the orbitals in the NH3 molecule, but this is not as easy as looking up the character table. You can see the 2px and the 2py orbitals in the ammonia molecule below (Fig. 2.3.53).
The two orbitals stand perpendicular to each other, but the NH3 molecule has C3 rotational symmetry. Therefore, why would the two orbitals be symmetry-degenerate? If they are degenerate, then there must be at least one symmetry operation that can interconvert the two orbitals. From the above depiction it not obvious what symmetry operation this could be.
Identifying the Double Degeneracy of 2px and 2py in NH3
We can do a little trick to see that a reflection operation is the one that can interconvert the two orbitals. The trick is to rotate the coordinate system by 45°. You can see the ammonia molecule in the bird’s perspective below (Fig. 2.3.54).
This means that we look perpendicular to the base of the pyramid. You can see the NH3 molecule with its three mirror planes above. Now we can choose our coordinate system so that the x and the y axes are rotated 45° with respect to one of the mirror planes. By definition, the 2px orbital must be oriented along the x-axis, and the 2py orbital must be oriented along the y-axis. You can see that one of the mirror planes, namely the one that oriented vertically, is able to interconvert the two orbitals. Upon reflection, the left and the right lobes of both orbitals will swap up their positions, so that after the symmetry operation is complete, the 2p orbitals are interconverted. Overall we can see that proving the degeneracy of two orbitals by inspection can be tricky. A quick look into the character table can tell us much more easily if two orbitals are degenerate.
Dr. Kai Landskron (Lehigh University). If you like this textbook, please consider to make a donation to support the author's research at Lehigh University: Click Here to Donate. | textbooks/chem/Inorganic_Chemistry/Inorganic_Coordination_Chemistry_(Landskron)/02%3A_Symmetry_and_Group_Theory/2.03%3A_Properties_and_Representations_of_Groups.txt |
Concept Review Questions
Section 1
1. What is the definition of symmetry, a symmetry operation, and a symmetry element?
2. There are five basic types of symmetry operations. Name them.
3. Of these five types of operations, two are not independent (i.e. they can be expressed by two of the three others). Name the two operations and explain why they are not independent.
4. What is the definition of the identity operation?
5. What is the definition of a proper rotational axis, and a proper rotation operation?
6. What is the definition of a mirror plane and a reflection operation?
7. What is the definition of an improper rotational axis and a rotation-reflection operation?
8. What is the definition of an inversion center and an inversion operation?
9. What is a principal axis?
10. What is the definition of a horizontal, vertical, and dihedral mirror plane, respectively?
11. What are the rules for giving axes and mirror planes primes?
12. How often do you have to carry out a proper rotation (order n), a reflection, an inversion, and a rotation-reflection (order n), respectively until you have reached the identity?
Section 2
1. What is a Platonic Solid?
2. Name the possible platonic solids.
3. Draw an icosahedron according to the rules that you have learned in class.
4. Which are the three low symmetry point groups? Which symmetry elements do they contain?
5. Name all high symmetry point groups.
6. What is a rotational subgroup?
7. What symmetry properties does a rotational point group have?
8. What symmetry properties does a dihedral point group have?
9. What different types of rotational point groups do you know?
10. What types of dihedral point group do you know?
Section 3
1. What is a matrix?
2. What are the multiplication rules for matrices?
3. Explain why transformation matrices represent symmetry operations?
4. What is an irreducible representation?
5. What is a reducible representation?
6. What is meant by an A and B symmetry type of an irreducible representation?
7. Explain briefly, why there are non-zero entries on positions other than those on the trace of the matrix for C3 symmetry operations?
8. The x and y coordinates are dependent for C3 rotations around the z axis in the point group C3v. Explain.
9. The irreducible representation of the type E contains characters other than +1 and -1. Explain how these characters are generated?
10. Two orbitals are degenerate when a symmetry operation can interconvert them. Show how the double degeneracy of the 2px and the 2py orbitals in the point group C3v can be demonstrated using this principle.
Dr. Kai Landskron (Lehigh University). If you like this textbook, please consider to make a donation to support the author's research at Lehigh University: Click Here to Donate.
Homework Problems Chapter 2
Homework Problems
Section 2
Exercise 1
Determine all symmetry elements and all unique symmetry operations of the following molecules:
a)
b)
c)
d)
e) bromine tetrafluoride
f) Boric acid
g) Dinitrogen tetroxide
h)
i)
j)
k)
l)
m)
Answer
a)
b)
c)
d)
e)
f)
g)
h)
i)
j)
k)
l)
m)
Exercise 2
When is a molecule chiral?
a) It has no mirror planes
b) It is has no inversion center
c) It has no principal axis
d) It has no rotation reflections (improper rotations)
Answer
d) It has no rotation reflections (improper rotations)
Exercise 3
If a molecule has a principal axis Cn , and n additional C2 axes standing perpendicular to Cn then it belongs to
a) A dihedral point group
b) A rotational point group
c) A low symmetry point group
d) A high symmetry point group
Answer
a) A dihedral point group
Exercise 4
Which of the following molecules are chiral:
a) CH4
b) CHCl3
c) HCFClBr
d) HOF
e) BHFCl
Answer
c) HCFClBr
Exercise 5
Determine the point groups of the molecules at symmetry.otterbein.edu/challenge/index.html until you feel that you can determine point groups effortlessly
Section 3
Exercise 1
Can the following matrices be multiplied and if so what is the product matrix?
a)
b)
c)
d)
Answer
a)
b)
c)
d)
Exercise 2
Determine the irreducible representations for the following orbitals in the point group D2h.
The z axis is defined as the axis of the principal C2 axis. C2’ is defined as the axis rotating around y. σv is defined as the xz plane.
Answer
Exercise 3
Determine the matrix representations of the symmetry elements of the following point groups:
a) D2
Define principal C2 axis as the axis running along z. C2 runs along x.
b) C3
If we define the principal C3 axis running along the z axis:
Answer
a)
b)
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Dr. Kai Landskron (Lehigh University). If you like this textbook, please consider to make a donation to support the author's research at Lehigh University: Click Here to Donate.
03: Molecular Orbitals
Covalent Bonding: Molecular Orbital Theory
Now that we have thoroughly studied symmetry, we can next apply symmetry to molecular orbital theory. Molecular orbital theory is a bonding theory that has been developed to explain covalent bonding, but as we will see in a bit, it can also make statements about ionic bonding. We will see that the application of symmetry to molecular orbital theory will greatly help us to understand molecular orbitals, in particular for more complex molecules. Before we apply symmetry to molecular orbital theory, however, let us briefly review the principles of molecular orbital theory.
Like all theories it is based on a few basic assumptions, also called axioms. The first assumption is the Born-Oppenheimer approximation. It says that the position of the nuclei are nearly fixed relative to electron motion (Fig. 3.1.1).
This is a good approximation because the nuclei are much more massive than the electrons. The second axiom is that molecular orbitals can be described as a linear combination of atomic orbitals (Fig. 3.1.2). Linear combination means a vectorial addition or subtraction.
Since orbitals are so-called vector functions they can be added and subtracted like vectors. Any orbital is a wave function that has a specific amplitude at a particular point in space. A point in space is defined by a vector that points from the origin to that point in space. Therefore, there is an amplitude associated with each vector that is associated with the wave function. In mathematical form we can say that a molecular orbital Ψab = N (caΨa ± cbΨb), whereby Ψa is the atomic orbital a of atom a, and Ψb is the atomic orbital b of atom b. The coefficients ca and cb determine how much the specific atomic orbitals a and b contribute to the molecular orbitals. The larger the coefficient, the greater the contribution of the particular atomic orbital to the molecular orbital.
N is the so-called normalization factor. The normalization factor has a value so that the probability to find the electron anywhere within the molecular orbital is 100%. Like in atomic orbitals, the square of the wave function for a molecular orbital reflects the probability to find the electron at a particular position, when we view the electron as a particle. Therefore, the integral of the square of the wave function over space must be 1, and the normalization factor is chosen so that it is 1. The number of orbitals that can be combined to molecular orbitals is not restricted to two. Also three, four or more orbitals can be combined. The number of molecular orbitals that results from a combination of atomic orbitals is always the sum of the atomic orbitals. So when two atomic orbitals are combined, two molecular orbitals must result, when three atomic orbitals are combined, three molecular orbitals must result, and so forth.
How can we qualitatively understand that the vectorial addition of atomic orbitals to form molecular orbitals explains covalent bonding? The nature of the covalent bond is that electrons are being shared between and that the bonds are directional. If we bring two atomic orbitals closer together they will start to interfere.
For example, two 1s orbitals of two hydrogen atoms will start to interfere as we bring the two atoms closer together. If that interference is constructive then the amplitudes of the two wave functions will add up and result in an increased amplitude between the atoms. This addition of amplitudes can be mathematically described by the vectorial addition of the atomic orbital amplitudes. An increased amplitude is associated with an increased electron density between the atoms. Because this electron density is located in between the atoms it can be interpreted as “shared” electron density explaining a directional, covalent bond. The molecular orbital can be called a bonding molecular orbital (Fig. 3.1.3).
However, we must also consider that negative interference can occur which can be described by vectorial subtraction of the amplitudes of the atomic orbitals. In this case the electron-density is depleted in between the atoms, and there is actually a node in between the atoms where the wave function of the molecular orbital changes its algebraic sign. This molecular orbital can be called an anti-bonding molecular orbital (Fig. 3.1.4).
The anti-bonding orbital has a higher energy than the bonding molecular orbital. This can be qualitatively understood from the fact that the energy of wave functions increases with the number of nodes. We have seen this principle before when we discussed atomic theory, and now we meet it again in molecular orbital theory. If a covalent bond forms will depend how many the electrons will be in the bonding or anti-bonding orbitals.
Electrons tend to occupy the lower energy states first, and therefore bonding orbitals will be filled first. However, the Pauli principle holds for molecular orbitals as it holds for atomic orbitals, and therefore we cannot fill more than two electrons in one molecular orbital. Once the bonding molecular orbital is filled, we must start to fill the anti-bonding orbital.
We can define a bond order in the molecule by subtracting the number of electrons in anti-bonding orbitals from the number of bonding orbitals, and divide the resulting number by two (Fig. 3.1.5).
If we stick to our example, the two H atoms, then we have two electrons overall to consider. We can fill both electrons into the bonding molecular orbital, and the anti-bonding molecular orbital remains empty. This gives a bond order of (2-0)/2=1. Therefore, we can say that one covalent bond has formed between the H atoms, have we have produced an H2 molecule.
Generally we can say that a molecule would expected to be stable when the bond order is larger than 0. That would mean that an H2+ molecular ion should be stable because its bond order would be (1-0)/2=½. The H2+ ion has only one electron which would be in the bonding orbital. However, this bond order is smaller than that of H2, and thus it should be less stable than H2. This is in accordance with experimental observations. Also an H2- ion should be stable. In this case, one of the three overall electrons would be in the anti-bonding orbital. The bond order would be (2-1)/2=1/2. An H22- anion, however would be expected to be unstable because there would be two electrons in bonding molecular orbitals, and two in anti-bonding ones, resulting in a bond order of (2-2)/2=0 (Fig. 3.1.6).
Molecular Orbital Diagram of H2
The covalent bonding in a molecule can be described by a molecular orbital diagram. Let us briefly review the principles of its construction for the example of the H2 molecule (Fig. 3.1.7).
Firstly, we write an arrow to the left. It indicates the relative energies of the orbitals, and is labeled with an E, standing for energy.
Next, we indicate the two 1s atomic orbitals by two horizontal lines, and give them appropriate labels. The lines must be at the same energy levels in the diagram, as both 1s orbitals have the same energy. If you know the exact energy of the orbitals you can write the exact energy next to the orbital name. In the case of the 1s orbitals this would be -13.6 eV.
Next we write the two molecular orbitals as horizontal lines at the appropriate energy levels into the middle of the diagram. The bonding orbital must have a lower energy than the atomic orbitals, and the anti-bonding orbital must have a higher energy. The energy difference between the molecular orbitals and the atomic orbitals must be approximately the same. Typically, the bonding orbital is slightly less bonding than the anti-bonding orbital is anti-bonding. When constructing qualitative molecular orbital diagrams, we do not know the exact energy values for the molecular orbitals, but we can estimate the relative energies of the orbitals according to the arguments just discussed. In the case of the H2 molecule the two 1s atomic orbitals overlap in σ-fashion, we can therefore denote the molecular orbitals with a σ symbol which we can write next to the lines for the orbitals. The anti-bonding MO gets a * in addition to indicate its anti-bonding nature. We connect the molecular orbitals with atomic orbitals by dotted lines to indicate that the molecular orbitals have been constructed from the 1s atomic orbitals.
In the last step, we fill the electrons into the atomic and molecular orbitals. Each hydrogen has one 1s electron, and we write the electrons as arrows in the 1s orbitals. One electron should be spin up and the other one spin down, because the electrons must have paired spins in the molecular orbital, and spin-reversal is quantum-mechanically forbidden. Lastly, we fill the two electrons into the molecular orbitals according to energy. This means we must write them with with paired spins into the bonding molecular orbital. Now our molecular orbital diagram is complete.
Factors Influencing the Degree of the Covalent Interaction
Now let us refine our understanding of molecular orbitals and molecular orbitals diagrams. Not all atomic orbitals can be combined to form molecular orbitals, and the degree of covalent interaction between two atomic orbitals can vary lot. What are the criteria according to which we can decide if covalent interaction between two atomic orbitals is possible, and if so how much? There are three criteria to consider.
The symmetry criterion, the overlap criterion, and the energy criterion. The symmetry criterion says that if there is a combination of atomic orbitals with bonding and the antibonding interactions that do not cancel out then there is a bonding interaction. We will discuss in a moment what this means.
Definition: Symmetry Criterion
If there is a combination of atomic orbitals with bonding and the antibonding interactions that do not cancel out then there is a bonding interaction.
The overlap criterion states that the better the atomic orbitals (of appropriate symmetry!) overlap the stronger the covalent interaction.
Definition: Overlap Criterion
The better the atomic orbitals (of appropriate symmetry!) overlap the stronger the covalent interaction.
The energy criterion states that the closer the orbitals are in energy the more covalent interaction between them.
Definition: Energy Criterion
The closer the atomic orbitals in energy the stronger the covalent interaction.
The Overlap Criterion
Let us now look at each criterion in more detail. Let us start with the one we can probably most easily understand, the overlap criterion. The greater the overlap the greater the covalent interaction. The overlap can be estimated according to three rules.
Rule 1
The first rule says that the overlap is the greater the smaller the distance between the two orbitals (Fig. 3.1.8).
This means that a small distance between the orbital leads to a strongly bonding and a strongly anti-bonding orbital, respectively while a large distance leads to a weakly bonding and a weakly anti-bonding orbital. When the distance is small then there is a large energy difference between the bonding and the anti-bonding molecular orbital, when the distance is large then the energy difference is small (Fig. 3.1.9).
Rule 2
Rule 2 states that a large “diffuse” orbital tends to overlap better (interacts more strongly) with another orbital when this orbital is also a large diffuse orbital. Small “little diffuse” orbital tend to interact more strongly with other little diffuse orbitals. If we combine a large orbital with a small orbital however, then this typically does not lead to good overlap and thus weak interaction (Fig. 3.1.10).
We can qualitatively understand this by looking at the image below (Fig. 3.1.11).
Only a small volume fraction of the large orbital can overlap with the small orbital due to the small size of the small orbital. Due that small overlap the bonding orbital is only weakly bonding, and the anti-bonding is only weakly anti-bonding. The energy difference between the bonding and the anti-bonding orbital is small. In the other two cases, the bonding orbitals tend to be strongly bonding, and the anti-bonding ones strongly anti-bonding. The energy differences between the orbitals tend to be large.
Rule 3
Rule 3 says that that orbitals that overlap in σ-fashion tend to interact more strongly than orbitals that overlap in π-fashion (Fig. 3.1.12).
One can see easily from the image below that two p orbitals that have the same distance d from each other overlap much more when they overlap in σ-fashion compared to π-fashion (Fig. 3.1.13).
This is because in the first case they point toward each other, and the orbital overlap is on the bond axis, while in the latter case they are oriented parallel to each other, and the orbital overlap is above and below the bond axis. This implies that the σ-overlap leads to more bonding and anti-bonding orbitals with a larger energy gap between them compared to the π-overlap.
The Energy Criterion
The energy criterion states that the covalent interaction is the larger the smaller the energy difference between the atomic orbitals. We can understand this qualitatively when considering that orbitals are waves, and waves of similar energy interfere more significantly with each other than waves with different energies. Just imagine two waves with very different wavelengths associated with very different energies. Would they interfere effectively? No, they wouldn’t. Rather, two waves with very similar wavelengths would interfere better. Because greater energy difference means less interaction, molecular orbitals that result from the interaction of two atomic orbitals with large energy difference are much more similar in shape, size, and location compared to molecular orbitals that result from atomic orbitals with similar energy.
The greatest covalent interaction is expected when the energy between the two orbitals is exactly the same. This is only possible when two, same orbitals A of the two same atoms overlap. In this case we form a perfect covalent bond with electrons exactly equally shared between the orbitals. The maximum of the amplitude of the bonding molecular orbital is exactly in the middle between the two atoms. In the molecular orbital diagram the energy difference between the bonding MO and the AOs is about the same as the energy difference between the anti-bonding MO and the AOs. Assuming that each atomic orbital is filled with one electron the two electrons are in the bonding molecular orbital where they are equally shared between the atoms (Fig. 3.1.14)
Now let us make the energy of the two atomic orbitals somewhat different. Because they are different we denote the atomic orbitals A and B now, whereby we choose the energy of orbital A to be somewhat higher than that of orbital B. Overlap between the atomic orbital still produces covalent interaction yielding a bonding and an anti-bonding molecular orbital. However, the energy difference of the molecular orbitals to the two atomic orbitals is no longer the same. The anti-bonding MO is now closer to the AO with the higher energy, and the bonding MO is now closer in energy to the AO with the lower energy. This has another consequence. The bonding molecular orbital is now localized primarily at atom B, and the anti-bonding orbital is located primarily at atom A. If we again assume that each AO contributed one electron to the covalent bond, then the two electrons will be in the bonding MO. Because the bonding MO is now primarily localized at atom B, the bonding electrons are primarily localized at atom B. This means they are no longer exactly equally shared, and we have a polar, covalent bond which his polarized toward atom B (Fig. 3.1.15).
Now let us make the energy difference between the two atomic orbitals of the atoms A and B very large (Fig. 3.1.16). In this case, the bonding MO is energetically very close to the AO of atom B, and is localized almost exclusively at atom B. Actually, the bonding MO closely resembles the AO of atom B in shape, size and localization. In other words, the AO of B has hardly changed due to the very weak covalent interaction resulting from the large energy difference between the atomic orbitals. Vice versa, the anti-bonding orbital is energetically very close to the AO of atom A, and is localized almost completely at atom A. The anti-bonding MO is very close to the AO in shape, size, and location. Due to the weak covalent interaction, there is almost no change to the atomic orbital of A. Assuming that each atomic orbital contributes two electrons, the bonding orbital will be filled. The two electrons will be almost exclusively located at atom B. This means that we have effectively transferred one electron from atom A to atom B in a redox process, and have produced an ionic bond. Atom A has now effectively a 1+ charge, and atom B has a 1- charge. This shows that molecular orbital theory, although designed for covalent bonding, can also make statements about ionic bonding. We can also see that a 100% ionic bond is not possible. To achieve 100% iconicity the energy difference between the two atomic orbitals would need to be infinitely large. This is practically impossible. However, a 100% covalent bond is possible because two electrons can be exactly equally shared between two atoms when the energy of the atomic orbitals is exactly zero.
Another conclusion that we can draw is that bonding electrons are located at the atom with the atomic orbital of lower energy, and anti-bonding electrons are located at the atoms with the atomic orbital of higher energy. Orbital energy is correlated with electronegativity. For orbitals of the same type and the same elements, orbitals with higher electronegativity have lower energy. For example, a 2s orbital of fluorine has a lower energy than a 2s orbital of oxygen because the electronegativity of fluorine is higher. Bonding electrons are therefore located primarily at the more electronegative atom, while anti-bonding electrons are located primarily at the less electronegative atom (Fig. 3.1.7). When there are enough anti-bonding orbitals occupied it is possible that the overall polarity in the molecule is such that the dipole moment points toward the more electro-positive atom. An example is carbon monoxide which is slightly polarized toward the carbon atom. We will discuss the MO diagram of the carbon monoxide in detail later.
The Symmetry Criterion
Lastly, let us look at the symmetry criterion. The symmetry criterion tells us if a covalent interaction between orbitals is possible based on the relative orientation of the orbitals. Only if bonding and anti-bonding interactions do not cancel out, a bonding interaction is possible, and we can construct molecular orbitals from atomic orbitals. Bonding and anti-bonding interactions cancel out when positive and negative interferences due to orbital overlap are exactly equal. This can be determined by inspection of orbital overlap.
For example, let us look at the orbital overlap between the 1s orbitals of hydrogen and the 2pz orbital of oxygen in the water molecule (Fig. 3.1.18).
In the water molecule the orbitals are oriented to each other in a specific way because of the bent structure of the water molecule. Due to the bent structure of the water molecule the 1s orbitals overlap differently with the two lobes of the 2pz molecule. The lobe that points downward overlaps more strongly than the lobe that points upward. The two lobes must have different algebraic sign. Now we choose the algebraic sings of 1s orbitals so that bonding is maximized. This means that we choose the algebraic signs so that they are the same as the those of the lobe that points downward. We can see that now the overlap of the 1s with the 2pz orbital produces more positive than negative interferences. The blue lobe is further away from the 1s orbitals than the orange lobe and thus positive and negative interferences do not cancel out. This is equivalent to saying that bonding and anti-bonding interactions do not cancel out. Therefore, the symmetry is “right”, we can construct molecular orbitals from this combination of atomic orbitals.
Can the 2px orbital of oxygen also combined with the 1s orbitals of the hydrogen atom to form molecular orbitals? The 2px orbital is oriented differently relative to the 1s orbitals in the H2O molecule (Fig. 2.1.19).
In this case, we must choose the algebraic signs of the two 1s orbitals to be different so that bonding interactions are possible. The bonding and the antibonding interactions only do not cancel out if the left 1s orbital has the same algebraic sign as the left lobe of the 2px orbital and the right 1s orbital has the same algebraic sign as the right lobe of the 2px orbital. If, for instance, we chose both 1s orbitals to be orange, then the bonding interactions between the left lobe and the left 1s orbital would be canceled out by the equally strong anti-bonding interactions between the right lobe and the right 1s orbital. Overall, we see however, that if we select the color (meaning algebraic sign) of the 1s orbitals appropriately then the symmetry is “right” and we can create molecular orbitals from the atomic orbitals.
What about the combination of the 2s of the oxygen with the 1s orbitals of the hydrogens (Fig. 3.1.20)?
We can see that is very easy to construct a combination in which there are bonding interactions. If we choose the algebraic sign of all atoms to be the same, then certainly bonding and anti-bonding interactions do not cancel out. The symmetry is “right”, and the construction of atomic orbitals from molecular orbitals is possible.
What about the interactions between the 1s orbitals and the 2py orbital (Fig. 3.1.21)?
In this case, the two 1s orbitals are in the paper plane, and the 2py orbital stands perpendicular to it. That makes a 1s orbital to overlap equally with both lobes of the 2py orbital. Because the two lobes of a 2py orbital must have different algebraic sign, the constructive and the destructive interferences will cancel out, no matter how we chose the algebraic signs of our 1s orbitals. That means there is no possibility to create orbital overlap in which bonding an anti-bonding interactions do not cancel out. Therefore, we cannot produce molecular orbitals from a combination of 1s and 2py orbitals. The 2py must remain non-bonding. You may be able to see the cancelation of the bonding and antibonding orbital overlap better if you choose your coordination system differently. Let us have the y-axis point up, and the x-axis point right (Fig. 3.1.21, bottom). Now we look at the H2O molecule from the bird perspective, and the 2py orbital is oriented vertically. The 1s orbitals are still on the x-axis. You can see the overlap between the 1s orbitals and the 2py orbital more clearly now. No matter how we choose the algebraic sign of our orbitals, the bonding and the anti-bonding interactions cancel out.
We have seen thus far that is is possible to decide about “right” and “wrong” symmetries by inspection, but we have noticed that this is not trivial. Generally, the more complex a molecule gets the more difficult it is to decide about “right” and “wrong” symmetry. As we will see in the following, group theory can greatly help us decide about “right” and “wrong” symmetry. It provides a formal pathway to unambiguously make such a decision.
Dr. Kai Landskron (Lehigh University). If you like this textbook, please consider to make a donation to support the author's research at Lehigh University: Click Here to Donate. | textbooks/chem/Inorganic_Chemistry/Inorganic_Coordination_Chemistry_(Landskron)/03%3A_Molecular_Orbitals/3.01%3A_Formation_of_Molecular_Orbitals_From_Atomic_Orbitals.txt |
Rules for the Symmetry-Adapted Linear Combination of Atomic Orbitals (SALC)
The construction of molecular orbitals using group theory follows a method called “Symmetry-adapted linear combination of atomic orbitals”. This method follows the following steps.
1. Firstly, we determine the point group for the molecule.
2. Secondly, we determine the axes of the coordinate system in a useful way.
3. Thirdly, we determine the ligand atomic orbitals that can be combined to form so-called ligand group orbitals (LGOs). These are usually orbitals of the same kind that a symmetry operation can interconvert. The number of LGOs is always the number of the ligand atomic orbitals.
4. In the next step, we determine the symmetry types of the ligand group orbitals, and we will talk in a moment how this works.
5. Finally, we determine the symmetry type of the atomic orbitals of our central atom, and combine ligand group orbitals and central atom orbitals of the same symmetry type. If they have the same symmetry type their symmetry is “right”. The MOs resulting from this combination will have the same symmetry type as the orbitals from which they have been made. Also the number of MOs must be the same as the number of LGOs and central atom AOs.
Now you can construct a molecular orbitals diagram using the MOs that you have constructed using SALC. After you have constructed the MO diagram, always check that the number of MOs is equal to the number of AOs. If not, you must have made a mistake and you have to go back and find the mistake.
Example - H2O
Let us carry out these steps using our water molecule as an example. As we determined previously, the point group of the water molecule is C2v. A sensible way to define the coordinate system is to have the x-axis point to the right, the z-axis point to the top, and the y-axis point out of the paper plane (Fig. 3.2.1). The water molecule would be in the xz plane pointing into z-direction. The next step is to define the ligand atoms and ligand orbitals. It is easy to see that the H atoms would be ligand atoms, and the O atom would be the central atom. The ligand atomic orbitals that would be combined to form ligand group orbitals would be the 1s orbitals of the hydrogen atoms. We would expect two ligand group orbitals because we combine two atomic orbitals.
Determination of LGO Symmetry Types
Next, we need to determine the symmetry types of the ligand group orbitals (LGOs). We can do this in two steps.
In the first step, we determine the reducible representation for the ligand group orbitals by a method called the “orbital swapping method”, and in the second step we determine the irreducible representations for the ligand group orbitals from the reducible representation. The irreducible representations will give us the symmetry type of the ligand group orbitals. Let us do the first step, and carry out the orbitals swapping method to determine the reducible representation. To do so, let us first denote the two 1s orbitals Ψ1 and Ψ2, respectively (Fig. 3.2.2).
Then, we decide if Ψ1 and Ψ2 get swapped up when a particular symmetry operation is carried out. If so, each gets a character zero, if not, they get a character +1 each. The sum of these characters give the character in the reducible representation for the particular symmetry operation. After we have applied orbital swapping for all symmetry operations, we know all the characters of the reducible representation. The E symmetry operation does not swap the orbitals, hence the character in the reducible representation is 1+1=2 (Fig. 3.2.3). The C2 symmetry operation does swap the orbitals, hence the character in the reducible representation is 0+0=0. The xz mirror plane leaves the orbitals at their positions, therefore the character in the reducible representation is 1+1=2. the yz mirror plane swaps the orbitals, hence the character is 0+0=0 (Fig. 2.3.2).
You may ask: Why does the orbital swapping method work? What deeper principle is behind it? The answer is. The orbital swapping method is a short-cut for getting the sum of the characters on the trace of a transformation matrix that converts the two wave functions. Before the symmetry operation is carried out the two wave functions can be described by a matrix of of the form below (Fig. 3.2.4).
When the orbitals are swapped, Ψ1 and Ψ2 on the matrix are swapped, and the matrix takes the form below (Fig. 3.2.5).
The transformation matrix when multiplied with the matrix in Figure 3.2.4 gives the matrix in Figure 3.2.5 which is the transformation matrix for the operation(s). This matrix has the form below (Fig. 3.2.6),
and we could use the multiplication rules to show that this matrix is the correct matrix. This means the transformation matrix for C2 and σyz is the matrix shown in Figure 3.2.6.
For the operations E and σxz the orbitals do not get swapped, and the final matrix is the same as the initial matrix in Figure 3.2.4. Thus, the transformation matrix has the form below (Fig. 3.2.8). For the matrix in Figure 3.2.6 the sum of the characters on the trace of the matrix is zero. For the matrix below
this sum is 2. These are the characters of the reducible representation that we previously determined using the orbital swapping method. Using the multiplication rules we can show that the matrix of Fig. 3.2.8 is the correct transformation matrix (Fig. 3.2.9)
In the next step we need to determine the irreducible representations from the reducible representation. Remember, a reducible representation is the sum of two or more irreducible representations. Thus, a reducible representation contains the information about the irreducible representations it is composed of, and we need to find a tool to extract that information.
The tool to do this is a formula from group theory called the reduction formula. You can see the reduction formula below (Fig. 3.2.10).
It says that the number of irreducible representations of a given symmetry type is one over the order of the point group times the sum of the products of the number of operations in a specific class in the character table of the point group times the character of the reducible representation for a given symmetry operation times the character of the irreducible representation for a given symmetry operation. The "sum of products" means the sum of the products for all classes of symmetry operations in the point group. The number of operations in a class and the character of the irreducible representation can be read from the character table for the point group. Here we see for the first time the utility of character tables for molecular orbital theory. The characters of the reducible representation can be read from the reducible representation previously determined by the orbital swapping method.
For clarity, let us use the reduction formula for the determination of the irreducible representations of the ligand group orbitals for the example of the water molecule. Our task is to find the number of irreducible representations of each possible symmetry type in the point group. In the point group C2v, these are the symmetry types A1, A2, B1, and B2. Let us start with A1. The number of irreducible representations of the type A1 is: A1 = 1/4 (1x2x1 + 1x0x1 + 1x2x1 + 1x0x1) = 1 (Fig. 3.2.12).
The result means that one of the two ligand group orbitals has the symmetry type A1. The factor ¼ is because the order of the point group is 4. We can determine the order just by counting the number of symmetry operations in the character table for C2v. The first product 1x2x1 (red) is the product for the identity E. The number of operations in the column for E is 1, there is never another conjugate operation in the class for the identity. Note that the number “1” is not explicitly spelled out. The number of operation within a class of a character table is only spelled out when the number is larger than 1. The character in the reducible representation for E is 2, and the character for the irreducible representation for the A1 symmetry type underneath E is also 1. The second product is 1x0x1 (blue) because the number of operations in the class of for C2 is 1, the character in the reducible representation for C2 is 0, and the character of the irreducible representation underneath C2 is 1. The third product for the σxz operation is 1x2x1 (green) because the character for the class σxz is 1, the character in the reducible representation for σxz is 2, and the character of the irreducible representation under σxz is 1. Lastly, the product for the operation σyz is 1x0x1 (black) because the number of operations in the class for σyz is 1, the character in the reducible representation under σyz is 0, and the character of the irreducible representation under σyz is 1 (purple).
Using the same procedure, we can also determine the number of ligand group orbitals that have the symmetry types, A2, B1, and B2.
The result (Fig. 3.2.13) is that the second ligand group orbital has the symmetry type B1, and the number of ligand group orbitals with the symmetry type A2 and B2 is 0. We have determined that the reducible representation is the sum of the two irreducible representations A1 and B2 (Fig. 3.2.14).
The result means that one ligand group orbital has A1 symmetry, the other B1 symmetry. The one with A1 symmetry reflects the combination of two 1s orbitals with the same algebraic sign, the one with B1 symmetry reflects the combination of two 1s orbitals with opposite algebraic sign.
We encountered the LGOs already previously by inspection. Here we see that group theory can formally derive them. We can also verify that the two LGOs have A1 and B1 symmetry, respectively, by carrying out the symmetry operations and see what they do to the respective LGO. For the LGO with A1 symmetry, no symmetry operation changes the LGO (Fig. 3.2.15). This is consistent with the symmetry type A1 for which all characters on the irreducible representation are 1.
For the LGO with B1 symmetry type we see that the C2 operation and the σyz operations reverse the algebraic sign of the wave function, while the identity and the σyz operations don’t (Fig. 3.2.16). This is consistent with the irreducible representation of the B1 symmetry type which has the characters +1 for E and σxz, and -1 for the operations C2 and σyz.
Symmetry Type Determination of Central Atom Orbitals
In the next step we have to find out the symmetry types of the central atom valence orbitals. In the case of the water molecule the central atom valence orbitals are the 2s, and the three 2p orbitals. We can easily read the symmetry types of these orbitals from the character table of the point group C2v (Fig. 3.2.17).
An s orbital is always of the totally symmetric symmetry type in the point group. This is the symmetry type with all characters having a +1 value. In the point group C2v this is the symmetry type A1. The symmetry type of the 2pz orbital is also A1. This is because the letter z is in the row for the symmetry type A1. Similarly, the 2px and the 2py orbitals have the symmetry type B1 and B2 respectively because the letters x and y are in the rows for the symmetry types B1 and B2, respectively.
Combination of AOs and LGO to form MOs
When central atom AOs have the same symmetry type then they have the “right” symmetry to be combined to form molecular orbitals. The molecular orbitals will have the same symmetry type than the AOs and LGOs from which they have been made. The number of MOs of a specific symmetry type must be equal to the sum of the LGOs and AOs that have the same symmetry type. When an orbital does not find a partner orbital of the same symmetry type then it is non-bonding.
Let us apply these ideas to the example of the water molecule. We saw that there were two AOs that had the symmetry type A1, namely the 2s and the 2pz orbital. There was also an LGO with A1 symmetry. We can therefore combine these three orbitals to form molecular orbitals. They also must have the symmetry type A1 (Fig. 3.2.18).
Next, let us think about orbitals of the symmetry type A2. No AO and no LGO has this symmetry type. Therefore, there are also no MOs of this symmetry type (Fig. 3.2.19).
For the symmetry type B1 we find that there is one AO, namely the 2px orbital and one LGO that have this symmetry type. This means that these two orbitals can be combined to form two MOs of the same symmetry type (Fig. 3.2.20).
For the B2 symmetry there is only the 2py AO, but no LGO it could be combined with. Thus, the 2py orbital remains non-bonding (Fig. 3.2.21).
Remember, we got the same result when we analyzed the orbitals by inspection. Here, group theory give us the same result through a formalism. This has the advantage that ambiguity and mistakes are avoided. Nonetheless, it should be kept in mind what the group-theoretical result means, and for that it is useful to have previously analyzed the interactions of the 2p orbitals with the 1s orbitals by inspection.
The MO Diagram of H2O
Now we can construct the molecular orbital diagram of H2O. As always we first draw a vertical arrow for the energy axis (Fig 3.2.22). Next we indicate the atom symbols on the left and the right side of the diagram. By convention, the the central atom is on the left side. After that we can write the atomic orbitals into the molecular orbital diagram according to approximate relative energy, and indicate their names and symmetry.
We know that the 2s orbitals of O must be lower in energy than the 2p orbitals of O, and we write them accordingly. We also know that the 2p orbitals all have the same energy, and so we write them together as close as possible. For the H orbitals we write the ligand group orbitals and indicate them LGO1 and LGO2. We do not know the exact energy of the LGOs relative to the 2p and 2s orbitals of oxygen, but we would expect that it is similar to the them, otherwise we could not make covalent bond. We also know that the O-H bonds are polarized toward O, so we would suspect that the LGOs are more similar to the 2p orbitals in energy compared to the 2s orbitals. We write the LGO2 somewhat above LGO1 to indicate the LGO2 has a slightly higher energy because it has a node. It is also ok though to write them out with the same energy, both methods are common in the literature. We indicate that LGO1 has the symmetry type A1 and LGO2 has the symmetry type B1.
Now we need to write out the molecular orbitals in the middle of the diagram. We go systematically through all the symmetry types as we construct the MOs. For example, we can start with the symmetry type A1. Because we combine three orbitals of this symmetry type we need to write three MOs into the middle of the diagram, and indicate the symmetry type. When we combine three orbitals the following approximation holds. There will be one bonding orbital of low energy, one anti-bonding orbital of high energy, and a third one of intermediate energy that is approximately non-bonding. We do not know if it is exactly non-bonding, slightly bonding or slightly anti-bonding. Exact quantum-mechanical calculations could determine this as well as the exact energy of the orbital, but here our focus is on the qualitative construction and understanding of molecular orbital diagrams. Therefore, we must leave this question unanswered. The bonding orbital should be written below the lowest energy atomic orbital which contributes to the MO, in this case the 2s orbital of O. The anti-bonding orbital should be drawn at an energy level above the highest-energy orbital that contributes to it, in this case the A1-type 2p orbital. You can draw the approximately non-bonding a1-type MO at approximately half the distance between the bonding and the anti-bonding MO into the diagram. The lowest energy orbital gets the label 1a1, the second-lowest, the label 2a1, and the highest energy orbital gets the label 3a1. Note that for the MOs we use lower case letters to indicate the symmetry type. The coefficients in front of the symmetry type numbers the orbitals according to increasing energy. In the last step we connect all A1-type AOs and LGOs with the a1-type MOs by dotted lines. This is indicated by the dotted, red lines in Fig. 3.2.22.
Next, we can draw the b1-type MOs. We expect two MOs, because there is one AO of this symmetry type, and one LGO of this symmetry type. One MO is expected to be bonding and should have low energy, the other one must be anti-bonding, and have high energy. The energy of the bonding MO should be lower than the energy of the lowest energy orbital that contributes to it, in this case the B1-type 2p orbital. The anti-bonding orbital should have a higher energy than the highest energy AO/LGO that contributes to it, in this case the LGO2. The bonding MO gets the label 1b1 and the anti-bonding orbital the label 2b1. We do not exactly know the energy of the b1-type MOs relative to the a1 type AOs when qualitatively drawing the MO diagram. For example, the anti-bonding 2b1 orbital is drawn with a higher energy than the anti-bonding 3a1 orbital, but we would not that for sure (only exact quantum-mechanical calculations could tell). We would however suspect, that the bonding 1b1 orbital has a higher energy than the bonding 1a1 orbital because the energy of the 2s orbital is significantly lower than the energy of the B1-type 2p orbital. We would also suspect that the 1b1 orbital is lower in energy than the 2a1 orbital because we know that the 1b1 is bonding, while the 2a1 is approximately non-bonding. Finally, we connect the AOs and LGOs with B1 symmetry with the MOs of b1 symmetry with dotted lines, indicated orange in Fig. 3.2.22.
Lastly, we still have to decide what to do with the 2py orbital that has the symmetry type B2. This orbital has no partner of the same symmetry type, and thus remains exactly non-bonding. Therefore, we write the 2py orbital as a b2-type orbital with unchanged energy in the middle of the MO diagram, and interconnect the two orbitals with a horizontal, dotted line, Fig. 3.2.22.
Now let us fill the electrons into the orbitals. The O atom has six valence electrons, two of them are in the 2s orbitals and four of them are in the 2p orbitals. The 2p orbitals are filled according to Hund’s rule. Because each H atom contributes one electron, the LGOs are both considered half-full with one electron each. Now we can fill the MOs according to increasing energy. Overall we have 6+2=8 electrons to fill into the MOs. Consequently, the 1a1 gets filled first, then the 1b1, then the 2a1, and finally the 1b2. The 1b2 is called the highest occupied molecular orbital, abbreviated HOMO. The next higher orbital, the 3a1 orbital is called the lowest unoccupied orbital, also called LUMO. HOMOs and LUMO are important for the chemistry of a molecule because the HOMO electrons are the most reactive electrons, and the LUMO is the orbital that easiest to fill with an electron coming from a co-reactant.
Now our MO diagram is complete. It is insightful to compare the MO diagram of water with the Lewis-dot structure of the water molecule to understand what additional information we can gain from the MO diagram compared to the Lewis dot structure. In the Lewis dot structure there, are two localized O-H bonds, so overall we have four bonding electrons. Can we we see these bonding electrons also in the MO diagram? Well, we recognize that there are two bonding molecular orbitals that are filled, the 1a1 and the 1b1. Therefore, also in the MO diagram there are four bonding electrons. You can see however, that the two MOs are not energetically equal, thus two electrons have a higher energy than the other two. This is something that we do NOT see in the Lewis-dot structure. According to the Lewis dot structure all four electrons are equivalent. The bonding MOs in water are delocalized over the entire molecule. In contrast to that in the Lewis dot structure two electrons are localized in the first O-H bond, and the other two in the second O-H bond. This is another difference between the Lewis-dot and the MO picture of the covalent bonding in water.
Next, let us see if there is an equivalent of the two electron lone pairs at O in the Lewis dot structure in the MO diagram. We can see that there are two filled, non-bonding MOs, the 1b2 is completely non-bonding, and the 2a1 is approximately non-bonding. We can argue that these non-bonding electrons are the counterparts of the four electrons in the two electron lone pairs. However, in the Lewis-dot structure the two electron lone pairs appear equivalent. In the MO picture we can see that the two non-bonding electrons have a higher energy than the other two.
Overall, we can say that the MO diagrams gives us a more refined picture of the covalent bonding compared to the Lewis dot structure. This however comes at the price of constructing a complicated MO diagram compared to a simple Lewis dot structure. Therefore, our decision to construct or not construct an MO diagram depends on how much detail we need to know in the context of a chemical problem that we want to solve. For example, light absorption due to electron transitions cannot be explained via a Lewis-dot structure, we need an MO diagram to understand that.
The MO Diagram of NH3
Let us do another, more complex example to enhance our understanding of MO theory, the example of the MO diagram of NH3. The NH3 molecule belongs to the point group C3v. The coordinate system can be chosen so that the z-axis points vertically, and the x-axis points to the right. The y-axis stands perpendicular to the paper plane. The NH3 molecule is oriented with the tip of the pyramid pointing up and one N-H bond being in the xz plane (Fig. 3.2.23).
We would choose the N atom as the central atom and the H atoms the ligand atoms. The relevant ligand orbitals for bonding would be the 1s orbital of the H atoms. We would group them to form ligand group orbitals. Three LGOs would be expected as three 1s orbitals would be combined.
The Reducible Representation of the LGOs of NH3
The next step is to determine the reducible representation of the LGOs using the orbital swapping method, Fig. 3.2.24.
We can see that the identity leaves all three orbitals unchanged, hence the character on the reducible representation for E is 1+1+1=3 (Fig. 3.2.24).
Next we need to see what a C3 operation does. It changes the position of all three orbitals, so the character on the reducible representation is 0+0+0=0.
Finally, we need to see what a vertical mirror plane does. We have three conjugate mirror planes and can select any one of the three. For example we can select the one that passes through orbital 1. Reflection at this plane does not change the position of orbital 1, but swaps up the positions of orbitals 2 and 3. Thus, orbital 1 gets a character 1, the other ones get the character 0. The character of the reducible representation is the sum of these three characters, and thus 1+0+0=1. Now, we have found all the characters of the reducible representation.
The Irreducible Representations for the LGOs of NH3
Next, we can determine the irreducible representations from the reducible representation using the reduction formula. In this case the number of A1 –type irreducible representations is 1, the number of E type irreducible representations is 1, and the number of A2-type irreducible representations are 0 (Fig. 3.2.25).
At first glance it looks as though we had found only two of the three expected irreducible representations. However, we need to consider that the symmetry type E is double-degenerate, so it counts like two irreducible representations. Consequently, two of the three ligand orbitals belong to this symmetry type, and are double-degenerated. The third one has the symmetry type A1.
Symmetry types of Central Atom Orbitals of NH3
Next, we need to determine the symmetry type of the central atom orbitals. A quick look into the character table can tell us that (Fig. 3.2.26).
The 2s orbital must have the totally symmetric symmetry type, thus it must have the symmetry type A1. The 2px and the 2py orbitals are found in the row for the symmetry type E. They are double-degenerated as one can see from the fact that the letters x and y are in parentheses. The 2pz orbital has A1 symmetry, because the letter z can be found the row of the symmetry type A1.
Combination of the AOs and LGOs of NH3 to form MOs
Now we must again combine central atom AOs and LGOs of the same symmetry type to form molecular orbitals. The molecular orbitals will have the same symmetry type than the AOs and LGOs from which they have been made. The number of MOs of a specific symmetry type must be equal to the sum of the LGOs and AOs that have the same symmetry type. When an orbital does not find a partner orbital of the same symmetry type, then it is non-bonding.
Let us apply these ideas to the ammonia molecule. We saw that there were two AOs that had the symmetry type A1, namely the 2s and the 2pz orbitals. There was also an LGO with A1 symmetry. We can therefore combine these three orbitals to form molecular orbitals. They also must have the symmetry type a1.
Next, let us think about orbitals of the symmetry type A2. No AO and no LGO has this symmetry type. Therefore, there are also no MOs of this symmetry type. For the symmetry type E we find that there are two AOs, namely the 2px and the 2py orbital and two LGOs that have this symmetry type. This means that these two orbitals can be combined to form four MOs of the E symmetry type (Fig. 3.2.28).
The MO Diagram of NH3
Now we can construct the MO diagram of NH3 (Fig. 3.2.29).
First, we draw the energy axis and indicate the atoms N and H on the left and the right side of the diagram. In the next step we draw the atomic orbitals for the N atom and indicate the symmetry type. The 2s orbital must be drawn below the 2p orbitals to indicate their lower energy.
Then, we can draw the ligand group orbitals on the right side of the diagram and indicate the the symmetry. We can draw the A1-type LGO somewhat below the E-type orbitals because it has a somewhat lower energy, but this is optional. We know that three MOs with a1 symmetry must form. When three MOs form then we can estimate that one will be a low-energy bonding one, one will be a high-energy anti-bonding one, and the third one will be an approximately non-bonding one of intermediate energy. Therefore, we will draw the bonding 1a1-orbital below the 2s orbital, and the anti-bonding 3a1 orbital above the A1-type LGO, and the 2a1 orbital about half-way in between. We also label the MOs according to the energy with the labels 1a1, 2a1, and 3a1, respectively. Lastly, we interconnect the AO and LGOs of the symmetry type A1 with the a1-type MOs through dotted lines. They are shown in red (Fig. 3.2.29).
Then, we draw the four E-type MOs. Because the symmetry type E implies that the orbitals are double-degenerate, we know that there must be two double-degenerate bonding MOs and two double-degenerate anti-bonding MOs of the symmetry type E. A valuable property of group theory is that double-degeneracy in symmetry also implies double-degeneracy in energy, and this means that we know that the two bonding MOs have the same energy, and the anti-bonding MOs also have the same energy. We must also draw them this way into the MO diagram. The bonding pair is labeled 1e and the anti-bonding pair is labeled 2e. We cannot know this for sure based on qualitative inspection, but we can suspect that the two bonding e-type orbitals have an energy in between the 1a1 and the 2a1 orbital. The 2a1 orbiital is higher than the 1e orbital because the former is approximately non-bonding, and the latter is bonding. The 1a1 orbital is likely lower in energy than the 1e orbital because the 1a1 orbital has a contribution from the low-energy 2s orbital, while the 1e orbital only has contributions from the energetically higher 2p orbitals. By similar arguments we can explain that the 2e orbital has a somewhat higher energy than the 3a1.
Now we still need to fill the electrons into the orbitals. The N atom has five valence electrons, two being in the 2s, and three being in the 2p orbitals. The three LGOs of H contain one electron each. This gives overall 5+3=8 electrons that we need to fill into the MOs according to energy. That fills the 1a1 orbital first, then the two 1e1 orbitals, and finally the 2a1 orbital. This makes the 2a1 orbital the HOMO in the molecule, and the 3a1 the LUMO in the molecule.
We can also make again a comparison of the MO and the Lewis-dot picture of the covalent bonding. We can view the non-bonding HOMO the equivalent of the electron-lone pair at N. There are six electrons in bonding MOs which is the equivalent of the six bonding electrons in the Lewis-dot structure. However, in the MO diagram we can see that the six electrons are not equivalent in energy. We are not able to see that in the Lewis-dot structure. Again, we can conclude that the MO diagram gives us more information about the bonding, which comes at the expense of the increased effort to construct an MO diagram. | textbooks/chem/Inorganic_Chemistry/Inorganic_Coordination_Chemistry_(Landskron)/03%3A_Molecular_Orbitals/3.02%3A_Molecular_Orbitals.txt |
Concept Review Questions
Section 1
1. What are the assumptions (axioms) of molecular orbital (MO) theory?
2. Explain qualitatively why the vectorial addition of atomic orbitals creates molecular orbitals (which underlying physical phenomenon is described by the vectorial addition?).
3. Explain qualitatively why molecular orbital theory is suitable to describe covalent bonding.
4. What are the three criteria that determine the degree of covalent interaction in MO theory?
5. Which three rules determine the degree of orbital overlap in MO theory?
6. Explain why the combination of a large, diffuse orbital and a small orbital produces only weak covalent interaction?
7. Explain why sigma-interactions between atomic orbitals typically produce larger orbital overlaps than pi-interactions?
8. Explain why the combination of orbitals of the same energy leads to the largest degree of covalent interaction?
9. What can be said about energy and location of a bonding and an anti-bonding molecular orbital that are made from atomic orbitals of large energy difference?
10. MO theory – even though designed for covalent bonding – can also make statements about ionic bonding. Explain why.
11. Explain the principles of SALC.
Dr. Kai Landskron (Lehigh University). If you like this textbook, please consider to make a donation to support the author's research at Lehigh University: Click Here to Donate.
Homework Problems Chapter 3
Homework Problems
Section 1
Exercise 1
What will lead most likely to the smallest covalent interaction:
a) Overlap of a small and a large orbital.
b) Overlap of two small orbitals.
c) Overlap of two large orbitals.
Answer
a) Overlap of a small and a large orbital.
Exercise 2
What will lead most likely lead to the largest covalent interaction:
a) orbital overlap in sigma-fashion
b) orbital overlap in pi-fashion
c) orbital overlap in delta fashion
Answer
a) orbital overlap in sigma-fashion
Exercise 3
Qualitatively construct the MO diagrams composed of
a) two 2s atom orbitals A and B of equal energy.
b) The orbital energy of atom A is significantly higher than that of B. Assuming both the bonding and the antibonding MO are filled with electrons: Where will bonding and antibonding electrons primarily be located. Explain briefly your decision.
Answer
Exercise 4
Decide by “inspection” which of the following combinations of orbitals have the “right” symmetries to form molecular orbitals.
a) The 2px orbital of the first N atom and the 2py orbital of the second atom in the molecule N2. The z axis is defined as the bond axis in N2.
b) The 2px (of F) and the 1s orbital (of H) in the HF molecule. The z axis is defined as the bond axis.
c) The 2pz orbital of F and the 1s orbital in the HF molecule: The z axis is defined as the bond axis.
Answer
a)
b)
c)
Exercise 5
The CH4 molecule belongs to the point group Td. You can find the character table of the point group in the internet.
a) Calculate the reducible representation for the ligand group orbitals (LGOs).
b) Calculate the irreducible representations of the ligand group orbitals (LGOs).
c) Draw a qualitative molecular orbital diagram for CH4.
Answer
Exercise 6
Which are the symmetry types of the central atom orbitals in the PCl5 molecule?
Answer
1. Determine point group of PCl5. --> D3h.
2. Decide what are the valence orbitals of the central atom: 3s, 3p
3. Look up the character table of D3h, eg. in the internet. You will find their symmetries to be: A1' (3s), A2'' (3pz), E'(3px, 3py)
Exercise 7
For the hypothetical BrKr+ molecule: Toward which atom is the HOMO polarized? Explain briefly why.
Answer
Exercise 8
Reconstruct the MO diagram for water and NH3 (repeat what we did in class without looking at your notes (only use the respective character tables).
Answer
Water
NH3
Dr. Kai Landskron (Lehigh University). If you like this textbook, please consider to make a donation to support the author's research at Lehigh University: Click Here to Donate. | textbooks/chem/Inorganic_Chemistry/Inorganic_Coordination_Chemistry_(Landskron)/03%3A_Molecular_Orbitals/Concept_Review_Questions_Chapter_3.txt |
Dr. Kai Landskron (Lehigh University). If you like this textbook, please consider to make a donation to support the author's research at Lehigh University: Click Here to Donate.
04: Acid-Base and Donor Acceptor Chemistry
A brief review of the Lewis acid-base theory
In this chapter we will discuss Lewis-acid base reactions from the perspective of molecular orbital theory. The Lewis acid-base concept has been developed by Gilbert Lewis (Fig. 4.1.1).
It is a very general acid-base concept, and includes the Broensted and Arrhenius acid-base concepts. It is very important for coordination chemistry. Let us briefly review what a Lewis acid, a Lewis base, and a Lewis acid-base reaction is.
A Lewis acid is defined as an electron pair acceptor.
Definition: Lewis Acid
A Lewis acid is an electron pair acceptor
A Lewis base is defined as an electron pair donor.
Definition: Lewis Base
A Lewis base is an electron pair donor
An example of a Lewis acid is BH3, and an example for a Lewis base is NH3. What happens in a Lewis-acid base reaction?
The Lewis base donates an electron pair to form a covalent bond with the Lewis acid (Fig. 4.1.2). A covalent bond formed in a Lewis acid-base reaction is usually called a dative bond because both electrons in the covalent bond come from a single partner. In a "conventional" covalent bond both partners contribute one electron to the covalent bond. There is however no fundamental difference between a “conventional” covalent bond and a dative covalent bond, it is just a matter of perspective. To indicate a dative bond one can draw an arrow pointing from the donor to the acceptor atom, instead of just a line. The reaction product of a Lewis acid-base reaction is called an adduct.
As mentioned previously, the Lewis acid-base concept is quite general and can explain the bonding in quite different compounds. It includes the Broensted acid-base concept, meaning that any Broensted acid is also a Lewis acid, and any Broensted base is also a Lewis base. However, the reverse is not true. Not every Lewis acid is a Broensted acid, and not every Lewis base is a Broensted base.
Here are a few examples that can illustrate the generality of the Lewis acid-base concept.
For example, an H+ ion is a Lewis acid and an OH- anion is a Lewis base, and the two can react to form water as a Lewis-acid base adduct. In this case one of the O-H bonds in the water molecule would be viewed as a dative bond, while the other one would be viewed as a “regular” covalent bond. Of course, there is no actual difference between the two bonds, we just have a different perspective on them. The same reaction could also be viewed as a Broensted acid base reaction with the OH- ion as the proton acceptor, and the H+ ion as the proton donor.
The Lewis acid-base concept also explains bonding in coordination compounds, and the formation of coordination compounds from metal ions and ligands. The ligand is the Lewis base and the metal ion is the Lewis acid, the coordination compound is the Lewis acid-base adduct. The bond between the metal ion and ligand is a dative bond pointing from the ligand to the metal.
For example, Zn2+ acts as a Lewis acid when reacting with 4 OH- as a Lewis base to form tetrahydroxo zincate (2-) anions (Fig. 4.1.4). The formed Zn-O bonds are dative bonds indicated as arrows pointing from the O to the Zn.
The Lewis acid-base concept can even be used to explain bonding in ionic crystals. In this case the anion would be the donor and the cation the acceptor.
For example Ca2+ can be viewed as a Lewis acid, and OH- as a Lewis base in a reaction that forms calcium hydroxide Ca(OH)2 as the Lewis acid-base adduct. The bonds between Ca2+ and OH- would be viewed as a dative bonds indicated as arrows pointing from O to Ca. In this case the ionic bonds would be interpreted as highly polar dative bonds. You may want to remember in this context that completely ionic bonds are not possible in general, and there must also be some degree of covalency due to the finite electronegativity difference between two elements.
MO theory and Lewis Bases
MO theory is a theory designed to explain covalent bonding. Because dative bonds are covalent bonds, MO theory should be able to explain dative bonds. So how can Lewis acids and bases and their reactions be viewed from an MO theory perspective? MO theory states that a molecule is a Lewis base when its HOMO is approximately non-bonding (Fig. 4.1.6).
This explains, for example, that NH3 is a Lewis base. Remember, its HOMO, the 2a1 orbital, is approximately non-bonding. Why does an approximately non-bonding HOMO make a molecule a Lewis base? Firstly, because the HOMO electrons are the highest energy electrons and thus the most reactive electrons, they get donated preferentially over all other electrons. The non-bonding nature of the HOMO is ideal, because if the HOMO was anti-bonding, the electrons would be so reactive, so that they would likely get completely transferred to the reaction partner. In this case we would not have a Lewis acid-base reaction, but a redox reaction. We would not form a covalent bond, but an ionic bond. If the HOMO was bonding, then the energy of the electrons would be too unreactive, simply no reaction would likely be observed.
MO theory and Lewis Acids
So what is then a Lewis acid according to MO theory? A Lewis acid, from the MO theory perspective, is a molecule that has an approximately non-bonding lowest unoccupied molecular orbital. The orbital needs to unoccupied, otherwise no electrons could be donated into it. For energy minimization arguments, electrons would be donated into the unoccupied orbital that has the lowest energy, which is the LUMO. So why is it ideal if the LUMO is non-bonding? If it was anti-bonding, its energy would likely be too high, and the donor would be unable to donate its electrons. There would not be a stabilization of electrons due to the high energy of the LUMO. The molecule would remain unreactive. If the LUMO was a bonding orbital, then its energy would likely be so low so that the electrons would likely be completely transferred to the Lewis acid. In this case we would not form a dative bond but an ionic bond via a redox reaction.
An example of a Lewis acid is the BeH2 molecule. It is a linear molecule belonging to the point group D∞h. The character table of this point group is a bit unusual because the principal axis has infinite order, and there is an infinite number of C2 axes and mirror planes (Fig. 4.1.7). Therefore, the symmetry types of this point group are not of the type we previously encountered. We do not want to understand the character table and its symmetry types in detail here, but just extract the information that is relevant for us to construct the molecular orbitals.
The central atom of the molecule is the Be atom, and its valence orbitals are the 2s and the 2p orbitals. According to the character table of D∞h the 2s orbital has the symmetry type Σg+, the 2pz orbital has the symmetry Πu, and the 2px and the 2py orbitals are double-degenerated, and have the symmetry type Σu+.
The two 1s orbitals of H make two ligand group orbitals. Determination of the reducible and the irreducible representation of the the LGOs via SALC would give the symmetry types of the LGOs which in this case are Σg+ and Σu+. We therefore expect that there is a bonding and an anti-bonding orbital of σg+ type, and a bonding and an anti-bonding orbital are of the σu+ type. The 2px and the 2py orbitals remain non-bonding. The Be atom has two valence electrons in the 2s orbital, and the 2 H atoms contribute two valence electrons each, so there are overall four electrons that we need to fill into the MOs. This fills the bonding σg+ and the σu+, making the σu+ the HOMO. You see that we have a bonding HOMO here, which would argue that BeH2 is not a Lewis base. However, we see that the LUMOs are the non-bonding 2px and 2py orbitals. This explains the Lewis-acidic character of the molecule.
Bonding in Lewis Acid Base Adducts
The greatest covalent interaction between two orbitals it is achieved when both orbitals have the same energy. We learned this previously in Chapter 3 when we discussed the energy criterion. The same holds for dative bonds. The more similar the energies of the donor HOMO and the LUMO acceptor, the greater the covalent interaction. Ideally, the energies are exactly same. In this case we form a perfect covalent bond with the electrons equally shared between the donor and the acceptor. We can graphically illustrate this in an MO diagram the following way (Fig. 4.1.9).
Let us assume a molecule A that acts as a donor and a molecule B that acts as an acceptor. The donor A has a HOMO and a LUMO that have a certain energy. The acceptor molecule B also has a HOMO and a LUMO with the LUMO having the same energy as the HOMO of A. The molecule B will also have HOMO which is energetically below that LUMO. Because their identical energies the combination of the HOMO of A with the LUMO of B leads to a bonding MO and an anti-bonding MO with equally shared electron density. Both MOs have equidistant energy from the HOMO of A and the LUMO of B. The electrons coming from the HOMO of A will be in the bonding MO. They will be equally shared, and the dative bond is an ideal covalent bond.
The exact match of the HOMO and LUMO energies of the donor and acceptor is rarely achieved. Let us consider a number of scenarios in which these energies are not the same, and what consequences this has for the dative bond.
Let us assume next, that the LUMO of the acceptor B is somewhat lower than the HOMO of donor A. In this case, we can still form a bonding MO and an anti-bonding MO due to covalent interaction between the HOMO of A and the LUMO of B. However, now the bonding MO will be localized primarily at B, and the anti-bonding MO will be localized mostly at A. The electrons from the HOMO of A will be in the bonding MO, and thus the bonding electrons in the dative bond will be localized mostly at B. This means that the dative bond is polar, and polarized toward the acceptor.
Let us consider next, what the bonding will be if the LUMO of the acceptor is much lower than the HOMO of the donor (Fig. 4.1.11). In this case, there is still the possibility to form a bonding MO and an anti-bonding MO, but the bonding MO will be localized practically exclusively at the acceptor, and the anti-bonding MO will be localized practically completely ar the donor. The electrons from the donor will be in the bonding MO, but because the bonding MO is located almost entirely at the acceptor, the electrons are transferred completely from A to B in a redox reaction, and the bonding will be ionic, and we have an ionic compound AB made of A2+ cations and B2- anions. This reaction would no longer be considered a Lewis acid-base reaction, and the reaction product will no longer be considered a Lewis acid-base adduct.
The next possibility to consider is that the LUMO of B is higher in energy than the HOMO of A (Fig. 4.1.12). In this case the bonding MO will be localized primarily at the donor A and the electrons in the dative bond will be predominantly at A. We have a polar, dative bond which is polarized toward A.
Next, let us raise the energy of the HOMO and the LUMO of B even higher. This leads to the fact that now the HOMO of B is energetically closer to the LUMO of A, compared to the energy difference between the HOMO of A and the LUMO of B. This results in the fact that the interaction will be mostly between the HOMO of B and and the LUMO of A. As a consequence, B will act now as the donor, and A will act as the acceptor. The dative bond may be polarized toward A or B, or not be polar at all depending on the relative energy of the HOMO of B and the LUMO of A.
If we raise the orbital energies of B even further, and the HOMO of B is much higher than the HOMO of A, then we will get a redox reaction. B gets oxidized, and A gets reduced. B2+ cations and A2- anions will form an ionic compound of the composition AB.
Examples
Generally spoken, the relative HOMO and LUMO energies of reaction partners decide if a Lewis acid-base, or a redox reaction takes place, and what the polarity of the bond is. Here are a few examples that will illustrate our general considerations (Fig. 4.1.15).
The relative orbital energies of the highest occupied atomic orbitals and the lowest unoccupied atomic orbitals of calcium, and the HOMO and LUMO energies of H2O are shown (Fig. 4.1.15). Can we predict the type of reaction? We can see that the highest occupied Ca orbital has a much higher energy than the LUMO of water. We would therefore expect that Ca gets oxidized, and water gets reduced. An ionic compound would be expected. This is what actually happens in experiment. The reaction of water and calcium yields calcium hydroxide and hydrogen gas.
Next, let us consider a possible reaction between water and chloride. We can see that the HOMO of Cl- is similar in energy compared to the LUMO of H2O. We would therefore expect a Lewis acid-base interaction with Cl- as the donor and H2O as the acceptor. Such an interaction indeed occurs in aqueous solutions containing Cl- in the form of weak hydrogen bonding between Cl- and H2O.
Next, let us consider the interactions between Mg2+ and H2O. In this case the LUMO of Mg2+ has about the energy of the HOMO of the water molecule. We would therefore expect that the water molecule acts as the donor and the Mg2+ acts as the acceptor. Indeed, Mg2+ forms a hexaaqua complex with water, which has the composition Mg(H2O)62+. The bonding should be very little polar.
Lastly, what are the interactions between F2 and H2O? The HOMO of H2O is much higher than the LUMO of F2. We would therefore expect a redox reaction in which F2 is reduced, and H2O is oxidized. In reality F2 can oxidize H2O to form OF2 and HF.
From the above examples it becomes also clear that we cannot necessarily predict the strength of the Lewis-acid base interactions. For example, the hydrogen bonding between H2O and Cl- is much weaker than the dative bonds between H2O and Mg2+. Other factors such as orbital overlap also need to be taken into consideration to make statements about the strength of the interactions.
Dr. Kai Landskron (Lehigh University). If you like this textbook, please consider to make a donation to support the author's research at Lehigh University: Click Here to Donate. | textbooks/chem/Inorganic_Chemistry/Inorganic_Coordination_Chemistry_(Landskron)/04%3A_Acid-Base_and_Donor_Acceptor_Chemistry/4.01%3A_Major_Acid-Base_Concepts.txt |
The Hard and Soft Bases Concept (HSAB)
The hard and soft acid and base concept (HSAB) can be conceived as a refinement of the Lewis-acid and base concept. Its most useful feature is that it can make predictions about the strength of the acid-base interactions. It can also estimate if the interactions are more ionic or more more covalent.
4.2.1 Visual representation of a unpolarized atom and an atom polarized in an electric field.
Let us first clarify what is meant by hard and soft, respectively. These terms refer to the polarizability of the electrons in an atom or a molecule (Fig. 4.2.1). An atom is soft when its electrons are easily polarizable. This means that the electron cloud easily deforms in an electric field. Easy deformation is consistent with the term “soft”. In an electric field the negatively charged electrons are attracted to the positive pole of the electric field. Therefore, the electron cloud distorts toward the positive end of the electric field, and the atom is polarized. Vice versa, when the electron cloud is not easily polarizable, we say the atom is hard. What is then meant by a hard and a soft acid or base? A hard or soft acid is a hard or soft Lewis acid, and a hard or soft base is a hard or soft Lewis base. As mentioned previously, the HSAB concept is useful because it make statements about the strength of the acid-base interactions, and thus the strength of the bonds. It can also make statements on whether the bonding is more covalent or more ionic (Fig. 4.2.2).
Generally, hard-hard interactions, meaning the interactions between a hard acid and a hard base, tend to be strong. The bonding is more likely ionic. Soft-soft interactions also tend to be strong, but they are more likely covalent. Hard-soft interactions tend to be weak.
Why are soft-soft and hard-hard interactions strong, while hard-soft interactions are weak? To answer this question, we must first understand what makes and acid or base hard or soft. Generally, we can say that the more delocalized the electrons are, the softer the species. For a single atom or ion this means that the larger the atom or ion is the softer the species. The larger the atom size, the more delocalized are its valence electrons. The size of a neutral atom is defined by its position in the periodic table. Generally, the higher the period, the softer the atom (Fig. 4.2.3).
For ions, also the charge plays a role. For cations, a higher positive charge makes a cation harder, for anions a higher negative charge makes the anion softer (Fig. 4.2.4).
The size of the donor/acceptor atom is not the only factor that determines the hardness. Also the ability of the species to make π-bonding is important. Species that have orbitals suitable for π-bonding tend to be soft even if size arguments suggest that they are hard. This is because π-bonding increases electron delocalization (Fig. 4.2.5).
For example, transition metals have d-orbitals available which are suitable for π-bonding with ligands, while alkali metals do not. Therefore, a relatively small transition metal cation such as Cu+ is softer than large alkali metal ions such as Cs+. For anions π-bonding is also important. For instance, CN- anions are soft bases even though the donor carbon atom is small because the CN- ligand has π-orbitals available for π-bonding with Lewis acids.
So back to the question: Why are soft-soft and hard-hard interactions strong, but hard-soft interactions weak? This essentially boils down to rule 2 of the orbital overlap criterion (see chapter 3 on MO theory). Two small orbitals have typically good overlap, and two large orbitals also have good orbital overlap, and thus the interaction is strong. However, large and small orbitals tend to have weak orbital overlap, and thus the bonding weak.
Now to the question of why soft-soft interactions tend to be more covalent, while hard-hard interactions tend to be more ionic. The answer has to do with the fact that in smaller atoms the energy differences between atomic orbitals tends to be larger compared to large atoms. As a consequence, it is statistically more likely that the HOMO and LUMO of two soft species have more similar energies compared to two hard species. When the HOMO and LUMO of a Lewis acid and a Lewis base are similar in energy, then the bonding is more covalent. When they are significantly different, then the bonding is more ionic.
Let us illustrate this by two qualitiative examples. Lithium oxide is made of O2- anions and Li+ cations. If we view the ionic bonding between the O2- and the Li+ ions as an extreme case of a polar, dative bond, then oxide anion acts as a Lewis base, and the Li+ ion acts as a Lewis acid. How can the high ionicity be explained? Both the Li+ and the O2- are small ions, thus they are both hard. The small size also implies that the energy differences between the atomic orbitals are large. Because of this, it is unlikely that the highest occupied atomic orbital of the O2- ion has a similar energy than the lowest unoccupied atomic orbital of the Li+ ion. Due to the large energy difference, the bonding MO will be localized almost completely at the O2- anion, and the bonding will be ionic (Fig. 4.2.7).
An example of a compound with strong soft-soft-interactions is silver iodide. While AgI is considered an ionic compound, the interactions have a significantly stronger covalent character. The soft nature of Ag+ is readily understood from the fact that Ag+ is a period 5 transition metal ion with low positive charge, and d-orbitals available for π-bonding. Iodine is a period 6 element, thus iodide is quite soft. Because Ag and I are elements of period 5 and 6, respectively, their orbital energy differences are significantly smaller than those of O and Li which are period 2 elements. Thus, the HOMO of I- and the LUMO of Ag+ are naturally closer in energy resulting in a more covalent interaction (Fig. 4.2.8).
We should be aware that we need to treat these considerations with caution. For example, B-O bonds are fairly covalent bonds despite the fact that B and O are both quite hard. The HOMO of the donor and the LUMO of the acceptor of hard species are not necessarily much different. Other parameters such as electronegativity differences also weigh in and must be taken into account to correctly predict the nature of the chemical bond.
Examples - Bases
To get a feeling for correctly estimating the hardness of a species let us consider a few examples. Let us start with some bases.
What statements about hardness can you make for the series F-, Cl-, Br-, I- (Fig. 4.2.9). They are all halogenide ions with the same negative charge. The size increases from fluoride to chloride, to bromide to iodide. Thus, the softness should increase in the same order. F- is a small ion with a small negative charge and should be be hard, I is a very heavy element therefore I- is very soft. Cl- and Br- are moderately hard, and soft ions, respectively.
Next, consider the series H2O, OH-, O2-, CH3O-, and PhO-. All of the species contain small O donor atoms, so all of them should be considered hard. The OH- ion is somewhat softer than the H2O because of the negative charge that increases the size of the donor O atom. For the same reason O2- is somewhat softer than OH-. CH3O- is somewhat softer than O2- because of the positive inductive effect of the methyl group. The phenolate ion is the softest because the negative charge at O can be delocalized in the aromatic ring. The delocalization of the negative charge leads to a greater polarizability, and thus softness. We could also think about if F- would likely be harder or softer than H2O. F is a little smaller than O, but F carries a negative charge, so the case is ambiguous.
Next, let us look at the series NH3, CH3NH2, and PhNH2 (Fig. 4.2.11). The N donor atom is a small, little polarizable atom, thus the species should be regarded hard. N is somewhat larger than O though, which means that NH3 is somewhat softer than H2O. CH3NH2 would be a bit softer than NH3 because the positive inductive effect of the methyl group, and aniline would be a bit softer than methyl amine because of possibility to delocalize the lone pair at N in the aromatic ring.
What about H2S, alkyl thiols and di alkyl thiols (4.2.12)? The S donor atom is in the 3rd period, and large enough to be considered soft. The softness would be expected to somewhat increase from H2S to R-SH to R-S-R because of the positive inductive effect of the alkyl group.
The next example is a phospine of the general formula PR3 (4.2.13). Like sulfur, phosphorus is a period 3 donor atom, and phosphines are generally soft.
Lastly, let us look at carbon monoxide and cyanide (Fig. 4.2.14). At first glance these species appear like hard bases because of the small carbon donor atoms. However, they are actually quite soft because of their ability to use their π and π* orbitals in π-bonding with Lewis acids, in particular transition metal ions. We will look closer at this issue later when we discuss the bonding in transition metal complexes in detail.
Example - Acids
Now let us think about the hardness of acids.
In the series H+, Li+, Na+, K+: Are these hard or soft acids and how does the hardness change within this series (Fig. 4.2.15)? The answer is: All alkali metal cations are considered hard acids, even the relatively large K+ cation in the 4th period. This is because alkali metal cations have only s-valence orbitals and thus a lack of orbitals suitable for π-bonding. Within the series H+ is the hardest and K+ is the softest because the ionic radius increases with increasing period.
Similarly, the earth alkaline metals Be2+, Mg2+, and Ca2+ are hard cations with the hardness decreasing from Be2+ to Ca2+. Due the the 2+ charge earth alkaline metal cations are expected harder than alkali metal cations of the same period, for example Be2+ should be estimated harder than Li+.
In the next series BF3, BCl3, B(CH3), and BH3 the hardness declines from BF3 to BH3 (Fig. 4.2.17). BCl3 is a softer than BF3 because of the smaller electronegativity of Cl versus F. The more electronegative F withdraws more electron density from the boron making it smaller, and thus harder. Because of the positive inductive effect, the B(CH3)3 is softer than BCl3. The BF3 and BCl3 molecules are considered hard acids overall, the B(CH3)3 is an intermediate case. At first glance, it would appear that BH3 is harder than B(CH3)3, nonetheless it acts more like a soft acid, possibly because of the hydride-like character of the compound.
Next let us consider the transition metal ion series Fe2+, Fe3+, Co2+, Co3+, Rh3+, Ir3+ (Fig. 4.2.18). These cations have the ability to make π-bonding, but because of the higher 2+ and 3+ charge respectively, none of them are soft. All period 4 cations with a 3+ charge, namely Fe3+ and Co3+ are hard acids, the Fe2+ and Co2+ ions are at the borderline between hard and soft due to their lower charge. Rh3+ and Ir3+ are also at the borderline. They have a higher positive charge, but are in period 5 and 6, respectively. Rh3+ would be expected to be harder than Ir3+ because it is in a lower period.
Ti4+ and Si4+ are both hard acids (Fig. 4.2.19). Generally, all ions with a charge of +4 or higher are hard acids.
What about the last series Cu+, Cd2+, Hg2+, Pd2+, and Pt2+ (Fig,. 4.2.20)? Think about it. They are all considered soft acids. Cu+ is a relatively low period 4 element, but has only a 1+ charge, and has d-orbitals for π-bonding. Hg2+, Pd2+, and Pt2+ have a somewhat higher 2+ charge, but are period 5 and 6 elements, and also have d-orbitals for π-bonding. Pt2+ would be expected softer than Pd2+ because of its higher period.
Quantitative Measures for Hardness
From the previous considerations we have seen that it is possible to make qualitative, and in some cases semi-quantitative estimates about the hardness of acids and bases, but they are not a quantitative measure for hardness. A hardness scale that allows for quantitative measure of hardness is Pearson’s concept of absolute hardness (Equ. 4.2.1).
Equation 4.2.1 Equation for the quantitative calculation of absolute hardness
It relates the hardness to the difference between the ionization energy and the electron affinity over 2.
Equation 4.2.2 Equation for the quantitative calculation of softness
The softness is then defined as the inverse of the absolute hardness (Eq. 4.2.2).
Equation 4.2.3 Equation for Mulliken's electronegativity
We can immediately see that the absolute hardness is related to Mulliken’s electronegativity scale which is the ionization energy + the electron affinity over 2 (4.2.3).
What is the idea behind this definition? We have previously qualitatively discussed that hard species tend to have large orbital energy differences, while soft species tend of have small orbital energy differences. Therefore, it makes sense to define the energy difference between the highest occupied atomic or molecular orbital and the lowest unoccupied atomic or molecular orbital a quantitative measure for the hardness of a species (Eq. 4.2.4).
Equation 4.2.4 Extended equation for the calculation of absolute hardness
The first ionization energy IE is minus the energy of the highest occupied atomic/molecular orbital: IE=-E(HOMO or HOAO) and the electron affinity is minus the energy of the lowest unoccupied molecular or atomic orbital: EA=-E(LUMO/LUAO). Therefore, the difference between the HOMO/HOAO and the LUMO/LUAO is the same as the difference between the ionization energy and the electron affinity.
Figure 4.2.21 illustrates the concept of absolute hardness for the example of the alkali metal cations. You can see that the lowest unoccupied atomic orbitals are fairly similar in energy, but the energy of the highest occupied atomic orbital increases significantly from the Li+ to the Cs+. Thus, the energy differences decrease from the Li+ to the Cs+, and the absolute hardness η is just half the value. The energy on the y-axis half-way between the HOMO and the LUMO energy is minus the energy associated with the Mulliken electronegativity.
In the table above (Fig. 4.2.22) you can see a number of acids and bases together with their hardness calculated from ionization energies and electron affinities. Let us check if the calculated values are in line with expectations and see what additional value the absolute hardness concept brings. For example, we can see that Li+ is harder than Na+ which is harder than K+. This is what we expected.
We can also see that we can determine relative hardness not possible by qualitative inspection. For example, we can see that Al3+ is harder than Li+. From atomic size perspective a neutral Al is larger than a neutral Li, and from that perspective the Li+ should be harder. On the other hand the positive charge is higher on Al compared to Li. From that point of view the Al should be harder. By qualitative inspection we could not tell which parameter dominates the overall hardness. The absolute hardness concept shows that (for this case) the charge is more important than neutral atom size. Similarly we could not decide by inspection that Mg2+ was softer than Li+ because charge arguments would suggest that Mg2+ is harder while neutral atom size arguments would say that Li+ should be harder. We can see that in this case neutral atom size has a larger impact, albeit only slightly. According to our expectations Mg2+ is harder than Na+ as both ions are neighbored in the same period, and thus very similar in atomic radius, but the Mg has the higher positive charge. We can also see that Ag+ and Au+ have much lower hardness than K+ which we would expect. We would also understand the Au+ has a lower value than Ag+ because these elements are in the same group, and Au+ is in period 6, while Ag+ is in period 5.
Below Au+ you can see a group of neutral molecules. Only the BF3 molecule is a Lewis acid, other molecules are Lewis bases. We can see that BF3 has a relatively high hardness, but is softer than K+. Of the bases, H2O is the hardest base, followed by NH3, followed by PF3 followed by PH3. This is what we expected. O is the smallest donor atom, followed by N, followed by P. PF3 is harder than PH3 because of the higher electronegativity of fluorine versus hydrogen.
The last group are the halogenide anions. According to expectations F- is the hardest and I- is the softest. We can see that Cl- has a lower hardness value than PH3. The chemical behavior of latter is that of a soft base, while the former is still regarded a relatively hard base. We can see that we also need to treat the absolute hardness values with some caution, it is not an omnipotent method, other factors but HOMO and LUMO energy values can also influence polarizability.
When comparing the three groups we see that the cations tend to have the highest hardness values, followed by the neutral molecules. The anions tend to have the lowest values. This means that a hard acid tends to have a higher absolute hardness value than a hard base. To interpret the values meaningful we should therefore only compare acids with acids and bases with bases.
The HSAB Concept and Solubilities
Because the HSAB concept can estimate the strength of the interactions between Lewis acids and Lewis bases, it can also estimate a number of other properties that derive from this strength of interactions. For example, it can be used to estimate solubilities. When the Lewis acid-base interaction between cation and anion is strong we would expect low solubility, when the interaction is weak then we would expect high solubility.
For example, it is know that the solubility of silver halogenides in water increases from AgI to AgBr to AgCl to AgF (Fig. 4.2.23). This is in accordance with the HSAB concept. Ag+ is considered soft, and thus it would make the strongest interactions with the softest anion, the iodide I-. Ag+ would make the weakest interactions with F-, because it is the hardest.
What would be our expectations for the lithium halogenides (Fig. 4.2.24)? In this case, we would expect the solubility to decline from LiI to LiBr, to LiCl, to LiF. Why? This is because Li+ is a hard cation, and thus the strongest interactions should result with F-. Consequently, LiF would have the lowest solubility. I- is the softest anion, thus it should make the weakest interactions with Li+.Consequently, the LiI would have the highest solubility. Is this what we observe experimentally? The experimentally greatest observed solubility is that of LiBr, followed by LiCl, followed by LiI. LiF has the lowest solubility. We can see that the LiI solubility is not what we expected. Instead of having the highest solubility, it has the second-lowest solubility. We can see here the limitations of the HSAB concept. There are also other factors that determine solubility, in particular solvation enthalpy. I- has a very low hydration enthalpy, which explains the lower than expected LiI solubility. In a way, the HSAB concept is able to explain the low hydration enthalpy of I- because it is based on the strength of interaction between I- and water. The water is a hard acid and therefore interacts only weakly with a soft base like I-. This can serve as an explanation for the low hydration enthalpy.
Another good example to illustrate the effects of solvation enthalpy on solubility is the solubility of the silver halogenides in liquid ammonia (not aqueous ammonia). It is actually reversed, the AgF has the smallest solubility , and the AgI has the highest solubility. The HSAB interaction between Ag+ and the halogenide ions are the same no matter of the solvent. Therefore, they cannot serve as an explanation. Rather, we can argue that the reverse enthalpies of solvation in liquid ammonia compared to liquid water are responsible for the inverse behavior. From the standpoint of HSAB, the ammonia molecules are already significantly softer than the water molecules, therefore interactions with the soft anions become significantly stronger.
The HSAB Concept and Stability
The HSAB concept can also be used to estimate thermodynamic stabilities of compounds, such as decomposition points, melting points etc. Generally, the greater the acid-base interactions the greater the expected thermodynamic stability. Let us do couple of exercises to practice this concept.
What order of thermodynamic stability would you expect for the alkali oxides (Fig. 4.2.25)? The answer is: The stability declines with increasing period of the alkali metal. Why? The oxide anion is considered a hard base due to its relatively small radius. Thus, the strongest interactions are expected with the Li+ which is the hardest alkali metal, and the weakest interactions would be expected for the Cs+ which is the softest alkali metal. This is in line with experimental observations. Only Li gives Li2O when burned in O2, Na gives sodium peroxide and the remaining alkali metals give superoxides.
We can ask the same question for the earth alkaline oxides (Fig. 4.2.26). Similarly, the stability of BeO is the highest because Be has the highest hardness. BaO has the lowest stability because Ba2+ is the softest earth alkali cation. Ba gives barium peroxide instead of barium oxide when burned in O2.
The HSAB Concept and Acidity
The HSAB concept can also explain Brønsted acidity. How?
Let us look at the series H2O, H2S, and H2Se (Fig. 4.2.27). The Brønsted acidity increases from H2O to H2S to H2Se. Why? H+ is a hard acid, and therefore the strongest interactions would be expected with the hardest base, the oxide ion, and the weakest interactions would be expected with the softest base, the Se2- anion. Therefore, H2Se loses a proton most easily, making it the strongest acid. For H2O the acidity is the smallest because the interactions between H+ and O2- are the greatest.
Let us go to a somewhat more complicated example (Fig. 4.2.28). The acidity of perchloric acid, chloric acid, chlorous acid, and hypochlorous acid declines from HClO4 to HClO3 to HClO2 to HClO. In this case all protons are bound to oxygen, so we cannot argue as before. However, we can argue that the negative charge in the anions of the acids is most delocalized in the case of the perchloric acid because the greatest number of resonance structures can be drawn for perchloric acid. Because the negative charge is most delocalized, the electron is most polarizable, and thus the softest. For the hypochloric acid we have the opposite case. There is no electron delocalization possible and only one resonance structure can be drawn for the hypochlorite anion. It is therefore the hardest, interacting the strongest with the proton.
Lastly, let us think about the relative basicity of NH3, PH3, and AsH3 (Fig. 4.2.29)? The N atom is the hardest base, and the interactions with protons are the strongest. Therefore, NH3 is the strongest base. AsH3 is the weakest base because As is the softest atom making the weakest interactions with protons.
Dr. Kai Landskron (Lehigh University). If you like this textbook, please consider to make a donation to support the author's research at Lehigh University: Click Here to Donate. | textbooks/chem/Inorganic_Chemistry/Inorganic_Coordination_Chemistry_(Landskron)/04%3A_Acid-Base_and_Donor_Acceptor_Chemistry/4.02%3A_Hard_and_Soft_Acids_and_Bases.txt |
Concept Review Questions
Section 1
1. What is the definition of a Lewis acid and a Lewis base?
2. What is a Lewis acid-base adduct?
3. What is a dative bond?
4. A Lewis base is typically a molecule with an approximately non-bonding HOMO? Explain why.
5. A Lewis acid is typically a molecule with an approximately non-bonding LUMO. Explain why.
Section 2
1. What is meant by a “soft” and a “hard” atom?
2. What is the influence of donor atom size on the hardness of a base?
3. What is the influence of acceptor atom size on the hardness of an acid?
4. What is the influence of positive charge on the hardness of an acid?
5. What is the influence of negative charge on the hardness of a base?
6. Explain the concept of Pearson’s absolute hardness.
7. The CH4 molecule does not make significant hydrogen bonding with Cl-. Explain this using Lewis acid/base and molecular orbital theory arguments.
Dr. Kai Landskron (Lehigh University). If you like this textbook, please consider to make a donation to support the author's research at Lehigh University: Click Here to Donate.
Homework Problems Chapter 4
Homework Problems
Section 2
Exercise 1
Decide if the following species are acids or bases according to Lewis theory. Order them with respect to their hardness and polarizability.
Ti3+, Ti4+, Zr3+, Hf3+
Answer
Hardness: Ti4+>Ti3+>Zr3+>Hf3+
Polarizability: Hf3+>Zr3+>Ti3+>Ti4+
Exercise 2
Decide if the following species are acids or bases according to Lewis theory. Order them with respect to their hardness and polarizability.
NH3, PH3, AsH3, SbH3.
Answer
Bases.
Hardness: NH3>PH3>AsH3>SbH3
Polarizability: SbH3>AsH3>PH3>NH3
Exercise 3
CsI is much less soluble in water than CsF. Why?
Answer
Cs-F hard-hard interactions are stronger than Cs-I hard-soft interactions. Nonetheless, CsI is less soluble because of the small solvation enthalpy for I- (weak hard-soft interaction between water and I-).
Exercise 4
Order the following species with regard to their expected solubility in water: ZnS, CdS, HgS.
Answer
ZnS> CdS>HgS
Exercise 5
AlF3 is insoluble in liquid HF but dissolves when NaF is present. When BF3 is added to the solution, AlF3 precipitates. Explain.
Answer
Hard-Hard interactions between H and F in H-F stronger than Hard-Hard interactions between Al3+ and F-. Consequence: no reaction occurs (no formation of AlF4-).
Hard-Hard interactions between Al3+ and F- are stronger than hard-hard interactions between Na+ and F-. Consequence: AlF3 dissolves to form AlF4-.
AlF3 precipitates upon addition of BF3 because Al-F hard-hard interactions are weaker than B-F hard-hard interactions (B3+ is harder than Al3+). AlF4- (aq) + BF3 (aq) --> BF4- (aq) + AlF3(s).
Exercise 6
Why were most of the metals used in early prehistory soft metals (in HSAB terminology)?
Answer
Because oxygen is the most abundant element in the earth crust. Thus most hard metals have formed oxides with oxygen. Soft metals do not easily combine with oxygen to form oxides. Therefore, they can be found more often in elemental form in nature. Because of that they have been used more often.
Exercise 7
Which of the following ions is the softest according to HSAB theory:
a) oxide (O2-)
b) peroxide (O22-)
c) ozonide (O3-)
Answer
c) ozonide (O3-)
Exercise 8
The oxidation of barium with oxygen gas yields barium peroxide while the oxidation of strontium with oxygen gas yields strontium oxide. Explain this from the standpoint of the HSAB theory.
Answer
Barium is softer, so it has a greater tendency to combine with the peroxide ion which is relatively soft compared to the oxide ion.
Exercise 9
A molecule with an electron pair in an antibonding HOMO will most likely undergo a reaction in which
a) this molecule will get oxidized
b) this molecule will be reduced
c) this molecule will act as a donor in a Lewis acid base reaction.
d) this molecule will act as an acceptor in a Lewis acid base reaction.
Answer
a) this molecule will get oxidized
Exercise 10
When a dative bond is polarized toward the acceptor this indicates that
a) the HOMO energy of the donor is higher than the LUMO energy of the acceptor.
b) the HOMO energy of the donor is lower than the LUMO energy of the acceptor.
c) the HOMO energy of the donor is equal to the LUMO energy of the acceptor.
Answer
a) the HOMO energy of the donor is higher than the LUMO energy of the acceptor.
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05: Coordination Chemistry I - Structures and Isomers
Coordination Chemistry
Now let us direct our focus toward actual coordination chemistry, also often called complex chemistry. In the following, we will apply the previously learned concepts about atomic theory, symmetry, molecular orbital theory, and acid-base chemistry to coordination compounds. Let us first ask: How are coordination compounds, or complexes defined? They are typically Lewis-acid base adducts between a metal atom or a metal ion as the Lewis acid and one or more ligands as a Lewis base. These ligands can be inorganic ligands such as halogenide ions, water, and ammonia molecules, or organic ligands like amines or alcohols. Coordination compounds are known since the antique due to their often intense color, and are used as a pigment or dye, for example Prussian Blue (KFe[Fe(CN)6] or tetrammine copper, both of which are intensely blue. You can see a few examples of coordination compounds below, namely the hexahydrate of copper (II) sulfate which is blue, iron (III) chloride, which is yellow, and nickel sulfate which is greenish-blue. Because of their intense color they have also attracted the attention of modern chemists from the early on, but the chemical bonding in these compounds remained a mystery for a relatively long time. The bonding in these compounds seemed more complex, hence the name complex compounds.
History on Coordination Compounds
What was the problem with the bonding in coordination compounds? Their empirical formulas could be easily determined by element analysis, but the results could not be explained with the concept of valence. In the early days of modern chemistry it was believed that the number of bonds in a compound could not exceed the valence. For example, in the compound of the formula Co(NH3)6Cl3 the valence of Co would be +3, therefore cobalt could not make more than three bonds. Assuming that the Co3+ ion would make three bonds to the 3 Cl- ions, how would one involve the six NH3 molecules in the bonding?
The chemist who solved this mystery was Alfred Werner. He correctly suggested that the number of bonds would not be restricted to the valence, but that more bonds would be allowed. It would be possible that the six NH3 molecules bind directly to Co3+ forming a so-called complex cation. They would be in the so-called first coordination sphere around the Co. The three chloride ions would then be bound loosely to the complex cation balancing the charge of the complex cation. They would be in the second coordination sphere.
Formula Writing in Coordination Chemistry
How can we write a formula of a coordination compound that contains complex ions?
For complex cations we write the element symbol for the metal ion first followed by the formula for the ligands. If there is more than one ligand, then we will place the ligands in parentheses, and indicate their number by a subscript behind the parentheses. The entire complex cation is placed in brackets. The formula for the anion in the second coordination sphere is placed behind the brackets and the number of anions is indicated by a subscript (Fig. 5.1.3).
For complex anions, we write the formula for the counter cation first, followed by the complex anion in brackets. You can see two examples above (Fig. 5.1.3 and 5.1.4). We have complex cations of Co with six NH3 ligands coordinated to it in the first coordination sphere. Three Cl- ions are in the second coordination sphere. Hence the formula is [Co(NH3)6]Cl3. The second example is the coordination compound with the complex anion having a 3- charge in which six cyanide anions bind to a central Fe3+ ion. The K+ ion in the second coordination sphere compensates the charge of the complex anion. Hence the formula is K3[Fe(CN)6].
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Classification of Ligands
Let us look a little closer at ligands and see how we can classify them. One way is to categorize them into monodentate and multi-dentate ligands.
Monodentate ligands have only one point of attachment to the metal ion.
Definition: Monodentate ligand
Monodentate ligands have only one point of attachment to the metal ion
Examples for such ligands are halogenide ligands, ammonia as ligand, and water as ligand. The molecules often have different names when acting as ligands, and you must know these names. For example, water as a ligand is called an aqua ligand, ammonia as a ligand is called an ammine ligand, chloride as a ligand is called a chloro ligand.
Ligands with more than one point of attachment are called multidentate ligands, or chelate ligands.
Definition: Multidentate ligands
Multi-dentate ligands have two or more points of attachment to the metal ion
Complexes with chelate ligands are called chelate complexes. Greek prefixes indicate how many points of attachment the ligand has (Fig. 5.2.1).
If there are two, then we have a bidentate ligand, when there are three, we have a tridentate ligand. We use the prefixes tetra, penta, and hexa to indicate four, five, and six points of attachment, respectively. Multi-dentate ligands with more than six points of attachment are rare.
The name chelate ligand comes from the Greek word chela, meaning great claw of the lobster. We can see that our lobster in Figure 5.2.2 happily chelates a Ni2+ ion with its two great claws!
Common Bidentate Ligands
A few common bidentate ligands are shown below.
The first one is ethylene diamine. It has two nitrogen donor atoms that can bind to the metal, and they are separated by an ethylene group (Fig. 5.2.3).
Another common ligand is the oxalate ligand. It has two O donor atoms separated by two carbon atoms. There are two negative charges that are delocalized over the four O atoms (Fig. 5.2.4).
As a third example you can see the bipyridyl ligand having two N donor atoms as part of two aromatic rings. The two N atoms are separated by two carbon atoms (Fig. 5.2.5).
Lastly, there is the acetyl acetonate ligand with two O donor atoms separated by three C atoms (Fig. 5.2.6). The acetyl acetonate carries a negative charge that is delocalized between the two O atoms. In this case the donor atoms are separated by three carbons.
Common ligands often have specific abbreviations. They are often used in the formulas of coordination compounds with these ligands. For example, the ethylene diamine ligand is abbreviated en, the oxalate ligand is abbreviated ox, the bipyridyl ligand is abbreviated bipy, and the acetyl acetonate ligand is abbreviated acac. There are not only bidentate ligands with O and N donor atoms, but also with others such as P and S.
Rings in Complexes with Chelate Ligands
In most chelating ligands the donor atoms are separated by two or three other atoms, mostly carbon atoms. This is because in this case the chelate ligands can form five- and six-membered rings with the metal ion (Fig. 5.2.7).
You can see that the trioxolato iron complex above has three five-membered rings containing one Fe, two O, and two C atoms. These ring sizes are particularly stable because they have the least ring strain. As a consequence, the respective chelate complexes are particularly stable.
Tridentate Ligands
Here are a couple of examples for tridentate ligands.
The first one is terpyridyl, abbreviated “Terpy”(Fig. 5.2.8), the second one is bisethylenetriamine, abbreviated “tris” (Fig. 5.2.9).
Both of them have three N-donor atoms separated by two carbon atoms. In the “tris”-ligand there are two ethylene groups between the N atoms, in the case of the “terpy”-ligand the N-atoms are part of three aromatic rings (Fig. 5.2.8. and Fig. 5.2.9)
Tetradentate Ligands
Two common tetradentate ligands are porphyrine and phtalocyanine (Fig. 5.2.9 and 5.2.10). Both are co-called macrocyclic ligands because they are large cycles. They both have four N-donor atoms pointing inside of the cycle. The phtalocyanine ligand has four additional N atoms connecting the five-membered rings via imine-linkages. Further the phtalocyanine has four benzene rings fused to the four five-membered rings. The porphyrine ligand is very important in biological systems. For example, it is part of the protein hemoglobin. In this case a an Fe2+ ion sits in the center of the porphyrin ring. It is also a component of chlorophyll in which case an Mg2+ ion sits in the center of the ring. Phtalocyanine ligands are important as components of dyes.
Hexadentate Ligand, EDTA
A very common hexadentate ligand is the ethylenediamine tetraacetic acid (EDTA) ligand. You can see its structure below (Fig. 5.2.11).
It has two N-donor atoms separated by an ethylene group. Each N-atom is further connected to two acetyl groups. The overall four acetyl groups carry four O-donor atoms. Overall, there are six donor atoms. The six donor atoms can coordinate octahedrally to a metal ion such as a Ca2+ ion (Fig. 5.2.12).
In coordinated form the EDTA ligand is deprotonated, and the O donor atoms carry a negative charge. Therefore an EDTA complex with a divalent cation such as Ca2+ has a 2- charge.
Nomenclature of Complexes
Now let us develop a nomenclature for coordination chemistry so that we can communicate them in an educated manner. One important aspect is that we name the number of ligands. To indicate the number of a particular ligand we use Greek prefixes (Fig. 5.2.13).
These are the same prefixes we got to know when we discussed multidentate ligands. If there are two ligands we use the prefix di, if there are three we use the prefix tri- and so fourth.
Nomenclature of Complexes with Anionic Coordination Spheres
We can now develop the full name of a coordination compound. Let us consider compounds with complex anions first. We can name them according to three steps.
In the first step we name the counter cation. We do not account for the number of counter cations in the name.
Next, we determine the name and the number of ligands. If the ligand is anionic it gets the suffix “o”. Note that sometimes for ease of pronunciation abbreviations are used. For example a Cl- is a chloro-ligand, and not a chlorido ligand. You must memorize these shorter forms. Generally, when an anion ends with “ide”, the “ide” is omitted, and replaced by “o”.
The third step is to name the metal ion and add the suffix “ate” to the name. You can either add the oxidation number of the metal in roman numerals or the charge of the complex anion in parentheses after the name of the metal. The first nomenclature is called the “Stock” system, the latter the “Ewing-Bassett” system. Note that if the element symbol of the metal is derived from a latin name then the latin name is used. For example if silver is the metal then the complex anion is an argentate, if lead is the metal the complex anion is a plumbate. Also here abbreviations are often used to make pronunciation easier. If the name ends with “um” that ending is replaced by “ate”.
An example for coordination compounds with a complex anion is shown above (Fig. 5.2.14). What is its name? There are three K+ cations, so the name starts with potassium. We realize next, that there are six cyanide anions as ligands, so the name continues “hexacyano”. The name of the metal is iron, but we use the latin name ferrum, and replace the ending “um” with the ending “ate”. The oxidation number of the iron is +3. We can see that from the fact that the complex ion has a 3- charge, and the six cyano ligands have a 1- charge each. If we used the Stock system we would therefore place the roman numerals for +3 in parentheses behind the name. If we used the Ewing-Bassett system, we would place the (3-) for the negative charge of the complex behind the name. So overall it would be either a potassium hexacyanoferrate (III) or a potassium hexacyanoferrate (3-).
Exercises
How would you name the following two compounds?
Let us look at the first example (Fig. 5.2.17). First, we need to name the cation. What is it? It is just “hydrogen”. Next we need to determine the name and number of ligands. We have six chloro ligands, so the name continues with “hexachloro”. The name of metal is platinum. We replace the ending “um” by “ate”. In the Stock system the roman numeral would be (IV) because the oxidation state of the Pt is +4. We can see this from the fact that the complex anion has a 2- charge, and the six chloro ligands have a 1- charge each. We must add +4 to -6 to get to -2. Therefore, in the Stock system, the name would be hydrogen hexachloroplatinate (IV), and in the Ewing-Bassett system it would be hydrogen hexachloroplatinate (2-).
The second example (Fig. 5.2.18) has three K+ cations, so the name starts with potassium. What is the name of the ligand? The name of the anion is thiosulfate. In this case we replace the ending “e” by the ending “o”. We have two ligands, so it is a “dithiosulfato”. The metal is silver, but we use the latin name “argentum”, and replace the ending “um” by the ending “ate”. So it is an “argentate” . Fusing the parts together gives “potassium dithiosulfato argentate”. The oxidation number of Ag is +1, the charge at the anion is -3. So put either (I) in roman numerals or 3- in parentheses behind the name.
Nomenclature of Complexes with Cationic Coordination Spheres
Now let us name complexes with complex cations. We name the complex cation first, and then the anion. Then, we determine the name and the number of the ligands and prefixes accordingly. If there is an anionic ligand we give it the suffix “o” again. Then, we name the metal. In this case, we always use English names. We place the oxidation number in roman numerals or the charge of the complex cation behind the name, depending on whether we want to use the Stock or the Ewing-Bassett system.
For example, how would you name the complex depicted above (Fig. 5.2.19)? There are two NH3 ligands which are neutral. We have to consider that NH3 as a ligand is called an “ammine” ligand. Note that it is spelled with two “m” in the middle. So the name starts “diammine”. The name of the metal is silver, and the oxidation state of silver is +1. This is because the complex has a 1+ charge and the ammine ligands are neutral. The anion is a chloride anion.
Therefore, the name in the Stock system would be “diamminesilver(I) chloride (Fig. 5.2.20).
In the Ewing-Bassett system it would be “Diamminesilver (1+) chloride” (Fig. 5.2.21)
Exercises
What would be the names of the two compounds listed below.
The first example, Fig. 5.2.22, has four NH3 ligands, so the name starts with tetraammine. Platinum is the metal, so the name continues “platinum”. The oxidation state of Pt is +2 because the complex cation has a 2+ charge, and the ligands are charge-neutral. The name of the anion is “choride”. So overall it is a “tetraammineplatinum(II) chloride” according to the Stock system or a “tetrammineplatinum(2+) chloride according to the Ewing-Bassett system.
The second compound (Fig. 5.2.23) has six water ligands, therefore the name starts with “hexaaqua” followed by the name of the metal with is “nickel”. The oxidation state of Ni is 2+ because the charge at the complex cation is +2, and the ligands are charge-neutral. The name of the anion is “chloride”. Therefore, the name is either hexaaaqua nickel (II) chloride or “hexaaquanickel (2+) chloride.
Nomenclature of Complexes with Cationic and Anionic Coordination Spheres
There is also the possibility that a coordination compound is made of a complex cation and a complex anion. In this case, the rules discussed previously hold, the only new thing to learn is that we name the complex cation first and the complex anion second.
In the compound depicted (Fig. 5.2.24) we have a diammine silver (I) cation and a hexacyanoferrate (III) anion.
Hence the name is diammine silver (I) hexacyanoferrate (III) in the Stock system or diammine silver(1+) hexacyanoferrate (3-), Fig. 5.2.24.
Nomenclature of Complexes with More Than One Ligand of the Same Kind
What if there are different ligands in a coordination compound? In this case, we name the ligands in alphabetical order, and give each ligand a prefix according to its number.
For example: What is the name of [Co(NH3)4Cl2]NO3 , Fig. 5.2.26? This is a compound with a complex cation containing ammine and chloro ligands. Because “a” comes before “c” in the alphabet we have to name the ammine ligand first. There are four ammine ligands and two chloro ligands. Therefore, we use the prefixes “tetra” in front of “ammine” and “di” in front of “chloro”. So the name starts “tetraamminedichloro”. Then, we name the metal which is cobalt. The oxidation number of cobalt is +3 because there are four charge-neutral ammine ligands, two anionic chloro ligands, and the charge at the complex cation is +1. +3 -2 = +1. So in the Stock system the compound is called tetramminedichlorocobalt(III) nitrate, in the Ewing-Basset system it is called tetramminedichlorocobalt(1+) nitrate.
Let us do one more example, Fig. 5.2.27. In the compound [Pt(NH3)BrCl(H2O)]SO4 there is a complex cation with four different ligands: an “ammine” ligand, a “bromo”-ligand, a "chloro"-ligand, and an "aqua" ligand. What is the order of them? According to the alphabet, “ammine” comes first, “aqua” is second, “bromo” is third, and “chloro” is fourth. They do not get a prefix because there is just one of them for each. The metal is platinum, and its oxidation state is +4 because the complex cation has a 2+ charge, and there are two neutral ligands, namely the aqua and the ammine ligands, and two anionic ones, namely the bromo and the chloro ligands: +4 - 2 = +2.
Therefore, the name is ammineaquabromochloroplatinum(IV)sulfate in the Stock system, and ammineaquabromochloroplatinum(2+)sulfate in the Ewing-Bassett system, 5.2.28.
Nomenclature of Complexes with Complicated Ligands in a Coordination Sphere
So far we only considered relatively simple ligands that were either monoatomic, or contained a few atoms only. However, many ligands, in particular chelating ligands, contain more atoms, and have more complex names. These names may already contain prefixes that we use to number ligands. For example the ethylenediamine ligand is a chelating ligand with a longer name that already contains the prefix “di”. In such cases, to avoid ambiguity, we put the ligand name in parentheses, and place a somewhat different prefix in front of it to account for the number of the ligands.
Instead of “di” we use “bis”, instead of “tri” we use “tris”, instead of “tetra” we use “tetrakis”, instead of “penta” we use “pentakis”, and so on, Fig. 5.2.29.
For example, what is the name of the complex cation depicted above (Fig. 5.2.30)? We first need to realize that there are two different ligands: chloro ligands and ethylenediamine ligands. We need to name the chloro ligands first, because they come first in the alphabet. Because there are two chloro ligands we use the prefix “di”. The ethylenediamine ligand is placed in parentheses, and the prefix “bis” is used instead of “di”. The metal is cobalt in the oxidation state +3 because the complex cation has a 1+ charge, and there are two chloro ligands with a 1- charge each, and two charge-neutral ethylenediamine ligands: +3 -2 = +1.
Therefore, the name is dichlorobis(ethylenediamine)cobalt(III) in the Stock system and dichlorobis(ethylenediamine)cobalt(1+) in the Ewing-Bassett system (Fig. 5.2.31).
Nomenclature of Complexes with Bridging Ligands
Ligands cannot only be terminal, they can also bridge two metal centers. To indicate that a ligand is a bridging ligand we give it a prefix μ. Bridging ligands are named before terminal ligands, and the name of the molecular fragments with the terminal ligands and metal ions is placed in parentheses after the name of the bridging ligand.
So what is the name of the complex cation depicted above (Fig. 5.2.32)? We can see that there are two different bridging ligands, one is a hydroxo-ligand OH-, and the other is called an amido ligand NH2-. These bridging ligands bridge two identical molecular fragments containing a Co atom and four ammine ligands each. The two bridging ligands need to be named first, and we have to name them according to alphabetic order. Hence, the name starts “ μ-amido- μ-hydroxo”. Then, we need to name the two fragments which are bridged. Because there is one Co and four NH3 ligands, the name of the fragment is “tetraamminecobalt”. Because this is a more complex name we need to put it into parentheses. There are two fragments, therefore we need to use the prefix “bis”. The oxidation state of Co is +3. This is because there are two anionic ligands, the hydroxo, and the amido ligand which have both a 1- charge. The other ligands are neutral. The complex cation carries a 4+ charge: +6-2=+4. Because there are two Co atoms, they have an oxidation number of +6/2=+3.
Hence, the overall name according to the Stock system is “ μ-amido-μ-hydroxobis(tetraamminecobalt)(III)”, Fig. 5.2.33.
In the Ewing-Bassett system it is μ-amido- μ-hydroxobis(tetraamminecobalt)(4+), Fig. 5.2.34. Now we have completed the chapter on nomenclature.
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Coordination Numbers (CN) and Structures of Complexes
Coordination compounds have many different structures or shapes, and therefore it is important that we are able to categorize the structures of coordination compounds, understand why a particular structure forms, and why certain structures are more common than others.
A central parameter that determines the structure is the coordination number. A coordination number is the number of points of attachment between the ligands and the metal.
Definition: Coordination Number
The coordination number is the number of points of attachment between the ligands and the metal
What are the main factors that are associated with the coordination number? Some of them are associated with the metal ion, and some of them are associated with the ligands. The shape and the size of a ligand greatly influences the coordination number. Generally, the bulkier the ligand the smaller than coordination number. With regard to the metal ion, the size of the metal ion plays an important role. The larger the metal ion, the more ligands fit around it, and the coordination number increases. In addition, the electron configuration of the metal ion plays a role. For certain electron configurations certain coordination numbers are preferred because these coordination numbers stabilize electron energies. Lastly, there are intermolecular interactions to consider. In solid state they occur between neighbored complexes, in solution they occur between the complex and solvent molecules. This can lead to different coordination numbers and structures in the solid state, and in solution, respectively.
Main Factors Determining CNs
1. Shape and size of ligands
2. Size of metal ion
3. Electron configuration of metal ion
4. Intermolecular interactions (solution vs. solid state)
As you can see, there are many factors that influence the coordination number, and the shape of the complexes. These factors are sometimes opposing, which makes the prediction of structures difficult. Nonetheless, there are a number of rules that allow for a fair accuracy in structure prediction, and we will go through these rules in the following.
Low Coordination Numbers (CN)
CN = 1
The coordination number 1 is very rare. When there is one point of attachment, then there are usually reactive coordination sites at the metal ion that lead to cluster formation. This reduces the energy and increases the coordination number. For example, in methyl lithium (MeLi) the Li appears to have the coordination number 1 because you may think that there is only a single, covalent Li-C bond. However, there is actually a tetrahedral cluster Li4Me4.
The four Li atoms build a tetrahedron and the methyl groups are placed above the faces of the tetrahedron (Fig. 5.3.1). One can also think of MeLi as a near cubic molecule in which the Li atoms occupy every other vertice of the cube, and methyl groups the remaining vertices of the cube (Fig. 5.3.2). The cluster formation avoids the coordination number 1, and stabilizes the methyl lithium. In addition to a tetrameric cluster there is also a hexameric cluster for Me-Li known.
.
The coordination number 1 is only possible when a large metal ion is surrounded by a very bulky ligand that suppresses cluster formation, and there is no coordination of solvent molecules to the metal.
An example the the cluster 2,6,-Trip2C6H3Tl. (Trip = 2,4,6, i-Pr3C6H2), Fig. 5.3.3. It has a Tl+ coordinated to a bowl-shaped, bulky ligand via a Tl-C bond. The Tl fits exactly into the bowl-shaped cavity in the ligand, thereby preventing cluster formation. The compound can be prepared from its lithium derivative and thallium chloride in diethyl ether at 0°C.
CN = 2
The coordination number 2 is also rare, but already a lot more common than the coordination number 1. Generally, d10 metal ions have a tendency to make structures with CN=2. These structures are linear structures. An example of a d10 ion is Ag+. Why is it d10? Ag is located in group 11, so it has 11 valence electrons. If we remove one, there are ten. The energy of the Ag+ ion is minimized when the 4d-subshell is full, and the 5s-subshell is empty. Ag+ makes many linear structures with ligands, example ammine ligands or cyano ligands. In an oversimplified picture, we may explain the structure using VSEPR arguments assuming there is an sp-hybridization of the 5s and one 5p orbital, and the two electron lone pairs at the two ligands get donated into the sp-hybridized orbitals. What other d10-ions can you think of? There would be the higher and lower homologue of the Ag+ ion, the Cu+, and the Au+ ion.
They make linear complexes such as CuCl2- and Au(CN)2-. Hg2+ is also a d10 ion that often makes linear complexes, for example Hg(CN)2-. This linear coordination is not only found in molecular compounds but also in extended solids. For example, HgO and HgS make zig-zag chains of linearly coordinated Hg2+ (Fig. 5.3.4). The bending of the chain occurs at the oxide anions, because of the electron lone pairs at O. Other metal ions but d10 ions make linear structures only when the ligands are very bulky. These examples are rare.
CN = 3
The coordination number 3 is also rare for coordination compounds. They are most common for d10 ions, especially when the ligands are bulky. In most cases the trigonal planar structure, or a structure close to that is adopted.
An example is the Cu(CN)32- anion in which Cu+ is surrounded by three cyanide anions in a trigonal planar fashion (Fig. 5.3.5).
CN = 4
The coordination number 4 is a very common coordination number. It is actually the second-most common coordination number, only surpassed by the coordination number 6. The by far most common structure associated with the coordination number 4 is the tetrahedral structure. Why? It is because the ligands have the greatest distance from each other, and the smallest steric repulsion. Generally, smaller ions and/or larger ligands favor the coordination number 4 over the coordination number 6.
This is the case in the permanganate anion and the chromate anion, for example. Here, the oxidation state of Mn and Cr are +7 and +6 respectively, making the ionic radius very small.
Another example is the tetrakis(pyridyl) copper(1+) ion. Here, the ligands are fairly bulky. The tetrahedral structure is particularly common for d0 and d10 ions, and mostly favored over the octahedral structure. All examples above have either d0 or d10 ions. Cr(VI) and Mn(VII) are d0 and Cu(I) is d10 thus fulfilling this requirement as well.
The second-most common structure after the tetrahedral structure is the square planar structure. It is only common for ions with d8 electron configuration. What are common d8 ions? Examples are the ions Ni2+, Pd2+, and Pt2+. We can find these elements in group 10 of the periodic table. Therefore, they must have ten valence electrons in the neutral state. Because of the 2+ charge, two electrons are removed leaving eight electrons. All eight electrons are in the d-valence subshell, therefore the ions are called d8 ions.
The arguably most well known square planar complex is cis-platinum. It has two chloro and two ammine ligands attached to Pt in cis-position. It is known as anti-cancer drug.
Ni(CN)42- is another example of a group 10 square planar complex.
Can you think of other d8 ions? The group 11 elements Cu, Ag, and Au have eleven valence electrons in the neutral state. In this case we need to remove three electrons to get to eight electrons, meaning that Cu3+, Ag3+, and Au3+ are d8 ions. Cu and Ag are not very stable in the oxidation state +3, therefore examples of square planer Cu(III) and Ag(III) are fairly rare.
Examples are AgF4- and CuF4-, Fig. 5.3.10. However Au3+ is more common and there is a relatively large number of Au(III) complexes, e.g. AuCl4-.
Because group 12 elements are not stable in the oxidation state 4, there are no group 12 d8 ions. However group 9 d8 ions are known. In this case the ions must be in the oxidation state +1 in order to have eight d electrons. Cobalt is not stable in the oxidation state +1, but Rh and Ir are, and square planar Rh(I) and Ir(I) complexes are common.
For example Rh makes a Rh(PPh3)3Cl complex (Fig. 5.3.11). It is a common hydrogenation catalyst. Vasca’s complex Ir(CO)(PPh3)2Cl is another example (Fig. 5.3.11). It has the property to reversibly bind oxygen.
The square planar coordination is possible also with metal ions other than d8 ions, but only when the ligand forces the metal ion into this coordination. Ligands that do this are for example the porphyrin ligand and the salen ligand. They are tetradentate ligands with donor atoms that form a square forcing the metal ion to adopt a square-planar coordination (Fig. 5.3.12).
A last possibility is the seesaw structure. It is know for main group elements, but not for transition metal elements.
An example is SF4 (Fig. 5.3.13). The seesaw structure is derived from the octahedral structure whereby two adjacent corners of the octahedron are occupied by electron lone pairs. Formally one could consider a compound like SF4 as a coordination compound in which four F- ligands bind to an S4+ cation. However, this view is commonly not adopted, and the S-F bonds are not viewed as dative covalent bonds.
CN = 5
For the coordination number 5, the two most common structures are the trigonal bipyramid and the square pyramid. They are about equally common because they have very similar energies. Which one is preferred, depends on the particular circumstances, the nature of the metal, the ligands, the solvent, etc. Many molecules have structures in between, mostly described as distorted trigonal bipyramids. Distortion is even more common in solid state due to packing effects. Packing effects minimize the void space in the crystal.
An example for a trigonal bipyramidal structure is CuCl53-, an example for a square-pyramidal structure is Ni(CN)53- (Fig. 5.3.14). The pentagon, another conceivable structure, is not known. Five-coordinate compounds are known for the full range of transition metals. A difference compared to the structures with the coordination numbers 2 to 4 is, that in the structures with the coordination number 5 not all ligands are symmetry-equivalent. For the trigonal bipyramidal shape we distinguish between axial and equatorial ligands. In square-pyramidal complexes, the fifth ligand at the tip of the pyramid is symmetrically different to the other four at the base of the pyramid.
Because the energy difference between the trigonal bipyramidal and the square pyramidal structure is small, and there is only a small activation barrier between the two structures, they are often fluxional. This means that they can dynamically interconvert.
This interconversion occurs according to a mechanism called the Berry pseudo rotation. In the Berry pseudo-rotation one 120° bond angle between two ligands in equatorial position increases until it is eventually 180°. The initial 180° angle between two ligands in axial position decreases until it is 120°. This means that the two previously equatorial ligands are now axial ligands, and the two axial ligands are now equatorial ligands. As the Berry pseudo rotations occurs, the complex moves from a trigonal bipyramidal structure through a square-pyramidal intermediate to another trigonal bipyramidal structure. The Berry pseudo rotation happens often very fast, therefore in many measurements the five ligands appear identical. For example, the PF5 molecule shows only one signal in the 19F NMR because the movement of F atoms from the equatorial to the axial position and vice versa is too fast for the NMR time scale. At low temperatures, the Berry pseudo rotation may be slow enough so that axial and equatorial positions can be resolved. In compounds with larger ligands the pseudo rotation may also be slow enough so that axial and equatorial positions are resolved in the NMR.
CN = 6
The most common coordination number is the coordination number 6. This coordination number exists for all transition metal electron configurations from d0 to d10. We can explain this by the fact that many metal ions have the right size to support six ligands around them, and there are many ligands that have the right size to surround a metal ion in the coordination number 6. The by far most common shape for the coordination number 6 is the octahedral shape. It is the shape for which the ligands have the largest distance from each other thereby minimizing steric repulsion. A second reason is that the orientation of atomic orbitals supports the octahedral shape, as many orbitals point along the x,y, and z axes, in particular the p orbitals, and the dz2 as well as the dx2-y2.
An example is the hexacyanoferrate(3-) anion in which six cyano ligands surround the metal ion octahedrally (Fig. 5.3.16).
Distortions of Complexes with CN = 6
Octahedral complexes can distort in two basic ways. The first common distortion is the so-called tetragonal distortion. Tetragonal distortion can be achieved either via elongating an octahedron along two opposite vertices, or compressing the octahedron along two opposite vertices.
In the first case, two opposite vertices are further away from the center of the octahedron compared to the other four corners (Fig. 5.3.17).
In the compressed octahedron two opposite corners are closer to the center compared to the four other corners (Fig. 5.3.18). Both the elongated and the compressed octahedron have the same symmetry, meaning they belong to the same point group. Which point group is it? We still have a C4 principal axis, but one only. It goes through the corners of the two opposite vertices along which we distorted. There is a horizontal mirror plane perpendicular to the C4 axis, as well as an inversion center. Thus, the point group must be D4h. In tetragonally distorted octahedra all faces are equivalent, but the distances of the ligands from the center of the octahedron are not the same.
When you elongate or compress the octahedron along to opposite faces you create a trigonal antiprism (5.3.19). It is called antiprism because the two opposite regular triangular faces, shown here in red are oriented in staggered fashion. This type of distortion is called trigonal distortion. The point group of a trigonal antiprism is D3d. The C3 axis goes through the two opposite regular triangular faces of the antiprism. There are three C2 axes standing perpendicular, and going through the centers of the six remaining distorted triangular faces. There are three vertical mirror planes also that bisect the angle between the C2 axes. In a trigonal antiprism the faces are not all equal, but the distance of the vertices are all the same.
When we rotate one of the two opposite triangular faces of an antiprism by 60°, the two faces become eclipsed and a trigonal prism results (Fig. 5.3.20). Trigonal prismatic compounds are mostly observed when there are three bidentate ligands connecting the top and bottom triangular faces. What is the point group? It is D3h. There is a C3 axes going through the two opposite triangles. 3 C2 axes are going through the centers of the three rectangles, and there is a horizontal mirror plane perpendicular to the C3 axis. Does an elongation or compression along the two opposite triangular faces change the symmetry? Think about it. No, it does not change it.
CN = 7
Coordination numbers higher than 6 are considered high coordination numbers. They are significantly less common than the coordination numbers, 4, 5, and 6. A simple explanation is that large coordination numbers require a very large cation or a very small anion of the combination of both to avoid steric repulsion. These conditions are statistically less likely. In addition, large coordination numbers are not so favorable from the stand point of orbital orientation.
Let us first look at the coordination number 7 in more detail. There are three structures possible which are about equally common in nature.
The first one is the pentagonal bipyramid of which ZrF73- is an example (Fig. 5.3.21).
The second one is the capped trigonal prism. TaF72- is an example for it (Fig. 5.3.22). The seventh ligand is placed above a square face of the prism, not a triangular face. Can you imagine why? Think about it for a moment. The answer is that the square face is larger than the triangular face, and thus there is less steric repulsion.
The third structure is the capped octahedron. An example is the tribromotetracarbonyltungstate(IV). In this compound three bromo ligands and three carbonyl ligands occupy the vertices of two opposite faces in the octahedron. The fourth carbonyl ligand is placed above the triangular face with the carbonyl ligands. You can see that like for the coordination number 5, the ligands are not symmetrically equivalent.
CN = 8
Now, let us go to the coordination number 8. What shapes could you imagine that adopt the coordination number 8? The most symmetric, and simple shape is the cube (Fig. 5.3.24).
However, complexes with cubic shapes are not observed. This is because there is another related structure which is energetically more favorable. It is the square antiprism (Fig.5.3.25). In a square antiprism, two opposite square faces are oriented in staggered fashion relative to each other. Relative to the cube one face of the cube is rotated by 45° relative to the opposite face, and the two opposite square faces are interconnected with each other to form triangular faces. The square antiprism is preferred over the square prism because the vertices where the ligands sit have a greater distance to each other compared to the cube.
An example of a coordination compound that makes a square antiprism is the tetraoxolato zirconate(4-) anion.
A structure related to the square antiprism is the dodecahedron. Note that we are not talking about the dodecahedron that is a platonic solid here. We can derive the dodecahedron by squeezing on the two opposite corners of the two squares. This distorts and bends a square to form two additional triangular faces. This produces the dodecahedron (Fig. 5.3.26).
A video of the square antiprism to dodecahedron transformation can be accessed through the link below:
https://www.youtube.com/watch?v=A8G7...ature=youtu.be
An example for this shape is the complex octacyanomolybdate (3-), (Fig. 5.3.26). As you can see from the examples, all metal ions are large metal ions, and the ligands a small, confirming that structures with large coordination numbers are favored by large metal ions and small ligands.
CN = 9
Finally, let us think about shapes associated with the coordination number 9.
One possible shape is the tricapped trigonal prism for which the nonahydridorhenate is an example (Fig. 5.3.27). Here three additional ligands are placed above the three rectangular faces of the trigonal prism. Unsurprisingly, the ligands are small, and the metal ion is large.
Another example is the tricapped antiprism. The nonaammine lanthanum(3+) cation is an example. In this case the three additional ligands are placed above three triangular faces of an antiprism. The coordination number 9 is not the upper limit for coordination numbers. Coordination numbers up to 16 have been observed, but they are rare.
Dr. Kai Landskron (Lehigh University). If you like this textbook, please consider to make a donation to support the author's research at Lehigh University: Click Here to Donate. | textbooks/chem/Inorganic_Chemistry/Inorganic_Coordination_Chemistry_(Landskron)/05%3A_Coordination_Chemistry_I_-_Structures_and_Isomers/5.03%3A_Coordination_Numbers_and_Structures.txt |
This section will be about isomerism in coordination compounds. You may know isomerism already from your organic chemistry classes. Here, we will apply isomerism to coordination chemistry. Some forms of isomerism in organic and coordination chemistry are the same, but there are some forms of isomerism that only occur in coordination chemistry.
Let us first briefly review the definition of isomerism: Isomerism is when two or more compounds have identical empirical formulas but different structures.
Definition: Isomerism
When two or more compounds have identical empirical formulas but different structures they are called isomers.
We distinguish between two basic types of isomerisms: Constitutional isomerism and stereoisomerism. What is the difference? In constitutional isomerism the bonds are not between the same atoms.
Definition: Constitutional Isomers
In constitutional isomers the bonds are not between the same atoms.
In stereoisomerism, the bonds are between the same atoms, but ligands are at different coordination sites.
Definition: Stereoisomers
In stereoisomers, the bonds are between same atoms, but the ligands are at different coordination sites.
Forms of Constitutional Isomerism
Hydrate Isomerism
First, let us look more closely at hydrate isomerism. In hydrate isomerism, in one isomer a water ligand is in the first coordination sphere, and in the other isomer it is in the second coordination sphere. A water molecule in the second coordination sphere is only defined for the solid state, but not in solution. This is because in solution the water molecule would become part of the solvent. Here are a few examples for hydrate isomers.
In the complex [Cr(H2O)6]Cl3 there are six aqua ligands in the first coordination sphere (Fig. 5.4.1). In one hydrate isomer, there are only five water molecules in the first coordination sphere, the sixth one is in the second coordination sphere. One chloride anion has moved from the second to the first coordination sphere. There is a third hydrate isomer which has four aqua ligands in the first coordination sphere and two water molecules in the second coordination sphere. Two chloro ligands are now in the first coordination sphere, and one in the second coordination sphere. More hydrate isomers would be possible theoretically, but for some reason nature does not make them.
Another example is [Co(NH3)4(H2O)Cl]Cl2. It has a hydrate isomer in which one aqua ligand has moved to the second coordination sphere, and one chloride anion has moved from the second to the first coordination sphere (Fig. 5.4.2).
Ionization Isomerism
A second form of isomerism is the ionization isomerism. In this case an ion moves from the first to the second coordination sphere and/or vice versa.
For example, in the compound [Co(NH3)5NO3]SO4 there is a nitrate ion in the first coordination sphere, and a sulfate ion in the second coordination sphere (Fig. 5.4.3). There is an ionization isomer to this compound, in which the sulfate ion is now in the first coordination sphere, and the nitrate is in the second coordination sphere.
Coordination Isomerism
Another form of constitutional isomerism is coordination isomerism. In this isomerism, ligands are bound to different metal ions. Naturally, this implies that this form of isomerism can only exist if at least one isomer has two distinguishable metal ions.
For example, the compound Pt(NH3)2Cl2 has two different coordination isomers. At first glance, they do not seem to be isomers at all. However, we can verify that they have the same empirical formula than the first compound (Fig. 5.4.4).
[Pt(NH3)3Cl][Pt(NH3)Cl3] has two Pt atoms, four NH3 units and four Cl atoms. That makes Pt2(NH3)4Cl4. This formula can be divided by 2 to to give Pt(NH3)2Cl2. We can now easily see that the two compounds are isomers. The third isomer also has two Pt atoms, four NH3 units, and four chloro atoms, so it must be an isomer as well.
Why are they coordination isomers? This is because in the first complex all ligands are bound to the same metal atom, whereas in the other two the ligands are bound to different metal atoms. In these compounds, one Pt atom belongs to a complex cation, and the other one to a complex anion. In the second isomer, three ammine ligands and one chloro ligand are bound to the first Pt atom, and one ammine and three chloro ligands are bound to the second Pt atom. In the third isomer, four ammine ligands are bound to the first platinum atom, and four chloro ligands belong to the second platinum atom.
Coordination isomerism is also possible for metal ions of different elements.
In the example shown (Fig. 5.4.5), the first isomer has ethylenediamine bound to Co and cyanide bound to Cr, whereas in the second isomer ethylenediamine is bound to Cr and the cyano ligand is bound to Co.
Further, in coordination isomers metal ions can be in different oxidation states. We can see that in the two depicted isomers there are complex cations, and complex anions with Pt in in different oxidation states (Fig. 5.4.6). In the first isomer Pt(II) makes the complex cation having four ammine ligands attached to it. Pt(IV) is part of the complex anion being surrounded by six chloro ligands. In the second isomer Pt(IV) forms the complex cation having four ammine and two chloro ligands in the first coordination sphere, and Pt(II) forms the complex anion having four chloro ligands.
Linkage isomerism
Linkage isomerism, also called ambidentate isomerism, is an isomerism that can be observed for ligands that have more than one reactive end. In two linkage isomers, the ligands will bind with different ends to the metal. Which end is reactive depends on the effective HOMO-LUMO interactions. Soft donor atoms tend to bind to soft metals, and hard donor atoms tend to bind to hard metals. Also the solvent can play a big role.
An example of an ambidentate ligand is the thiocyanate anion (Fig. 5.4.7). It can bind either with the sulfur or with the nitrogen end to a metal ion. When it binds with the S-end it is called the thiocyanato-ligand, when it binds with the N-end it is called the isothiocyanato ligand. Which atom binds to the metal can depend on the solvent. In the example shown, thiocyanate binds with S to Pd in polar solvents, but with N in apolar solvents. We could try to rationalize why. A possibility is that in polar solvents the more electronegative N atom can engage in hydrogen bonding which is not possible in apolar solvents. Steric arguments could also play a role. You can see that the triphenyl arsine ligands are fairly bulky. When the ligand binds with the nitrogen, then it binds in linear fashion avoiding steric interference with one of the arsine ligands. So it may be that in apolar solvents steric interactions dominate the behavior, while in polar solvents solvent-ligand interactions are in control.
It is even possible that two, same ambidentate ligands bind with opposite ends to the metal in one and the same molecule.
An example is the complex shown (Fig. 5.4.8). In this molecule, there is a thiocyanato and an isocyanato ligand binding to Pd. What arguments would we have to explain this ambidentate isomerism? Think about it for a moment. We can see that the two methyl groups are far less bulky than the two phenyl groups. When thiocyanate binds with the S-atom, then it can bend away from the two bulky phenyl groups. The second thiocyanate anion binds with the N atom because there is no significant interference between the methyl groups and the linear isothiocyanate ligand. This behavior indicates that thermodynamically, the Pd-N interaction is stronger, but only a little bit stronger because other factors such as steric interference can easily reverse the behavior.
Thio- and isothiocyanato ligands are not the only examples of ambidentate ligands.
Another example is the nitrite anion. It can either bind with the N- end or the O-end to a metal. In the first case it is called a nitroisomer, in the latter it is called a nitrito isomer (Fig. 5.4.9). Nitritoisomers are usually more stable.
Stereoisomers (Configuration Isomers)
Now let us discuss the second major type of isomerism: Stereisomerism. As mentioned previously, in stereoisomerism the bonds are between the same atoms, but the positions at which the ligands bind, the coordination sites, are different. There are two basic types of stereoisomerism: diastereomerism, and enantiomerism (Fig. 5.4.10).
In diastereomerism, the diastereomers are not mirror images to each other.
Definition: Diastereomers
Diastereomers are stereoisomers that are not mirror images of each other. They can be chiral or not chiral.
In enantiomerism, the enantiomers are mirror images to each others. Enantiomers are always chiral molecules.
Definition: Enantiomers
Stereoisomers that are mirror images of each other. They are always chiral.
Diastereomers can be chiral, but do not have to be chiral. Remember, a molecule is chiral when it does not have an improper rotational axis.
Example - Diastereomers and Enantiomers
Let us take an example from organic chemistry to illustrate the difference between enantiomerism and diastereomerism.
Above (on the top left of Fig. 5.4.11) you see the Fischer projection an isomer of a compound called tartaric acid, it is called L-tartraric acid. The first OH group from the top points to the left, and the second one points to the right. This is the natural form of tartaric acid. The real structure of L-tartaric acid is depicted below (bottom left). You can see the the OH group that points to the right in the Fischer projection points toward us, while the other one, that points to the left in the Fischer projection, points away from us.
In the D-tartaric acid isomer, the first OH group in the Fischer projection points to the left, and the second one points to the right. For the real structure this means that the first OH-group points away from us, and the second one points toward us. These two isomers are mirror images to each other, and thus they are enantiomers. You may not immediately see in the real structure that the two are mirror images, but you can see it when you rotate the molecule around the C-C bond axis so that the carboxylic acid groups point to the the left (Fig. 5.4.11). Can you see it now?
There is a third isomer which is the so-called mesotartraric acid. We can draw its Fischer projection with both OH groups pointing to the right side (Fig. 5.4.11). In the real structure now both hydroxo groups point to the front. Is this molecule a mirror image of one of the two previous molecules? No, it is not! Therefore, it is a diasteromer relative to the other two. At first glance it seems that we can draw another molecule which is the mirror image of the mesotatraric acid. It would have two hydroxo groups pointing to the left in the Fischer projection, and two hydroxo groups pointing away from us in the real structure. However, the two molecules can be rotated so that they superimpose, and therefore they are not isomers, but identical molecules.
Cis-Trans Isomerism
Let us now discuss some common forms of stereoisomerism. The cis-trans isomerism is one very common stereoisomerism. It occurs when two, same ligands are in adjacent or opposite positions. For example, in a square planar complex two ligands can be adjacent or in opposite positions. When in adjacent position, the bond angle is 90° and we have a cis-isomer, when in opposite position, the bond angle is 180° and we have a trans-isomer.
The probably most well known example of a cis-isomer in coordination chemistry is cis-platinum which is an anti-cancer drug (Fig. 5.4.12). Its trans-isomers does not have these pharmaceutical properties showing that cis-trans isomerism can have a profound impact on the properties of a molecule. Overall cis-trans isomerism in Pt(II) complexes have been most intensely studied, but cis-trans isomerism is also known for other d8 metal ions in square planar complexes. We can also ask if cis- and trans isomers are diastereomers or enantiomers. Let us look at the example of cis- and trans-platinum to answer this question. Clearly, these two isomers are not mirror images to each other, so they must be diasteromers. Generally, cis- and trans- isomers are diastereomers.
Cis-trans isomerism extends beyond square planar complexes, and is also known for other shapes, for example, the trigonal bipyramidal shape, and the octahedral shape. In the cis-isomer of an octahedral complex two ligands occupy positions on the same face of the octahedron, whereas in the trans-isomer they occupy opposite position of the octahedron.
For example, in the complex diaquabromochlorooxalato cobalt(1-) there are cis and trans isomers known (Fig. 5.4.13). In the trans-isomer the two aqua-ligands stand in opposite position, and there is a 180° angle between them. In the cis-isomer they are in adjacent position, and the angle is 90°. We can see that the two ligands are on the same triangular face of the octahedron, shown in red.
Are there rules that can help us to decide if a cis- or a trans- complex will form? As you might suspect, the largest ligands usually go in trans-position due to steric repulsion arguments. Bidentate ligands usually form the cis-isomer because bidentate ligands are usually designed to make five- or six-membered rings with the metal ion. Are cis-trans isomers chiral? If the ligands are simple, then they are usually not, but if they are more complicated, then they can be. In this context it needs to be said that ideal structures are rarely observed. For example, a square planar complex is rarely an ideal square plane e.g. because the bond angles may be somewhat distorted from the 90° angle due to the fact that the ligands have unequal steric requirements. A slight distortion actually can remove one or many symmetry elements, leading to chirality, however, such small deviations are usually ignored.
Fac-mer Isomerism
Another common type of stereoisomerism in coordination chemistry is fac-mer isomerism. Fac stands for facial and mer stands for meridional. In a fac-isomer the same ligands are on a common face of a polyhedral complex, in the mer isomer they are on a plane that bisects the polyhedron. This kind of isomerism is very common for octahedral complexes, but not restricted to those.
For example the complex triammine trichloro cobalt(III) has a fac- and a mer-isomer. You can see that in the fac-isomer the identical ligands are on two opposite triangular faces of the octahedron. In the mer-isomers they lie on two planes that bisect the octahedron. We can again ask if they are diastereomers or enantiomers? The answer is: They are not mirror images to each other, so they are not enantiomers, but diastereomers.
Fac-mer Isomerism with Tridentate Ligands
Fac-mer isomerism is also common for octahedral complexes with tridentate ligands, for example the ligand diethylenetriamine (Fig. 5.4.15).
In this ligand two NH2 groups and one NH group are interconnected by two ethylene units. For clarity reasons, in complexes with this ligand, often only the N donor atom of the NH2 and NH groups is depicted, and the ethylene diammine units are simplified as bent lines interconnecting the N-atoms. We will use this simplification in the following (Fig. 5.4.16).
Here is an example of bis(diethylenetriamine) cobalt complexes that are fac-mer isomers (Fig. 5.4.16). One ethylenediamine ligand is shown in green, the other one is shown in blue. You can see that the left isomer is the fac-isomer with all three donor atoms of a ligand placed on the same face of the octahedron. On the right side is the mer-isomer, with all donor atoms of the same ligand placed on planes that bisect the octahedron. Also in this case the two isomers are not mirror images and are diasteromers to each other.
Triethylenetetramine Complexes
There is isomerism similar to fac-mer isomerism in octahedral complexes with the tetradentate ligand triethylenetetramine (Fig. 5.4.17).
In this ligand, two NH and two NH2 groups are bridged by four ethylene groups. We can again depict the ligand in a simplified way by just showing the N donor atoms, and by simplification of the ethylene groups as bent lines.
One possibility to realize an octahedral complex with triethylenetetraamine ligands is to orient all three rings in the same plane, we can also say we have an octahedral complex with three coplanar rings. This is shown below (Fig.5.4.18).
The three ethylenediamine units make three five-membered rings which are co-planar to each other. The N-atoms occupy overall four vertices of the octahedron. There are two other generic ligands X located at to opposite vertices of the octahedron. Therefore, this isomer is called the trans-isomer.
We can ask if this molecule is chiral? The answer is no, because there is a mirror plane which is co-planar with the three five-membered rings. In addition to this trans-isomer, there are also two cis-isomers, one is called the α-isomer, the other one is called the β-isomer. In the β-isomer, there are now only two co-planar rings, the third one is now out of plane. The two ligands X are in cis-position relative to each other. Is the trans-isomer a diastereomer or enantiomer to the β -isomer? They are clearly not mirror images to each other, therefore they must be diasteromers. Is this β-isomer chiral? Yes, it is! There is no improper rotational axis. Therefore, an enantiomer must exist, and this enantiomer must be a mirror image of the β-form. Here it is! Naturally, it must also have two co-planar rings, and two ligands X in cis-position. It is therefore also considered a β-isomer.
Now let us discuss the α-isomerism. In an α-isomer no two ligands are co-planar any more. This isomer is a diasteromer to the trans- and the β-form because the isomers are not mirror images. Also the α-isomer is chiral. Therefore, there must be another α-isomer which is the mirror image to the first α-isomer. It is shown in the bottom right of Figure 5.4.18.
Nomenclature for Propeller Complexes
Complexes with two or more non-coplanar rings that are non-adjacent are always chiral and are called “propeller complexes”. There is a special nomenclature for propeller complexes that we will briefly discuss in the following.
Let us first understand why these complexes are propeller complexes? They are called like that because the non-coplanar rings are oriented relative to each other similarly to propeller blades. A propeller must have at least two blades, but can also have more than that. We distinguish between so-called left handed propellers and right-handed propellers. A left-handed propeller has the property to move away from you in a medium like air or water when rotated counter-clockwise.
Definition: Left-Handed Propeller
Counterclockwise rotation moves a left-handed propeller away
A right handed propeller moves away from you when rotated clock-wise.
Definition: Right-Handed Propeller
Clockwise rotation moves a right-handed propeller away
The tip of one blade of a left-handed propeller describes a left-handed helix as it moves away, the tip of a blade of a right-handed propeller describes are right-handed helix as it moves away.
Determining the Handedness of "Propeller Compexes"
How can we determine the handedness of a propeller molecule? Let us determine it using the example of the tris(oxalato) ferrate(III) anion. Fig. 5.4.19.
It is a three-bladed propeller molecule because it has three five-membered rings between the three oxolato ligands and the iron center. Each ring is considered a blade. None of the rings are co-planar and none of them are adjacent, meaning that the rings do not share a donor atom.
In order to determine the handedness, we can first rotate the molecule so that one of the rings is oriented horizontally and points to the back. Unless we have a three-dimensional model, we need to do this in our mind. Once that is accomplished we can draw a horizontal line between the two donor atoms of that ring. Next we can draw a line between two donor atoms of another ring. If necessary, we extend the line so that the line crosses the horizontal line. Next, we determine the smallest angle between the two lines, and rotate the non-horizontal line until it becomes horizontal. We can do this just in our mind. If we need to rotate anti-clockwise, the propeller is considered left-handed, we call it a $\Lambda$ -isomer. If we need to rotate clock-wise, the the propeller is right-handed, we call it $\Delta$-isomer.
Determining Handedness for Propeller Molecules Requiring More Than One Label
In some cases coordination compounds have multiple non-adjacent, non-coplanar ring combinations, and the handedness must be determined for each combination. Examples are octahedral EDTA complexes.
In those complexes there are overall five different rings which we can label R1 to R5. You can see the different rings in different colors (Fig. 5.4.20) For clarity we have abbreviated the linkers between the donor atoms by bent lines in different colors. Each color represents a different linker making a different ring.
We realize that ring R3 is shared with all other rings, therefore it is adjacent to all other rings and does not need to be considered. R1 is adjacent to R2, and R3, but not adjacent and not co-planar with R4 and R5. Therefore we need to consider the R1-R4 and R1-R5 combinations. Further, R2 is coplanar with R3 and R4, and adjacent to R3, but is not coplanar and not adjacent to R5. Therefore, we need to consider the R2-R5 combination as well. R4 and R5 are adjacent, so we do not need to consider this combination. We have found all combinations we need.
We now need to rotate the complex so that one relevant ring is oriented horizontally and points to the back. We can for example choose the ring R4. Luckily, is already correctly oriented. Next, we can connect the donor atoms of the ring R4 to produce a horizontal line. Now let us draw a line between the donor atoms of the ring R1. We see that we need to rotate the horizontal line counter-clockwise to make it parallel to this line, thus the R1-R4 ring combination is a $\Lambda$-configuration.
We can next determine the R1-R5 handedness by first reorienting the molecule so that R5 points to the back. R1 could also be chosen. However, R5 is part of two relevant ring combinations, and so we can kill two bird with one stone if we choose R5. We can move ring R5 to the back by rotating 90° counter-clockwise around the blue axis shown in Fig. 5.4.20. In goes through an O atom, the cobalt atom, and an N atom. This axis stands perpendicular to the square indicated by dotted lines. The square has four donor atoms on the four vertices, and these donor atoms move 90° counter-clockwise as the rotation is carried out. The orientation of the rings follow the movement of the donor atoms. The O-donor atom of the ring R5 moves so that R5 is now horizontal, and points to the back. Next we can interconnect the donor atoms of R5, and the donor atoms of R1 by lines. We can see that the horizontal line must be rotated clockwise so that it becomes parallel with the other one. Therefore, the R1-R5 configuration is a $\Delta$-configuration.
For the determination of the combination R2-R5 the complex does not need to be reoriented again. We can directly draw a line through the donor atoms of R2, and determine the direction of rotation. This time we need to rotate counter-clockwise again, meaning the R2-R5 combination is in $\Lambda$-configuration.
Overall, we have a ΛΔΛ-(ethylenediamine tetracetato)cobaltate(III). Note that the order of designation is arbitrary, we could have named the molecule a ΛΛΔ-complex or a ΔΛΛ-complex also.
Exercise
Let us practice this by one more example.
What is the handedness of the complex shown? First of all: Does it meet the criteria for a propeller complex? The answer is yes. There are two rings, R1 and R2, and they are both not coplanar, and non-adjacent.
To determine handedness we need to first move the molecule so that one ring is oriented horizontally and points to the back. We could choose any of them, but it seems easier to rotate the ring R1 that points to the front and is oriented horizontally by 180° around an axis that is defined by one N and on Cl atom (shown in blue in Fig. 5.4.21)), and is oriented vertically. This rotation moves one of the donor atoms of the other ring by 180°, which defines the new orientation of the ring. Now we can draw a horizontal line through the donor atom of ring 1, and a line through the donor atoms of ring R2. We can see that we must rotate clock-wise, and the propeller complex is right-handed.
Ring Conformations
In addition to isomerism, also conformerism has an influence on the structure of a coordination compound. There may be two or more conformers possible for a particular isomer. As an example, let us consider a ring between a metal ion and an ethylenediamine ligand (Fig. 5.4.22).
In contrast to simplified depictions often used, it is not planar, because of the sp3-hybridization of the carbon and nitrogen atoms. There are two different conformations possible that differ in the orientation of the C-C bond within the ring. They are shown above (Fig. 5.4.22). In the conformation shown left we must rotate a thought horizontal line going through the two nitrogen donor atoms counterclockwise so that it becomes parallel to the C-C bond. For the other conformer, we need to rotate the thought horizontal line clockwise. We call the first conformer the $\lambda$-conformer, and the second one the $\delta$-conformer. In this case we use lower case letters to indicate conformerism and not isomerism.
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Chapter 5
Section 1 & Section 2
1. Explain how the name “chelate” ligand originates.
Section 3
1. What are the main factors determining coordination numbers?
2. Is the coordination number 1 common? Explain.
3. For which electron configuration are the coordination numbers 2 and 3 common?
4. What is the structure of methyl lithium?
5. What electron configuration is common for square planar shapes?
6. What is the most common shape for the coordination number 4?
7. Which are the most common shapes for the coordination number 5?
8. What is the Berry pseudo rotation? Describe its mechanism.
9. What are the most common shapes of coordination compounds with the coordination number 7? What are the point groups associated with these shapes.
10. Explain briefly, why molecular coordination compounds with cubic shapes are not known, while square anti-prisms are relatively common.
11. Explain why coordination compounds with high coordination numbers (7 and larger) are not very common.
12. Name an example of a coordination compound with the coordination number 9.
Section 4
1. What is the difference between a constitutional isomer and a stereoisomer.
2. Name four types of constitutional isomerism.
3. What is the definition of hydrate isomerism?
4. Why is hydrate isomerism only possible in solid state?
5. What is the definition of ionization isomerism?
6. What is the definition of coordination isomerism?
7. What is the definition of linkage (ambidentate) isomerism?
8. What is the difference between a diasteromer and an enantiomer?
9. What is the definition of cis-trans isomerism?
10. What is the definition of fac-mer isomerism?
11. What is the definition of a propeller complex?
12. What is the definition of a left-handed/right-handed propeller?
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Homework Problems Chapter 5
Homework Problems
Exercise 1
Name the following compounds:
1. [(NH3)5Cr-O-Cr(NH3)5 ]4+
2. Na[Fe(NH3)3Cl3]
3. VOCl3
4. Na2PdCl4
5. [Ni(NH3)3(NMe3)](PF6)2
Answer
a) μ-oxo bis(pentaamminechromium (III)) (or 4+)
b) sodium triammine trichloroferrate (II) (or (1-))
c) trichlorooxo vandadium(V) (or (0))
d) sodium tetrachloropalladate (II) (or (2-))
e) triammine (trimethylamine) nickel (II) hexafluorophosphate or triammine (trimethylamine) nickel (2+) hexafluorophosphate
Exercise 2
Write a plausible structural formula for
1. pentaaquathiocyanatoiron(III).
2. dicyanoargentate (I)
3. Tris(triphenylphosphine)gold(I)
Answer
a)
b)
c)
Exercise 3
What other stereoisomers can you draw for the complexes depicted below. Are they diasteromers or enantiomers? Determine the handedness of all isomers.
Answer
Exercise 4
What type of isomerism is present between the following molecules:
Answer
Ionization isomerism
Exercise 5
How many hydrate isomers are theoretically possible for [Al(H2O)6]Cl3. Write down their formulas.
Answer
[Al(H2O)5Cl]Cl2 x H2O
[Al(H2O)4Cl2]Cl x 2 H2O
[Al(H2O)3Cl3] x 3 H2O
Exercise 6
What isomerisms do you see between the following two compounds:
[M(H2O)4Cl2]Br2 and [M(H2O)3ClBr2]Cl x H2O.
Answer
Hydrate isomerism and ionization isomerism
Exercise 7
What distortions are common for the octahedral shape?
Answer
Tetragonal and trigonal distortions
Exercise 8
Which coordination number and shape would you most likely expect for a complex between Ag+ and S2O32-?
Answer
CN = 2, linear shape.
Exercise 9
Which of the following complexes are most likely square planar:
1. AuCl4-
2. ZnCl42-
3. TiCl4
4. FeCl42-
Answer
a) AuCl4-
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06: The 18 Electron Rule
Electron Counting In Transition Metal Complexes
In this chapter we will learn how to count valence electrons in coordination compounds. Electron counting is important because the number of electrons in a complex can tell us a lot about the stability and reactivity in a coordination compound. In addition, it allows us to predict and understand structures to a certain extent. Electron counting sounds trivial, but it is not as trivial as it seems, actually there are even two different methods for electron counting. Each method leads to the same result. Which method you prefer is “personal taste”, but each method is about equally common in the literature, so you need to know both of them.
The Neutral Atom Method
The first method is called the “neutral atom method”. As the name suggests, we will break up the complex into neutral fragments, and count the electrons that contribute to the bonding in each of the fragments. The neutral atom method is carried out according to the following three steps. First, we count the number of the valence electrons of the metal. We consider the metal as neutral atom. The number valence electrons is the same as the group number of the transition metal in the periodic table. For transition metals the group number varies from 3 to 12. In the second step, we account for the ionic charge of the complex, if the complex is not neutral. This will reduce the number of electrons for a complex cation, and increase the number of valence electrons for a complex anion. In the third step, we need to determine how many electrons are contributed by each ligand. This is the most complicated step. To determine the number of electrons must cleave each metal-ligand bond so that a ligand fragment results that is neutral. We then count the number of electrons at the ligand that contributed to the bond. The number of electrons contributed by each ligand is then summed up. This is sum is then added to the number of electrons determined by the previous steps. This is gives the overall number of valence electrons.
Let us apply these rules by an example, the cis-platinum complex (Figure \(1\)). We first need to count the number of electrons of the metal. The metal is platinum which is located in group 10 of the periodic table. Therefore, a neutral platinum atom has ten valence electrons.
Next, we look at the charge of the complex. In this case, there is no charge, and therefore no electrons are added or subtracted.
Lastly, we count the electrons of the ligands. There are two types of ligands, the chloro-ligands and the ammine ligands. Now, our task is to cleave the Pt-ligand bond so that neutral ligand fragments result. We can see that for the chloro-ligands we must cleave the Pt-Cl bond, homoleptically, meaning in the middle, assigning one electron to the Pt and one electron to Cl, because doing so creates a neutral chlorine atom. The fact that we cleaved the bond homoleptically, means that the chloro ligand contributed one electron. Because we have two chloro ligands, there are overall two electrons.
Now, let us think about how many electrons the ammine ligands contribute. In this case we need to cleave the metal-ligand bond heteroleptically to produce a neutral ligand fragment. Both bonding electrons are assigned to the ligand. This produces a neutral NH3 molecule. This means that each ammine ligand contributes two electrons. Overall, that makes four electrons, because we have two ammine ligands.
Finally, we need to sum up the electrons from all three steps. That is ten electrons from Pt, zero electrons due to charge, two electrons from the chloro ligands, and four electrons from the ammine ligands equaling 16 valence electrons total. This is the final result.
"Oxidation State" Method
The oxidation state method is also comprised of three steps. The first step is the same as in the neutral atom method. We determine the number of electrons of the neutral metal which the same as its group number in the periodic table. The next step is different though. It requires the determination of the oxidation state of the metal. How can we determine it? First, we cleave the metal-ligand bonds heteroleptically so that all bonding electrons are assigned to the ligands. Then we determine the charge of the ligands. Ligands can either be neutral or negatively charged. We determine the overall number of charges at the ligands. The difference between that number and the charge of the complex is the oxidation state of the metal. We either add or subtract electrons depending on the oxidation state of the metal. If the oxidation state is positive we subtract electrons, if it is negative, which is rare, then we add electrons. The third step counts the number of the electrons contributed by the ligands. Because we cleaved all bonds heteroleptically, all bonding electrons are considered to be contributed by the ligands, and we count them accordingly. Finally, we sum up the electrons of all three steps which gives the total number of electrons.
Let us apply the method to the previous example cis-platinum. Applying the first step gives us 10 valence electrons for the platinum.
Next, we need to determine the oxidation state of Pt. To do so, we must now cleave all metal-ligand bonds heteroleptically, so that all bonding electrons are assigned to the ligand. By the way, this is equivalent to saying that we treat all the bonds as dative bonds with all electrons coming from the ligands as the donors. When we do this for the chloro ligands we see that this created chloride anions with a 1- charge. Cleaving the Pt-N bonds heteroleptically leads to neutral NH3 molecules. Therefore the overall number of charges at the ligands is 2x(-1)+2x0=-2. The charge at the complex is zero, therefore the oxidation state of Pt is 0-(-2)=+2. We must therefore subtract two electrons from the 10 electrons of the platinum.
Now we must determine the number electrons coming from the ligands. Because all bonds are considered as dative bonds, the chloro ligands contribute two electrons each, and the ammine ligands contribute two electrons each. That makes overall eight electrons.
In sum, 10 electrons from the neutral Pt atom minus two electrons due to the +2 oxidation state of Pt plus 2x2=4 electrons from Cl plus 2x2=4 electrons from NH3 gives 16 electrons total. We can see that we have arrived at the same results as in the case of the oxidation state method.
We can discuss the advantages and disadvantages of both methods, also. The neutral atom method has the advantage that we do not have to think about charges at ligands and oxidation states. However, we have to think about how to cleave bonds to create neutral fragments. We may need to cleave bonds in an way that is not reflecting the donor-acceptor nature of a coordination compound. The oxidation method does account for the donor-acceptor nature of a coordination compound because the bonds are considered dative bonds, and the electrons are assigned to the ligands and metals accordingly. We do not need to think how to cleave bonds, because we cleave the bonds always heteroleptically. However, it requires us to think about charges at the ligands to determine oxidation states which is an additional, non-trivial step.
Counting Electrons: Ligand Contributions
The most difficult step in electron counting is usually the determination of the number of electrons a ligand provides. Therefore, let us practice this by a few examples.
Let us first look at an hydrido ligand which is for example present in the nonahydridorhenate(2-) complex anion (Figure \(3\)). What is the charge at the ligand in the two methods? In the neutral atom method, the charge is always zero, this is a no-brainer. In the case of the oxidation statement method, we need to treat the bond as a dative bond, and that means that we must cleave the bond heteroleptically, so that both bonding electrons can be assigned to the ligand. An H atom with two electrons is a hydride anion with a -1 charge. Next, let us think about the number of electrons donated. In the neutral atom method we need to produce neutral ligand fragments. To do so we must cleave the M-H bond homoleptically, because this will create a neutral hydrogen atom. How many electron will it contribute? It will contribute one electron, because we cleaved the bond homoleptically assigning only one of the two bonding electrons to H. In the oxidation state method, the hydrido ligand contributes two electrons because the bond was considered dative and therefore cleaved heteroleptically. Both bonding electrons were assigned to the ligand, therefore the ligand contributes two electrons.
Next, let us consider a halogenide ligand (Figure \(4\)). What is the charge at the ligand? For the neutral atom method, the answer is trivial, the charge is always zero. In the oxidation state method both bonding electrons in the metal-ligand bond get assigned to the ligand. This gives the ligand a -1 charge. What is the number of electrons contributed? In the neutral atom method we need to think again how to cleave the metal-ligand bond to create a neutral ligand fragment. We need to recognize that we must cleave the bond homoleptically to produce that fragment. Cleaving the bond homolopetically means that the ligand has contributed one electron. In the oxidation state method, we cleave the bond always heteroleptically so that all bonding electrons are assigned to the ligand. Thus, the ligand contributes two electrons.
A halogenide anion as a ligand cannot only be terminal, but also bridging. An example is the μ-dichloro bis(tetraethylene rhodium(I)) complex in which two chloro-ligands bridge two rhodium atoms. What is the charge in the neutral atom method? Of course, it is zero. What is the charge in the oxidation state method? We can see that if we consider both metal-ligand bonds dative bonds, and cleave the bonds heteroleptically, that the Cl atom is surrounded by eight unshared electrons, which gives it a -1 charge. How, many electrons are contributed in the neutral atom method? To answer this question, we need to decide if we have to cleave the bonds homo- or heteroleptically to produce a neutral Cl atom. Can you see it? The answer is: We must cleave one bond homoleptically, and the other one heteroleptically. How many electrons are then contributed by the chloro ligand? It is the two electrons from the heteroleptically cleaved bond, and one electron from the homoleptically cleaved bond. So overall it is three electrons. What about the oxidation state method? In this case both bonds are cleaved heteroleptically, and this means that overall four electrons are contributed.
What are the charges and electrons contributed by an alkoxy ligand (OR), Figure \(6\)? An example for a complex with such as ligand is hexaaphenoxy ferrate (3-). The charge according to the neutral atom method is zero. In the oxidation state method we cleave the bonds heteroleptically, and our ligand becomes an alkoxide anion. This anion has a 1- charge. How many electrons does the ligand contribute? To get a neutral fragment we must cleave the bond homoleptically. This actually produces an alkoxy radical. This radical contributes its radical electron, thus there is one contributed electron. In the oxidation state method we treat the bond as a dative bond and cleave the bond heteroleptically. Therefore two electrons are contributed according to the oxidation state method.
Next, let us apply electron counting to the carbonyl ligand. For example, five carbon monoxide molecules form an iron pentacarbonyl complex with iron. The charge at the ligand in the neutral atom method is zero. In the oxidation state method, we assign both bonding electrons to the ligand, and that produces a neutral carbon monoxide molecule. Note that the carbon atom is formally negatively charged, because it is surrounded by 5 electrons, but the oxygen atom is positively charged because it is surrounded by five electrons. So overall, the molecule is neutral. What is the number of electrons contributed in the neutral atom method? To produce a neutral ligand we must cleave the bond heteroleptically. This produces a neutral carbon monoxide molecule. Because we cleaved the bond heteroleptically, the ligand contributes two electrons. In the oxidation state method we always cleave the bonds heteroleptically, and thus two electrons come from the ligand, too.
The last ligand we discuss here is the isonitrile ligand. An example is the pentakis-(tert-butyl isonitrile) iron molecule. The charge of the ligand is zero in the neutral atom method, but what is it in the oxidation state method? Let us see what happens as we assign both bonding electrons to the ligand. We we see that the carbon is now surrounded by five electrons. Three are in the carbon-nitrogen triple bond and the other two come from the electron lone pair at the carbon atom. This means that the carbon atom has a -1 formal charge. Now, let us look at the N atom. We see that it is surrounded by four electrons, three coming from the C-N triple bond, and one from the C-R bond. That means that the N atom has a +1 charge. Overall, the molecule is neutral and does not carry a charge. What about the number of electrons contributed? We can see that we must cleave the bond heteroleptically to produce a neutral isonitrile, there are two electrons are contributed in the neutral atom method. The oxidation state method must cleave the bond heteroleptically, therefore, the number of electrons contributed is also two.
The 18 Electron Rule
Electron counting is important in the context of an important rule in coordination chemistry: The 18 electron rule. The 18 electron rule states that for d-block elements normally complexes with 18 electrons in the shell (ns2(n-1)d10np6 configuration) are most stable. If this number is not reached, the species is coordinatively unsaturated and tends to add more ligands. It also tends to be reduced because adding electrons brings the complex to, or at least closer to 18 electrons. Coordinatively unsaturated complexes therefore tend to have a higher reactivity.
Definition: Coordinatively Unsaturated Complexes
A complex is coordinatively unsaturated when the 18 electrons are not reached in the (ns2(n-1)d10np6 configuration) shell. It tends to add more ligands, and tends to be reduced. It is associated with higher reactivity.
If a species has more than 18 electrons it is coordinatively oversaturated and tends to lose ligands. It is usually easily oxidized. Both loss of ligands and oxidation reduces to the number of electrons to or at least closer to 18.
Definition: Coordinatively Oversaturated Complexes
A complex is coordinatively oversaturated when it has more than 18 electrons in the shell (ns2(n-1)d10np6 configuration). It tends to lose ligands and tends to get oxidized.
The 18 electron rule has many exceptions, and therefore needs to be applied with caution. In particular, group 3, 4, and 10 complexes deviate often from the 18 electron rule.
For illustration purposes, let us count the number of electrons of the tetrahedral tetrabenzyltitanium(0) complex by the oxidation state method. We could also use the neutral atom method, which would give the same results. This complex is a group 4 complex because titanium is in group 4. How many electrons will the titanium contribute? Because the number of electrons is always the same as the group number, it will contribute four electrons. Next, what is oxidation state of Ti? To determine it we must determine the charge at the ligands. To do that we cleave the bonds heteroleptically. This will give benzylate anions with -1 charge. There are four of these ions, and therefore there will be four negative charges overall. The complex is charge-neutral, and thus the oxidation state is +4 because -4+4=0. Therefore, we need to subtract four electrons. Because we cleaved the bond heteroleptically, each ligand contributes two electrons, giving overall eight electrons coming from the four ligands. This means that we have overall eight electrons, or an 8-electron complex. This is far, far away from 18 electrons. Nonetheless, the complex is stable. How can we explain this? The answer is that in order to achieve 18 electrons it would need to add five additional ligands if each ligand is considered a 2-electron donor. This would increase the coordination number to 9 which is too high to produce a stable complex. In order to reduce the complex to an 18 electron complex, 10 electrons would need to be added. This would produce a complex with a -10 charge which is way to high to be stable. The arguments are generalizable for group 3 and group 4 complexes. Because these elements only have a few d electrons, the ligands would need to contribute a lot of electrons to produce an 18 electron complex. This would require just too many ligands to add. The coordination numbers would get too high. If electrons are added instead of ligands, the negative charge at the complex would be to high to be stable based on electron-electron repulsion arguments.
These arguments cannot be applied for group 10 elements, because these elements have many d electrons. The explanation in this case is that these elements like to make square planar complexes when in the oxidation number is +2. Square planar complexes prefer 16 instead of 18 electrons. We will learn later, when we discuss bonding in coordination compounds, why this is. You can see that the square planar diamminedichloro palladium complex shown is square planar and has sixteen electrons. There are 10 electrons coming from Pd. If we use the neutral atom method, no electrons need to be added or subtracted due to the charge at the complex. The complex is charge-neutral. To assess how many electrons come from the ligands we need to cleave the bonds so that neutral ligands are produced. The Pd-Cl bonds need to be cleave homoleptically, the Pd-N bonds need to be cleave heteroleptically. Therefore, the two chloro ligands are 1e donors, and the two ammine ligands are 2e donors. This gives 10+4+2=16 electrons.
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Concept Review Questions Chapter 6
Concept Review Questions
Section 1
1. Describe the principles of the neutral atom method.
2. Describe the principles of the oxidation state method.
3. What does the 18 electron rule state?
4. What is a complex with 18 electrons called?
5. Why do group 3 and group 4 metal complexes often show deviations from the 18 electron rule?
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Homework Problems Chapter 6
Homework Problems
Section 1
Exercise 1
What is the number of valence electrons of the following complexes? Determine the number of valence electrons using both the neutral atom method as well as the oxidation state method:
1. Ni(CO)4
2. RhI2(PPh3)- (Ph = phenyl)
3. [Au(CN)2]-
4. PtCl4
5. [Fe(CN)6]3-
6. [Fe(H2O)5(SCN)]2+
7. CH3Mg(THF)2Cl (THF = tetrahydrofurane)
8. VOCl3
Answer
a) 18
b) 14
c) 14
d) 14
e) 17
f) 17
g) 8
h) 10
i) 18 | textbooks/chem/Inorganic_Chemistry/Inorganic_Coordination_Chemistry_(Landskron)/06%3A_The_18_Electron_Rule/Chapter_6.1%3A_The_18_Electron_Rule.txt |
Bonding in Coordination Compounds
This chapter is devoted to bonding theories for coordination compounds. Let us first think about, what a good theory should be able to do in general. The answer is, that it should be able to make many correct explanations for experimental observations based on a few, sensible, assumptions. In addition, it should be able to predict experimental observations. The more the theory can explain and predict, and the fewer the necessary assumptions, the better the theory. What does this mean for a bonding theory? What would a good bonding theory for coordination compounds be able to do? It should certainly be able to explain and predict the number of bonds and the shape of a molecule. In addition, it should be able to explain the magnetism of molecules, in particular dia- and paramagnetism. Remember, a molecule is diamagnetic when it has no unpaired electrons. It is paramagnetic when there are unpaired electrons. A diamagnetic molecule is repelled by an external magnetic field. A paramagnetic molecule is attracted by an external magnetic field. It should further be able to explain the stability and reactivity of complexes, as well as the optical properties of complexes. Optical properties of compounds are linked to bonding because they are related to electronic states.
Valence Bond Theory
There are essentially three bonding concepts that are used to describe the bonding in coordination compounds. The first one is the valence bond theory. The valence bond concept was introduced by Linus Pauling in 1931 to explain covalent bonding in molecules of main group elements.
The basic idea is to overlap half-filled valence orbitals to form covalent bonds in which the two electrons are shared between the bonding partners (Fig. 7.1.1). These orbitals can either be atomic orbitals, or hybridized atomic orbitals. The concept works very well to explain the shapes of molecules of main group elements. The valence bond concept in its original form assumes that each bonding partner contributes one electron to the covalent bond. This is not consistent with the dative bonding in coordination compounds where it is assumed that one partner donates an electron pair and the other partner accepts it. To adapt valence bond theory to suit coordination compounds, Pauling suggested that a dative bond is formed via the overlap of a full valence orbital of the donor and an empty valence orbital of the acceptor. We will see that this concept can explain the shapes of coordination compounds in some cases, but overall it does not work very well. We will also see that valence bond theory can explain magnetism in some cases, but also here the valence bond theory has significant deficits. By its nature, valence bond theory cannot explain optical properties. Overall, valence bond theory is far more suitable for main group element molecules compared to transition metal complexes.
Crystal Field Theory
The second major theory is the crystal field theory. It is actually not a bonding theory because it is based on repulsive electrostatic interactions. It was originally developed to explain color in ionic crystals. Later, it was found that it can also explain colors in molecular coordination compounds, and is suitable to explain shapes and magnetism of complexes. However, because it is based on repulsive electrostatic interactions it cannot actually explain what holds the atoms in a molecule together. However, the crystal field theory is quite simple and convenient to use, and there is a lot of practicality to it.
Ligand Field Theory
The third theory is the ligand field theory. It is the most powerful theory, but also the most complicated one. Basically, it is molecular orbital theory applied to coordination compounds. It can make detailed statements about the number of bonds and shapes of molecules, and can explain the magnetism and optical properties of coordination compounds.
Valence Bond Theory for Coordination Compounds
Octahedral Complexes
Let us have a closer look at the valence bond theory, and assess valence bond theory for complexes by a number of examples.
The first example is the hexaammine chromium (3+) cation (Fig. 7.1.2). From experiment we know that it has an octahedral shape, with six dative Cr-N bonds. Can valence bond theory explain the six bonds and the octahedral shape satisfactorily? In order to explain the six dative Cr-N bonds we would need to overlap six empty chromium valence orbitals with six filled valence orbitals of N. We can see that the six ammine ligands have one electron lone pair each that can serve as the valence orbitals. Does chromium have six empty valence orbitals? In order to assess this, we first need to know the oxidation state of the chromium. It is +3 because the ligands are all neutral when the bonds are cleaved heteroleptically, and the complex cation has a 3+ charge. Therefore, the chromium is a Cr3+ cation.
Next, we need to know the electron configuration of the Cr3+. A neutral Cr atom has the electron configuration 4s13d5. When a transition metal loses electrons to form a cation, it always loses its two valence electrons first, and then its d electrons. For chromium this means that we must remove the one 4s electron, and two of the five 3d-electrons. The three remaining 3d electrons are expected to be spin up in three different d orbitals according to Hund’s rule. How many empty valence orbitals remain? These would be two 3d and the 4s orbitals. In addition, it would also be justified to consider the three 4p orbitals as valence orbitals because the 4p orbitals are energetically only slightly higher than the 4s orbital. That means that we would have the six valence orbitals that we would need to explain the six bonds. There is, however, a complication. The six bonds in the complex are not distinguishable, but the six valance orbitals in the Cr3+ ion are distinguishable, for example, the 3d orbitals have different shape and energy than the 4s orbital, which is different from the 4p orbitals. Therefore, if we overlapped these orbitals with the electron lone pairs at N, the bonds would not be equivalent, or indistinguishable. We would have difficulty to explain the highly symmetric octahedral shape of the molecule. To go around this issue, valence bond theory uses the concept of hybridization. In this concept we mathematically mix the wave functions of the valence orbitals to form hybridized orbitals. In our example we would mix the two empty d-orbitals, the 4s orbital, and the three 4p orbitals to form six so-called d2sp3 hybridized orbitals. They have the same shape and size, and their lobes point toward the corners of an octahedron. Therefore, we can now create overlap between these six orbitals, and the six electron lone pairs at N to form six equivalent, indistinguishable Cr-N bonds. We conclude that we have now satisfactorily explained the bonding and the shape of the complex.
Can we also explain its magnetism? From experiment we know that the complex is paramagnetic, and that there are three unpaired electrons. Does valence bond theory predict the same? Yes, it does. There are three unpaired electrons in the three half-filled d-orbitals.
Tetrahedral Complexes
Our next is example is a tetrahedral complex, the tetrahydroxo zincate (2-) complex anion, Fig. 7.1.3.
When viewing it as a Lewis-acid base complex with dative bonds it can be thought as an adduct of a Zn2+ and four hydroxide anions. One of the three electron lone pairs at the hydroxide ions would donate its electrons into empty Zn valence orbitals. That means we would need overall four empty Zn valence orbitals to explain the four Zn-O bonds. A neutral zinc atom has the electron configuration 4s2 3d10. We can derive this from the fact that zinc is in group 12 of period 4 in the periodic table. A Zn2+ ion has two electrons less. Because we must remove s electrons before we remove d electrons, the Zn2+ has the electron configuration 3d10. Like in the previous example we can justifiably consider the 4p orbitals as additional valence orbitals. We can see that we have four empty orbitals available to make the four bonds, namely the 4s and the 4p orbitals, but these orbitals are not equivalent, and do not have the correct orientation to explain the tetrahedral shape of the complex. There is a 90° angle between the p-orbitals which is smaller than the 109.5° tetrahedral bond angle in the molecule. However, we can solve this problem by hybridizing the 4s and the three 4p orbitals to form four sp3-hybridized orbitals. These hybrid orbitals have the property that their lobes point toward the corners of a tetrahedron. Thus, they are suitable to explain the tetrahedral shape of the molecule. We can place the ligands around the Zn2+ ion and approach the ligands on the bond axes to create orbital overlap between the empty sp3-hybridized orbitals and one electron lone pair at the oxygen atom. This produces the tetrahedral tetrahydroxo zincate (2-) anion.
Can we also explain the magnetism of the molecule? What magnetism would valence bond theory predict? We can see that there are no unpaired electrons in any of the metal valence orbitals. Thus, the complex should be diamagnetic. This is also what we find experimentally. Thus, valence bond theory is able to explain the magnetism of this complex anion.
Square Planar Complex
Now let us see if the valence bond theory can also explain a square planar complex such as tetracyanonickelate (2-).
In the valence bond picture we view the Ni-CN bonds as dative bonds, and the complex is considered an adduct of Ni2+ and CN-. To explain the four bonds, the Ni2+ ion would need to have four empty valence orbitals. Ni is a group 10 metal and a neutral Ni atom has the electron configuration 4s23d8. To create a Ni2+ ion we must remove the two 4s electrons, and thus the Ni2+ has the electron configuration 3d8. Do we have four empty orbitals available? Yes, the 4s and the three 4p orbitals are empty but again they are not equivalent and thus not suitable to explain four equivalent Ni-C bonds. Can we hybridize these orbitals? Yes, we can, but the resulting four sp3 hybridized orbitals would not be suitable to explain the square planar shape, only the tetrahedral shape. What valence bond theory suggests in this case is to reverse the spin of one of the unpaired d electrons and move it into the other half-filled d-orbital. This produces an empty d-orbital that we can now hybridize with the 4s and two of the 2p orbitals to four dsp2-hybridized orbitals. These four orbitals have the property that their lobes point toward the vertices of a square, thus they are suitable to explain the square-planar shape. We can approach the ligands now on the bond axes to create orbital overlap between the empty dsp2 Ni and the electron lone pairs of the ligands. We can also say that the ligands donate their electron lone pairs into the hybridized metal orbitals. This produces the four covalent bonds that we need and yields a molecule of a square planar shape.
We can see that the valence bond theory can still explain the square planar shape, but only with the help of the additional assumption that one of the d-electrons gets spin-reversed and moves into another d-orbital. An assumption a theory makes should always be reasonable, so let us critique how reasonable this assumption is. Firstly, is the spin-reversal reasonable? Spin-reversal is a quantum-mechanically forbidden process, and thus it is questionable to assume that it happens. Secondly, there is no good explanation for why the electron moves. The energy of two spin-paired electrons in the same orbital is actually higher than that of two spin-paired electrons in different orbitals. So overall, we see that valence bond theory has difficulties to explain the square planar shape. It must make assumptions that are not very plausible.
Octahedral d5 High and Low Spin Orbital Complex
The valence bond theory has also difficulties to explain so-called high spin and low spin octahedral complexes. For example, it is known from magnetic measurements for 3d5 transition metal ions that they can make octahedral complexes with either one unpaired electron or five unpaired electrons, depending on the ligand. In the first case, the number of paired electrons in the d-orbitals is maximized, and we have a low-spin complex, in the other case the number of unpaired electrons is maximized, and we have a high spin complex. What approach does valence bond theory take to explain this phenomenon?
In the case of a low spin-complex, valence bond theory assumes a so-called inner orbital complex. Like in the square planar complex it is assumed that unpaired electrons reverse their spins and move into other half-filled d-orbitals so that spin-pairing is maximized. In the case of a d5 ion, two electrons reverse their spin, and move into two other half-filled orbitals. This leaves one unpaired electron. We see that due to the movement of the two electrons two 3d-orbitals are empty now, and so are the 4s and the 4p orbitals. The six empty orbitals can now be combined to form d2sp3-hybridized orbitals that can explain the octahedral shapes. Approaching the ligands overlaps the electron lone pair at the ligand with the empty hybrid orbitals to form a dative, covalent bond. We can also say the ligands donate electron lone pairs to form six covalent bonds. We can again criticize that spin-reversal is forbidden and spin-pairing is energetically unfavorable making the approach valence bond theory takes to explain the low-spin complex unsatisfactory.
What about the 3d5 high spin complex (Fig. 7.1.6)? In this case we cannot pair spins to create empty d-orbitals because we need to explain five unpaired electrons. Now, valence bond theory makes another new assumption. It assumes that the outer 4d orbitals get involved in the bonding. These orbitals are empty and available for hybridization. We can therefore hybridize two 4d, the 4s, and the three 4p orbitals to form d2sp3 hybridized orbitals. In the last step we can approach the ligands, and the ligands can donate their electron lone pairs into the transition metal d-orbitals. Now we have explained the six bonds, the octahedral shape, and the five unpaired electrons.
We can again critique the valence bond approach. What justification is there to assume that the 4d orbitals are involved. The answer is: Very little. These orbitals are just too high in energy to be considered valence orbitals. It is not reasonable to assume that they are involved in the bonding. Therefore, again, we see that valence bond theory has difficulties to explain the properties of a complex. Valence bond theory also does not explain distortions of octahedral complexes due to the Jahn-Teller effect.
Octahedral d7High and Low Spin Orbital Complex
High-spin and low-spin complexes are not only observed for octahedral complexes of d5-ions, but for example also for octahedral d7 ion complexes. A low-spin complex has three unpaired electrons and a high-spin complex has one unpaired electron. We will see that valence bond theory has even greater troubles to explain these compounds.
In a d7 ion there are four paired and three unpaired electrons according to Hund’s rule (Fig. 7.1.7). We can reverse the spin of one unpaired electron and pair it with an unpaired electron in another half-filled orbital to reduce the number of unpaired electrons to one. However, this gives us only one empty 3d orbital available for d2sp3-hybridization. In this case we cannot produce more empty 3d-orbitals by reversing the spin. Therefore, we must make again the questionable assumption that outer orbitals are involved in the bonding such as the 4d orbitals. Valence bond theory now suggests to move the unpaired electron from the 3d to the 4d orbital. This is simply done to create another empty 3d orbital that we need for d2sp3-hybridization. However, why would the 3d electron just go into another orbital of much higher energy? If we make this questionable assumption though, we have indeed six orbitals available for hybridization, and we can let the ligands donate an electron pair into the empty hybrid orbitals.
Finally, let us discuss an octahedral d7 high-spin outer orbital complex (Fig. 7.1.8). In this case we cannot pair any spins in the 3d orbitals. Therefore we assume again that the 4d orbitals get involved in the bonding, and hybridize two of them with the 4s and the three 4p orbitals. The six ligands can then donate six electron pairs into the orbitals thereby creating six bonds and explaining the octahedral shape.
Overall, we see that in the valence bond theory we move around electrons as we please in order to explain shapes and magnetism of complexes without good justification. Therefore, the valence bond theory, while extremely valuable for main group compounds, is only of limited use for transition metal complexes.
Crystal Field Theory
Now let us discuss the second bonding theory for coordination compounds, the crystal field theory. It is actually not a bonding theory because it is based on repulsive electrostatic interactions. Nonetheless, it has many features of a bonding theory in the sense that it can explain many phenomena that a bonding can explain, in particular molecular shape, magnetism, and optical properties.
What are the principles of crystal field theory? Crystal field theory assumes that the electrons in the metal d-orbitals are surrounded by an electric field which is caused by the ligand electrons. This electric field is called the crystal field. The name crystal field comes from the fact that this principle was first applied to transition metal ions surrounded by anions in crystals, and was only later extended to transition metal ions surrounded by ligands in molecular coordination compounds. The assumption that ligands surrounding a transition metal ion produce an electric field makes sense because the ligands contain electrons that are associated with an electric field. It is further assumed that the crystal field raises the energy of the metal-d-orbitals because of electrostatic repulsion between the ligand electrons and the metal electrons.
Let us first assume the hypothetical case that the ligand electrons surround the metal d-orbitals exactly spherically (Fig. 7.1.9, bottom left). In this case the electric field is completely isotropic, and this means that the energy of all five metal d-orbitals increases to the same extent. Now let us consider the practical case that the ligands surround the metal in octahedrally, which is the case in an octahedral complex. We can say we have an octahedral crystal field. The electric field will now not be spherical any more, it will be the strongest where the ligands are, namely on the vertices of the octahedron, and less strong elsewhere. The vertices of the octahedron lie on the x, y, and z axes of the coordinate system. Thus, the crystal field is the strongest on the axes, and less strong elsewhere. What consequence does this have on the metal d-orbital energies relative to the spherical crystal field? Orbitals that have their electron density mostly on the axes will experience a greater electrostatic repulsion from the crystal field, and therefore will be higher in energy. Orbitals that will have their electron density mostly elsewhere, meaning not on the axes, will experience smaller repulsion, and thus the energy will be smaller compared to the spherical crystal field. Which are the orbitals that have their electron density mostly on the axes? These are the dz2 and the dx2-y2 orbitals. The dz2 orbital has its energy density mostly on the z-axis, the dx2-y2 – orbital has its energy mostly on the x and the y axes. The energy of both orbitals is increased by exactly the same amount. This is not obvious given that the orbitals have very different shapes. To understand this, it helps to remember that when orbitals are symmetrically degenerate, they also must be energetically degenerate.
An octahedral complex belongs to the point group Oh and in the point group Oh the dz2 and the dx2-y2-orbitals are degenerate and have eg symmetry. Therefore, the dx2-y2 and the dz2-orbitals in an octahedral crystal field are also often just called the eg-orbitals. The remaining d-orbitals, the dxy, the dyz, and the dxz orbitals have their electron density mostly in between the axes, therefore their energy is lower compared to the spherical crystal field. The energy of all three orbitals is reduced by exactly the same amount. We can again understand this when considering that these orbitals are triple-degenerate in the point group Oh and have the symmetry type t2g. For that reason the dxy, the dyz, and the dxz in an octahedral crystal field are also often called the t2g-orbitals. The energy difference between the t2g and the eg orbitals is called Δo. The energy of the t2g orbitals is decreased by 2/5 Δo relative to the spherical crystal field, and the energy of the eg-orbitals is increased by 3/5 Δo relative to the spherical crystal field. Where do the factors 2/5 and 3/5 come from? They are due to the law of the conservation of energy. The overall energy reduction due to the energy decrease of the t2g-orbitals must be equal to the overall energy increase due to the energy increase of the eg orbitals: ΣE(t2g)=-Σ(E(eg). Because there are three t2g orbitals but only two eg orbitals this equation only holds when the energy of the eg orbitals is increased by 3/5 Δo and the energy of the t2g orbitals is decreased by 2/5Δo, or 3 x 2/5 Δo = 2 x 3/5 Δo. Note that the energy of all orbitals will be greater in comparison to the case of no electrical field existing, but the energy is increased to a greater extent for the eg-orbitals compared to the t2g orbitals.
The Tetrahedral Crystal Field
What about a tetrahedral complex with a tetrahedral crystal field?
In this case, the ligands do not approach on the axes (Fig. 7.1.10). We can understand this when we consider that we can inscribe a tetrahedron in a cube. If we connect every other corner of a cube then we obtain a tetrahedron. We can define the coordinate system so that the three axes go through the centers of the square faces of the cube. We can see that the axes do not point toward the ligands, thus the ligands do not approach on the axes. Therefore, the orbitals that have their electron density mostly on the axes have a decreased energy relative to the spherical crystal field. These are the dz2 and the dx2-y2 orbitals. Their energy is the same despite the fact that they have quite different shape. We can explain this again with symmetry arguments. In the point group Td the dx2-y2 and the dz2 – orbitals are double-degenerate and have the symmetry type e. Because they are symmetrically degenerate, they are also energetically degenerate. The energy of the dxy, the dyz, and the dxz orbitals have most of their energy density in between the axes. Therefore, their energies are increased relative to the spherical crystal field. They are increased by the same amount because the orbitals are triply degenerated in the point group Td and have the symmetry type t2. The energy difference between the e and the t2 orbitals is called Δt. The energy of the t2 orbitals is decreased by 2/5 Δt. The energy of the e orbitals is increased by 3/5 Δt. This is again the due to the law of the conservation of energy. The total amount of decreased energy must equal the total amount of increased energy. The tetrahedral crystal field energy is smaller than that of the octahedral field because the octahedral field interacts more strongly with the d-orbitals compared to the tetrahedral field. One can calculate that it is actually just 4/9 of the octahedral field. This is mainly because the lobes of the t2 orbitals to not point exactly toward the vertices of the tetrahedron, while the lobes of the eg orbitals do point exactly toward the vertices of the octahedron.
Tetragonally Distorted Octahedral and Square Planar Crystal Fields
One nice feature of crystal field theory is that can can readily explain distortions such as the tetragonal distortion of octahedral complexes (Fig. 7.1.11).
In a tetragonally distorted complex there is a tetragonally distorted crystal field. In an elongated octahedron two ligands are further away from the metal than the four others. Let us assume these two ligands are on the z-axis. Then, the crystal field is weaker on the z-axis. To keep the overall crystal field constant we must bring the other four ligands closer to the metal center. That means that we compress along the x and the y-axis. What will happen to the energy of the orbitals as the octahedron distorts? Because we elongate in the z-direction, the dz2 orbital, that has most electron density on the z-axis goes down in energy. Because we compress on the x and the y-axis, the energy of the dx2-y2 orbital increases. What about the t2g orbitals? The dxy orbital goes up in energy because it has electron density in the xy plane, but not along the z-axis. The dxz and dyz decrease in energy because they have a significant electron density in z-direction, and the electron density in z-direction is the same for both orbitals. We can now also think about, if the crystal field theory can explain why tetragonal distortion occurs preferentially for certain electron configurations of the metal. For example, it is known that metal ions with d9 electron configuration often make octahedral complexes with tetragonal distortions. For instance, the hexaaqua copper (2+) complex is an example of a tetragonally distorted complex with a Cu2+ ion that has a d9 electron configuration. We can understand that the tetragonal distortion occurs when comparing the energy of the electrons in the undistorted vs the distorted octahedron. In the undistorted octahedron we have three electrons in the degenerated eg orbitals. As we distort, we can move two electrons in the energetically lower dz2-orbital and fill the third one into the energetically higher dx2-y2-orbital. Thus, overall the electrons have a lower energy explaining the distortion. This is an example of the Jahn-Teller effect. In general, the Jahn-Teller effect can occur when there are partially occupied degenerate orbitals. In this case a molecule can lower its energy through distortion. Note that not only partial occupation of the eg-orbitals, but also partial occupation of the t2g-orbitals can cause the Jahn-Teller effect, although the effect is typically smaller. For example in complexes with metal ions the d4-electron configuration, all four electrons can be stabilized through Jahn-Teller distortion. It should also be noted that in addition to an elongation, there is also the possibility of compressing the octahedron along the z-axis. In this case the order of energy of the dz2 and the dx2-y2 orbital reverses, and the order of energy of the dxy, as well as the dxz and dyz reverses as well.
Finally, let us look at the square planar crystal field in square planar complexes. To understand the square planar crystal field it helps to understand the square planar shape as an infinitely elongated octahedron. If we move the two ligands along the z-axis infinitely far away from the metal ion, then we have created a square planar structure. How will the orbital energies change compared to an elongated octahedron? The dx2-y2 orbital will have an even higher energy due to the necessity to compensate for the decreased field associated with the ligands on the z-axis by further compressing along the x and the y-axis. The dz2-orbital is even further decreased in energy because the ligands along the z-axis are now completely gone. The dxy orbital is increased in energy because of the enhanced field in the xy-plane. It is now higher than the dz2-orbital. The dxz and the dyz orbitals are further decreased in energy because they have significant electron density in z-direction.
Viewing the square planar shape as an extreme case of a tetragonally elongated octahedron also lets us understand why the square planar shape is so often adopted by d8-metal complexes. We can see that the stabilization energy, and thus the tendency to distort is the greatest when two electrons are in the metal eg orbitals. In this case two electrons lower their energy through distortion and no electron has an increased energy. Thus, we would understand that the distortion becomes so great, so that the octahedral complex eventually loses two ligands and adopts the square planar shape. This is a nice example how crystal field theory can explain shapes and the number of bonds in a complex without actually being a bonding theory.
High Spin and Low Spin Complexes
One of the greatest strength of crystal field theory is that it can explain high-spin and low spin octahedral complexes in a simple way. The basis for that is the assumption that different ligands produce crystal fields of different strengths and that the differently strong crystal fields produce different Δo values. This assumption is plausible because it can be expected that different ligands interact differently with a metal ion, for example, the bond length or the bond strength may be different. If the ligands produce a large crystal field then we would expect a large Δo, when the crystal field is small, then we would expect a small Δo. The size of the Δo determines if we get a high spin or a low spin complex. If the Δo is larger than the spin pairing energy, then a low spin complex is favored, if Δo is smaller than the spin-pairing energy, then the high spin complex is favored.
For example in a d4-metal complex with small Δo all electrons are unpaired (Fig. 7.1.12, left). Three of them are in the t2g-orbitals, and the fourth one is in the eg-orbital. Now let us assume a different ligand that can produce a Δo that is large enough to overcome the spin pairing energy. In this case the fourth electron would pair the spin of one of the three electrons in the t2g-orbitals (Fig. 7.1.12, right). We would obtain a low spin d4-complex.
In the case of a d5-metal high spin complex there are five unpaired electrons, and all orbitals are half-filled. If the the ligand produces a crystal field large enough, the spin pairing energy is overcome and there are two paired and one unpaired electron in the t2g orbitals (Fig. 7.1.13).
For a d6-metal high spin complex with a weak crystal field there are two unpaired electrons in the t2g and eg orbitals respectively. In the case of a strong field ligand and a low spin complex all electrons are in the t2g orbitals and all spins are paired (Fig. 7.1.14).
In the case of a d7-high spin complex there are two electron pairs and one unpaired electron in the t2g orbitals, and there are two unpaired electrons in the eg orbitals. For the low-spin complex the spin pairing energy is overcome. However, one electron must remain unpaired in the eg-orbitals because the t2g-orbitals are fully occupied with six electrons. The number of unpaired electrons in high and low spin complexes predicted by crystal field theory is what is experimentally observed. Therefore we can state that crystal field theory can quite elegantly explain high and low spin complexes.
We can also understand why there are no d1, d2, d3, d8, d9, and d10 low and high spin complexes. In the electron configurations d1-d3 all electrons are unpaired in the t2g orbitals. Therefore, there is no electron that could be moved from an eg to a t2g orbital. For the configurations d8-d10 all t2g orbitals are necessarily full. Therefore, there is no possibility to move an electron from an eg to a t2g orbital regardless the crystal field strength.
Further, we can explain why there are no low spin tetrahedral complexes. We have learned previously that a tetrahedral crystal field has only 4/9 of the strength of an octahedral field. Because it is so much weaker, no ligand is able to produce a field strong enough to overcome the spin pairing energy. High spin and low spin complexes are possible though for square planar complexes.
Strong and Weak Field Ligands: The Spectrochemical Series
Crystal field theory cannot only explain magnetism well, it can also make statements about the optical properties of a coordination compound. A complex can absorb light when an electron is excited from a d orbital of lower energy to a d orbital of higher energy. The larger the Δ, the smaller the wavelength of the absorbed light. In the case of an octahedral complex, absorption of light would excite an electron from a t2g to an eg orbital.
For example, an octahedral [Cr(H2O)]62+ complex has a smaller Δo compared to an octahedral [Cr(CN)6]4- complex (Fig. 7.1.16), and thus the aqua complex absorbs light of longer wavelength compared to the cyano complex. Vice versa, measuring the absorption spectrum of the complexes, allows us to make statements about the relative crystal field strength of the ligands. By measuring the absorption spectrum of many complexes with a variety of ligands we can develop a so-called spectrochemical series that orders ligands according to their field strength.
You can see such a series containing a non-exhaustive number of ligands in Fig. 7.1.17. You can see that the iodo ligands is on the very left side, and is the weakest field ligand, the carbonyl ligand is on the very right hand side, and is the strongest field ligand. All others are in between. We can see for example, that an aqua ligand is a stronger field ligand compared to a fluoro ligand, but a weaker ligand than an ammine ligand. Generally, ligands at the lower end of the series produce weaker fields, smaller Δs, and absorb light light with longer wavelengths. Ligands at the upper end of the series produce stronger fields, create larger Δs, and absorb light of shorter wavelengths. However, crystal filed theory cannot explain different ligand strength. Why does one ligand produce a stronger field than another? To answer this question, we need the ligand field theory.
Ligand Field Theory
Ligand field theory is the most powerful bonding theory for coordination compounds. It is essentially molecular orbital theory applied to coordination compounds. It can explain covalent bonding in coordination compounds, it can explain their shapes, it can explain their magnetism, and their electronic spectra. It can also explain the stability of coordination compounds, in particular the 18e rule and their exceptions. It is, however, more complicated than other bonding theories.
Just like in molecular orbital theory we can apply symmetry adapted linear combination of atomic orbitals to construct molecular orbitals. However, slightly modified rules apply to optimize molecular orbital theory for coordination compounds. These modifications are useful due to the greater complexity of coordination compounds. In the first step we determine the point group of the molecule and assign axes in a useful way. In the second step, we determine the valence orbitals, also called frontier orbitals of the metal. For example, for a transition metal of the 4th period we would consider the 4s, the 4p, and the 3d orbitals as the frontier orbitals. Next, we determine the symmetry of these orbitals. We can do this by just looking into the character table of the respective point group. In the following we select the highest-energy ligand orbitals that are suitable for σ-bonding. For ligands that are molecules or polyatomic ions, these orbitals are the HOMOs suitable for σ-bonding. For simple ions like chloro-ligands these are the highest occupied atomic orbitals. Next, we group the selected ligand orbitals to form ligand group orbitals (LGOs), and determine their symmetry types. To do so, we determine the reducible representation, and then the irreducible representations of the LGOs. This gives us the symmetry types of the ligand group orbitals. We then combine metal frontier orbitals and ligand group orbitals of the same symmetry type to form molecular orbitals. Now we have constructed all molecular orbitals suitable for σ-bonding, and can draw a molecular orbital diagram for the σ-bonding in the coordination compound.
Next, we look for ligand orbitals that are suitable for π-bonding with the metal. We then determine the symmetry types of their ligand group orbitals. Ligand group orbitals and metal orbitals of the same symmetry will then be combined to form molecular orbitals. These MOs represent the π-bonding in the molecule. We add these molecular orbitals to the molecular orbital diagram. Finally, we check if there are ligand orbitals suitable for $\delta$-bonding with the metal. If so, we also form ligands group orbitals for these, determine their symmetry types, and combine ligand group orbitals and metal orbitals of the same symmetry to form molecular orbitals. These orbitals will then also be added to the molecular orbital diagram.
Applicable rules for the construction of molecular orbital diagrams using ligand field theory
1. Determine point group of the complex and assign axes.
2. Determine which are the relevant metal frontier orbitals for bonding.
3. Determine symmetry types of these metal orbitals.
4. Select ligand HOMOs (suitable for σ-bonding) if the ligand is a molecule. If a simple ion, select the highest occupied atomic orbital.
5. Form ligand group orbitals (LGOs) from selected ligand orbitals and determine their symmetry types.
6. Combine metal orbitals and ligand group orbitals of appropriate symmetries to form molecular orbitals.
7. Select ligand orbitals for $π$ and $σ$ bonding if applicable, determine their symmetry and combine them with appropriate metal orbitals.
Ligand Field Theory for an Octahedral Complex of a 4th Period Transition Metal
Let us apply the ligand field theory to a 4th period octahedral transition metal complex first. According to the rules we must first determine the point group and define the coordinate system. The point group is obviously Oh.
We can define the coordinate system so that the ligands approach on the x, y, and z axes, respectively. Next, we need to see what the frontier orbitals are. These would be the 4s, the 4p, and the 3d orbitals (Fig. 7.1.18). What symmetry types do they have? We can see this by looking into the character table for the point group Oh. An s orbital always has the totally symmetric symmetry type which is always listed first in the character table. In the point group Oh this is the symmetry type A1g. What about the 4p orbitals? We can see that the letters x,y, and z are in parentheses in the irreducible representation of the symmetry type T1u. This means that the three 4p orbitals are triply degenerate and have the symmetry type T1u. Finally, we need to determine the symmetry types of the 3d orbitals. We find xy, xz, and yz in parentheses in the irreducible representation of the symmetry type T2g. Thus, these orbitals have the symmetry type T2g. We further find 2z2-x2-y2 and x2-y2 in the line for the symmetry type Eg , thus the dz2 and the dx2-y2 orbitals are degenerate and have the symmetry type Eg. Remember here from the chapter about atomic structure that 2z2-x2-y2 mathematically describes a cone, and that the dz2 orbital has a conical node. We have now found all the symmetry types of the frontier orbitals.
Next, we need to direct our attention to the ligand and find the highest occupied molecular or atomic orbital suitable for σ-bonding. Of course, it depends now what the ligand is.
As an example let us choose the common and interesting carbonyl ligand (Fig. 7.1.19). To determine its HOMO suitable for σ-bonding with the metal, we will first need to know the molecular orbital diagram for the carbon monoxide molecule. The carbon monoxide molecule is a linear, polar molecule that belongs to the point group C∞v.
The character table of this point group is somewhat hard to work with because of the infinite order of the principal axis, and the infinite number of vertical mirror planes. We will therefore use the subgroup C4v instead (Fig. 7.1.20). A subgroup of a group is a group that results when we remove certain symmetry elements from the original group. We should remove symmetry elements so that degeneracies in the molecular orbitals are not overlooked, which can happen when you reduce the symmetry too much. The point group C4v is the point group with the lowest symmetry we can choose without overlooking degeneracies. Essentially, this is because the atomic orbitals of C and O are only 2s and 2p, and the 2p-orbitals perpendicular to the C-O bond axis can be rotated by 90° to make the orbitals interconvert. This requires a rotational axis of the order 4. If we chose the point group C2v, which has even lower symmetry, we would still be able to construct a molecular orbital diagram, but we would overlook that the 2px and the 2py orbitals are degenerate. We see that in the case of a diatomic linear molecule there is no central atom and no ligands. Therefore, we cannot apply the SALC method exactly as we previously learned it. Therefore, we treat the C and O atoms like central atom orbitals that interact with each other. To determine their symmetry types we can just look into the character table for C4v. We find that the 2s orbitals and the 2pz orbitals have the symmetry type A1 and the 2px and the 2py orbitals are double-degenerate and have the symmetry type E (Fig. 7.1.20). Now we can just combine the atomic orbitals to form molecular orbitals.
To construct a molecular orbital diagram we must consider that O is considerably more electronegative than C, and therefore, the 2s orbital of O is lower in energy compared to the 2s orbital of C. The 2p orbitals of O are lower in energy than those of C. Now we can label our orbitals with their symmetry types, and combine orbitals of the same symmetry type to form molecular orbitals (Fig. 7.1.21). We can start, for instance, with the orbitals of the A1 symmetry type. There are two on each side, so we will have overall four. Our qualitative assumption would be that there should be a strongly bonding and a strongly anti-bonding orbital, and weakly bonding and a weakly anti-bonding orbital in addition. The strongly bonding one would have the lowest energy and be labeled 1a1, the weakly bonding one would have higher energy and is labeled 2a1, the weakly anti-bonding one has an even higher energy and is labeled 3a1, and the strongly anti-bonding one would have the highest energy of all the four, and has the label 4a1. The strongly anti-bonding orbital must be above the 2pz orbital of C, and the strongly bonding one must be below the 2s orbital of O. We can write the 2a1 and the 3a1 orbital at energy levels so that the energy differences between all four a1 orbitals are about the same. We can now also connect the MOs and AOs of a1 symmetry with dotted lines. Now we can turn our attention to the orbitals with E symmetry. There are overall four orbitals of E symmetry, therefore, we expect two bonding double–degenerate orbitals with e-symmetry, and two anti-bonding double-degenerate orbitals with e-symmetry. The bonding MOs must be below the energy level of the 2p orbitals of O, and the two anti-bonding orbitals must be written above the energy levels of the 2p orbitals of C. We can now connect the atomic and molecular orbitals with dotted lines. Finally, we still need to fill in the electrons. There are four electrons coming from the carbon and six electrons coming from the oxygen, which gives ten electrons overall. This means that the 1a1, the 2a1, the 1e1 and the 3a1 molecular orbitals are full. This makes the 3a1 orbital the HOMO. It is suitable for σ-bonding with the metal because it has been created through σ-interactions between the 2pz and the 2s orbitals of O and P respectively.
It is again interesting to compare the MO picture of the bonding in CO with the Lewis-dot structure (Fig. 7.1.19). In the Lewis dot structure we have a triple bond with six bonding electrons. They correspond to the bonding 1a1 and 1e electrons. In the Lewis dot structure all six electrons are energetically indistinguishable, but in the MO diagram we can clearly see that two electrons have a lower energy than the other four which are energetically degenerate. The 1a1 MO is a σ-orbital while the other two are π-orbitals because the 2px and the 2py orbitals interaction in π-fashion. The 2a1 and 3a1 orbitals can be approximated as non-bonding MOs representing the two electron lone pairs at C and O respectively. Because the 3a1 is slightly anti-bonding it has its electron density mostly at C, while the 2a1 is slightly bonding and therefore its counterpart is the electron lone pair at O. Interestingly the dipole moment of CO is slightly polarized toward C by 0.1 Debye despite the fact that O is the more electronegative atom. This can be attributed to the fact that the HOMO as a weakly anti-bonding orbital is primarily located at C, In addition, the 2a1 orbital is fairly close in energy to the 2s of C, therefore the 2s of C contributes significantly to this orbital. This leads to the fact that there is a significant amount of electron density located at C as well.
In the next step, we need to group the six HOMOs to form ligand group orbitals, and determine their symmetry types. This is being done by first determining the reducible representation of the orbitals using the orbital swapping method, and then determining the number of irreducible representations of a given type using the reduction formula from group theory. We will not go through the details of the calculations here, because there is nothing really new to learn, but just discuss the outcome. The outcome is that there is one ligand group orbital with the symmetry type a1g, two doubly degenerated ligand group orbitals with the symmetry type eg, and three triply degenerated ligand group orbitals having the symmetry type t1u. The a1g orbital does not have a node because it is totally symmetric. The t1u orbitals have one node, and the eg-orbitals have two nodes (Fig. 7.1.22). This is not a result of the reduction formula, but one could show that by actually computing the ligand group orbitals using a formula called the projection operator (which we will not discuss in detail here).
Because we now know our the symmetry types of our metal frontier orbitals and our ligand group orbitals we can construct a qualitative molecular orbital diagram (Fig. 7.1.23). To do so we can plot the frontier orbitals on the left side of the molecular orbital diagram, and the ligand group orbitals on the right side of the molecular orbital diagram.
For a 4th row transition metal the sequence of energy is 3d<4s<4p. It makes sense to assume that the ligand group orbitals have about the same energy as the 3d orbitals of the metal, and we can plot them accordingly. Next, we can assign the orbitals their previously determined symmetry types. Then, we can start to combine orbitals of the same symmetry types to form molecular orbitals. We can start for instance with the orbitals having the symmetry type A1. The 4s orbital has that symmetry type. Also one ligand group orbital is of this type. Therefore, we would expect one bonding and one anti-bonding molecular orbital. We can label them 1a1g and 2a1g, respectively, and connect them with the A1g atomic and ligand group orbitals by dotted lines. Next, for example, we can consider the Eg orbitals. There are two metal d orbitals and two ligand group orbitals of that symmetry. We therefore produce two doubly degenerate bonding, and two doubly degenerate anti-bonding molecular orbitals. We can label them 1eg and 2eg respectively. We can further see, that there are the three T1u metal 4p orbitals that we can combine with the ligand T1u orbitals. This gives three triple-degenerated bonding orbitals and three triple-degenerated anti-bonding orbitals with t1u symmetry. Lastly, there are the metal T2g orbitals. There are no ligand group orbitals with the same symmetry, and therefore the T2g orbitals remain non-bonding. We have to write them with unchanged energy into the molecular orbital diagram.
Now we are finished with the construction of molecular orbitals, but still need to fill the electrons into them. We consider the metal-ligand bond a dative bond, with electron pairs being donated by the ligand’s HOMO. Therefore, we consider all ligand group orbitals to be full with electrons. That means that we have overall 6x2=12 electrons to consider. What about the metal electrons? Well, it depends now which metal ion we have. Let us assume here that we have a d0 metal ion with no d electrons. That means that we overall have 12 electrons that we need to fill into the molecular orbitals according to energy. This fills the 1a1g, three t1u, and the two 1eg orbitals. Now let us assume that we would not have a d0 metal ion, but a metal ion with d electrons. There could be up to ten d electrons. Where would they go? They would go into the five orbitals with the next highest energies. Which ones are these? Well, these are the t2g and the 2eg orbitals. The t2g orbitals are the non-bonding metal dxz, dxy, and dyz orbitals, therefore these orbitals are located only at the metal. The 2eg orbitals are anti-bonding molecular orbitals that have been constructed from the dz2 and the dx2-y2 orbitals, and are similar in energy to the dx2-y2 and dz2 orbitals. We can therefore say that the t2g and the 2eg orbitals are the d orbitals under the influence of an octahedral ligand field. Due to the presence of the ligand field the energies of the metal d-orbitals split and the difference in energy is the octahedral ligand field energy $\Delta$o. Now we can see that there is an interesting analogy to the crystal field theory. Like the d-orbitals split in energy under an octahedral crystal field into t2g and eg orbitals, the d-orbitals split in energy in an octahedral ligand field into t2g and eg orbitals. Like in crystal field theory the energy of the dz2 and the dx2-y2 is raised. The energy of the other three d-orbitals is unaffected similar, but not exactly the same as the dxy, dyz, and the dxz orbitals in an octahedral crystal field.
Now we have understood the σ-bonding in an octahedral complex, next let us consider the π-bonding. That means that we have to think about what orbitals of the ligand has the right symmetry for π-bonding with the metal. They should also have an energy similar to the metal orbitals so that significant covalent interaction can occur. We shall stick with our carbonyl ligand, thus we need to look into the molecular orbital diagram of the CO molecule again, and see if there are molecular orbitals suitable for π-bonding (Fig. 7.1.24). The CO molecule has the 1e and the 2e orbitals, that are bonding and anti-bonding π-orbitals, respectively. We can understand this when considering that they are constructed from the 2px and the 2py orbitals that overlap in π-fashion to make the π-bonding in the molecule. Each ligand has two 1e and two 2e orbitals which gives overall four orbitals. These orbitals are energetically similar to the HOMO, thus we can expect that they are energetically close enough to the energy of d-metal orbitals, and can get involved in bonding. We have six ligands meaning that we have 6x12=24 orbitals overall. The twelve 1e orbitals are bonding π-orbitals, and the twelve 2e orbitals are anti-bonding π*-orbitals.
Why are these orbitals suitable for π-bonding with metal d-orbitals? Let us look at their shape, and how they can overlap with a metal d-orbital (7.1.25).
Look for example at a 1e orbital constructed from two 2px orbitals, and how it is oriented relative to the metal-ligand bond axis which we shall define as the z-axis. Next, let us consider a metal dxz orbital orbital. We can see that it is in plane with the 1e orbital, and that the overlap between the dxz and the 1e orbital occurs in π-fashion. Now let us consider the analogous π*-orbital. We can see that it has an additional node, but it can also overlap with a dxz orbital in π-fashion. The ligand does not only have π and π*-orbitals through overlap of two 2px orbitals, but also π and π*-orbitals that result from the overlap of two 2py orbitals. Those orbitals can overlap in π-fashion with a metal dyz orbital. So far, we considered the interactions of a single ligand with the metal only. There are five other ligands, one also approaching from the z-axis, and the other four approaching from the x and y axes. They also have the π and π*-orbitals that interact with the dxz, the dyz, or the dxy orbitals in π-fashion, depending on the direction of approach.
So what do we do with all these orbitals? We group the twelve bonding ones to form a set of ligand group orbitals, and group the twelve anti-bonding ones to form another set of ligand group orbitals. We determine the symmetry types of each set by first determining the reducible representation, and then the irreducible representations using the reduction formula. The result of this process is that the twelve bonding ligand group orbitals have T1g, T2g, T1u, and T2u symmetry. This means that there are three triple-degenerated ones that have T1g symmetry, three other triple-degenerated ones that have T1u, symmetry, three more triple-degenerated ones that have T1u symmetry, and another three triple-degenerated ones that have T2u symmetry. The twelve anti-bonding ligand group orbitals have the same symmetry types. Three have T1g symmetry, three have T2g symmetry, three have T1u symmetry, and another three have T2u symmetry.
It should be mentioned that there are a number of other orbitals that can make π-interactions when the ligand is not a CO ligand. For example halogenide ligands have p-orbitals that have suitable orientation to overlap with metal d-orbitals in π-fashion (Fig. 7.1.26). When there are complexes with metal-metal bonds then there is also the possibility of metal d-orbitals to overlap with other metal d-orbitals in π-fashion.
Now that we have determined the symmetry types of the ligand group orbitals available for π-bonding, we need to select those ligand group orbitals that have suitable symmetry to form molecular orbitals with metal d-orbitals in the octahedral ligand field. These are the 2eg and the t2g MOs that resulted from the σ-interactions. For simplicity sake we will call the 2eg molecular orbitals just the metal eg orbitals. The t2g MOs are identical to the metal T2g atomic orbitals because the metal T2g atomic orbitals remain exactly non-bonding with respect to σ-interactions. The ligand group orbitals have T1g, T2g, T1u, and T2u symmetry respectively. That means that we can combine the t2g metal orbitals and the T2g ligand group orbitals to form molecular orbitals. The metal eg orbitals and the ligand T1g, T1u, and T2u orbitals remain non-bonding because they do not find a partner (Fig. 7.1.27). The three bonding T2g LGOs will form six MOs of the same symmetry type with the three metal t2g orbitals. In addition, we need to consider that there are also three anti-bonding T2g* LGOs that were formed from the anti-bonding π-orbitals. They can also interact with the metal t2g d-orbitals. Because the number of MOs of a given symmetry type is always the sum of the atomic orbitals + the LGOs of that symmetry type this adds three MOs to the six MOs giving overall nine t2g MOs. There are also the T1u*, the T1g*, and the T2u* orbitals. They just remain non-bonding because they do not find a partner.
We can now think of two extreme cases for the combination of the metal t2g with the ligand T2g and T2g* orbitals. In the first case the T2g LGOs are much closer in energy to the metal t2g orbitals, and the T2g* LGOs are energetically much higher than the metal t2g orbitals. In this case we can neglect the covalent interactions between the T2g* orbitals and the metal t2g orbitals, and the T2g* orbitals remain effectively non-bonding. We would consider only the interactions between the T2g LGOs and the metal t2g orbitals to form three bonding and three anti-bonding MOs of t2g symmetry (Fig. 7.1.28). Now we need to consider the electrons. The bonding T2g LGOs are full, and therefore there are overall six electrons to consider. These six electrons would go into the three bonding t2g MOs. Now can also have up to ten metal d electrons. Six of them can be accommodated by the t2g metal orbitals, the other four would be in the eg orbitals. Upon the formation of the π-bond, the t2g metal electrons will be in the anti-bonding t2g orbitals. The eg electrons will simply remain non-bonding. We can see that the π-interactions lower the energy of the ligand electrons, but increase the energy of the metal electrons. If we get a net stabilization of electron energies will depend on how many d electrons we have. As long as there are less than six d electrons we will see a stabilization, if there are more there will be an overall destabilization. We can also ask what impact the π-bonding has on the magnitude of the Δo. We can see that the π-bonding decreases the Δo. It is larger before the bonding compared to after the π-bonding.
We call a ligand that has T2g orbitals of similar energy to the metal t2g orbitals, and T2g* orbitals of much higher energy compared to the metal t2g orbitals a π-donating ligand, or a π-donor (Fig. 7.1.28). This is because before the π-bonding the ligand electrons are localized exclusively at the ligand, and after the π-bonding they are in the bonding t2g MOs which are shared between the metal and the ligand. Thus, a partial electron transfer has occurred from the ligand to the metal. An example for π-donating ligands are halogenide ligands such as a bromo ligand.
Now let us consider the opposite case in which the anti-bonding T2g* ligand orbitals are energetically close to the metal t2g orbitals and the bonding T2g ligand orbitals are energetically too low in order to significantly interact with the metal t2g orbitals. This means that the T2g-orbitals remain practically non-bonding. The T2g* ligand group orbitals and the metal t2g orbitals form three triply degenerated bonding molecular orbitals and three triply degenerated anti-bonding molecular orbitals. Because the T2g* ligand group orbitals are empty, no ligand electrons are filled into the new molecular orbitals. This means that up to six metal d electrons can be filled into the bonding molecular orbitals of t2g symmetry. Any remaining d electrons in the eg orbitals just remain in the eg orbitals and do not change in energy. We can see that in contrast to the previous case, we can lower the energy of the metal d electrons through π-interactions. Because the bonding t2g electrons are shared between the metal and the ligand, electron density has been transferred from the metal to the ligand. Therefore, a ligand that primarily utilizes its anti-bonding T2g*-orbitals for π-bonding is called a π-acceptor ligand, it accepts d-electron density from the metal. An example for a π-acceptor ligand is the carbonyl ligand (Fig. 7.1.29). It is also important to understand the influence of a π-acceptor ligand on the size of Δo. Because the anti-bonding t2g MO is higher in energy than the non-bonding eg orbitals, the Δo is now defined by the energy difference between the bonding t2g MOs and the eg orbitals (Fig. 7.1.29). We can see that Δo is increased when π-acceptor interactions are taken into account.
We have now discussed the two extreme cases, however there are many ligands that are actually in between these two extremes, and there is a continuous spectrum from strongly π-donating, to weakly π-donating, to weakly π-accepting, to strongly π-accepting ligands. It is also possible that π-donor and π-acceptor effects cancel out. This is the case when the the T2g and T2g*-ligand group orbitals are energetically about equidistant to the metal t2g orbitals. Some ligands also do not have orbitals suitable for π-bonding at all, and there are neither π-donating nor π-accepting effects.
The effect of π-bonding on Δo can nicely explain the spectrochemical series. Because π-accepting ligands increase Δo, these complexes absorb light of shorter wavelength and of higher energy. π-donating ligands decrease Δo and and thus lead to the absorption of light having longer wavelengths. We see here that the ligand field theory is more powerful than the crystal field theory. The latter was not able to explain why different ligands produce different Δo values.
Ligand field theory is also able to nicely explain the magnetism of coordination compounds, and high and low spin complexes in particular. It is also able to explain why certain ligands tend to produce low spin complexes, while others tend to form high spin complexes. According to ligand field theory π-acceptors tend to make low spin complexes, and π-donors tend to make high-spin complexes. This is in accordance with experimental observation.
Octahedral Complexes and The 18 Electron Rule
Another great feature of ligand field theory is that it can explain the 18 electron rule, and exceptions from the 18 electron rule. For example, the octahedral hexacarbonyl chromium complex is an 18 electron complex. Let is construct a qualitative molecular orbital diagram and see if the MO diagram supports the stability of the complex. The MO diagram considering only the σ-interactions is shown in Fig. 7.1.30. We can see that all the twlelve ligand electrons are in the bonding molecular orbitals 1a1, 1tu, and 1eg. In addition the chromium has six valence electrons. These electrons remain non-bonding when considering σ-interactions only. However, this changes, when we consider π-interactions. The CO ligand is a strong π-acceptor ligand, therefore we consider only its T2g* π-ligand group orbitals for bonding. They must be located energetically above the LGOs for σ-bonding. The interaction of the metal t2g orbitals creates three bonding t2g MOs and the three anti-bonding ones. Because we can fill metal d electrons into the bonding MOs the status of the d-electrons has changed from non-bonding to bonding. We can see now that all 18 electrons are in bonding MOs, and that no electron is in-non-bonding or anti-bonding MOs. When all electrons are in bonding MOs then this is the ideal situation for complex stability. This is explains why the 18 electron complex is stable. If we analyzed the MO diagrams of many other stable 18 electron complexes then we would also mostly find that all bonding MOs are filled, and all other MOs are empty. This explains the 18 electron rule.
Next, let us construct a qualitative molecular orbital diagram of WCl6. This is not an 18 electron complex, it has only twelve electrons coming from the six chloro ligands. W is in the oxidation state +6 and is a d0 species contributing no electons. Can ligand field theory explain this exception from the 18 electron rule? Let us again start with the MO diagram considering σ-bonding only. The twelve ligand electrons go into the bonding 1a1g, 1tu, and 1eg orbitals. The non-bonding t2g and the anti-bonding 2eg orbitals remain empty due to the absence of metal d electrons. We can see that all bonding molecular orbitals are full and all others are empty, explaining the stability of the molecule, and thus the exception from the 18 electron rule. Now let us consider π-bonding in addition. A chloro ligand is a typical π-donor which uses its 3p electrons that are suitably oriented for π-bonding. Therefore, we only consider the T1g ligand group orbitals for bonding here. These orbitals are full with electrons because a chloride anion has a full 3p subshell. The interaction of the T2g LGOs with the metal t2g orbitals creates a bonding t2g MO and an anti-bonding t2g MO. We can see that the ligand π-electrons now have a lower energy than without the π-interactions of the metal. Therefore, the π-bonding has further stabilized the complex. In a way, we can now even say that we have an 18 electron complex because when we add th 6 π electrons to the 12 σ electrons we get 18 bonding electrons overall. These additional 6 electrons are not accounted for in electron-counting because electron-counting treats the W-Cl bond as a single bond and only considers the σ-interactions between W and Cl.
Tetrahedral complex of a 4th period Transition Metal
Next, let us consider a tetrahedral complex and see if ligand field theory can explain it well. The point group for a tetrahedron is Td, and we will need the character table of this point group (Fig. 7.1.32)
We choose the coordinate system like in the crystal field theory. We inscribe the tetrahedron into a cube by occupying every other vertice of the cube with a ligand (Fig. 7.1.34). Then, we let the axes run perpendicular to the faces of the cube. Next, we need to think about the symmetry types of the metal frontier orbitals. For a fourth period transition metal these are the 3d, the 4s, and the 4p orbitals. Looking into the character table of the point group Td gives us their symmetry types (Fig. 7.1.32 and Fig. 7.1.33). The 4s orbital has the totally symmetric symmetry type A1, We find the letters x,y, and z in parentheses in the irreducible representation of the type T2, and this means that the 4p orbitals are triply degenerated and have the symmetry type T2. The 3dxy, the 3dyz, and the 3dxz orbitals are found in the same irreducible representation, are also triple-degenerated, and have T2 symmetry. The 3dx2-y2 and the 3dz2 orbitals have the symmetry type E according to the character table for Td.
Now we need to think about the ligands. Because we need to consider σ-bonding first, we have to find the HOMO of the ligand suitable for σ-bonding. We can choose again a carbonyl ligand as an example ligand, and in this case the HOMOs of the CO ligands would be used for σ-bonding. This is actually not immediately obvious. If we chose the coordinate system of the metal and ligands to be same same, then then the CO bond axis, which we previously defined as the z-axis, would not point toward the metal, and no σ-overlap with a metal orbital would be possible. Therefore, we must give each ligand a different coordinate system with the z-axes pointing toward the metal (Fig. 7.1.34). Only then, the CO molecule would point toward the metal, and would be oriented to make σ-bonding with the metal. Generally, when constructing MOs, the ligands should always be oriented to that bonding is maximized. This is usually the case when orbital overlap for σ-bonding is maximized.
Because we have four ligands we will have four ligand HOMOs that we would group to form four ligand group orbitals. Which symmetry types do they have? In order to obtain the symmetry types we would need to determine the reducible representation first, and then the irreducible representations this reducible representation is composed of. We won’t do this explicitly here and not go through the entire mathematical process, and only consider the results.
The result is that one ligand group orbital has the symmetry type A1 and the other three have the symmetry type T2 (Fig. 7.1.35).
We can now construct the molecular orbital diagram for σ-bonding (Fig. 7.1.36). First, we can plot the metal frontier orbitals according to the energy on the left side of the diagram, and label the orbitals according to their symmetry types. On the right side we plot the four ligand group orbitals, and label them according to their symmetry type. Their energy should be approximately that of the metal d-orbitals. Now we can construct molecular orbitals by combining atomic and ligand group orbitals of the same symmetry type. The 4s orbital and one LGO have A1 symmetry, and therefore we can combine them to form a bonding 1a1 and an anti-bonding 2a1 orbital. Next, let us direct our attention to the t2 symmetry type. The 4p and three metal d orbitals have that symmetry type, and so have three LGOs. That is overall nine orbitals, so we must get nine MOs with t2 symmetry. Because of triple-degeneracy, there must be three sets of triple-degenerate MOs. We can estimate that one set will be bonding, one will be approximately non-bonding, and one will be anti-bonding. We can label them 1t2, 2t2, and 3t2 respectively. The 2t2 is actually somewhat anti-bonding in nature. Now only the metal e-orbitals are left. They do not find a partner and remain non-bonding.
We still need to fill the electrons into the MOs. The four ligand HOMOs are considered full, which gives 4×2=8 electrons. The electrons occupy the bonding 1t2 and the 1a1 orbitals explaining the four dative metal-ligand bonds. All metal d electrons, which could be up to ten, would need to go into the e and/or the t2-orbitals. The e-orbitals are the non-bonding dz2 and dx2-y2 orbitals, and the 2t2 orbitals are only weakly anti-bonding and have strong d-metal orbital character because they are have been constructed from the dxy, the dyz, and the dxz orbitals, and are fairly similar in energy to those orbitals. We can therefore say that the the e and the 2t2 orbitals are the metal t2 and eg orbitals in a tetrahedral ligand field. The energy difference between the e and the 2t2 orbitals is the tetrahedral ligand field energy ΔT. You see here again the relationship between ligand and crystal field theory. You can also see that ligand field theory can explain why crystal field theory works as a bonding theory even though it is not an actually bonding theory.
π-bonding in a tetrahedral complex
What is the π-bonding in a tetrahedral complex?
First, we need to decide if there are ligand orbitals that are suitably oriented to overlap with metal orbitals in π-fashion. We can see that no ligand orbital is overlapping with a metal d-orbital exactly in π-fashion, however, the e π and 2e π*-orbitals of the ligands still overlap so that at bonding interaction is created, and this orbital overall occurs similarly to π-overlap. Therefore, we can still approximate the bonding interactions as π-bonding. We need to consider though that because of smaller orbital overlap, the π-bonding in tetrahedral complexes is weaker than in octahedral complexes. How many ligand orbitals will we need to consider? Assuming that we will stick with CO as the ligand, there will be four per ligand, and thus there will be 4×4=16 overall. Of these, there will be eight bonding e π-orbitals and eight anti-bonding e π*-orbitals. We group the bonding ones to form a set of eight ligand group orbitals, and group the anti-bonding ones to form another set of eight anti-bonding ligand group orbitals. What will be their symmetry types? We can determine the reducible representations and irreducible representations to find this out. We are not going through the exact process here, but only look at the results. The result is that each set of ligand group orbitals has the two E-type orbitals, three T1-type, and three T2-type ligand group orbitals.
Since we now know the symmetry of the ligand group orbitals we can combine them with same-symmetry metal d-orbitals in the tetrahedral ligand field to form π-molecular orbitals. Let us do this for the example of nickel tetracarbonyl, a tetrahedral complex of Ni in the oxidation state 0 and four carbonyl ligand. We can use the MO diagram for σ-bonding which we constructed previously as a starting point, and modify it so that it accounts for π-bonding (Fig. 7.1.38).
First, we need to add the new ligand group orbitals to the MO diagram. We only need to consider the E and T2 LGOs and not the T1 LGOs because no metal orbital has t1 symmetry. Next, we need to take into account that the CO ligand is a strong π-acceptor ligand. This means that we only need to consider the LGOs formed from the anti-bonding π*-orbitals. We must draw these orbitals above the LGOs for σ-bonding into the MO diagram because the LGOs for σ-bonding have been formed from the CO HOMOs while the anti-bonding LGOs for π-bonding have been formed from the CO LUMOs. We can now combine the π-LGOs and metal d-orbitals of the same symmetry to form π-MOs. We see that we can combine the e-type LGOs with the non-bonding e-type metal dz2 and dx2-y2 orbitals to form a pair of bonding MOs of e-symmetry, and an anti-bonding pair of e-symmetry, which we can label 1e and 2e, respectively. Now what is the effect on the T2-type π-LGOs? We first need to realize that we have already three sets of triple-degenerated t2-MOs generated through σ-interactions. The interaction of the T2-LGOs with the σ-t2-MOs must create another set of triply-degenerated orbitals so that the overall number of orbitals with t2-symmetry remains the same. The interaction will occur mostly between the 2t2 MO and the T2-LGOs because the 2t2 MOs have the most similar energy to the T2-LGOs. This leads to the lowering of the energy of the 2t2 MOs, so that these formerly weakly anti-bonding orbitals become weakly bonding. The additionally created t2-orbitals are weakly anti-bonding. We can label them 3t2. The former anti-bonding 3t2 –orbital will be re-labeled 4t2. We can double-check that we have constructed the t2-MOs correctly by verifying that the number of T2-LGOs, and that includes σ and π, plus the number of metal T2 orbitals equals the number of t2 molecular orbitals. We see that the number of metal T2 orbitals + the number of T2 LGOs is 6+6=12. The number of t2 MOs is 4×3 which is also 12.
Finally, we still need to fill the electrons. As previously mentioned, the π-LGOs as empty and therefore the ligand does not contributed any electrons to the π-bonding. The Ni is in the oxidation state 0 and thus contributes 10 electrons because Ni is in the 10th group of the periodic table. Because a neutral Ni atom has the electron configuration 3d84s2 we can draw eight electrons in to the 3d orbitals and two electrons into the 4s orbitals. The 1t1 and the 1a1 MOs are already full with ligand electrons due to σ-bonding. Thus, the metal electrons must go into the 2e and the 2t2 MOs. Both the e-orbitals and the 2t2-orbitals are bonding, and thus we can say that the metal d-electrons have been stabilized due to the π-acceptor nature of the ligand. We see here an analogy to the octahedral ligand field. Like in the octahedral ligand field π-acceptor ligands tend to lower the energy of the metal d-electrons. We can also ask: What is the effect of a π-acceptor ligand on ΔT? The answer is: Like Δo gets larger in the octahedral ligand field, ΔT gets larger in a tetrahedral ligand field. However, the increase is much smaller compared to the octahedral field. Why? this is because in the octahedral ligand field, the energy of the eg orbitals is not affected through the π-acceptor interactions, and only the t2g orbitals are lowered in energy. In the case of the tetrahedral ligand field both the energy of the e and the t2 orbitals get lowered, and the energy of the e-orbitals gets only slightly more lowered than the energy of the t2 –orbitals. Therefore, the ΔT only slightly increases. This fact can also serve as an additional explanation why tetrahedral complexes never make low-spin complexes, even with strong π-accepting ligands. The effect of the ligand on ΔT is simply still too small because both the energy of the e and the t2 orbitals have been lowered.
Finally, let us discuss the results in the context of the 18 electron rule. We can see that the number of electrons involved in σ-bonding are in bonding molecular orbitals. Thus, it is justified to say that the ligand field theory is able to explain the 18 electrons rule.
Now let us think about how a π-donating ligand influences the magnitude of ΔT. In this case we only need to consider the bonding E and T2 ligand orbitals if the ligand has π-bonds. Those orbitals will be energetically below the LGOs for σ-bonding. If the ligand is a simple ion like chloride, then we would consider LGOs from the filled p-orbitals that that have suitable symmetry for π-overlap. Those LGOs would have the essentially the same energy as the σ-LGOs. Let us construct a qualitative MO diagram of TiCl4 which has chloro ligands which are typical π-donating ligands. When constructing the MO diagram for π-bonding we can again start with the MO-diagram for σ-bonding, and then modify this diagram. We can first plot the filled π-T2 and Eg LGOs with similar energy as the σ-LGOs on the right side of the diagram. The E ligand group orbitals will now interact with the e-metal orbitals. That interaction leads to a pair of bonding MOs of e-symmetry, and a pair of anti-bonding MOs of e-symmetry. This means that effectively, the formerly non-bonding e-metal orbitals become anti-bonding MOs, and move up in energy. The bonding e-type MO is created in addition and must have lower energy than the E-type π-LGOs. The interaction of the T2-LGOs with the metal t2-orbitals creates an additional set of bonding t2-orbitals. The 2t2 orbitals move up in energy, and become more anti-bonding, as a consequence of that. We can just renumber the t2 and the e MOs. Now we still need to fill the electrons into the MOs. The π-LGOs have overall eight electrons. These electrons will go into the newly formed bonding 1e and 1t2 molecular orbitals. We see that this leads to a stabilization of these electrons. This is an analogy to the octahedral ligand field. When we have a π-donor ligand, then that the π-ligand electrons. What about the metal electrons. In TiCl4 the Ti is in the oxidation state +4, therefore it is formally d0, and it does not have any electrons. This explains the stability of the TiCl4 as complex that does not obey the 18 electron rule. It has only 8 electrons, but the bonding situation is nonetheless ideal. All bonding MOs are full, and all others are empty. If the Ti had d electrons they would need to go into the 2e and the 3t2 orbitals which are both anti-bonding. This would destabilize the complex. In general π-donating ligands increase the energy of metal d-electrons, and this is another analogy to the octahedral ligand field. Now to the question, what is the effect of a π-donating ligand on ΔT? Both the metal e and t2 frontier orbitals have increased in energy but the e-orbitals more so than the t2 orbitals. That means that the ΔT has overall decreased. This is a further analogy to the octahedral ligand field. π-donating ligands lead to a smaller ΔT.
Overall the ligand field theory can again provide an explanation for the spectrochemical series. π-donors lead to smaller ΔT ligands, and thus the complex absorbs light of larger wavelengths, π-acceptors lead to larger ΔT and those complexes absorb light of smaller wavelengths.
Square Planar Complex of a 4th Period Transition Metal
As a last example we will discuss now the molecular orbital diagram of a square planar complex. This will be the most complicated MO diagram we will discuss. The greater complexity stems from the lower symmetry in a square planar complex. It belongs to the point group D4h whereas tetrahedral and octahedral complexes belong the high symmetry point groups Td and Oh.
The lower symmetry leads to a reduction in the degeneracy of the molecular orbitals. This creates more energy levels and hence a more complicated molecular orbital diagram. Let us think next about the definition of the axes. I would be sensible to define the xy plane as the plane of the molecule with the x and the y-coordinates going through the bonds. The z-axis would stand perpendicular to the plane of the molecule. What are the symmetry types of the metal frontier orbitals. Assuming a 4th period transition metal the 3d, the 4s, and the 4p orbitals would be the frontier orbitals. Looking into the character table of the point group D4h (Fig. 7.1.10) shows that the symmetry of the 4s orbital is A1g, the symmetry of the 4p orbitals is A2u for the 4pz orbital and eu for the 4px and the 4py orbitals. The d orbitals have the symmetry types A1g, B1g, B2g, and Eg for the 3dz2, dx2-y2, 3dxy, and 3dxz/3dyz, respectively.
Next, we to determine the ligand HOMOs suitable for σ-bonding. There are no square planar carbonyl complexes, therefore we will not choose the carbonyl ligand as a example ligand here. Instead we will use the related cyano ligand. The cyano ligand is isoelectronic with the carbonyl ligand. Formally, the O-atom in CO is replaced by N. Because N has one electron less than O, we must add one electron, giving the cyano ligand a 1- negative charge. Like in CO, the cyano ligand has a triple bond between C and N, and there is an electron lone pair at C, and one at N. The MO diagram is similar to that of CO, just the energy difference between C and N orbitals is somewhat smaller than that of the energy difference between C an O orbitals (Fig. 7.1.41). The number, symmetry, and energy order of the MOs of CN- and CO is the same. Therefore, the CN- has the same HOMOs suitable for σ-bonding, and we can select it for the construction of σ-MOs. Each ligand has one HOMO, therefore we have overall four HOMOs.
We can group them to form LGOs and determine the symmetry types of the LGOs. The result is that there is on ligand group orbital with A1g symmetry, there are two with Eu symmetry, and there is one with B1g symmetry (Fig. 7.1.42).
Square Planar Complex: σ-bonding
We have now the all information necessary to construct a qualitative molecular orbital diagram for σ-bonding (Fig. 7.1.43). As usual we first plot the metal frontier orbitals on the left and the ligand group orbitals on the right side of the MO diagram, and label the symmetry types. Then, we need to combine metal frontier and ligand group orbitals of the same symmetry type to form molecular orbitals. We can for instance begin with the symmetry type A1g. The 4s orbital and the 3dz2 orbitals have this symmetry type, and so has one ligand group orbital. We will therefore get three molecular orbitals of that symmetry type, a bonding one, an approximately non-bonding one, and an anti-bonding molecular orbital. We plot them according to their expected energy into the MO diagram, and label the MOs 1a1g, 2a1g, and 3a1g, respectively. Next, we can look at the B1g orbitals. The 3dx2-y2 orbital and one LGO of that symmetry type, and thus we can form a bonding and an anti-bonding MO from this combination. We can further see that we can combine the 4px and the 4py orbitals that have Eu symmetry with the remaining two LGOs of the same symmetry. This gives two double-degenerated bonding MOs of eu symmetry, and two anti-bonding bonding ones of the same symmetry. Are there any orbitals with other symmetry types left? Yes, there are the metal dxy, dxz, and dyz orbitals that have B2g, and Eg symmetry, respectively, In addition, there is the pz orbital that has a2u symmetry. We can see that these metal orbitals do not find a partner. There is no LGO of these symmetry types. We therefore have to draw these orbitals as non-bonding orbitals into the MO diagram.
Now it is the time to fill the electrons into the molecular orbitals. The LGOs are occupied with electrons because they have been formed from the HOMOs of the ligand. Because we have four LGOs we have eight electrons to consider. These electron will therefore fill the 1a1, the 1b1g, and the 1eu molecular orbitals. These orbitals are all bonding molecular orbitals. Thus, ligand field theory is able to explain the four bonds and the square planar shape of the molecule. Where would the metal d-electrons go? We could have up to ten d-electrons depending on the metal ion, and these ten electrons would go into the five MOs of the next higher energy. These are the non-bonding b2g and eg orbitals, the approximately-non-bonding 2a1g orbital, and the anti-bonding 2b1g orbital. We can consider these orbitals as the metal-d-orbitals in a square planar ligand field. The b2g orbitals is the dz2 orbital in the square planar ligand field, the eg orbitals are dxz and dyz in a square planar ligand field, the 2a1g orbital is the dz2 orbital in a square planar ligand field and the 2b1g orbital is the dx2-y2 orbital in a square planar ligand field. We can see that these results are similar yet not completely analogous to those we obtained in the crystal field theory. Remember, that in the crystal field theory the dx2-y2 orbital had also had the highest energy. However, the dz2 was lower in energy than the dxy, and the dxy was not degenerate with the dxz and the dyz orbitals. The difference it due to the fact that in ligand field theory the dxy orbital is considered non-bonding, not interacting with the ligands at all, whereas in crystal field theory a strong electrostatic repulsion between the dxy and the ligands is assumed because the dxy orbital is in the plane of the molecule.
Square Planar Complex π-bonding
Now let us look at the π-bonding of a square planar transition metal complex (Fig. 7.1.44). If we assume a CN- ligand, then the orbitals that have suitable symmetry to overlap with the metal frontier orbitals in π-fashion are the 1e and 2e π and π* orbitals from the MO diagram of CO. There are two π and two π* orbitals per ligand. Because we have four ligands there are overall eight π and eight π* orbitals. Now we need to consider that four of the π-orbitals are in the xy plane, while the other four are above and below the xy plane. We call the former the parallel orbitals, and the latter the perpendicular orbitals, because they are oriented parallel and perpendicular to the xy plane, respectively. We must distinguish them because they have a different symmetry and overlap differently with the metal d-orbitals. This is illustrated for the bonding π-orbitals in Fig. 7.1.44. The anti-bonding ones behave analogously. Therefore we group the orbitals to form four sets of ligand group orbitals π , π||, π* , and π*||. Our next task is to determine the symmetry types of these orbitals. We again omit the exact process, but only look at the results. Both the π and the π* orbitals have one A2u, two Eg and one B2u ligand group orbitals. The π|| and π*|| orbitals have one A2g, two Eu, and one B2g ligand group orbitals.
We now need to add the π-molecular orbitals to the molecular orbital diagram that we had constructed previously to account for σ-bonding. First, we need to consider which π-LGOs have the right symmetry to produce π-MOs with metal d-orbitals (Fig. 7.1.45). We can see that among the parallel LGOs the A1g and the B1g orbitals find partners, but not the Eu orbitals, For the perpendicular LGOs the Eg orbitals find partners, but not the A2u and the B2u orbitals. We therefore only need to consider the A1g, the B1g and the Eg LGOs. To make matters more complicated, to get a realistic MO diagram for a square planar complex with cyano ligands it is necessary to consider the interactions of the 4s and the 4p orbitals with the π-LGOs as well. The symmetry of the 4s was A1g and the symmetry of the 4p orbitals was A2u and Eu, respectively. The 4s still does not find a partner, but the 4p orbitals do.
MO Diagram of a Square Planar Complex With a Cyano Ligand
Let us now construct the MO diagram of a square planar complex with a cyano ligand considering both σ and π-bonding for the example of the Ni(CN)42- complex anion. Because of the complexity of the diagram we will not modify the MO diagram for σ-bonding, like we did previously, but construct the complete MO diagram from scratch (Fig. 7.1.46). Like previously we can plot the metal 3d, 4s, and 4p orbitals on the left side of the diagram. On the right side of the diagram are now the LGOs for σ as well as π-bonding. For the Ni(CN)42- complex the σ-LGOs are energetically somewhat below the metal d-orbitals, the bonding π-LGOs have about the same energy as the metal d-orbitals, and the anti-bonding π*-LGOs have an energy significantly higher than the 4p orbitals. We can again construct MOs symmetry type by symmetry type until the MO diagram is complete. We can start with the symmetry type A1g. We see there is only one σ-LGO that interacts with the 4s and the 3dz2 which also have A1g symmetry. That leads to a bonding, an approximately non-bonding, and anti-bonding a1g MO which we can label 1a1g, 2a1g, and 3a1g, respectively. Next, we can for instance look at the B2g orbitals. There is one d-orbital that has that symmetry, as well as two LGOs. The interaction is almost only with the bonding B2g LGO because the energies are similar. This leads to a bonding and an anti-bonding π-MO with B2g symmetry. The anti-bonding LGO is too high in energy and it does not significantly interact. Therefore we can draw it with the same energy into the MO-diagram. Now let us go to the B1g orbitals. There is a metal d-orbital with B1g symmetry, and a σ-LGO with B1g symmetry. This produces a bonding and anti-bonding MO with B1g symmetry. Now let us go to the Eg orbitals. There are two Eg metal d-orbitals, two Eg π-LGOs and two Eg π*-LGOs. Again, the π*-LGOs are too far off in energy, so that the bonding interaction is mostly between metal orbital the π-LGOs. This leads to a pair of bonding MOs and a pair of anti-bonding MOs. The π*-LGOs can be drawn with unchanged energy into the MO-diagram. The 4p orbitals of the metal have symmetry types, we have not considered yet: A2u and Eu. Let us begin with the Eu orbitals. There are two σ-Eu LGOs, two π-Eu-LGOs and two π*-Eu LGOs. Overall we have four pairs of EU-orbitals, and we would therefore expect four pairs of eu-MOs. We can make the approximation that there will be a bonding one, a weakly bonding one, a weakly anti-bonding one, and a strongly anti-bonding one. We can label them 1eu, 2eu, 3eu, and 4eu, respectively. The 4eu has almost only contributions from the anti-bonding π*-EU LGOs due to energy arguments, therefore it is only connected to these via a dotted line, and not to the other Eu orbitals. Now to the A2u orbitals: There is a π and a π*-orbital of A2u symmetry that can be combined with the 4p A2u orbitals. This leads to three A2u MOs overall, one bonding, one approximately non-bonding, and one anti-bonding one. Now we have used all metal d-orbitals. We notice that there are still some LGOs that we have not used. These are the π and π* A2g and the B2u LGOs. They remain non-bonding can be drawn accordingly in the MO diagram.
We now need to fill the electrons into the MO-diagram. Firstly, there are the σ-electrons to consider. There are overall eight electrons stemming from the four occupied σ-LGOs. We can fill these eight electrons into the four MOs with the lowest energy. These are the 1a1, the 1eu and the 1b1. These are all bonding MOs explaining the four σ-bonds within the complex. We can call these four orbitals the σ-bonding orbitals. Next we need to consider the electrons in the π-LGOs. There are overall eight of these orbitals containing 16 electrons. We must fill these electrons into the eight MOs of the next-higher energies. These are the 1b2g, the 1eg, the 1a2u, the 2eu , the 1b2u, and the 1a2g orbitals. We see that these orbitals are either bonding or non-bonding, and thus overall the ligand π-electrons get stabilized due to the π-interactions with the ligand. We can call these orbitals the π-bonding molecular orbitals. Finally we need to look at the metal d-electrons. We assume Ni as the metal with Ni in the oxidation state +2 and d8 electron configuration. We need to fill these eight electrons into the four next higher-energy MOs. This fills the 2b2g, the 2eg, and the 2a1g, orbitals which are either non-bonding or anti-bonding. This means that the metal electrons are overall destabilized. This behavior is consistent with the cyano-ligand acting as a π-donating ligand. We understand from this why square planar complexes prefer 16 over 18 electrons and disobey the 18 electron rule. If we had two more electrons then these would need to go into the energetically much higher 2a2u orbitals. The energy gap between the 2a2u and the 2a1g orbital is significantly higher than that between the 1a1g, the 2b2g, and the 2eg orbitals. Overall we can call the 2au, the 2a1g, the 2eg, and the 2bg orbitals the metal d-orbitals in a square planar ligand field. In a square planar ligand field we have three ∆s. ∆3 is the energy difference between the 2b2g and the 2eg orbitals, ∆2 is the energy difference between the 2eg and the 2a1g orbitals, and ∆1 is the energy between the 2a1g and the 2a2u orbitals. ∆1 is much larger than the other two. It also almost always larger than the spin-pairing energy, therefore square planar complexes are almost always low-spin complexes.
Overall, we see that an MO diagram can become very complex even for a relatively simple molecule of relatively high symmetry. It should be noted that for MO diagrams as complex as this one, it is not possible any more determine the exact order of the energy levels just by qualitative inspection, and without exact computations. Nonetheless, you can see that it is possible to readily understand a complex MO diagram by following the symmetry-adapted linear combination of atomic orbitals approach. We just need to go through the symmetry types step by step and construct MOs by combining orbitals of the the same symmetry type until the process is completely. We have now reached the end of the chapter on ligand field theory.
Dr. Kai Landskron (Lehigh University). If you like this textbook, please consider to make a donation to support the author's research at Lehigh University: Click Here to Donate. | textbooks/chem/Inorganic_Chemistry/Inorganic_Coordination_Chemistry_(Landskron)/07%3A_Coordination_Chemistry_II_-_Bonding/7.01%3A_Theories_of_Electronic_Structure.txt |
Concept Review Questions
Section 1
1. What should a good theory be able to do?
2. What should a good bonding theory be able to do?
3. What is the major conceptual difference between valence bond theory of main group compounds with regular covalent bonds vs. coordination compounds with dative bonds?
4. Valence bond theory cannot explain optical properties of molecules. Explain why.
5. Explain the concept of crystal field theory.
6. Explain why the energy of the dz2 and the dx2-y2 orbitals in the octahedral crystal field is higher than that of the dxz, the dyz, and the dxy orbitals.
7. Explain why the the dz2 and the dx2-y2 orbitals have the exactly the same energy in the octahedral crystal field.
8. Explain why the energy of the dxz, the dyz, and the dxy orbitals are decreased by exactly 2/5 Δo.
9. Explain why the energy of the dxz, the dxy, and the dyz orbitals are higher than the energy of dz2, dx2-y2 orbitals in the tetrahedral crystal field.
10. Explain why the energies of the dxz, the dxy, and the dyz orbitals are exactly the same in the tetrahedral crystal field.
11. Explain why the tetrahedral crystal field is smaller than the octahedral crystal field (ΔT = 4/9ΔO).
12. Which orbital energies increase and which ones decrease when an octahedral crystal field tetragonally distorts (elongated octahedra)? Explain why.
13. Explain why octahedral complexes of d9 ions frequently distort tetragonally.
14. Explain the relationships between a tetragonally distorted octahedral crystal field and a square planar crystal field.
15. Explain why crystal field theory can explain why complexes with d8 ions are often square planar.
16. Explain how crystal field theory explains high-spin and low-spin complexes.
17. Explain why crystal field theory can make statements about the electronic spectra of complexes.
18. What is meant by the spectrochemical series?
19. What is the relationship between molecular orbital theory and ligand field theory?
20. Describe the steps involved to construct a molecular orbital diagram in ligand field theory?
21. Why can ligand field theory explain the magnetic properties of coordination compounds?
22. Why is ligand field theory able to explain electronic spectra?
23. What are the relationships between crystal field theory and ligand field theory?
24. What is a π-donating ligand and what is a π-accepting ligand?
25. How do π-donating ligands change ΔO in octahedral complexes? Explain briefly.
26. How do π-accepting ligands change ΔO in octahedral complexes? Explain briefly.
27. How does ligand field theory explain the spectrochemical series?
28. How does ligand field theory explain that there are no low spin complexes for tetrahedral compounds?
29. What is the strength of π-bonding in tetrahedral complexes vs. octahedral complexes? Explain why.
Dr. Kai Landskron (Lehigh University). If you like this textbook, please consider to make a donation to support the author's research at Lehigh University: Click Here to Donate.
Homework Problems Chapter 7
Homework Problems
Section 1
Exercise 1
Decide if valence bond theory is able to explain shapes of the following species. Explain your decision using appropriate electron box diagrams.
1. PdCl42- (square planar)
2. NiCl42- (tetrahedral)
3. Cr(CO)6 (octahedral)
4. ZnCl42- (tetrahedral)
5. Ag(NH3)2+ (linear)
6. Au(PPh3)3+ (trigonal planar)
7. Cu(NH3)42+ (tetrahedral)
Answer
a) No. One would need to move an electron from one half-filled d-orbital into the other half-filled d-orbital. This would require to reverse the spin of the electron which is quantum-mechanically not allowed.
b) Yes, because s and p orbitals are empty and can be sp3 hybridized:
c) No, because one would need to move and spin-pair one s and two d electrons under spin-reversal.
d) Yes, because empty s and p orbitals are available for hybridization
e) Yes, because empty s and p orbitals are available for sp hybridization which is able to explain linear shape.
f) Yes, because empty s and p orbitals are available for sp2 hybridization which is able to explain trigonal planar shape.
g) Yes, because empty s and p orbitals are available for sp3 hybridization which is able to explain tetrahedral shape.
Exercise 2
We assume a hypothetical cubic crystal field. Predict how the energies of the metal d-orbitals split in this crystal field? Explain your decision briefly. What is the symmetry of these orbitals?
Answer
Exercise 3
The octahedral crystal field ΔO for a d4 metal complex is larger than the spin pairing energy. How many unpaired electrons would you expect in this complex?
Answer
I would expect two unpaired electrons.
Exercise 4
Verify by applying the symmetry adapted linear combination of atomic orbitals (SALC) method that the ligand group orbitals suitable for sigma bonding have the symmetry types Ag, Eg, and T1u for an octahedral complex.
Answer
Exercise 5
Can the 4p orbitals of a period 4 transition metal also make pi-bonding with the ligands in an octahedral complex? Explain your decision briefly?
Answer
Yes, the 4p orbitals have T1u symmetry which is suitable to make pi-bonding with the T1u pi-ligand group orbitals.
Exercise 6
Which of the following ligands would you expect to be π-acceptors? Explain your decision briefly.
a) NO+
b) H-
c) Cl-
d) CH3-
Answer
Only NO+, because it is the only ligand that has π*-orbitals.
Exercise 7
A general trigonal planar complex of the composition M(CN)3 is given. How many π- and π* orbitals of the ligands will be involved in the π-bonding?
Answer
3*4=12 orbitals
Exercise 8
The energies of the metal d orbitals of the above complex are -24 eV. The energies of the π-orbitals of the ligand are -27 eV, the energies of the π*-orbitals are -10 eV. Would you expect the ligand to act as a π-donor or as a π-acceptor? Why?
Answer
It would act as a pi-donor, because the energy of the pi-orbitals is much closer to the energy of the metal d orbitals.
Exercise 9
Two octahedral complexes MX6 and MY6 are given. The energy of the ligand HOMOs suitable for sigma-bonding is about the same, but the orbital overlap between the metal and the ligand Y is significantly greater than the orbital overlap between the ligand X and the metal. For which complex would you expect a larger ΔO?
Answer
ΔO will be greater for the orbitals with the ligand Y. This is because the stronger interactions raise the energy of the eg* orbitals, and thus the energy difference between these orbitals and the t2g orbitals becomes larger.
Exercise 10
Construct a qualitative molecular orbital diagram of a square pyramidal complex using the symmetry adapted linear combination of atomic orbitals approach. Consider sigma-bonding only.
Answer
Exercise 11
Strong π-donating ligands lead to a decrease of the octahedral ligand field Δo. Illustrate this by sketching the relevant part of MO diagram.
Answer
Dr. Kai Landskron (Lehigh University). If you like this textbook, please consider to make a donation to support the author's research at Lehigh University: Click Here to Donate. | textbooks/chem/Inorganic_Chemistry/Inorganic_Coordination_Chemistry_(Landskron)/07%3A_Coordination_Chemistry_II_-_Bonding/Concept_Review_Questions_Chapter_7.txt |
A closer look at electronic spectra
Let us take a closer look at optical absorption spectra, also called electronic spectra of coordination compounds. We have previously argued that ligand field theory can predict and explain the electronic spectra. However, only in relatively few cases ligand field theory (LFT) is sufficient to explain the spectra. For example the hexaaqua nickel (2+) complex ion is an octahedral d8-complex ion. According to ligand field theory the metal d-orbitals in an octahedral field are the t2g and the eg–orbitals (Fig. 8.1.1). Six electrons are in the t2g orbitals, and two electrons are in the eg orbitals. Ligand field theory would predict that there is one electron transition possible, namely the promotion of an electron from a t2g into an eg orbital. This process would be triggered by the absorption of light whereby the wavelength of the light would depend on the Δo between the t2g and the eg orbitals. Overall, this should lead to a single absorption band in the absorption spectrum of the complex. We can check this prediction by experimentally recording the absorption spectrum of the complex (Fig. 8.1.1).
What we find is that the absorption spectrum is far more complex than expected. Instead of just a single absorption band there are multiple ones. Obviously, ligand field theory is unable to explain this spectrum. The question is why? The answer is: LFT assumes that there are no electron-electron interactions. However, in reality there is repulsion between electron in d-orbitals and this has an effect on their energy.
Effect of Electron-Electron Interactions
Let us look closer at this. To simplify the problem let us consider a d2 ion in an octahedral ligand field first (Fig. 8.1.2).
According to ligand field theory, the two electrons would be in the t2g orbitals when in the ground state. For instance, they could be in the xy, and the xz orbitals. This is called a microstate. It is called a microstate of a state because there are other combinations of orbitals possible. For example, the ground state would also be realized if the two electrons were in the xz and the yz orbitals. This combination would be another microstate associated with the ground state electron configuration. Upon excitation using light the electron in the dxy orbital could be promoted into an eg orbital, which could be either the dz2 or the dx2-y2. Those two possibilities reflect to different microstates associated with the excited state (Fig. 8.1.2). Let us assume the electron goes into the dz2 orbital. In this microstate one electron would be in the dxz orbital and the other one in the dz2 orbital. Now there us another possibility how to excite an electron. We could assume that not the dxy electron is promoted into the dz2 orbital, but the dxz electron gets promoted. In this case, we would realize a microstate in which one electron in the dxy orbital and the other one in the dz2 orbital.
Now let us compare the two cases. Ligand field theory would argue that both excited microstates have the same energy. However, in fact they do not. Why? That is because the electrons in the two excited microstates interact differently. This becomes plausible when considering the orbital shapes (Fig. 8.1.3). The dxz orbital has electron density on the z-axis, while the dxy orbital has not. Therefore, electrons in these orbitals would be expected to interact differently with an electron in a dz2 orbital which has most of its electron density on the z-axis. As a result, the two excited microstates do not have the same energy. This means that in order to promote the electron into the two different microstates we need different amounts of energies. Thus, the complex would absorb light with more than just one wavelength. This is in contrast to what ligand field theory predicts. What will we need to do in order to correctly predict how many absorption bands we get? The answer is, we must find all possible microstates for the d2 electron configuration and group those together which have the same energy. A group of microstates with the same energy is called a term. The number of electron transitions can then be predicted from the number of terms.
Russell Saunders Coupling of Electrons in Free Ions
Our task is now to find all the terms for the octahedral complex with d2 electron configuration. To simplify matters let us first look at a free d2 ion. A free ion is an ion which is not coordinated by any ligand.
The energy of the microstates depends on the quantum numbers ml and ms of the electrons (Fig. 8.1.4 and Fig. 8.1.5). This is because these quantum numbers of determine how the electrons interact. This is also called the Russell-Saunders coupling of electrons. Let us first see how the quantum number ml of the electrons influences the term symbol. The term is determined by the microstate with the highest quantum number ML, whereby ML is the sum of the quantum numbers ml of the individual electrons. The maximum ML value is also called the total orbital angular momentum quantum number L. The L value also tells you how many microstates belong to a term due to these magnetic interactions. The number of microstates is 2L+1. This is because ML can vary between –L and +L. So for instance if L=2, ML can adopt the values -2, -1, 0, +1, and +2, and that translates to 2L+1=5 microstates.
The quantum number L alone does not define the term yet. In addition, we must also consider the interaction of the spins of the electrons (Fig. 8.1.5).
The spin quantum number ms of a single electron can be either +1/2 or -1/2. When the spins of the electrons interact, their spin quantum numbers add up, defining the total spin angular momentum Ms. The microstate with the maximum Ms value defines the total spin angular momentum quantum number S. The number of microstates due to the spin-spin interactions that belong to the term is 2S+1. This is also called the spin multiplicity (Fig. 8.1.6).
Overall the term has (2L+1)(2S+1) microstates. The total orbital angular quantum number L and the total orbital spin quantum number S define the term symbol (Fig. 8.1.7).
When L=0 we have an S term, when L=1, we have a P term, when L=2, we call it a D term, and when L=3 we call it an F term. These symbols are chosen in analogy to the symbols s, p, d, and f for orbitals. The spin multiplicity 2S+1 is then written as a superscript in front of the letter defined by L. This gives the overall term symbol for a term. We call a term with a spin multiplicity of 1 a singlet term, when 2S+1=2, we call the term a doublet term. A term with a spin multiplicity of three is called a triplet term, one with a spin-multiplicity of 4 a quartet term, and so forth.
Microstates for a free d2 ion
What are the possible microstates and the overall number of microstates for a free ion with d2-electron configuration? We can express this in a so-called microstate table (Fig. 8.1.8).
You can see the microstate table for the d2 electron configuration depicted above. Each column represents a possible value ML. For the d2 configuration, ML can adopt values from -4 to +4, hence there there are 9 columns. Each row represents possible Ms value. For the d2 electron configuration, Ms can vary between +1 and -1. +1 means both electrons have the spin +1/2, 0 means one electron has the spin +1/2 and the other one has the spin -1/2. The Ms value of -1/2 is adopted when both electrons have the spin -1/2. Now we can combine each Ms value with each ML value which defines a particular field in the table. You can see that some fields are empty due to the Pauli principle. For example, the field for ML=-4 and Ms=-1 is not filled because in this case both electrons would have the same quantum numbers, namely ml=-2, and ms =-1/2. We can further see that for some fields only one combination of electrons is possible, while for others there are multiple. For example, for the field with ML=4 and Ms=0 there is only one combination of electrons possible. You can see the symbol “2-2+” in this field. This means that the first electron has an ml value of 2 and and a spin of -1/2 and the second electron also has an ml value of 2 and an ms value of +1/2. The most populated field is the field for ML=0 and Ms=0. There are overall five microstates with that combination of ML and Ms values. If we count the overall number of microstates in the table then we arrive at the number 45. This is consistent with what we would expect according to the formula #microstates = (2L+1)(2S+1). In our example L=4 and S = 1, and thus the number of microstates is ((2x4)+1)((2x1)+1)=45. In general for a dn electron configuration with n d-electrons the number of microstates is (10!)/((10-n)!n!).
One useful property of a microstate table is that we can derive the terms and term symbols for a particular electron configuration from it. For simplification sake, we indicate a possible microstate in the table just by an “x”. Then we draw the largest possible rectangular box into the microstable that contains only fields with at least one possible electron combination (Fig. 8.1.9).
You can see that the largest possible box is the red box drawn (Fig. 8.1.9). This box contains the microstates that belong to the first term of the electron configuration d2. The number of microstates is equal to the number of fields within the red box. That makes 7x3=21 microstates. What is the term symbol for this term? In order to answer this question we need to find the microstate in the red box that has the highest ML and the highest Ms value. You can see that it is the one with ML=3 and Ms = 2. Thus, L=3 and S=1. This defines the term as a 3F term because 2S+1 = 3, and L=3 corresponds to the term symbol F. This term is a triplet term. Note that a triplet term also includes microstates with paired spins. Overall seven of the 21 microstates have Ms=0.
Next, we look what is the next-largest rectangular box we can draw, and which contains microstates we did not consider yet. We can see that the blue box contains nine fields equaling nine microstates. The microstate within this box that has the highest ML and Ms values is the one with ML=4 and Ms=0. This defines a 1G term. This is a singlet term. Note that a singlet term ONLY has microstates with paired spins (Ms=0).
There is another rectangular box which also contains nine microstates. It is the green one. The microstate with the highest ML and Ms values is the one with ML=1 and Ms=1. Thus, L =1, and 2S+1=3 which defines a 3P term. Note that despite this being a triplet term, the term also contains three microstates with paired spins.
The next-largest box is the purple one containing five fields and five microstates. What is the term with the highest ML and Ms values? It is the one with ML=2 and Ms=0. This means that L=2 and S=0 which defined a 1D term. This is a singlet term. A singlet term contains only microstates with paired spins. There is one microstate left we did not consider thus far. It is one with ML=0 and Ms = 0. These are the highest ML and Ms values because there is no other microstate. Thus, this term contains only one microstate and the term symbol is 1S. Now we have found all terms and term symbols.
The Energy of Terms
What is the relative energy of the terms? First we need to consider Hund’s rule which states that the higher the spin multiplicity, the lower the energy of a term.
We therefore first need to rank the terms with regards to the spin multiplicity. Secondly, we need to consider the value of the total orbital angular momentum quantum number L. The higher the value the lower the energy for the term. This means that the energy of terms decreases from S, to P, to D, to F.
Note that this is contrary to orbital energies which increase from s, to p, to d, to f. Therefore, what would be the expected energy sequence for the terms associated with a d2 free ion? It should be: 3F<3P<1D<1S. However, the actual energy sequence is 3F<1D<3P<1G<1S. This means that, unexpectedly, the 1D term is energetically lower than the 3P term. We can understand from this is why Hund’s rule is a called a rule and not a law because there are exceptions. In most but not all cases a triplet term has a lower energy than a singlet term.
Free-Ion Terms of Other d Electron Configurations
What are the term of the other d-electron configurations (Fig. 8.1.10)?
You can see above that there is only a single 2D term for a d1 free ion. How can we explain this term? Because there is no other electron, there are no electron-electron interactions, and thus every possible microstate has the same energy. This explains why there is only one term. Why is it 2D? The term symbol is determined by the microstate with the highest ML value and the highest Ms value. This is achieved when we fill the electron spin-up into the d-orbital with ml=2 (Fig. 8.1.10). Because there is only one electron, ml=ML=L=2. This makes the term a D term. It is a doublet term because Ms is maximized when the electron is spin-up or ms=+1/2. Thus, ms=Ms=S=1/2. 2S+1=2×1/2+1=2, thus we have a doublet state. How many microstates are associated with this term? (2L+1)(2S+1)=5×2=10. What does that mean? This means that we can fill the one electron that we have either spin-up or spin down in any of the five d-orbitals. All the 10 microstates are energetically equal.
When you go from d2 to d3 to d4 to d5 you can see that the number of the terms increases dramatically. This is because with increasing electron numbers the number of possible permutations increases, and thus the number of possible microstates. As a consequence, there are more terms. From d6 to d10 the number of terms decreases again. The terms for d6 are the same as the terms for d4, the d7-terms are identical to those for d3, the d8-terms are the same as the d2 and the d9-terms are the same as terms for the electron configuration d1. Why is that? This is because unoccupied states in orbitals permute the same way as occupied ones do. For instance, in the d3 electron configuration has three occupied states and seven unoccupied ones, while the d7 electron configuration has seven occupied states and three unoccupied ones. The d10 electron configuration has also only one term with the term symbol 1S (Fig. 8.1.10). Can we explain this? In the d10 electron configuration, there is no way to permute the electrons, hence there is only one microstate possible. In this microstate L=0. Why is that? When we sum up the individual ml-values of the ten electrons, then this sum is ML = L = 2×(-2)+2(-1)+2×(0)+2×(+1)+2×(+2)=0. The spin multiplicity of the term is 1 because all electrons are paired (2S+1=(2×0)+1=1). Hence, we have a singlet 1S term.
Spin-Orbit Coupling
We have still not considered all electron-electron interactions. In addition to the interaction between the angular magnetic momenta, and the magnetic interactions between the spins, there are also magnetic interactions between the angular momenta and the spins. This is called the spin-orbit coupling. It is expressed by an additional quantum number J. J can run from L+S to L-S.
The quantum number J is added as a subscript behind letter describing the term. So a full term symbol is described as 2S+1LJ. Spin-orbit coupling may lead to additional terms depending on L and S (Fig. 8.1.11).
For instance, for a 3P term, S=1, and L=1. Thus J can adopt values between L+S=1+1=2 and L-S=1-1=0. Thus, overall J can adopt the values, 0, 1, 2. This means a 3P term splits into three terms due to the spin-orbit coupling. Their term symbols are 3P0, 3P1, and 3P2. Their energy increases with increasing quantum number J. 1S and 1D terms on the other hand to do not split into additional terms. For a 1S term L+S=0+0=0, and L-S=0-0=0. The full term symbol is 1S0. For the 1D term, L+S=2+0=2 and L-S=2-0=2, thus the full term symbol is 1D2. Overall, the energy differences between terms due to J are very small, in the order of tens of wavenumbers, and orders of magnitudes smaller than energy differences due to RS-coupling, which are typically in the order of tens of thousands of wavenumbers. Thus, we usually ignore the effect of J when interpreting electronic spectra.
Dr. Kai Landskron (Lehigh University). If you like this textbook, please consider to make a donation to support the author's research at Lehigh University: Click Here to Donate. | textbooks/chem/Inorganic_Chemistry/Inorganic_Coordination_Chemistry_(Landskron)/08%3A_Coordination_Chemistry_III_-_Electronic_Spectra/8.01%3A_Quantum_Numbers_of_Multielectron_Atoms.txt |
Term Splitting of Terms in an Octahedral Field
Thus far we have only considered free ion terms, that means terms without the presence of a ligand field. Let us next think about the influence of an octahedral field on a term. Terms are wavefunctions, just like orbitals, and therefore they behave like orbitals in a ligand field. D-orbitals split into t2g and eg orbitals in an octahedral ligand field. A D-term behaves similarly. It splits into T2g and Eg terms. P-orbitals are triple-degenerate having T1g symmetry in the point group Oh and do not split in energy. P-terms have the same symmetry and also do not split. We can call a P-term in an octahedral ligand field a T1g term. Following analogous arguments, S terms become A1g terms. F terms do split in energy like f-orbitals and become T1g, T2g, and A2g terms. Overall, the presence of the octahedral field increases the number of terms from four to seven (Fig. \(1\)). It is easy to see that the ligand field leads to many states, and many potential electron transitions. Thus, we would expect quite complicated spectra. For other ligand fields, the terms also behave analogously to orbitals. For, instance in a tetrahedral field D-terms split into E and T2 terms, and so forth.
Term Splitting for octahedral d2 metal complexes
Now let us think about how the term energies of our free d2-ion changes when placed in an octahedral ligand field, depending on the ligand field strength. We can express this by a correlation diagram (Fig. 8.2.2). In a correlation diagram, we plot the energies of the terms relative to the field strength.
On the left side we plot the terms without any field according to their energies. In the case of a d2 ion, the energies are 3F<1D<3P<1G<1S. Next, we plot the relative energies in a weak octahedral ligand field, and label the terms according to their symmetry. We can see that the D, F, and G terms split in energy, while the S and P terms do not. Because of the weak field, energy differences are very small. Now let us increase the ligand field strength continuously, until we have reached a very strong ligand field. We see that some of the terms move up in energy, while other terms move down as the ligand field increases. For example, two of the three terms resulting from the F-terms increase in energy while one decreases. It is also possible that a term does not change its energy. For example, the 3T1g term from the 3P term does not change its energy. In very strong ligand field there are three groups of terms that have similar energy. In the hypothetical case of an infinitely strong ligand field, the terms that belong to a particular group become identical in energy. In this case, there are only three states for the electrons possible. The field is considered so strong so that the energy associated with electron-electron interactions become negligible compared to the energy of the field. The electrons behave as though there were no electron-electron interactions. Therefore, we can call the lowest energy state the t2g2 state. It is equivalent to the state of the two electrons being in the t2g orbitals. The second state is called the t2geg state. It is equivalent to the state of one electron being in the t2g and one electron being in the eg orbitals. The third state is called the eg2 state, which is equivalent to the state with both electrons in the eg2-orbitals.
Selection Rules
In most cases, the ligand field strength is in between the very weak and very strong case, and thus, we could expect very complicated spectra. Fortunately, nature does not make things quite as complicated, because not all possible electron transitions are quantum-mechanically allowed. The allowed transitions are defined by two rules: The spin selection rule and the Laporte rule.
The spin selection rule states that only transitions are allowed in which the total spin quantum number S does not change. When S does not change also the spin multiplicity does not change. Thus, electron transitions are only allowed for transitions that do not involve a change in spin multiplicity. For example, it would be allowed to excite an electron from a triplet term to another triplet term, but not from a triplet term to a doublet or singlet term.
The Laporte rule state that transitions are only allowed when there is a change of parity. This means a transition from a gerade (g) to an ungerade (u)-term and vice versa is possible, but not a transition from a g-term to another g-term, or the transition from a u-term to another u-term. For example, the transition from a T2g to a T1u term would be allowed, but not the transition from a T2g to a T1g term.
Tanabe-Sugano diagram of a d2 octahedral complex
A particular correlation diagram is the Tanabe-Sugano diagram (Fig. 8.2.3). It is illustrated here for an octahedral d2 complex. The difference to the previously discussed correlation diagram is that the energy of the ground state is plotted horizontally, and the energy of all other terms are plotted relative to that. In the case of the d2 electron configuration, the 3T1g term is the ground term and is plotted as a horizontal line. We can see that the ligand field strength on the x-axis is given in units of B, and the energy of the terms is also given in units of B. B is a so-called Racah parameter, that is a quantum-mechanical energy unit for the electromagnetic interactions between the electrons. It is chosen because it provides “handy” numbers.
You can see that some lines in the diagram are bent, and some are straight. Bending of lines occurs when two terms interact with each other because they are close in energy and have the same symmetry. This is again an analogy to orbitals. Like orbitals interact when they have the same symmetry type and similar energy, also terms interact when they have the same symmetry and the similar energy. Without taking their interactions into account, their energies can cross when the energy of term A declines and the energy of term B increases with increasing field strength (Fig. 8.2.4). The closer the terms come to the point where they cross, the stronger their interactions, because their energies become more and more similar. The interactions lead to the fact that the terms "bend away" from each other, leading to bent curves. This means that curves for two terms of the same symmetry type will bend away in a Tanabe-Sugano diagram and never cross. For example the terms for the two 1A1g terms bend away from each other and do not cross.
Next, let us think about which electron transitions would be allowed under the consideration of the spin selection and the Laporte rule. We notice that in the symmetry types the “g” for gerade has been omitted (Fig. 8.2.3). This is a common simplification made in the literature. We have to remind us that all the terms in the diagram are “g” terms. What does this mean for the allowance of electron transitions? It means that no electron transition would be allowed, and that would imply that the complex could not absorb light. The Laporte selection rule however does not hold strictly. It only says that the probability of the electron-transition is reduced, however, not forbidden. This means that an absorption band that disobeys the Laporte rule will have lower intensity compared to one that follows the Laporte rule, but it can still be observed. The spin-selection rule, however, holds strictly, and transitions between terms of different spin multiplicity are strictly forbidden, meaning that they have near zero probability to occur. Overall, we can therefore excite an electron from the 3T1 ground state to other triplet terms, namely the 3T2 term, and the 3A2 term (Fig. 8.2.3).
Tanabe-Sugano diagram of d3 octahedral complexes
Now, let us have a look at the Tanabe-Sugano diagram of a d3 ion in an octahedral ligand field (Fig. 8.2.7). What is the ground term? We can see the term designation on the horizontal line reads “4A2”, therefore this term is the ground term. How many electron transitions from the ground state should we expect? To answer this question we need to count the number of other quartet terms. There is the 4T2, the 4T1, and another 4T1. Thus, there are overall three electron transitions possible. Can we understand why the ground state is a quartet term?
It helps to consider how we would fill the electrons into the d-orbitals for the electron configuration d3. All three electrons would be filled spin-up in the t2g orbital following Hund’s rule (Fig. 8.2.8). Because each electron has the spin +1/2 the total spin of all three electrons is 3x1/2=3/2. Thus, the spin multiplicity is ((3x3/2)+1)=4. Note that the microstate we have drawn is actually only one of the (2L+1)(2S+1) microstates. S=3x1/2=3/2, but what is L? You can see one the left side of the diagram that the 4A2 term originated from a 4F term. This means L=3, and (2L+1)((2S+1)=7x4=28. This means that there are actually 27 other microstates that have the same energy as the microstate that we drew. Why did we draw this microstate in favor of the others? This is because this microstate is the state with the maximum ML (=L) and Ms (=S) values determining the term symbol.
Tanabe-Sugano diagram of d4 octahedral complexes
Now let us look at the Tanabe-Sugano diagram of a d4 octahedral complex. You can see that this diagram is separated into two parts separated by a vertical line. The line indicates the ligand field strength at which the complex changes from a high spin complex. At lower ligand field strengths, the ground term is a 5E term. At higher field strength the ground term is a 3T term.
We can rationalize this again by drawing the orbital box representation of the d-orbitals in the octahedral ligand field. In the high spin state, there are four unpaired electrons, thus S=4x1/2=2, and 2S+1=5. In the low spin state, there are two unpaired electrons, and thus S=2x1/2=1, and 2S+1=3. This explains the quintet and the triplet nature of the high and low spin ground terms. Note again, that the two microstates represented by the orbital box diagrams (Fig. 8.2.10) are not the only microstates that have the respective energy. They are only the "representative" microstates because the have the maximum ML and MS values.
How many electron transitions are possible from the ground term? For a high spin complex there is only one because the is only one other quintet term, namely the 5T2 term. For the low spin complex, there are five transitions because there are five other triplet terms.
Tananbe-Sugano diagram of d5 octahedral complexes
Also the Tanabe-Sugano diagram of a d5 octahedral complex is divided into two parts separated by a vertical line (Fig. 8.2.11). The left part reflects the high spin and the right part the low spin complex.
The high spin ground state is a sextet term, and the low spin ground state is a doublet term. We can understand the sextet and doublet nature of the terms when considering that the associated electron box diagram have five and one unpaired electrons respectively. S=5x1/2=5/2 and 2S+1=6 for the high spin term, and S=1x1/2=1/2 and 2S+1=2 for the low spin term. What are the possible electron transitions from the ground state? For the high-spin complex there is no other sextet term, meaning that there is no electron transition possible. Hence, high-spin octahedral d5-complexes are colorless. An example is the hexaaqua manganese (2+) complex. A solution of this complex is near colorless, only very slightly pinkish. The slight color is because also spin-forbidden transitions can occur, albeit at a very low probability. For a d5-low spin complex there are three additional doublet states, and thus there are three electron transitions possible.
Tanabe-Sugano diagram of d6 octahedral complexes
The next diagram is the one for the d6 electron configuration (Fig. 8.2.13). Again, the diagram is separated into parts for high and low spin complexes. You can see dotted lines in the diagram. They indicate the terms that have a different spin multiplicity than the ground term. This way we can more easily see how many electron transitions are allowed.
The ground term for the high-spin complex is the quintet 5T2 term. It is a quintet term because four electrons in the d-orbitals are unpaired, and two are paired. The value for S is thus 4x1/2=2, and the spin multiplicity is 2S+1=5. The ground term for the low spin complex is a 1A1 term. It is a singlet term because all electrons are paired, and thus S=0, and 2S+1=1. How many electron transitions are there for the high-spin complex? There is only one because the 5E term is the only other quintet term. There are five transitions possible for the low-spin case because there are five additional singlet terms.
Tanabe-Sugano diagram of d7 octahedral complexes
Next, let us look at the Tanabe-Sugano diagram of a d7 octahedral complex (Fig. 8.2.15). In this case, the high spin complex has a 4T1 ground term, and the low spin complex has a 2E ground term.
The microstate that "represents" the high spin ground term has three unpaired electrons, hence the spin quantum number S=3/2 and the spin multiplicity is 2S+1=4. The microstate that represents the low spin ground term has one unpaired electron, an S value of ½, and a spin multiplicity of 2. There are three other quartet terms, and four other doublet terms, hence there are three electrons transitions for the high-spin complex, and four for the low spin complex.
Tanabe-Sugano diagram of d8 octahedral complexes
Now let us look at an octahedral complex with d8 electron configuration. For this electron configuration, there are no high and low spin complexes possible, therefore, the Tanabe-Sugano diagram is no longer divided into two parts (Fig. 8.2.17).
There is a single ground term of the type 3A2. It is a triplet state because the microstate representing the term has two unpaired electrons in the eg orbitals (Fig. 8.2.18). Thus, S=2x1/2=1, and the spin multiplicity is 2S+1=2. How many electron transitions would you expect? There are three other triplet states, namely the 3T2 and two 3T1 terms. Therefore, there are three electron transitions possible.
We could also ask: Are there Tanabe-Sugano diagrams for d1, d9, and d10? For, d1 there are no electron-electron interactions, thus the simple orbital picture is sufficient. The 2D term splits into T2g and Eg terms, and there is only one electron transition possible. The d9 electron configuration is the “hole-analog” of the d1 electron configuration. It has also just one 2D term which splits into a T2g and an Eg term in the octahedral ligand field. Therefore, also in this case there is only one electron transition from the T2g into the Eg term possible. In the case of d10 all microstates are filled with orbitals, and there is only the 1S term which does not split in an octahedral ligand field. Therefore, there are no electron transitions in this case.
Finally, it should be mentioned that it is also possible to construct Tanabe-Sugano diagrams for other shapes such as the tetrahedral shape, but we will not discuss these further here.
Charge Transfer Transitions
We are still not done with our electronic spectra. Thus, far we have only considered transitions of d-electrons between d-orbitals, and their terms. They are called d-d transitions. However, there are also so-called charge transfer transitions possible, that are not d-d transitions. We can easily see that there must be other transitions but d-d transitions when we look at the color of d10 and d0 ions. For those, the are no d-d transitions possible. Therefore, they all should be colorless. However, that is not always true. Some of these ions are indeed colorless, but some are not (Fig. 8.2.19). For example, Zn2+, a d10 ion is colorless in complexes, but not Cu(I) which is also d10. While tetrakis(acetonitrile)copper (+) is colorless, bis(phenanthrene) copper(+) is dark orange. Similar is true for d0 ions. While TiF4 and TiCl4 are colorless, TiBr4 is orange, and TiI4 is brown. Some d0 species are even extremely colorful, for example permanganate with Mn7+ which is extremely purple, and dichromate with Cr(VI) which is bright orange.
The explanation of these phenomena are charge-transfer transitions (Fig. 8.2.20). There are two types of charge-transfer transitions, the ligand-to-metal (LMCT) and the metal-to-ligand (MLCT) charge transfer transitions. For the ligand-to-metal transitions, electrons from bonding σ and π-orbitals get excited into metal d-orbitals in the ligand field, for example the t2g and the eg orbitals in an octahedral complex. If the energy difference between the σ/π-orbitals and the d-orbitals is small enough, then this electron-transition is associated with the absorption of visible light. The transition is called a ligand-to-metal transition because the ligand σ/π-orbitals are mostly located at the ligands, while the metal-d-orbitals in a ligand field are mostly located at the metal. Vice versa, the metal-to-ligand transition involves the transition of an electron from metal d-orbitals in a ligand field to ligand π*-orbitals. This essentially moves electron density from the metal to the ligand, hence the name ligand-to-metal-charge transfer transition. If the energy-difference between the ligand π* and the metal orbitals is small enough, then the absorption occurs in the visible range. Charge-transfer transitions are usually both spin- and Laporte allowed, hence if they occur the color is often very intense. How can we distinguish between d-d and charge transfer transitions? Charge transfer transitions often change in energy as the solvent polarity is varied (solvatochromic) as there is a change in polarity of the complex associated with the charge transfer transition. This can be used to distinguish between d-d transitions and charge-transfer bands.
LMCT Transitions
Can we predict when the energy windows between the bonding molecular orbitals and the metal d-orbitals are small enough so that LMCT transitions in the visible can take place? Generally, it would be desirable if the energy of the metal orbitals was as low as possible and the energy of the bonding ligand orbitals are as high as possible. The energy of metal d-orbitals decreases with increasing positive charge at the metal because the effective nuclear charge on the metal increases. This means that very high metal oxidation states favor an LMCT transition. The d-orbitals should have few or no electrons, so that electrons can be promoted into the orbitals, and orbital energy decreases because electron-electron repulsion is minimized. Examples are Mn(VII), Cr(VI), and Ti(IV). The energy of MOs from bonding ligand orbitals increases when the ligand orbitals have high energy this is typically the case for π-donor ligand with negative charge (Fig. 8.2.21).
Examples of ligands are oxo- and halo ligands. This explains for example the LMCT transitions in permanganate. The Mn is in the very high oxidation state +7, and the ligands are are oxo-ligands wich are π-donors with a 2- negative charge. The transitions are both Laporte and spin-allowed leading to very high intensity of light absorption, and thus color (Fig. 8.2.21).
MLCT Transitions
What are favorable metal ion and ligand properties for a metal-to-ligand transition, then? In this case we would like to keep the energy of the metal orbitals as high as possible so that the energy difference between a metal d-orbital and a π*-orbital is minimized. This is accomplished when the positive charge at the metal ion is small, and there are many d-electrons that can repel each other, thereby increasing orbital energies, for examples Cu(I), Fig. 8.2.22.
The ligand should be a π-acceptor with low-lying π*-orbitals, for example phenanthroline, CN-, SCN-, and CO. For instance, the bis(phenanthroline) copper(+) is dark-orange and has a MLCT absorption band at 458 nm. Also, the MLCT transfer is both spin and Laporte-allowed.
It should be mentioned that some complexes allow for both metal-to-ligand and ligand to metal transitions. For example, in the Cr(CO)6 complex the σ-orbitals are high enough and the π*-orbitals are low enough in energy to allow for light absorption in the visible range. Finally, also intraligand bands are possible when the ligand is a chromophore.
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Concept Review Questions
Section 1
1. Explain the difference between a microstate and a state.
2. Explain what is meant by a term? Why do we need terms to explain electronic spectra?
3. What is meant by “Russell-Saunders coupling”?
3. What is the definition of the total angular momentum quantum number L?
4. What is the definition of the quantum number Ml?
5. What is the definition of the quantum number ml?
6. What is definition of the total spin quantum number S?
7. What is the definition of the quantum number Ms?
8. What is the definition of the quantum number ms?
9. How many microstates are associated with a term defined by a quantum number L and S respectively?
10. Which formula can be used to calculate the number of microstates with a specific d electron configuration?
11. What is meant by the spin-orbit coupling?
12. Why is the spin-orbit coupling usually neglected for electronic spectra?
13. What is the dependency between the quantum number L and the energy of a term?
14. What is the dependency between the quantum number S and the energy of a term?
15. What is meant by a “free ion term”?
Section 2
1. How do D, P, and S terms split in an octahedral and tetrahedral ligand field, respectively?
2. What is a correlation diagram?
3. What is meant by t2g2, t2geg, and eg2 terms in octahedral complexes? Explain their relationship to molecular orbitals.
4. What is the major difference between a Tanabe-Sugano diagram and a correlation diagram?
5. Why do terms with the same symmetry never cross in Tanabe-Sugano diagrams?
6. When are lines in Tanabe diagrams straight and when are they bent?
7. What is a metal-to-ligand charge transfer (MLCT) transition?
8. What is a ligand-to-metal charge (LMCT) transfer transition?
9. Under which conditions does a MLCT transition occur in the visible range? Name one example of a complex that does MLCT.
10. Under which conditions does a LMCT transiton occur in the visible range? Name one example of a complex that does LMCT?
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Homework Problems Chapter 8
Homework Problems
Section 1
Exercise 1
How many microstates exist for the electron configurations
a) p3
b) f2
c) d3
Answer
a) p3 - 6!/((6-3)!3!) = 20
b) f2 -14!/((14-2)!2!)=91
c) d3 -10!/((10-3)!3!)=120
Exercise 2
What is the term symbol for the one term associated with the following electron configurations?
a) p6
b) p1
c) f14
d) s1
Answer
a) p6 - 1S
b) p1 - 2P
c) f14 - 1S
d) s1 - 2S
Exercise 3
What is the number of microstates for a
a) 5D term
b) 4G term
c) 2F term
Answer
a) (2L+1)(2S+1) = 5 x 5 = 25
b) (2L+1)(2S+1) = 9 x 4 = 36
c) (2L+1)(2S+1) = 7 x 2 = 14
Exercise 4
In a triplet term there are
a) only microstates with unpaired electrons.
b) only microstates with paired electrons.
c) microstates with paired and unpaired electrons.
Answer
c) microstates with paired and unpaired electrons.
Exercise 5
In a singlet term there are
a) only microstates with unpaired electrons.
b) only microstates with paired electrons.
c) microstates with paired and unpaired electrons.
Answer
b) only microstates with paired electrons.
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09: Coordination Chemistry IV - Reaction and Mechanisms
Chemical Reactivity and Reactions of Complexes
In this chapter we will discuss the chemical reactivity and reactions of coordination compounds. First, let us briefly review what are the basic parameters that define if a reaction can be observed or not. The first parameter is the Gibbs free energy of reaction ΔG. The Gibbs free energy defines the thermodynamic driving force for the reaction. The more negative ΔG is, the greater the thermodynamic driving force. The Gibbs free energy of reaction is a measure for the difference in the stability of the reactants and products. The less stable the reactants, and the more stable the products, the more negative ΔG. The second parameter is the kinetics of a reaction. The kinetics is a measure of the speed at which a reaction occurs. The kinetics is independent from the thermodynamics. This means that even if there is a strong thermodynamic driving force for the reaction, the rate of the reaction may be extremely slow, and practically no reaction is observed. For example, graphite is thermodynamically more stable than diamond at ambient pressure and temperature, but the kinetics of the conversion at room temperature is so slow, so that practically no reaction is observed. The rate of the reaction is determined by the the activation energy that needs to be overcome to make the reaction happen. The lower the activation energy, the faster the reaction. In coordination chemistry the activation energy and the reaction rates are associated with the inertness and the lability of complexes. The more inert and the less labile a complex is the higher the activation barrier and the more slowly it reacts. The lability is the opposite of the inertness and is defined by the half-life time of a dative bond. The greater the half-life time, the more inert, and the less labile the complex.
Inert and Labile Complexes
The greater the half-life time of a dative bond, the more inert, and the less labile the complex. Where do we actually draw the borderline between a labile and an inert complex? A commonly accepted definition is that complexes with metal-ligand bond half life times of <30 s at standard conditions are considered labile complexes with a low activation energy. Those complexes undergo fast substitution reactions, and the chemical equilibrium is reached fast (Fig. 9.1.1).
An example is the substitution of a thiocyanato ligand in a pentaaquathiocyanato iron (2+) complex by a fluoro ligand. This substitution reaction is fast because the half-life time of the metal-ligand bond is smaller than a second.
If the half-life time is greater than 30s, then the complex is considered inert. For example, the hydrolysis of hexaammine cobalt (3+) is slow because the half-life time of the Co-N bond is larger than a day. The chemical equilibrium is reached only slowly (Fig. 9.1.2).
Can we predict which complexes are labile, and which ones are inert? Many factors determine this, but there are some trends we can keep in mind. For transition metals, octahedral d3, low spin d4-d6, and square planar d8 complexes tend to be inert. A second trend is that the lability decreases with the period within a group. For instance, we would expect that in group 10 the lability decreases from Ni to Pd to Pt. This is because covalency of the bonds tends to increase with increasing period due to the softer nature of higher period metals. A third trend is that higher positive charges at the metal ion lead to a decrease in liability. Metal ions with higher oxidation numbers also tend to make more covalent bonds because the high positive charge at the metal is stabilized by covalent interactions (Fig. 9.1.3).
We can also compare the d-block metals with the other blocks. S-block metal complexes tend to have the greatest lability, followed by the f-block, followed by the d-block. P-block metal complexes can widely vary in lability (Fig. 9.1.4).
Also the nature of the ligand influences the lability. Generally, when the ligand is a good leaving group, then the bond tends to be labile. A ligand is usually a good leaving group when it is stable in the uncoordinated form. For instance, a chloro ligand is a good leaving group because chloride anions are quite stable in their non-coordinated form. An alkyl ligand on the other hand would not be considered as a good leaving group because the an alkylide anion is not a very stable species in the free form.
Stability of Complexes
Now let us have a closer look at the stability of complexes. The more negative the Gibbs free energy of formation, the greater the equilibrium constant of formation K. The larger the equilibrium constant K of formation, the more stable the complex. This can be expressed by the van’t Hoff equation ΔG = -RTlnK.
For the formation of a single dative bond M-L the constant of formation is given by K=[ML]/[M][L] (Fig. 9.1.5). The greater K, the farther the equilibrium is on the side of the product. K is usually reported as logK because K can vary over 35 orders of magnitude.
Rationalizing the Different Stability of Complexes
How can we rationalize the different stabilities of coordination bonds, and coordination compounds? Here, the HSAB concept is very useful. Hard-hard and soft-soft interactions are typically strong and lead to high constants of formation and stable complexes. Hard-soft interactions lead to small constants of formation and less stable complexes.
Let us understand this better by the example of the interaction of Ag+ with different ligands (Fig. 9.1.6). You can see in the table the logK values for the different ligands. Br- has the largest one, followed by NH3 and Cl-. F- has the smallest one. Can we explain this with the HSAB concept? Ag+ is a soft cation, thus it would be expected to make the strongest interactions with the softest ligand. Bromide is the softest of the four, and hence HSAB correctly predicts the stability. F- is the hardest ligand and would be expected to make the weakest complex. This is in accordance with the small logK value for F-. NH3 and Cl- have similar intermediate hardness, which is consistent with the similar, intermediate logK values.
Stabilities of Multi-ligand Complexes
How does the stability of a complex change as more ligands are added? Let us take a look at the constants of formation to answer the question. The addition of an additional ligand is associated with an additional chemical equilibrium which relates to an additional constant of formation. The number of equilibrium constants is equal to the number of ligands. For instance, when we have a complex ML4 with four ligands, then for the addition of the first ligand we get a constant K1= [ML] / [M] [L] , for the second ligand we get a constant K2=[ML2] / [ML] [L] , for the third ligand there is a constant K3=[ML3] / [ML2] [L], and for the fourth ligand there is a constant K4=[ML4] / [ML3] [L]. When we analyze the K-values, we notice that they decline from K1 to K4. For a general complex MLn we find that the constants decline from K1 to Kn. How can this be explained? The explanation is that the more ligands are getting added, the less many free ligands are available that could be further added. Thus, the probability that further ligands are added decreases, and this results in a smaller equilibrium constant. So it is a statistical effect that explains the phenomenon.
Then what is the overall stability constant K of the complex associated with the equation M + 4 L → ML4 and more general M + nL → MLn? It is the product of the constants associated with each step. For the complex ML4 it is K=K1K2K3K4. For the general complex MLn it is K=K1K2.........Kn (Fig. 9.1.8).
There are exceptions from the rule that the stability constants of complexes associated with the addition of one ligand decreases with the number of ligands. If there is an exception, then this is because the ligand addition causes a major change in the electronic structure of the complex. Here are a couple of examples.
In the first example (Fig. 9.1.9) the aqua ligands of a hexaaqua iron (2+) complex are substituted by bipyridyl ligands. The bipyridyl ligand is a bidentate ligand, and so one bipydridyl ligand substitutes two aqua ligands. In the first substitution, the two aqua ligands get replaced by one bipyridyl ligand to form a tetraaqua(bipyridyl) iron (2+) complex. This step is associated with a logK1 value of 4.2.
In the second substitution step, two aqua ligands of the tetraaqua(bipyridyl) iron (2+) complex get replaced by another bipyridyl ligand to form a diaqua bis(bipyridyl) iron(2+) complex. As expected this step is associated with a smaller logK2 value of 3.7.
In the third step, the last two aqua ligands get substituted by another bipyridyl ligand forming a tris(bipyridyl) iron(2+) complex. What is the logK3 value? It is 9.3! This is much larger than logK1 and logK2, so this result is unexpected. A major change in electronic structure must have occurred. What is it? In this case it is a change from a high-spin to a low spin complex. The bipyridyl ligands are strong π-acceptors that increase Δo. Substitution of the first four aqua ligands increases ΔO but the increase is not large enough yet so that it induces a switch from a high-spin to a low-spin complex. Only the third bipyridyl ligand does that. As a consequence, the formation constant logK3 is unusually large.
In the second example (Fig. 9.1.10) we subsequently substitute aqua ligands by chloroligands in a hexaaqua mercury(2+) complex. The first substitution leads to a pentaaquachloro mercury(+) complex. This reaction is associated with a logK1 value of 6.74. The second substitution leads to a tetraaquadichloro mercury(0) complex, and this reaction is associated with a smaller logK2 value of 6.48. The third substitution reaction leads to an aquatrichloro mercury(1-) complex. The logK3 value for this reaction is 0.85. We see that this value is smaller than logK2, and this is what we expected, but the difference between logK3 and logK2 is much larger than that between logK1 and logK2. The increased difference is too large to be just caused by a statistical effect, but must be induced by a change in electronic structure. In this case it is a change in coordination number. As the third chloro ligand is added, the complex does not lose only one but three aqua ligands. This reduces the coordination number from six to four.
Stability of Chelate Complexes
The Thermodynamic Chelate Effect
Chelating complexes tend to be more stable than complexes with monodentate ligands. This is called the “thermodynamic chelate effect”. The effect deserves an explanation. The explanation is the increase of entropy that occurs when two or more monodentate ligands are replaced by a chelating ligand. The entropy increases because the overall number of particles increases as the substitution takes place.
For example, the substitution of six ammine ligands in the hexaammine nickel (2+) complex by three ethylenediamine chelating ligands increases the number of molecules from four to seven, and hence the entropy increases, in this case by 88 J K-1 and mol-1 (Fig. 9.1.11)
The Kinetic Chelate Effect
In addition to the thermodynamic chelate effect, there is the kinetic chelate effect. Chelate complexes are frequently more inert than complexes with monodentate ligands. Chelate complexes are more inert for two reasons (Fig. 9.1.12).
Firstly, the whole ligands needs to rotate and bend in order to cleave the first metal-ligand bond. This requires time and slows the kinetics of the bond cleavage. The second reason is that the detached donor atom cannot leave the proximity of the complex because the ligand is still attached via the other donor atom. This increases the probability of the re-formation of the metal-ligand bond which decreases the probability of both bonds being cleaved.
Mechanism of Ligand Substitution Reactions
Now let us talk about the mechanism of ligand substitution reactions. We know two basic mechanisms according to which a ligand substitution can occur. One is called the intimate mechanism, and the other one is called the stoichiometric mechanism.
The intimate mechanism goes through a single transition state as the reaction proceeds.
The stoichiometric mechanism goes through an intermediate. The intermediate is a shallow local minimum on the reaction coordinate. The transition state is the maximum on the reaction coordinate. For both mechanisms the principle of microscopic reversibility holds. This means that the reverse reaction follows the same free energy path as the forward reaction. For both mechanisms the free Gibbs energy of reaction is given by the difference between the free energy of the products and the free energy of the reactants. The reaction can only spontaneously occur if the free energy of the products is smaller than the free energy of the reactants. The activation energy is the free energy difference between the transition state and the reactants.
Classifications of Mechanisms
There are two forms of the stoichiometric mechanism: The dissociate mechanism D and the associative mechanism A. In the dissociative mechanism (Fig. 9.1.15), an old ligand X is lost first, and the intermediate has a lower coordination number. In the associative mechanism (Fig. 9.1.16), the new ligand Y is added first and the intermediate has a higher coordination number. The intermediates must be detectable, thus must be present in relatively high concentration to measurable with an analytical technique. This requires are relatively deep local thermodynamic minimum for the intermediate. In practice, many local minima are too shallow to allow for the clear detection of the intermediate. We call such a mechanism an Interchange mechanism. There are two forms of interchange mechanisms: Ia and Id. Ia stands for associative interchange mechanism and ID stands for dissociative interchange mechanism. Overall we can have a continuous range of mechanisms that range from A to Ia to intimate to Id to D.
For a dissociative and the Id mechanism the dissociation is the rate limiting step. This is because the activation energy is associated with the dissociative step. We need to go strongly uphill to achieve dissociation, and go into the local thermodynamic minimum. From, there we only need to go slightly uphill to add the new ligand.
For the associative and the Ia mechanism (Fig. 9.1.16), the associative step is the rate limiting step because the activation energy is associated with the associative step. In this case we first need to go energetically strongly uphill to achieve association and reach the local thermodynamic minimum. We only need to go slightly uphill from the local thermodynamic minimum to lose the old ligand.
How to Distinguish Between D/Id and A/Ad
We can use the relationships between the rate determining step and the mechanism to experimentally distinguish between associative A/Ia and dissociative D/Id mechanisms. If the reaction rate strongly depends on the new ligand, then we likely have an associative or Ia mechanism. This is because the associative step is the rate determining step. If the reaction rate does not strongly depend on the new ligand, then the mechanism is likely dissociative D or Id. This is because in the dissociate mechanism, the addition of the new ligand is not the rate-determining step.
For example, we can substitute a chloro ligand in a [PtCl(dien)]+ complex by iodo and bromo ligands, respectively (Fig. 9.1.17). What we measure is that the reaction rates are by a factor of 100 different. This means that the reaction rates does depend on the new ligand and this indicates an associative or Ia mechanism.
Steric arguments can help to predict if a mechanism is associative or dissociative. Lower coordination numbers and little steric crowding tends to favor A and Ia mechanisms, while higher coordination numbers and greater steric crowding tend to favor the dissociative mechanism.
Dr. Kai Landskron (Lehigh University). If you like this textbook, please consider to make a donation to support the author's research at Lehigh University: Click Here to Donate.
Concept Review Questions Chapter 9
Concept Review Questions
Section 1
1. When in general does a chemical reaction occur?
2. What is meant by an inert and a labile complex?
3. For which d electron configuration would you expect inert complexes?
4. How does the leaving group effect influence the inertness of a coordination compound?
5. With which concept can the stability of coordination compounds be estimated?
6. What is the linear free energy relationship? Explain its origins qualitatively.
7. Explain why the constant of formation for multi-ligand complexes decreases with the addition of additional ligands? Under which circumstances are there exceptions?
8. What is the “thermodynamic chelate effect”?
9. What is the “kinetic chelate effect”?
10. What is meant by an “intimate” and a “stoichiometric” mechanism respectively?
11. What is the difference between a transition state and an intermediate?
12. What is the difference between an associate mechanism A and an interchange mechanism Ia?
13. What is the difference between a dissociative mechanism D and an interchange mechanism Id?
14. How can one experimentally distinguish between an associative and a dissociative mechanism?
15. Which factors favor associative mechanisms and dissociative mechanisms respectively?
Dr. Kai Landskron (Lehigh University). If you like this textbook, please consider to make a donation to support the author's research at Lehigh University: Click Here to Donate.
Homework Problems Chapter 9
Homework Problems
Section 1
Exercise 1
An associative mechanism is favored when
a) the oxidation state of the metal is high
b) the coordination number of the metal is low
c) the complex has bulky ligands
d) the metal –ligand bonds are labile.
Answer
b) the coordination number of the metal is low
Exercise 2
Which of the following metal-ligand bonds would you expect to have the highest thermodynamic stability?
a) Hg-S
b) Hg-O
c) Hg-N
d) Hg-F
Answer
a) Hg-S
Exercise 3
Chelate complexes are thermodynamically particularly stable because
a) The substitution of a simple ligand by a chelating ligand is enthalpically particularly favorable.
b) The substitution of a simple ligand by a chelating ligand is entropically particularly favorable.
c) Chelating ligands are more inert than simple ligands.
Answer
b) The substitution of a simple ligand by a chelating ligand is entropically particularly favorable.
Exercise 4
What is true about intimate mechanisms?
a) The intimate mechanism involves an intermediate that is detectable
b) The intimate mechanism involves an intermediate that is not detectable.
c) The intimate mechanism involves a transition state.
Answer
c) The intimate mechanism involves a transition state.
Exercise 5
What is true about complexes that undergo metal to ligand charge transfer?
a) The metal ion has no or few d electrons
b) The ligands are good pi-acceptors
c) The metal ions have a high charge.
Answer
b) The ligands are good pi-acceptors | textbooks/chem/Inorganic_Chemistry/Inorganic_Coordination_Chemistry_(Landskron)/09%3A_Coordination_Chemistry_IV_-_Reaction_and_Mechanisms/9.01%3A_Substitution_Reactions.txt |
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10: Organometallic Chemistry
Organometallic Chemistry
In this chapter you will get an introduction in a subfield of coordination chemistry: Organometallic chemistry. How is an orgamometallic complex defined?
Definition: Organometallic Complex
A complex with bonding interactions between one or more carbon atoms of an organic group or molecule and at least one metal atom is called an organometallic complex.
It is defined as a complex with bonding interactions between one or more carbon atoms of an organic group or molecule and at least one metal atom. It is important to understand that just the presence of an organic ligand is not sufficient to define an organometallic compound yet. There must be interactions between a carbon and a metal atom. If an organic ligand is present but no donor atom is a carbon atom, then we call this complex a metal-organic complex.
Let us practice this by one example: which of the following molecules are metal-organic and organometallic respectively: tetramethyl tin and tetramethoxy tin (Figure $1$). The former is organometallic while the latter is metal-organic. Why? This is because in tetramethyl tin there are Sn-C bonds, while in tetramethoxy tin there are Sn-O bonds.
Industrial Importance of Organometallic Compounds
Organometallic compounds are an industrially very important class of compounds. For example, aluminum and tin alkyl compounds are produced at amounts in the order of tens of thousand of tons per year. Even more important however is the use of organometallic compounds as catalysts to produce other compounds. This ranges from commodity polymers like polypropylene and polyethylene to simple organic molecules like acetaldehyde and acetic acid. These compounds are produced at a scale in the order of millions of tons per year.
History of Organometallic Chemistry
Let us take a historical approach to organometallic chemistry and see how the field has evolved. The arguably first organometallic compound was kakodyl oxide, an organo-arsenic compound.
It was accidentally produced by the French chemist Louis Cadet in 1760 when he was working on inks. He heated arsenic oxide and potassium acetate and obtained a red-brown oily liquid, known as Cadet’s fuming liquid.
It consists mostly of cacodyl and cacodyl oxide (Figure $2$). In cacodyl, there is an As-As bond and two methyl groups attached to each As atom. In cacodyl oxide, there is an O atom in between the two As atoms, and each As is bound to two methyl groups. The names of these compounds come from the greek word “kakodes”=bad smell. Indeed, they have a very intense, garlic-like smell. The reaction can be used to identify arsenic in samples. For instance, if you suspected your food was poisoned with arsenic, you could heat a sample together with potassium acetate. If a bad, garlic-like smell evolved, then this would indicate that there is arsenic in your sample.
Another important milestone in coordination chemistry was the discovery of Zeise’s salt by the Danish chemist William Zeise in 1827 (Figure $3$). Zeise’s salt was the first olefin complex in which an olefin was bound side-on to a metal using its π-electrons for σ-bonding. Zeise observed that when sodium hexachloroplatinate (2-) was heated in ethanol, a compound with the composition Na[PtCl3(C2H4)] could be isolated. The chemical composition of the compound suggested that there was a C2H4 organic fragment bonded to the platinum, but it was not clear how.
This question was only answered more than 100 years later in 1969 with the crystal structure analysis of Zeise’s salt. The crystal structure revealed that an ethylene molecule was bound side-on to the platinum (Figure $4$). The two carbon atoms had about the same distance to the Pt arguing that they were equally strongly involved in the bonding with Pt. The first C had a distance of 216 pm while the second carbon had a distance of 215 pm. The ethylene molecule was oriented perpendicular to the plane made of the three chloro ligands and the platinum atom. The Cl-Pt-Cl bond angles were near 90°. Overall, the structure could be described as a structure derived from a square planar structure with the ethylene as the fourth ligand that would stand perpendicular to the plane in order to minimize steric repulsion with the chloro ligands.
An important question was how to describe the bonding in the compound. It was noteworthy that the C-C bond length in Zeise’s salt was significantly longer (144 pm) than that in a free ethylene molecule (134 pm). Based on the bond-strength bond-length concept this argued that the bond was weaker than a regular C=C double bond, and that the bond order was smaller than 2. What could explain this lower bond order and the side-on coordination of the ethylene in Zeise’s salt?
Bonding in Zeise's Salt
The answer is that the ethylene molecule uses its π-electrons for σ-bonding with a metal d-orbital (Figure $5$). You can see above that the lobe of a dx2-y2 orbital has the correct orientation to overlap with the bonding π-orbitals of the ethylene ligand in a σ-fashion. Through this interaction electron density gets donated from the bonding ligand π-orbital into the metal-d-orbital. This results in a lower electron density for π-bonding within the ligand, and thus the bond order is decreased. In addition, there is another effect that lowers the bond order. A metal d-orbital can interact with the π*-orbitals in π-fashion. In this process, electron density can be donated from the metal d-orbital into the empty π-orbitals of the ligand. The increased electron density in the π-orbitals further decreases the bond order.
In 1890, a further milestone in organometallic chemistry was reached with the synthesis of the first carbonyl complex, the nickel tetracarbonyl. Nickel tetracarbonyl is a colorless liquid that boils at 43°C.
The compound was prepared by the German chemist Ludwig Mond (Figure $6$). Nickel tetracarbonyl forms spontaneously from nickel metal and carbon monoxide at room temperature.
The reaction is the basis of the Mond-process that is used up to date in order to purify nickel (Figure $7$). Carbon monoxide gas can be streamed over impure Ni metal at temperatures of 50-60°C to form Ni(CO)4 in gaseous form. Impurities are left behind in solid form.
Then, the nickel tetracarbonyl is thermally decomposed at ca. 220° to form nickel and carbon monoxide. The process can be varied to produce nickel in powder form, as spheres (Figure $8$), and as coatings. The Mond process is used up to date despite the very high toxicity of nickel tetracarbonyl.
In 1900 the first Grignard reagents were discovered. Victor Grignard (Figure $9$) was an enthusiastic young French chemist who discovered how to make organomagnesium halides (RMgX) while working for his Ph.D.
His supervisor, Sabatier, had been trying this chemistry for some time, but Victor was the genius who solved the problem. Organomagnesium halides form from magnesium and organic halides in ethers. This discovery in 1900 changed the course of organic chemistry and won Grignard and Sabatier the Nobel Prize in 1912.
In 1917, the first lithium alkyls were prepared by Wilhelm Schlenk (Figure $11$). Like Grignard reagents, lithium alkyls are very valuable reactants in synthetic organic chemistry. Lithium alkyls are very air-sensitive compounds, and some do even self-ignite in air, for example tert-butyl lithium. Therefore, they need to be handled carefully under inert gas.
In this regard, Wilhelm Schlenk made an important invention: The Schlenk-lines (Figure $12$). The Schlenk lines allow to evacuate reaction flasks and back-fill them with an inert gas. A Schlenk line has a dual manifold with several ports. One manifold is connected to a source of purified inert gas, while the other is connected to a vacuum pump. The inert-gas line is vented through an oil bubbler, while solvent vapors and gaseous reaction products are prevented from contaminating the vacuum pump by a liquid nitrogen or dry ice/acetone cold trap. Special stopcocks or Teflon taps allow vacuum or inert gas to be selected without the need for placing the sample on a separate line.
The discussed discoveries were important milestones in the development of organometallic chemistry, but the field expanded only rapidly with the discovery of the first metallocene: The ferrocene. Like often in science, ferrocene was discovered through an accident.
In 1951 Peter Pauson and Thomas Kealy attempted to synthesize the organic compound fulvalene through oxidative coupling of cyclopentadienyl magnesium chloride with iron (III) chloride. However, instead of obtaining fulvalene they observed the formation hydrofulvalene together with an orange powder with “remarkable stability”. Analysis of the powder showed that the chemical compound contained two cyclopentadienyl rings and one iron atom. According to the knowledge on bonding in organometallic compounds that was known at the time Kealy and Pauson suggested a structure for the molecule in which two cyclopentadienyl group were bound to a single Fe atom via two Fe-C bonds (Figure $13$).
However, the remarkable stability of the compound was in contradiction to this structure. The structure would be a 10 electron complex which would be highly coordinatively unsaturated. This would be inconsistent with the observed stability. We can briefly practice our electron counting skills again in order to prove that there are only 10 electrons (Figure $14$).
For example we can use the oxidation state method for electron counting. Fe in the neutral state has eight valence electrons. We would cleave the bonds heteroleptically which would create cyclopentadienyl anions with a 1- charge. This would mean that Fe is in the oxidation state +2, and thus we would subtract two electrons. The cyclopentadienyl anions would contribute two electrons each giving ten electrons overall. In contrast to expectations, the compound was stable in air, and could be sublimed without decomposition. Further the double-bonds in the cyclopentadienyl rings resisted hydrogenation. In addition, the structure was inconsistent with spectroscopic observations. Only one C-H stretch vibration was observed in the IR and only one signal was observed in the 1H NMR. Kealy and Pauson’s structure would have been consistent with three C-H stretch vibrations and three NMR resonances.
For these reasons, Kealy’s and Pausons structure was soon questioned. Ernst Otto Fischer at University of Munich and Sir Geoffrey Wilkinson at Harvard University (Figure $15$) suggested an alternative structure that was very different from all known organometallic structures. The bonding in this structure was completely different from all bonding concepts considered thus far for organometallics.
What structure did they propose? They proposed a sandwich structure in which the two aromatic cyclopentadienyl anions would sandwich an Fe2+ ion. What would be the bonding in this structure? The cylopentadienyl anion would donate its six π-electrons into the valence orbitals of the metal. Thus, all five carbon atoms would be equally involved in the bonding, the ligand would act as a ƞ5-ligand. Because of the donation of π-electrons in the cycopentadiene unit, the structure was called “ferrocene”. The ferrocene structure would explain the stability and low reactivity of the compound.
Electron-counting gives an 18 electron complex. Let us verify this using the oxidation state method. Fe would contribute eight electrons in the neutral state. The ligands would be considered as cyclopendienyl anions with a 1- charge. This would give an oxidation state of +2 for Fe reducing the number of electrons from eight to six. The two ligands would contribute six electrons each giving twelve electrons. 12+6=18. Hence, the structure fulfills the 18 electron rule. The structure is also consistent with the spectroscopic observation of one NMR resonance and one C-H stretch vibration. All these were excellent arguments to support the ferrocene sandwich structure, however they were not an absolute proof. The proof finally came with the X-ray crystal structure determination which unambiguously confirmed the sandwich structure.
Bonding in Ferrocene and MO Theory
Can the stability of the ferrocene structure also be explained by molecular orbital theory? Let us check! First, we need to decide on the point group. We will make the simplification that the two Cp cyclopentadienyl rings are in eclipsed formation, and then the point group is D5h. Actually, the rings are in staggered confirmation and the point group is D5d, but the energy difference is minimal, and there is a very small activation barrier between the two conformers.
\begin{array}{|c|rrrrrrrr|cc|}
\hline \bf{D_{5h}} & E & 2C_5 & 2C_5^2 & 5C_2 & \sigma _h & 2S_5 & 2S_5^2 & 5 \sigma_h & h=20 & \
\hline A_{1}’ & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & & x^2+y^2, \; z^2\A_{2}’ & 1 & 1 & 1 & -1 & 1 & 1 & 1 & -1 & R_z & \
E_{1}’ & 2 & 2cos(72 ^{\circ}) & 2cos(144 ^{\circ}) & 0 & 2 & 2cos(72 ^{\circ}) & 2cos(144 ^{\circ}) & 0 & (x, \; y) & \
E_{2}’ & 2 & 2cos(144 ^{\circ}) & 2cos(72 ^{\circ}) & 0 & 2 & 2cos(144 ^{\circ}) & 2cos(72 ^{\circ}) & 0 & & (x^2-y^2,\; xy) \
A_{1}” & 1 & 1 & 1 & 1 & -1 & -1 & -1 & -1 & & \
A_{2}” & 1 & 1 & 1 & -1 & -1 & -1 & -1 & 1 & z & \
E_{1}” & 2 & 2cos(72 ^{\circ}) & 2cos(144 ^{\circ}) & 0 & -2 & -2cos(72 ^{\circ}) & -2cos(144 ^{\circ}) & 0 & (R_x,\;R_y) & (xz,\; yz) \
E_{2}” & 2 & 2cos(144 ^{\circ}) & 2cos(72 ^{\circ}) & 0 & -2 & -2cos(144 ^{\circ}) & -2cos(72 ^{\circ}) & 0 & & \
\hline \end{array}
We can define the z-axis standing perpendicular to the Cp rings, and the xy plane to be coplanar with the Cp rings.
The Fe 3d, 4s and 4p orbitals will be our frontier orbitals and we can read their symmetry types from the character table of D5h. The 3dxz and 3dyz have E1’’ symmetry, the 3dx2-y2 and the 3dxy have E2’ symmetry, the 3dz2 has A1’ symmetry, the 4s has A1’ symmetry, and the 4px and the 4py orbitals have E1’ symmetry, and the 4pz has A2’’ symmetry.
Next, we need to determine the ligand group orbitals. We know that the π-ligand molecular orbitals are the ones that donate the electrons into the metal-orbitals, thus we have to have a closer look at these orbitals. The π-MOs of the ligand are made from the five 2pz orbitals of the carbon atoms that stand perpendicular to the ring. This means that there are 5 MOs to consider.
You can see the five MOs and their relative energy above. One is strongly bonding and has no node. Then, there are two double-generate weakly bonding ones that have one node, and finally there are two anti-bonding ones that have two nodes. Because the cyclopentadienyl anion has six π electrons, the bonding MO and the two weakly bonding MOs are full, the anti-bonding MOs are empty.
Because we have two Cp- anions to consider we have overall 10 MOs to combine. We would therefore expect ten ligand group orbitals (LGOs). We can determine their symmetry type by determining the reducible and irreducible representations. The results are shown below.
We can generally divide the LGOs into three groups with a different number of nodes. There are two 0-node LGOs with A1’ and A2’’ symmetry. They are made from the 0-node MOs of the ligand. There is a bonding an anti-bonding combination possible. In the bonding combination the 0–node ligand MOs have the lobes with the same algebraic sign pointing toward each other. The anti-bonding combination has the lobes with opposite algebraic sign pointing toward each other.
There are four 1-node LGOs with E1’ and E1’’ symmetry. They are constructed from the two 1-node ligand MOs. Like for the 0-node orbitals, there is a bonding and an anti-bonding combination possible.
Finally, there are four 2-node LGOs with E2’ and E2’’ symmetry. They are made from the 2-node ligand MOs and there is also a bonding and an anti-bonding combination possible.
The Molecular Orbital Diagram of Ferrocene
Now we have all the information to draw the molecular orbital diagram of ferrocene. As usual we plot the metal frontier orbitals on the left and label their symmetry. We plot the Cp LGOs on the right and also label the symmetry. We can order the LGOs according to energy with the 0-node LGOs having the lowest energy and the 2-node LGOs having the highest energy. Then, we can start to combine orbitals of the same symmetry type to form MOs. There are two metal AOs of the A1’ type and one A1’ LGO giving three MOs of this symmetry type, one bonding, one approximately non-bonding, and third one anti-bonding. We can connect AOs, LGOs, and MOs with dotted lines. Next, we can combine the A2’’ AO and the A2’’ LGO to form a bonding and an anti-bonding MO. We again connect the AOs, LGOs, and MOs with dotted lines. Then, we can produce two bonding e1’’ and two anti-bonding e1’’ MOs from the E1’’ metal AOs and the E1’’ LGOs and connect the orbitals with dotted lines. There are two E1’ LGOs and two E1’ AOs that can be combined to two bonding 1e1’ and two anti-bonding 2e1’ MOs and we again connect the orbitals with dotted lines. The two E2’ LGOs can be combined with the two E2’ d-orbitals to form a pair of bonding 1e2’ and anti-bonding 2e2’ molecular orbitals. Lastly, we notice that the E2’’ LGOs do not find a partner, and we have to write them as non-bonding with the same energy into the MO diagram.
Now we need to fill the electrons into the orbitals. The ligands have twelve electrons overall. They can be filled into the six orbitals of the lowest energy. This fills the 1a1’, the 1a2’’, the 1e1’’, the and 1e1’ orbitals. We notice that all MOs are bonding which supports the stability of the molecule. We still need to consider the metal d-orbitals. We have an Fe2+ ion and thus six metal d-electrons. They would go into the 1e2’ orbital which is bonding and the 2a1’ orbital which is weakly bonding. The 2a1 is the HOMO, and the next higher 2e1’’ is the LUMO. We can see that we can fill all metal electrons into bonding MOs. The LUMO is an anti-bonding orbital, and thus overall all bonding MOs are filled, and no non-bonding and anti-bonding orbitals need to be filled. This is the ideal situation for a stable molecule. We can also see that the MO diagram explains the 18 electron rule. All 18 electrons are in bonding MOs.
We can consider the 1e2’, the 2a1’, and the 2e1’’ metal d-orbitals in the ligand field produced by the Cp-ligands as these 5 orbitals can hold the maximum possible number of 10 d-electrons, have similar energy then the d-orbitals, and have contributions from them.
Metallocenes with other cyclic π-ligands
Is it possible to make metallocenes with other π-conjugated rings but the cylopentadienyl anion?
The answer is yes, for instance benzene is known to act as a ligand bis-(benzene) chromium (0). In this case the ligand acts as a ƞ6-ligand because all six carbon atoms are involved in the bonding. Why does chromium give stable metallocene complex with benzene? We can explain this again with the 18 electron rule. In bis-(benzene) chromium (0), chromium is in the oxidation state 0 because the benzene ligand has no charge. Thus, chromium contributes six electrons. Adding the 12 π-electrons from the two benzene ligands gives 18 electrons.
Other cyclic π-ligands
Also cyclobutadiene can act as a cylic π-ligand in complexes. The cyclobutadiene is is different from the cyclopentadienyl anion and the benzene ligand in two ways. Firstly, it has much more ring strain then the previous two, and secondly it is not an aromatic, but an anti-aromatic ligand. Remember, we have an aromatic ring when there are 4n+2 π-electrons, whereby n is an inter number. This means that rings with two (n=0), six (n=1), and ten (n=2) π-electrons are aromatic. Anti-aromatic rings are those that have 4n electrons, such as four (n=1), eight (n=2) and so forth. Cyclobutadiene has four electrons, and thus it is anti-aromatic. Anti-aromatic rings are less stable than aromatic ones because not all π-electrons are in bonding molecular orbitals. Let us illustrate this by constructing the MO diagram for the π-system of the cyclobutadiene molecule (Figure $25$)
The π-system is made of four carbon atoms contributing a half-filled 2pz orbital each (if we define the plane of the molecule as the xy plane). That makes four p-orbitals with four electrons that give four molecular orbitals. There is a bonding MO with no node, two doubly-degenerate non-bonding ones with one node, and one anti-bonding one with two nodes (Figure $25$). There are four electrons. We can fill two electrons into the bonding MO, but the other two must go into the two non-bonding ones under obedience of Hund’s rule.
Jahn-Teller distortion in cyclobutadiene
One may think that the cyclobutadiene is a square planar molecule because of complete delocalization of the π-electrons, but that is actually not the case. There are two shorter double-bonds and two longer single-bonds, and the shape of the molecules is a rectangle. This means that the two double bond are localized. We can view this effect as a Jahn-Teller distortion.
The distortion of the square to form a rectangle is energetically favorable because it lowers the energy of the two non-bonding electrons. Why? Let us look at the non-bonding orbital 1 first and elongate in x-direction, and squeeze in y-direction (Figure $26$). We can see that we bring the p-orbitals with bonding interactions further apart, and bring those with anti-bonding interactions closer together. This means that the energy of this orbital goes up and the orbital becomes anti-bonding. Let us do the same with non-bonding orbital 2. We see that the opposite happens. The bonding interactions are enhanced and the anti-bonding interactions are weakened. Therefore, this orbital becomes bonding and the energy goes down. We can now fill the two electrons into the bonding MO. We see that we have lowered the energy of the electrons through the distortion.
We could also have squeezed in x-direction and elongated in y-direction. In this case orbital 1 would have become bonding and orbital 2 anti-bonding. However, this distortion is symmetry-equivalent to the previous one, and does not produce a new molecule. Both molecules can be superposed by a simple 90° rotation.
Cyclobutadiene as ƞ4-ligand
Cyclobutadiene (Cb) in its free form is not stable because of the high ring strain and the anti-aromaticity, but complexes with cyclobutadiene as a ligand are stable. We would expect that 18 electron complexes are most stable.
What would be an 18 electron sandwich complex with two cyclobutadiene ligands? Because each cyclobutadiene contributes four electrons, ten electrons would need to come from the metal, and we would expect a nickel cyclobutadienyl complex Ni(Cb)2. This complex is not known, but derivatives are. For example a Ni complex with two tetraphenyl butadiene ligands are known (Figure $27$). Another example is the butadienyl tricarbonyl iron (0) complex. Interestingly, when cyclobutadiene acts as a ƞ4-ligand, then it is not distorted, but square planar. An explanation is that we can formally treat the cyclobutadiene ligand as an Cb2- ligand binding to a metal cation. For instance in FeCb(CO)3 the Fe would be an Fe2+, in the nickel complex, the nickel would be a Ni4+. The Cb2- anion would formally aromatic because it has 4n+2=6 electrons. The additional two electrons would be in the non-bonding orbitals. Because the two non-bonding orbitals are completely filled now, there would be no longer a driving force for the distortion. Note though that is is a formal view only, and there are arguments that speak against this view. One is that the addition of the two electrons to the ligand should further destabilize the ring because the added electrons are non-bonding. Aromaticity would rather be achieved by removing two electrons to form a Cb2+ cation that would have only two electrons in the bonding orbital.
Cyclooctetraene (cot) as a ligand
Also metallocenes with cyclooctatetraene (cot) acting as ƞ8-ligand is known. However, because of the large ring size only metals with large atomic radii, can make metallocenes with this ligand. For example uranium makes a uranocenes with two cyclooctatetraene ligands (Figure $28$).
Like cyclobutadiene, cyclooctatetraene is an anti-aromatic ligand with 4n=8 π-electrons (n=2). In the free cyclooctatetraene molecule is non-planar and the π-electrons are localized (Figure $29$).
In metallocenes however, the ring becomes planar (Figure $28$). One can again formally explain that by assuming an aromatic cot2- ligand that binds to a metal 2+ cation, however one should keep in mind again that this is a formal view and not necessarily reflects the bonding situation in the compound.
Metals having smaller atomic radius can bind to cot in ƞ2, ƞ4, and less commonly in ƞ6-mode. The 18 electron rule holds in most cases. For instance, cot can make a ƞ4-complex in tricarbonyl cyclooctatetraene ruthenium (0) (Figure $30$).
Also two metals can bind to a single cot-ligand. This is for instance realize in μ-cyclooctetraene bis(tricarbonyl ruthenium (0) (Figure $31$).
Dr. Kai Landskron (Lehigh University). If you like this textbook, please consider to make a donation to support the author's research at Lehigh University: Click Here to Donate. | textbooks/chem/Inorganic_Chemistry/Inorganic_Coordination_Chemistry_(Landskron)/10%3A_Organometallic_Chemistry/10.01%3A_Historical_Background.txt |
Introduction into Carbonyl Complexes
In this chapter we will look closer at carbonyl complexes, often just called carbonyls. Why? They are interesting for a number of reasons. Firstly, they are a quite extensive class of compounds with a diverse coordination chemistry. We will learn that the CO ligand can bind to a metal in various, sometimes non-obvious ways. Further, carbonyls are frequently used as starting materials for other coordination compounds. This is because the carbonyl ligand has no charge and carbon monoxide is a gas. Because of that a ligand substitution reaction can be easily driven to the right side by purging CO out of the reaction vessel. CO is a C1 unit, and this is frequently used in catalysis with carbonyls. Carbonyls are often intermediates in reactions that add a single carbon atom to a hydrocarbon chain. A fascinating fundamental property of carbonyls is that its $\pi$-accepting properties stabilize metals in low, sometimes even negative oxidation numbers.
Bonding in CO
The carbon in the CO molecule is the more reactive end, and thus CO prefers to bind with the carbon to a metal, and not with oxygen. This is not obvious because the carbon is less electronegative than oxygen. The reason is that the HOMO of the CO molecule is an approximately non-bonding orbital which is primarily localized at the carbon (Figure $1$). In the valence bond picture it is resembled by the electron-lone pair at the carbon atom. The electron lone pair at the oxygen is resembled by the 2a1 molecular orbital which has a significantly lower energy, therefore it is rarely used in bonding. Both the $\pi$ and $\pi$*-orbitals are relatively close in energy to the HOMO, and thus these orbitals can also be used for bonding in carbonyls. This explains the diverse coordination chemistry of carbonyls.
σ-binding Modes of The Carbonyl Ligand
In the most simple case the CO uses its electron lone pair at the carbon to bind to single metal atom M (Figure $2$). In this case we call CO a terminal carbonyl ligand. In addition, it is possible that the electron lone pair is shared between two or even three metals that are interconnected via metal-metal bonds. In these cases, we say that the CO acts as a $\mu$-bridged, and $\mu_3$-brigded ligand, respectively. If two metals held together by a metal-metal bond are different, then the interaction of CO with one metal may be stronger than with the other. In this case the CO ligand acts as a semibridging ligand. In all these cases CO acts as a 2-electron donor because it donates its two electrons in the electron lone pair in the carbon atom.
In addition to the electron lone pair at the carbon the CO ligand can also use its binding $\pi$-orbitals for electron-donation into metal orbitals (Figure $3$). However, these electrons can only be used in conjunction with the electron lone pair at C because the electron pair has a higher energy than the $\pi$-electrons, and electrons of higher energy are always used first. For steric reasons the electron lone pair and the $\pi$-electrons are always donated to different metals that are held together by metal-metal bonds.
In the most simple case the CO ligand binds in $\mu_2$-$\eta_2$-mode which implies that the electron lone pair binds to one metal and two $\pi$-electrons bind to another. In this this case the CO ligand acts as a 4-electron donor. In addition, it is possible that all four $\pi$-electrons are involved involved in the bonding in addition to the electron lone pair. In this case three metal atoms are involved. The first one interacts with the electron-lone pair, the second one with the two $\pi$-electrons, and the third one with the other two $\pi$-electrons. Because three metals are bridged and both atoms of the ligand are involved in the bonding with the metal we can say that CO binds in $\mu_3$-$\eta_2$-fashion and acts as a 6-electron donor. It is also possible that the electron-lone pair is being shared between two metals, and two $\pi$-electrons are donated to the third metal. Also in this case the ligand binds in $\mu_3$-$\eta_2$-fashion, but acts as a 4-electron donor. Another way the CO ligand can act as 4-electron donor is when it acts as an isocarbonyl ligand. In this case it uses its electron pairs at both the C and the O-atoms. This is only possible when steric circumstances favor the utilization of the O-electron lone pair over the energetically higher $\pi$-electrons. This is rare.
$\pi$-binding Modes of The Carbonyl Ligand
The carbonyl ligand can use its $\pi$*-orbitals for bonding with metal d-orbitals in $\pi$-fashion. The ligand acts as a $\pi$-acceptor. There are two possibilities for the binding (Figure $4$).
The first one occurs when the CO-ligand acts as a terminal ligand binding end-on. In this case the two lobes of the $\pi$*-orbitals at the carbon interact with the the lobes of a metal d-orbital. Also the bonding $\pi$-orbitals have the right symmetry to overlap with the metal d-orbital in this mode, but their energies are much further away from those of the d-orbitals, so that these interactions can be neglected. The second possibility is that the CO binds side-on to the a metal-d-orbital. In this case one lobe at C and one lobe at O interacts with the metal d-orbital. These interactions are less strong compared to the end-on interactions because the orbital overlap is less efficient due to the unequal size of the lobes of the $\pi$*-orbital. The $\pi$*-orbital is mostly localized at the carbon-atom because the carbon atom is the more electropositive atom. Note that the bonding $\pi$-orbitals do not have the right symmetry to overlap with a metal-d orbital in $\pi$-fashion, they can only overlap in σ-fashion when the binding is side-on.
The Dewar-Chatt-Duncanson model
The σ-donor and the $\pi$-acceptor interactions in carbonyl complexes synergistically reinforce each other. This synergistic effect is called the Dewar-Chatt-Duncanson model for carbonyls. How can we understand the synergistic interactions. Let us consider a carbonyl ligand that binds end-on to a metal.
Let us first look at the σ-donor interactions. The electron lone pair at the carbon donates electron density into empty metal d-orbitals and a dative bond is formed between the metal and the carbon. The donated electron density enhances the energy of the metal d-electrons due to increased electron-electron repulsion. Because of their increased energy the d-electrons get more easily accepted by the carbonyl ligand through the $\pi$-acceptor interactions. The $\pi$-acceptor interactions increase the bond order between the metal and the carbon bond. At the same time the bond order between carbon and oxygen gets decreases because electron density has been transferred from the metal into the $\pi$*-orbitals (Figure $5$).
The strength of the $\pi$-acceptor interactions can differ significantly in carbonyls. One can draw three different structures for weak, intermediate, and strong $\pi$-acceptor interactions (Figure $6$ to $8$).
When only weak interactions are present we can represent the M-C bond as a single bond, and the C-O bond as a triple bond.
When the interactions have intermediate strength we can represent both the M-C bond and the C-O bond as a double-bonds.
When the $\pi$-acceptor interactions are strong then the M-C bond becomes a triple bond and the C-O bond becomes a single bond.
On what does the strength of the $\pi$-acceptor interactions depend? It depends mostly on the charge on the carbonyl. Carbonyl cations with a charge of +2 or higher tend to have weak $\pi$-acceptor interactions. Neutral carbonyls or carbonyls with a +1 or -1 charge have intermediate $\pi$-acceptor interactions. and those with a negative charge of 2- or higher have strong $\pi$-acceptor interactions. One can see from this that the smaller the positive charge and the higher the negative charge, the higher the metal-carbon bond-order, and the stronger the metal-carbon bond. As a consequence negative charges tend to stabilize carbonyls, while positive charges destabilize them. Therefore, carbonylate anions tend to be more stable compared to carbonyl cations. The stability of neutral carbonyls is intermediate.
Homoleptic Carbonyls of The Transition Metals
After having understood the principles of the bonding in carbonyls let us next think about what structures transition metal carbonyls make. The structures follow mostly the 18 electron rule. For the most simple homoleptic carbonyls, in which the carbonyl ligand binds end-on to the transition metal acting as a two-electron donor, we would just assemble as many ligands as needed until we have 18 electrons. This would give us the coordination number in the carbonyl from which we could deduct the structure. This works well for transition metals with an even number of valence electrons.
In this case the number of ligands x is just 18 minus the number of metal electrons divided by 2. We would therefore expect a mononuclear carbonyl of the type M(CO)x (Figure $9$). For metals with an odd number of electrons, the situation is more complicated because an odd number times two multiplied with an integer number never gives 18. In this case, as many ligands x as needed to make a 17 valence electron complex are added to the metal, and then the 17 valence electron complex dimerizes to form a dinuclear carbonyl of the composition M2(CO)x. An exception is the vanadium. It makes a 17 valence electron vanadium hexacarbonyl that does not dimerize because a coordination number of 7 is not favorable.
Homoleptic carbonyls of the 6th group (M = Cr, Mo, W)
Now let us look closer at the carbonyls with metals with an even number of electrons. We can start out with group 6 which contains the elements chromium, molybdenum, and tungsten. Our task is to determine the composition and structure of the carbonyls.
According to the 18 electron rule we need 18 electrons overall. How many are contributed by the metal? Because the metals are in group 6, they all have six electrons. The carbonyls carry no charge, so there are no electrons to be subtracted or added. How many ligand electrons do we need to get to 18? Well, that is 18-6=12 electrons. How many carbonyl ligands are needed then? Because each ligand is considered a two electron donor, we need 12/2=6 ligands. The composition of the carbonyl will therefore be M(CO)6. The carbonyl adopts octahedral shape in order to maximize the distance of the ligands from each other. The group 6 carbonyls are colorless, crystalline, and sublimable solids.
Charged Octahedral Carbonyls with 18 Electrons
Are there also other 18 e carbonyls of the type M(CO)6 with other metals but group 6 metals? Yes, but the charge at the carbonyl needs to be adjusted so that the complex has 18 electrons. For instance, we can replace Cr by V, but then the we need a 1- charge which results in a carbonylate of the formula V(CO)6-. Why? The vanadium is in group 5 and has one electron less than chromium. For that reason, we must add an electron which gives the complex a 1- charge. Also the higher homologues Nb(CO)6- and Ta(CO)6- are known. What if we go further to the left in the periodic table? Titanium is left to the vanadium, and has four valence electrons. Therefore, the 18 valence electron hexacarbonyl of titanium has a 2- charge, and the formula is Ti(CO)62-. Again, the higher homologues of Ti(CO)62- , the Zr(CO)62- and the Hf(CO)62- are also known. Can we also go to the right in the periodic table? Manganese sits to the right of the chromium. The manganese has one electron more than the chromium, therefore the 18 electron hexacarbonyl of manganese must be a carbonyl cation with a 1+ charge, and the formula is Mn(CO)62+. Of the higher homologues, both the Tc(CO)6+ and the Re(CO)6+ are known, too. To the right of the Mn, there is the Fe which has 8 valence electrons. Therefore the 18 electron iron hexacarbonyl must have a 2+ charge, and the formula is Fe(CO)62+. The higher homologues Ru(CO)62+ and Os(CO)62+ are also known. Can we go even further to the left, to the Co? The expected formula for Co carbonyl would be Co(CO)63+, but it is not known, and neither is the higher homologue, the Rh(CO)63+. Only the Ir(CO)63+ has been isolated. This reflects a general trend in carbonyl chemistry. Positive charges destabilize the carbonyl, and carbonyls with high positive charges are frequently not stable. On the other hand, negative charges stabilize a carbonyl and even carbonylates with high negative charges are stable. We can easily explain this by the fact that the bond order between the metal and the carbon decreases with increasing positive charge, and increases with increasing negative charge. Increasing bond order strengthens the stability of the carbonyl. We see from this that the $\pi$-acceptor properties are quite important for the stability of carbonyls. We also see that in carbonylates the oxidation numbers of the metals are negative. Negative oxidation numbers are quite unusual for metals. We conclude that the carbonyl ligand has the unusual property to stabilize the metal in negative oxidation states.
Is there a way to easily measure the bond order of an M-C bond? We can do this indirectly by measuring the IR-spectrum of the carbonyl. Because the bond order of the C-O bond decreases with increasing bond order for the M-C bond, the wavenumber for the C-O stretch vibration can be used as a measure for the M-C bond order. The lower the wavenumber, the lower the C-O bond order, and the higher the M-C bond order. When we look at the numbers for the different carbonyls within a period we see that as expected the wavenumbers increase with increasing group number indicating a decreasing bond order. We can also look down the groups. We see that in this case, the numbers hardly change, meaning that the bond order is hardly affected by the period of the metal.
Homoleptic Carbonyls of the 8th Group (Fe, Ru, Os)
What charge-neutral carbonyls would we expect for the group 8 elements Fe, Ru, and Os?
Overall, these carbonyl need to have 18 electrons. In this case the metals have 8 electrons. The charge is 0. This means that we would have 18-8=10 electrons that would need to come from the ligand. The number of ligands needed would therefore be 10/2=5. Thus, we would expect pentacarbonyls of the composition M(CO)5 (Figure $12$). These carbonyls adopt the trigonal-bipyramidal shape. The group 8 carbonyls are all liquids. The iron pentacarbonyl has a light-orange color while the ruthenium and the osmium carbonyls are colorless.
Charged Trigonal-Bipyramidal Carbonyls with 18 electrons
We can again ask if there are charged isosteric pentacarbonyls with 18 e of metals other than group 8 metals. The answer is yes. The manganese is left to the iron in the periodic table and has one electron less. Therefore its pentacarbonyl needs to have a 1- charge and the composition is Mn(CO)5-. Also the higher homologues, the Tc(CO)5- and the Re(CO)5- exist. The Cr has one electron less than than the Mn, therefore its pentacarbonyl has a 2- charge and the formula is Cr(CO)52-. Mo(CO)52- and W(CO)52- are also known. The vanadium has another electron less because it is in the 5th group, therefore its pentacarbonyl has a 3- charge, and the formula is V(CO)53-. Again, the higher homologues, the Nb(CO)53- and the Ta(CO)53- are also stable. Also for the pentacarbonylates high negative charges are no problem for the stability of the complexes due to the stabilizing effect of the $\pi$-acceptor interactions. What about carbonyl cations with trigonal bipyramidal shape? As we go from Fe to Co, we need a 1+ charge at the carbonyl to achieve 18 electrons. The respective Co(CO)5+ cation is stable, and so are their higher homologues, the Rh(CO)5+ and the Ir(CO)5+. However, the group 10 pentacarbonyl cations Ni(CO)52+, Pd(CO)52+, and Pt(CO)52+ are not stable. The 2+ positive charge weakens the metal-carbon bond too much. This behavior is consistent with the weakening of the M-C bond with increasing positive charge due to weaker $\pi$-acceptor effects.
Homoleptic carbonyls of group 10 (Ni, Pd, Pt)
What about the group 10 carbonyls?
In this case the metals have 10 electrons. There is no charge. The number of ligand electrons needed is 18-10 = 8 electrons. Thus, the number of ligands is 8/2=4. Therefore, the composition is M(CO)4 (Figure $14$). The shape is tetrahedral because we have an 18 and not a 16 valence electron complex. Nickel tetracarbonyl is a volatile, colorless liquid. The higher homologues, the Pd(CO)4 and the Pt(CO)4 are not stable. We can explain this when remembering that the $\pi$-orbital overlap in tetrahedral complexes is generally weak. Therefore, the M-C carbon bond is less effectively stabilized by the $\pi$-acceptor interactions compared to octahedral and trigonal-bipyramidal carbonyls. This effect is even more pronounced for the Pd and Pt carbonyls because they have larger orbitals that can overlap even less effectively with the relatively small $\pi$*-orbitals of the CO ligand. In this case the $\pi$-acceptor interactions are so weak so that the entire molecule is no longer stable.
Charged Tetrahedral Carbonyls with 18 electrons
What about charged 18 electron tetrahedral carbonyls with other metals? Again, negatively charged carbonylates are stable, even with high negative charges. Co(CO)4-, Fe(CO)42-, Mn(CO)43-, and Cr(CO)44- are all stable and so are their higher homologues. The Cu(CO))4+ which has a 1+ positive charge is also known, but its higher homologues, the Ag(CO)4+ and the Au(CO)4+ are not.
Overall Stability Trends
Overall the stability of carbonyls tends to decrease from group 6 to group 8 to group 10 (Figure $16$). This trend is because the octahedral shape allows for the best orbital overlap for $\pi$-acceptor interactions, followed by the trigonal bipyramidal, followed by the tetrahedral shape. The stability also tends to decline from the 4th to the 6th period. This trend can be explained by less efficient orbital overlap due to increasingly different orbital size between the ligand and the metal orbitals. These effects are sufficiently large for Pd(CO)4 and Pt(CO)4 to make them instable.
Group 4 and Group 12 Carbonyls (non-existing)
We have discussed group 6, 8, and 10 carbonyls, but not group 4 and group 12. Why? Because no carbonyls of these groups are stable. What is the reason? For group 4 carbonyls, we would need seven carbonyl ligands. This would lead to a coordination number of seven which is unacceptably high for carbonyls. For group 12 carbonyls, we would only need three carbonyl ligands. This would lead to an unacceptably low coordination number of 3.
Homoleptic carbonyls of the 7th group (M = Mn, Tc, Re)
Now let us think about what structures the homoleptic carbonyls with metals having an odd number of electrons form. Let us start with the 7th group which consists of the metals Mn, Tc, and Re. Our approach is to see how many ligands we need to produce a 17 valence electron fragment, and then dimerize that fragment.
The group 7 metals have seven valence electrons. Again, we will assume that there is no charge on the carbonyl. The number of ligand electrons to produce the 17 valence electrons fragment will therefore be 17-7 = 10. The number of ligands needed is therefore 10/2=5 assuming that the carbonyl ligand is a 2-electron donor binding end-on to the metal. This gives an M(CO)5 fragment that dimerizes to form an M2(CO)10 carbonyl with a metal-metal bond. The structure can be thought to be derived from an octahedral structure in which each metal is surrounded octahedrally by five CO ligands and an M(CO)5 fragment acting as the sixth ligand. Each M(CO)5 unit has one axial ligand and four equatorial ligands, whereby the axial ligand is co-directional with the metal-metal bond. The four equatorial ligands of the first metal are in staggered conformation relative to the four equatorial ligands of the second metal due to steric repulsion arguments.
Charged Isoelectronic Carbonyls of the Type M2(CO)10
Can we make charged isoelectronic carbonyls of the type M2(CO)10 with other metals? The answer is yes, but only carbonylate anions with low negative charge, and no carbonyl cations. The formation of carbonyl cations is prohibited because of the weakening of the $\pi$-acceptor interactions that result from the positive charge. High negative charges are not possible because too many negative charges on the metal atoms lead to electrostatic repulsion and the destabilization of the metal-metal bonds.
There are therefore only Cr2(CO)102-, and the higher homologues Mo2(CO)102- and W2(CO)102-, but no other M2CO10 type carbonyl ions.
Homoleptic carbonyls of the 9th group (M = Co, Rh, Ir)
What are the expected structures for the homoleptic carbonyls of group 9?
In this case, the metal contributes 9 electrons, and this means that 17-9=8 electrons need to come from the carbonyl ligand. We therefore need 8/2=4 CO molecules to produce an M(CO)4 17 valence electron fragment. Two of the fragments then dimerize to form a dinuclear carbonyl of the composition M2CO8. In nature, however, only the Co-carbonyl is realized, the Rh and the Ir analogs are unstable. The structure of the Co2(CO)8 carbonyl can be derived from an M(CO)4L trigonal-pyramid, whereby L is the second M(CO)4 unit (Fig. 10.2.19). According to that, the coordination number is 5. There are two axial and six equatorial ligands whereby the equatorial ligands are oriented in a staggered conformation. In solution this structure is in dynamic equilibrium with a second structure with C2v symmetry in which two CO ligands are bridging. Due to this equilibrium there is constant ligand scrambling, meaning that the CO ligands constantly move from axial to equatorial positions, and migrate from one metal atom to the other. In solid state the C2v-type structure is present exclusively indicating that it is slightly more stable.
Charged Isoelectronic Carbonyls of the Type M2(CO)8
The charged, isoelectronic carbonyls of the type M2(CO)8 behave similarly to that of the type M2(CO)10. There are no highly charged binuclear carbonylate anions, and no carbonyl cations.
There are only the weakly negatively charged [Fe2(CO)8]2- , [Ru2(CO)8]2-, and [Os2(CO)8]2-. The weakly negative charge stabilizes the metal-carbon bond without destabilizing the metal-metal bond too much. We can further see from this that a weakly negative charge is better for the stability of the species than no charge at all.
Homoleptic Carbonyls of the 11th Group (M = Cu, Ag, Au)
The group 11 metals Cu, Ag, and Au have eleven valence electrons.
Therefore, six ligand electrons are needed to achieve 17 electrons (Figure $21$). This means that 6/2=3 CO ligands are required to produce the 17 valence electron fragment. The dimer of it has the composition M2(CO)6 and its structure can be derived from a tetrahedral M(CO)3L structure in which L is the second M(CO)3 fragment. Neither the Cu, nor the Ag, nor the Au carbonyl are known due to the weak $\pi$-overlap in tetrahedral coordination. However, the weakly negatively [Ni2(CO)6]2- ion is known which may be explained by the stabilizing effect of one negative charge on the M-C bond.
Overall Stability Trends
Overall, we see the same stability trends for the dinuclear carbonyls as compared to the mononuclear ones. The stability decreases with the group number because $\pi$-overlap becomes smaller with smaller coordination numbers. Also, we see that the stability decreases with the period because of decreasing match of orbital sizes.
Homoleptic carbonyls of the 5th group (M = V, Nb, Ta)
The 17 valence electron fragment of the carbonyl of vanadium has the composition V(CO)6. It does not dimerize because the coordination number of 6 is much more favorable compared to the coordination number 7. V(CO)6 is a dark violet radical of octahedral shape. The radical electron is sterically inactive. V(CO)6 can be used as an oxidant because it can be reduced easily to the 18 valence electron species V(CO)6-. The higher homologues Nb(CO)6 and Ta(CO)6 are not known. This may be explained by the weaker orbital overlap between the relatively small $\pi$*-orbitals of the carbonyl ligand and the large metal d-orbitals of Nb and Ta.
Dr. Kai Landskron (Lehigh University). If you like this textbook, please consider to make a donation to support the author's research at Lehigh University: Click Here to Donate. | textbooks/chem/Inorganic_Chemistry/Inorganic_Coordination_Chemistry_(Landskron)/10%3A_Organometallic_Chemistry/10.02%3A_Ligands_in_Organometallic_Chemistry.txt |
Isolobality
In the previous chapter we discussed the most simple homoleptic carbonyls. We still need to discuss more complex carbonyls, but before we do this, we will learn about the concept of isolobality. This isolobality concept greatly helps to understand and rationalize complex carbonyls, but its applications are not limited to carbonyl chemistry. It is generally very useful to predict the stability of organic and inorganic compounds. This makes it helpful in synthesis, because in synthesis it is important to be able to estimate if the target molecule is stable or not.
The concept of isolobality was developed by Roald Hoffmann, who won the Nobel prize in chemistry in 1981. It comes from the greek words “isos” meaning “similar” and “lobos” meaning lobe. Hence, “isolobal” means “similar lobes”. In his Nobel speech he defined isolobality as follows: “Molecular fragments are isolobal if the number, symmetry properties, approximate energy and shape of the frontier orbitals and the number of electrons in them are similar - not identical, but similar. “ When molecular fragments are isolobal then they can likely be combined to form a stable molecule. Thus, isolobal fragments are compatible building blocks for the construction of stable molecules. We can also say that the concept of isolobality helps us to predict when a stable bond forms. Whenever two molecular fragments are isolobal, then they will likely form stable bonds between them.
The Concept of Isolobality
Are there simple means how to estimate if two molecular fragments are isolobal? In main group chemistry the octet rule is a very helpful guide, especially for the period 2 elements for which the octet rule holds fairly strictly. The difference between the number "8" and the number of valence electrons (VE) of the molecular fragment is equal to the number of frontier orbitals and the number of electrons in it: # frontier orbitals = number of electrons in the frontier orbitals = 8 – VE(fragment). Thus, when two fragments have the same number of valence electrons, they can be considered isolobal.
For example, the CH3 fragment has 7 VE, therefore it has one frontier orbital with one electron it (Fig. 10.3.2). The same is true for the NH2 and the OH fragments. The NH2 fragment has 5+2=7 VE, and the OH fragment has 6+1=7 VE. Therefore, these three fragments are isolobal. Isolobality between two fragments is symbolized by an arrow with a loop, and therefore we can write two such arrows between the three isolobal fragments. It should be possible to combine these isolobal fragments to form stable molecules. Let us test this. Two CH3 fragments can be combined to form an ethane molecule which is known to be stable. The combination of CH3 and NH2 gives methylamine which is stable, and the combination of CH3 and OH gives methanol, which is also stable. Two NH2 fragments give hydrazine, H2N-NH2 which exists, the combination of NH2 and OH gives the known hydroxylamine molecule NH2OH. The combination of two OH fragments gives hydrogen peroxide H2O2.
The Concept of Isolobality
As another example, the CH2 fragment is a 6 VE fragment, thus there are 8-6=2 frontier orbitals with overall two electrons in them (Fig. 10.3.3). An NH-fragment and an O-atom also have 6 VE, and are therefore isolobal to CH2. In NH there are 5+1=6 valence electrons, and an oxygen atom has 6 valence electrons. We can combine two CH2 fragments to form ethene H2C=CH2. Combining CH2 with NH and O, respectively gives methylene imine H2CNH, and formaldehyde H2C=O, respectively. Methylene imine is stable in the gas phase, and oligomerizes in higher concentrations to form a hexamer, called urotropine. Other oligomers and polymers are also known. The combination of two NH fragments gives the known molecule diazene HN=NH, and the combination with an O atom gives H-N=O, which is known as nitroxyl or azanone, and is stable in the gas phase. The combination of two oxygen atoms gives the well known O2 molecule.
Lastly, let us look at a few 5 VE fragments (Fig. 10.3.4). The CH fragment has 4+1=5 VE, and an N-atom has 5 VE as well, thus they can be considered isolobal. Two CH fragments give acetylene C2H2, and two N atoms give the dinitrogen molecule N2. The combination of a CH fragment with an N fragment gives H-CN, well known as hydrogen cyanide.
In the way the octet rule can help to predict the number of frontier orbitals and the electrons in them for main group element fragments, the 18 electron rule can be used to predict the number of frontier orbitals and electrons for organometallic fragments, including carbonyl fragments. The number of frontier orbitals and the number of electrons in them is 18 minus the number of valence electrons the organometallic fragment has: # frontier orbitals = number of electrons in them = 18-VE. For instance for a 17 VE fragment such as Mn(CO)5 there is one frontier orbital with one electron, for a 16 VE fragment such as Fe(CO)4 there are two frontier orbitals with one electron in each of them, and for a 15 VE fragment such as (CO)3Co there are three frontier orbitals with overall three valence electrons (Fig. 10.3.5). This implies that a 17 VE carbonyl fragment is isolobal to a 7 VE organic fragment such as CH3. Similarly, a 16 VE carbonyl fragment is isolobal to a 6 VE fragment such as CH2, and a 15 VE fragment is isolobal to a 5 VE fragment such as CH (Fig. 10.3.5).
It should be possible to combine the isolobal fragments to form stable molecules. Let us check how well this works (Fig. 10.3.6). It should be possible to combine the two 17 VE electron fragments such as Mn(CO)5 to form (CO)5Mn-Mn(CO)5. This is a known molecule. Remember we encountered it previously when we discussed the homoleptic carbonyls of metals with an odd number of electrons. Combining this fragment with a 7 VE CH3 fragment leads to (CO)5Mn-CH3 which is also stable. Can we also combine two 16 VE electron fragments to form (CO)4Fe=Fe(CO)4 with an Fe=Fe double bond? The answer is no. Metal-metal double bonds, and also metal-metal triple bonds in carbonyl complexes are not formed. Instead clusters with single bonds are realized. The number of single bonds a metal makes is equal to the number of its frontier orbitals. In the case of 16 VE fragments trimeric clusters with single bonds are formed. In this cluster the two frontier orbitals of each fragment make two single bonds to two other fragments. In contrast, a 16 VE carbonyl fragment such as (CO)4Fe can be combined with a 6 VE fragment to form a compound with a Fe=C double bond. A compound with a metal-carbon double bond is called a carbene complex. Similarly, the combination of 15 VE fragments of the type Co(CO)3 does not lead to a stable (CO)3Co≡Co(CO)3 molecule with a Co≡Co triple bond. Instead, nature realizes a tetrameric tetrahedral cluster in which the three frontier orbitals of each 15 VE fragment make three single bonds to the other three 15 VE fragments. In clusters the CO ligands may not only be terminal, they can also be bridging. In the tetrameric Co-cluster, there are are nine terminal and three bridging CO-ligands. The three bridging CO ligands are connecting the three Co atoms at the base triangular face of the tetrahedron. What about the combination of a 15 VE fragment with an organic 5 VE fragment? The combination of a 15 VE fragment such as Co(CO)3 with a 5 VE organic fragment such as CH to does yield stable complexes such as the (CO)3Co≡CH complex with a Co≡C triple bond. Complexes with metal-carbon triple bonds are called carbine complexes.
Synthesis of Terameric Cluster Carbonyls
How can we make a carbonyl cluster like ((Co(CO)3)4?
It can be prepared by heating the Co2(CO)8 to a temperature above 54°C. Above this temperature the reactant loses four CO ligands and rearranges to form the cluster (Fig. 10.3.7). Interestingly, the higher homologues of the Co-cluster, the Rh4(CO)12 and the Ir4(CO)12 form spontaneously from the elements. Remember, that we previously determined that the Rh2(CO)8 and the Ir2(CO)8 are not stable. Rh and Ir favor the tetrameric clusters with 12 COs over the dimer with 8 CO ligands.
Structures of Co4(CO)12, Rh4(CO)12 and Ir4(CO)12
Like in the Co-cluster, there are 9 terminal and three bridging CO ligands in the Rh4(CO)12 cluster. By contrast, there are only terminal CO ligands in the iridium cluster (Fig.10.3.8) . This may be explained by the larger Ir-Ir bond length in comparison to the Co-Co and Rh-Rh bond lengths. The observation reflects the general rule that only 3d and 4d elements have bridging CO ligands, while only terminal ligands are observed in metals with 5d electrons.
Isoelectronic Charged Carbonyl Clusters
The previously discussed clusters were charge-neutral. Can we construct charged isoelectronic ones with different metals, and what is their stability? Let us start with the tetrameric Co-cluster made of four 15 VE fragments. The element left to the Co is the Fe. It has one electron less, therefore Fe(CO)3- is the 15 VE electron fragment which is isoelectronic to the neutral Co(CO)3 fragment. Its tetramer would be a Fe4(CO)124- cluster. This cluster is not known. The cluster already carries a 4- charge which is too high to support stable Fe-Fe bonds. Each Fe carries formally a 1- negative charge and there is too much electrostatic repulsion between the Fe atoms. Another possibility to realize a 15 VE fragment with Fe is a (CO)4Fe+ fragment. Here an additional ligand is added contributing two electrons, and one electron is removed resulting in a fragment with a 1+ charge. The fragment could be combined to form an Fe4(CO)164+ cluster. However, such a cluster is also not stable. This is because the coordination number would be 7, which is too high for a carbonyl. In addition, there is the destabilizing effect of the positive charge on the Fe-C bond. Similarly, tetramers of the 15 VE fragments Mn(CO)4, Cr(CO)4- and (CO)5Cr+ are also not known. Also in these cases, the coordination numbers would be too high. The Mn(CO)4 and the Cr(CO)4- would lead to a coordination number of 7, in the case of the Cr(CO)5+ fragment the coordination number would be even 8. We see here the limitations of the concept of isolobality. The presence of isolobality is not always sufficient to make a stable molecule, also other factors like coordination numbers and charge needs to be considered.What about charged isoelectronic clusters made from 16 VE? We saw previously that the neutral 16 VE Fe(CO)4 fragment gave a Fe3(CO)12 trimer. If we go from the Fe to the Mn, the isoelectronic 16 VE fragment would be a Mn(CO)4- fragment because Mn has one electron less than Fe. This fragment can indeed be trimerized to give a stable Mn3(CO)123- cluster with a 3- charge. This charge is not too high yet to destabilize the Mn-Mn bond and the coordination number of 6 is ideal for the stability of carbonyls. We could add a ligand and remove two electrons to produce a 15 VE Mn(CO)5+ fragment. However, in this case the coordination number would become 7 upon trimerization, which is too high to produce a stable cluster. In addition, the positive charge destabilizes the Mn-C bond. Replacing Mn by Cr in Mn(CO)5+ would give a neutral 16 VE Cr(CO)5 fragment, but its trimer is not stable because also in this case the coordination number becomes too high. Replacing Fe in Fe(CO)4 by Co would require a 1+ charge giving a Co(CO)4+ 16 VE fragment. Its trimer would have the coordination number 6, but it would carry a 4+ charge which is too high. A Co(CO)3- fragment would also have 16 VE, but its trimer is also not known, possibly because the relatively small coordination number of 5, which is less favorable than a coordination number of 6. Overall we can say as a rule for the stability for carbonyl clusters that there should not be positive charges, and no high negative charges. There should not be high coordination numbers.
Clusters with 16 VE and 6 VE Fragments
The cluster chemistry of carbonyls is even more versatile due to the possibility to substitute 16 and 15 VE fragments by 6 and 5 VE fragments, respectively. For instance, we can replace one 16 VE Fe(CO)4 fragment by a 6 VE CH2 fragment in the trimeric triiron dodecacarbonyl Fe3(CO)12 cluster (Fig. 10.3.9). In the resulting cluster a methylene group bridges two Fe atoms, therefore the complex can be regarded a μ-methylene complex. The substitution reduces the number of cluster valence electrons (CVE) from 48 to 38. A second substitution of another 16 VE tetracarbonyl iron fragment by another 6 VE methylene fragment gives a cluster with two Fe-C bonds and one carbon-carbon bond having 28 cluster valence electrons. This complex can be regarded as a ethene complex in which an ethene molecule binds side-on to a 15 VE Fe(CO)4 fragment. Upon binding, the two Fe-C bonds are formed at the expense of the C-C π-bond and the bond order in the C-C bond is reduced from 2 to 1. If we substitute the last tetracarbonyl iron fragment by a third CH2 unit, then a completely organic molecule, cyclopropane, results. All the molecules are known to be stable molecules, and their stability can be nicely understood by using the concept of isolobality.
Similarly, we can substitute 15 VE tricarbonyl cobalt units by 5 VE CH fragments in tetracobaltdodecacarbonyl Co4(CO)12. A first substitution gives a 50 VE cluster with one CH fragment that makes three bonds to three Co(CO)4 fragments (Fig. 10.3.10). Because the CH group bridges three metal atoms it is regarded a μ3-methine complex. A second substitution produces a 40 VE cluster with an organic CH-CH unit that makes four bonds to two Co atoms of two 15 VE Co(CO)3 fragments. The CH-CH fragment can be considered an ethine molecule binding side-on to a dicobalthexacarbonyl fragment, whereby the two π-bonds in ethine are expended to form four Co-C single bonds. The cluster can be considered a μ22-ethine complex because the ethine bridges two Co atoms and both carbons are involved in the bonding with Co. A third substitution leads to a 30 CVE cluster with three CH units binding to one Co(CO)3 fragment. There are three C-C single bonds and three Co-C bonds in the cluster. The three CH units form a cyclopropenylium complex that binds side-on to a Co(CO)4 fragment. A free cyclopropenylium molecule is a 3-ring with three π-electrons that are delocalized in the ring. Therefore, the bond order in a free cyclopropenylium molecule is 1.5. In the complex the three π-electrons are being used to make the three single bonds with the cobalt atoms. The bond order in cyclopropenylium is thereby reduced from 1.5 to 1. Finally, the fourth substitution of a tricarbonyl cobalt fragment by a methine fragment yields the completely organic tetrahedrane molecule. This molecule and the others discussed before are stable. The isolobality concept lets us rationalize these complex structures easily and understand their stability.
Synthesis of Cluster Complexes From Alkynes
The view that the 40 VE clusters are alkyne clusters is not just a formal view. They can be synthesized from alkynes. For instance alkyne-dicobalt clusters are accessible from dicobalt octacarbonyl and alkynes (Fig. 10.3.11).
The cluster forms via the donation of the four π electrons of the alkynes into the metal d-orbitals of the dimeric cobalt carbonyl. This leads to a loss of of two CO ligands, and the formation of four Co-C bonds. Under harsher conditions the C-C triple bond can also cleave and this can give access to methine complexes.
Carbonyl Hydrides
Now let us leave the carbonyl cluster compounds and talk about another interesting class of carbonyls: carbonyl hydrides. They can be rationalized by the concept of isolobality as well. A hydrogen atom can be conceived as a species with one frontier orbital containing one electron. In this case the frontier orbital is simply the 1s orbital of the hydrogen. Thus, it should be possible to combine an H atom with a 17 VE carbonyl fragment like tetracarbonyl cobalt.
Indeed, one can combine such fragments to form stable carbonyl hydrides such as tetracarbonylhydrido cobalt (0). This molecule can be synthesized by reduction of bis(tetracarbonyl cobalt) (0) with dihydrogen (Fig. 10.3.12). Counter-intuitively, this molecule is a strong acid, it has an acidity similar to sulfuric acid. One would think that the Co-H bond would be polarized toward H based on electronegativity arguments. However, this is not the case. The carbonyl fragment has a frontier orbital which is energetically higher than that of H, and thus the bond is polarized toward H. In another view we can explain the high acidity be the fact that the loss of the proton leads to the very stable 18 VE tetrahedral Co(CO)4- anion. Thus, loss of the proton occurs easily.
Another 17 VE electron fragment is the Mn(CO)5 fragment. Also this fragment can be combined with an isolobal H fragment to form a stable molecule, which has the composition HMn(CO)5 (Fig. 10.3.13). It can be prepared from bis(pentacarbonyl manganese) and dihydrogen at 200 bar and 150°C. This carbonyl hydride is a weak acid, it has a similar acidity as H2S. We could argue that this lower acidity may be because the loss of the proton reduces its coordination number from 6 to 5. The coordination number of 6 is the preferred coordination number for carbonyls and thus the tendency to lose the proton is relatively small. The different metal and the number of carbonyl ligands will likely lead to a different energy of the frontier orbital compared to the previous example, which leads to a different polarity.
Overall, one can tune the properties in carbonyl hydrides from highly acidic to hydridic by choice of the metal the coordination number, and also by the choice of additional ligands L other than carbonyl. The electronic and steric properties of the ligands have an influence on the energy of the HOMO of the fragment, and thus on the polarity of the metal-hydrogen bond, and the acidity.
Ligands related to CO
The cyano ligand (CN-)
Now let us leave the CO ligand, consider a number of related ligands, and discuss similarities and differences compared to the CO ligand. Let us start with the cyano ligand CN-. This ligand is isoelectronic to the CO ligand. The N atom has one electron less than O, but the negative charge at the cyanide ligand compensates for that. One question we can ask is: What is the reactive end? The answer is: In analogy to the carbonyl ligand the reactive end is the carbon atom. We can explain this by the fact that the MO diagram is similar to that of CO, only the differences in atomic orbital energies are smaller due to the smaller electronegativity difference between C and N compared to C and O. Therefore, like in CO, the HOMO is represented by the electron lone pair at the C, which makes C the more reactive end. Due to the smaller electronegativity difference, the difference in energy between the lone pair at C and the lone pair at N is smaller, therefore, in contrast to CO, the cyano ligand acts far more often as a bridging ligand between two metals using both of its electron lone pairs.
Would we expect the cyano ligand to be a stronger or weaker σ-donor compared to the CO ligand? Think about it for a moment. The answer is: It is a stronger σ-donor because of its negative charge. The negative charge at the ligand increases electrostatic repulsion between the electrons, and this increases the orbital energies. Therefore, there is a stronger tendency to donate the electrons. Our next question is: Is the CN- ligand a stronger or weaker π-acceptor than CO? The energy of the π*-orbitals is higher compared to CO because of the negative charge at the ligand. Because of that, electrons from the metal cannot be as easily accepted by the ligand. Therefore, the cyano ligand is a weaker π-acceptor than the carbonyl ligand. Our last question is: Are cyano complexes more stable with metals in high or low oxidation states. Because of electrostatic arguments a cyanide anion interacts more strongly with a metal cation rather than a metal in a zero or negative oxidation state. Therefore, unlike CO, CN- does not stabilize metals in low oxidation states. It prefers to make complexes with metal in high, positive oxidation numbers.
The Nitrosyl Ligand NO
The nitrosyl ligand NO is another ligand similar to CO. It has one more electron that CO because N has one more electron than C. The additional electron makes NO an “odd” molecule with a radical electron. Like in CO and CN-, the more electropositive element is the reactive end. In the case of NO it is the N atom. The radical electron is the most reactive electron that can be most easily donated, however the electron lone pair at N may be donated in addition. In the former case, the NO is a 1-electron donor, in the latter it is a 3-electron donor. How can we tell if one or three electrons have been donated? When only one electron is donated, the electron lone pair at the nitrogen is sterically active and leads to a bent structure (Fig. 10.3.14).
An example is the trans-bis-(ethylenediamine)chloronitrosyl cobalt (1+) cation. Generally, the O-N-M bond angle in nitrosyl complexes with NO as a 1e-donor can vary between 119 and 140°. We can also identify a bent nitrosyl ligand in the IR spectrum. Typical wave numbers are 1520-1729 cm-1.
When the nitrosyl ligand donates all three electrons, then it binds to the metal in a linear fashion (Fig. 10.3.15). The electron lone pair is no longer sterically active because it is involved in the bonding. An example is the nitroprusside anion [Fe(CN)5(NO)]2-. It is used as a medication for the treatment of high blood pressure. The bond angles in complexes with NO as the 3-electron donor are often not exactly 180°, but vary between 165 and 180°. It can also be identified in the IR because it has a characteristic band between 1610-1680 cm-1.
There is a also a different view on nitrosyl complexes with linear and bent structures. In bent structures we can also consider the ligand as an NO- anion that donates two electrons. The NO- anion has two electron lone pairs at N. When it donates two electrons then one sterically active electron lone pair remains at the nitrogen atom. In linear structures we can regard the ligand also as an NO+ cation that donates two electrons. An NO+ cation has one electron lone pair at N, and when it donates that lone pair then there is no sterically active electron at the nitrogen left, and thus the ligand binds in a linear fashion.
We could also ask: What can neutral NO not be a 2-electron donor with the radical electron left at N? The answer is: The radical electron is the highest energy electron, and is always used first in interactions with a metal.
Phosphine Ligands
Phosphines are another class of important ligands in coordination chemistry. To approach the coordination chemistry of phosphines let us have a look at the MO diagram of the PH3 molecule and compare it with the NH3 molecule. As one would expect, the MOs are overall similar, but there is one important difference. While the LUMO in NH3 is the anti-bonding 3a1 orbital, the LUMOs in the PH3 molecule are the anti-bonding 2e orbitals. The relative energies of the 3a1 and 2e orbitals in the PH3 and NH3 molecules are swapped up. This can be attributed to the fact that the the P atom uses the 3s and the 3p orbitals as valence orbitals, while N uses the 2s and 2p orbitals. The 3s and the 3p orbitals are larger and overlap less effectively with the small 1s orbitals of the hydrogen. They also have a higher energy making the P-H bonds less polar than the N-H bonds. The energy of the PH3 HOMO is higher than that of the NH3 HOMO. Both the HOMO and the LUMO of PH3 are more diffuse and polarizable than the respective orbitals in NH3.
The higher energy of the HOMO in PH3 makes it a better donor than NH3. In addition, the PH3 has π-acceptor properties because the 2e LUMO are relatively low-lying anti-bonding 2e orbitals and have suitable shape for π-overlap with metal d-orbitals. NH3 does not have these π-acceptor properties because its LUMO is the 3a1 orbital, and not the 2e orbitals. The 2e orbitals in NH3 are energetically too high in order to allow for significant π-acceptor interactions.
Bonding in Phosphine Complexes
The σ-donor and π-acceptor properties of the phosphine ligand can be modified by substituting H by other groups. Generally, more electron donating groups increase the energy of the HOMO and the LUMO. This strengthens the σ-donor and reduces the π-acceptor properties. Vice versa, more electron withdrawing groups decrease the energy of the HOMO and the LUMO. As a consequence, the ligand becomes a weaker σ-donor and a stronger π-acceptor (Fig. 10.3.17).
The table above shows the HOMO and LUMO energies of three phosphines. As expected the HOMO and LUMO energies decrease from PMe3 to PH3 to PF3 due to the increasingly electron-withdrawing nature of the substituent. As a consequence, the σ-donor properties weaken and the π-acceptor properties strengthen from to PMe3 to PH3 to PF3.
Not only the energies, but also the orbital overlap is important for the strength of the π-acceptor properties of the phosphine ligands. The more strongly electron-withdrawing the group is the more the anti-bonding e-type LUMO orbitals are located at the P atom. The more these orbitals are localized at P, the better they can overlap with the ligand. Vice versa the localization of the HOMO shifts toward the group as the group becomes more electron-withdrawing, thereby weakening the σ-donor properties.
Phosphine ligands with strongly electron withdrawing groups such as PF3 have π-acceptor properties strong enough to stabilize metals in low oxidation numbers, similar to CO. For example, the PF3 ligand forms stable complexes with Pd and Pt atoms in the oxidation number 0, while the respective Pd and Pt carbonyls are not known.
Dihydrogen Complexes
Even a simple dihydrogen molecule can serve as a ligand. We call the respective complexes dihydrogen complexes. When dihydrogen acts as a ligand, it binds side-on to the metal (Fig. 10.3.18). It uses its bonding σ-orbital for σ-donation. In addition, it uses its σ*-orbital as π-acceptor. The combination of the σ-donor and π-acceptor interactions leads to the lowering of the bond order and the lengthening of the H-H bond in the dihydrogen molecule. When these interactions are strong enough, then the H-H bond can cleave and two, separate hydrido-ligands bind to the metal via two metal-hydrogen single bonds. The π-acceptor interactions tend to be strong enough to cleave the H-H bond when the metal is relatively electron-rich. Dihydrogen complexes are therefore usually observed when the metal has a high oxidation number and/or when it has other strongly π-accepting ligands such as CO. For instance, the Mo(PMe)5H2 complex is a dihydride while the Mo(PR3)(CO)3(H2) complex is a dihydrogen complex because the latter contains three strongly π-accepting CO ligands, while the former does not.
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Concept Review Questions
Section 1
1. What is the difference between an organometallic compound and a metal-organic compound?
2. What is kakodyloxide?
3. What is Zeise’s salt and what is unusual about its structure and bonding?
4. What is the Mond process?
5. Explain why the physical and chemical properties of ferrocene are inconsistent with the monohapto coordination of Fe by cyclopentadienyl ligands.
6. Describe qualitatively how the cyclopentadienyl ligands are bonded to Fe in ferrocene.
7. Explain why the cyclobutadiene is not square planar in free state but is square planar when coordinated to transition metals in Ƞ4 fashion?
8. Explain briefly why cyclooctatetraene makes sandwich complexes (metallocenes) only with very large metal ions?
Section 2
1. Explain why carbonyls are interesting and important.
2. Which molecular orbitals of CO can be involved in bonding with transition metals? Explain why these orbitals can be involved.
3. Which types of binding do you know for carbonyls in which CO acts as 2-electron donor?
4. Which types of binding do you know for carbonyls in which CO acts as 4- and 6- electron donor?
5. Which types of binding do you know for carbonyls in which CO acts as an electron acceptor? Which of these binding types leads to a stronger interaction? Explain this.
6. What is the Chatt-Dewar-Duncanson model for carbonyls?
7. What is meant by pi-backbonding in carbonyls?
8. Explain what the strength of the pi-backbonding in carbonyls depends on?
9. Explain how one can derive the structure of the most simple carbonyls?
10. How can one experimentally measure the strength of the pi-backbonding in carbonyls?
11. What is unusual about the structure of Co2(CO)8? Explain.
12. What is unusual about the structure of V(CO)6? Explain.
13. Why do the most simple group 4 and group 12 carbonyls not exist?
14. Why don’t highly positively charged carbonyls exist?
15. Why don’t highly negatively charged dimeric carbonyls exist?
Section 3
1. What is meant by the concept of isolobality?
2. Explain how the concept of isolobality is helpful for the prediction of stable structures?
3. Explain how one can explain the structure of Co4(CO)12 using the concept of isolobality.
4. What is the structural difference between Co4(CO)12 and Ir4(CO)12. Explain this difference.
5. The stability of the most simple monomeric carbonyls decreases from group 6 to group 10. Explain.
6. The stability of the most simple monomeric carbonyls decreases from period 4 to period 6. Explain.
7. Explain why the following complexes are considered ethene and ethine complexes, respectively.
8. What is a carbonyl hydride?
9. Explain why HCo(CO)4 is more acidic than HMn(CO)5?
10. Why is the CN- ligand a stronger sigma-donor and a weaker pi-acceptor compared to the CO ligand?
11. Explain why the CN- ligand tends to make complex cations, while the CO ligand prefers to make complex anions?
12. Explain why the carbon atom is the reactive end in cyanide complexes?
13. What are the two binding modes in nitrosyl (NO) complexes?
14. Why is a neutral NO ligand never a two electron donating ligand?
15. How can one explain the two different binding modes in NO complexes?
16. Why is phosphine a pi-accepting ligand while ammonia is not?
17. Why is PF3 a stronger pi-acceptor than PH3?
18. Why is PMe3 is stronger sigma-donor than PH3?
19. Why is H2 a pi-accepting ligand?
20. Explain how the pi-accepting properties lead to the dissociation of the H-H bond in the H2 molecule when H2 acts as a ligand.
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Homework Problems Chapter 10
Homework Problems
Section 1
Exercise 1
Predict the charge n of cobaltocene (Co(Cp)2)n. Cp=cyclopentadienyl. Explain briefly.
Answer
+1, because the 18 electron rule is fulfilled.
Exercise 2
Bis-benzene chromium (Ƞ4-C6H6)Cr is a sandwich compound having two parallel benzene rings in an eclipsed conformation. Sketch the ligand group orbitals for this molecule.
Answer
Section 2
Exercise 1
The MO diagram of the CO molecule is given below. Identify the molecular orbitals that can act
a) As sigma-donors
b) Pi-acceptors in carbonyls.
Answer
a) 5 σ, 4 σ, bonding π orbitals
b) anti-bonding π orbitals
Exercise 2
Which carbonyl will most likely exhibit the highest frequency for the C-O stretch vibration? Give a brief explanation for your decision.
a) V(CO)6-
b) Cr(CO)6
c) Mn(CO)6+
Answer
c) Mn(CO)6+
Carbonyl cations have the weakest pi-backbonding, therefore the bond order for the C-O bond is the highest and thus the frequency of the C-O stretch vibration is the highest.
Exercise 3
Predict the right charge of the most stable pentacarbonyl of
a) Iron
b) Mn
c) Cr
Answer
a) 0
b) 1-
c) 2-
Section 3
Exercise 1
What structure would you expect for a carbonyl made of the following fragments according to the concept of isolobality? Predict if the structure would likely be stable.
a) Cr(CO)5+
b) Cr(CO)4-
c) Mn(CO)4-
Answer
a) 15VE -> tetrahedral cluster (not stable b/c CN too high)
b) 15 VE -> tetrahedral cluster (not stable b/c CN too high)
c) 16 VE -> trimer (stable b/c CN = 6, charge not too high).
Exercise 2
Which of the following fragments are isolobal to the CH2 fragment?
a) NH2
b) NH
c) OH
d) Fe(CO)3
e) Co(CO)4+
Answer
b) NH
Exercise 3
Which of the following fragments are isolobal to the Mn(CO)5 fragment?
a) CH3
b) CH2CH3
c) OH
d) F
e) NH
Answer
a) CH3
b) CH2CH3
c) OH
d) F | textbooks/chem/Inorganic_Chemistry/Inorganic_Coordination_Chemistry_(Landskron)/10%3A_Organometallic_Chemistry/Concept_Review_Questions_Chapter_10.txt |
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11: Complexes with Metal-Metal Bonds
Molecular orbital considerations in Dinuclear Metal Complexes with Multiple M-M Bonds
Let us begin a new chapter and think about dinuclear transition metal complexes with multiple metal-metal bonds. What is the maximum bond order that we could expect? The transition metals have five d-orbitals available, and in order to determine the maximum possible bond order we need to see how they overlap to form molecular orbitals (Fig. 11.1.1).
If we define the M-M bond axis as the z-axis, then two dz2-orbitals can overlap in σ-fashion to form a bonding and an anti-bonding σ-molecular orbital. A dxz, and a dyz can overlap with another dxz and another dyz in π-fashion to form two degenerated bonding π and two degenerated anti-bonding π*-orbitals. π-overlap is smaller than σ-overlap, therefore the split in energy between the bonding and the anti-bonding π-orbitals is smaller than the split between the bonding and the anti-bonding σ-orbitals. The dxy orbital can overlap with another dxy orbital in $\delta$-fashion, and so can the dx2-y2 with another dx2-y2. This gives two bonding $\delta$-orbitals and two anti-bonding $\delta$-orbitals. The $\delta$-overlap is even smaller than the π-overlap, therefore the energy split between the bonding and the anti-bonding $\delta$-orbitals is even smaller than those for the π and π*-orbitals. So what would be the maximum bond order that could be achieved? The maximum bond order would be achieved when all bonding MOs were full, and all anti-bonding MOs were empty. We have overall five bonding MOs that we could fill with ten electrons from two metal atoms with d5-electron configuration. The maximum bond order BO would therefore be BO=5. However, in practice only bond orders up to 4 are well known. This is because the dx2-y2 orbital is usually too involved in the bonding with the ligands thereby becoming unavailable for metal-metal interactions. The dx2-y2 orbital makes the strongest interactions with the ligands because most ligands approach on or near the x and the y-axes.
Electron Configurations and Multiple M-M Bonds
We can easily predict now which electron configuration leads to which metal-metal bond order. The bond order increases from 1 to 4 with the electron configuration changing from d1 to d4. At d4 all bonding MOs are full with eight electrons.
An example of a complex with a bond order of 4 is the tetraacetatodiaquadichromium complex shown (Fig. 11.1.2). Cr is in the oxidation state +2, which makes the chromium a d4 species. We can quickly show this by counting the valence electrons. A neutral Cr atom has 6 VE, and an electron configuration 3d54s1. There are four acetate ligands having a 1- charge each which gives four negative charges overall. The complex is overall neutral which means that each Cr must have formally a 2+ charge, and the electron configuration is 3d4.
With even more electrons in the metal d-orbitals the bond order begins to decrease again as anti-bonding MOs need to be filled (Fig. 11.1.3). The combination of two metals with d5 electron configuration leads to a triple bond, two d6-metals give a double bond, and two d7 metals give a single bond. A metal-metal bond should not exist for two d8-metals because then the bond order is zero.
Evidence for M-M Multiple Bonds
What experimental evidence can support the existence of a particular bond order (Fig. 11.1.4)? One argument is the bond length which can be obtained through the crystal structure determination of the complex. The shorter the bond, the higher the bond order. For instance, in the four tungsten complexes shown the bond lengths decrease from 272 pm, to 248 pm, to 230 pm to 221 pm corresponding to a single, double, triple, and quadruple bond, respectively.
Another hint can be the conformation of a molecule (Fig. 11.1.5). For instance, the two square-planar units of the Re2Me82- complex anion show eclipsed conformation. Steric repulsion arguments would favor the staggered conformation, so there must be a reason why the two ReMe4 units are eclipsed. The rhenium is in the oxidation state +3, thus it is a d4 species, and we would argue that there may be a Re-Re quadruple bond. This quadruple bond can only form when the dxy orbitals are in eclipsed conformation, and this is only possible when the two ReMe4 fragments are in eclipsed conformation. The very short bond length of 218 pm further supports the existence of the quadruple bond.
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Concept Review Questions Chapter 11
Concept Review Questions
Section 1
1. What experimental evidence exists for the presence of metal-metal multiple bonds?
2. Why is the bond order in metal-metal bonds limited to four?
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Homework Problems Chapter 11
Homework Problems
Section 1
Exercise 1
Which bond order would you expect for metal-metal bond when the metals have
a) d1
b) d4
c) d5
d) d7
electron configurations?
Answer
a) d1 --> single bond
b) d4 --> quadruple bond
c) d5 --> triple bond
d) d7 --> single bond
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12: Organometallic Reactions and Catalysis
Introduction
In this chapter we will discuss common reactions and catalysis of organometallic compounds. We can classify reactions into two main categories. The first one includes reactions with gain or loss of ligands. You will hear about four reactions that belong to this category: ligand substitution reactions, oxidative additions, reductive eliminations, and nucleophilic displacement. The latter three can involve a change in coordination number while ligand substitution reactions do not.
The second main category are reactions that involve the modification of ligands. There are two reactions you will learn about: The migratory insertion and the β-hydride elimination.
Oxidative Addition Reactions
Let us start with the oxidative addition. How is it defined? An oxidative addition is a reaction in which one or more ligands add to the metal center of a complex and oxidize it. This addition can occur in a cis- or trans-fashion. A cis-addition means that two ligands are added so that they have cis-orientation to each other after the reaction is complete. A trans-addition implies that two new ligands are in opposite, trans-position after they have been added. We can further distinguish between mononuclear and dinuclear additions. Mononuclear addition involves a complex with a single metal. Dinuclear additions are additions to a metal-metal bond. The addition cleaves the metal-metal bond and adds one ligand to each fragment.
Here are a few of examples. The first example is a mononuclear cis-addition (Fig. $1$:). In the reaction we add dihydrogen to a square planar iridium complex. After the reaction there are two additional hydrido ligands, and the coordination number has increased from 4 to 6. We can see that the two hydrido ligands are oriented in cis-fashion relative to each other. We can verify that the addition of the hydrido ligands oxidized the metal by determining the oxidation number of the metal before and after the reaction. There are three neutral ligands and one negatively charged chloro ligand. There is no overall charge at the complex. This means that the oxidation number of Ir is +1. After the reaction, there are the two hydrido ligands that have both a -1 charge. Remember, we get the charge at a ligand by cleaving the metal-ligand bond so that all electrons in the bond get assigned to the ligand. This means the oxidation number must be increased by +2, and is +3 after the reaction.
We can rationalize why the addition occurs in cis-fashion and not in trans-fashion. The two hydrido ligands are the smallest ligands, therefore, steric repulsion is minimized in the complex is minimized when the two hydrido ligands are in cis-position.
Fig. $2$: Mononuclear oxidative trans-addition (Attribution: A. Vedernikov, U Maryland (modified))
The second example is a mononuclear trans-addition (Fig. $2$:). In this reaction CH3Br is added to the iridium complex. In this case the two new ligands, a CH3 and the Br group are in opposite position. Again, the coordination number increases from four to six. We can again verify that an oxidation has occurred by analyzing the oxidation number of Ir. As previously determined, the oxidation number of iridium in the reactant complex is +1. Let us analyze the product complex. The chloro ligand and the bromo ligand have a 1- charge. In addition to that also the CH3 ligand has a 1- charge assuming that we treat the Ir-C bond as a dative bond. Thus, the oxidation number of Ir must be +3 to give a neutral complex.
The third example is a dinuclear addition (Fig. $3$:). This addition is neither cis nor trans because it cleaves a metal-metal bond, and the one new ligand is added to each fragment. In the example the Co-Co bond of the dicobalt octacarbonyl gets cleaved upon the reaction with dihydrogen, and two HCo(CO)4 complexes with Co-H bonds form. In this case the coordination number does not increase. Both the reactant and the products have the coordination number 5. We can see that an oxidation of Co has occurred by comparing the oxidation numbers. In the reactant the oxidation number of Co is zero because all CO ligands are neutral and the complex is overall neutral. In the carbonyl hydride product, Co has the oxidation number +1, because the hydrido ligand has a 1- charge.
Oxidative Additions with Intact Ligands
In the previous examples, the ligands that reacted with the complexes lost their integrity. The H-H bond in the H2 molecule cleaved to form the hydrido ligands, and the C-Br bond in CH3Br was cleaved to form separate Br and methyl ligands. Not all ligands lose their integrity upon their addition, in particular when they contain multiple bonds. These ligands are called intact ligands. In these cases the bond order within the ligand is reduced, but the number is still greater than zero.
Examples for intact ligands are are alkenes, alkynes, O2, etc. In an oxidative addition with an alkyne for example, the alkyne binds side-on to the transition metal under the formation of two metal-C single bonds, and the reduction of the bond order within the alkyne from 3 to 2 (Fig. $4$:). In the shown reaction, a diphenyl ethine molecule adds to tris(triphenyl phosphine) platinum under the formation of two Pt-C single bonds. The C-C triple bond in the ligand becomes a double bond. This reaction occurs under loss of one of the three triphenyl phosphine ligands. The coordination number at the metal increases from 3 to 4. We can again show that the reaction is oxidative by comparing the oxidation numbers of the metal. In the reactant, Pt has the oxidation number 0 because the three triphenyl phosphine ligands have no charge and the complex has overall no charge. After the reaction, the oxidation number of Pt is +2. This is because when we assign the four electrons of the two Pt-C single bonds to the ligand, the ligand becomes a (Ph-C=C-Ph)2- ligand with each C atom carrying a 1- charge. Because the product complex is overall neutral, the oxidation number of Pt is +2.
Reductive Elimination Reactions
The reductive elimination reaction is the reverse of the oxidative addition. The reductive elimination can only occur when the two ligands to be eliminated are in cis-position.
For instance the previously discussed oxidative cis-addition of H2 to the iridium complex is reversible, and the reverse is called the reductive elimination (Fig. $5$:). The cis-orientation of the two hydrido ligands is necessary to form an H-H bond. The reaction can be thought of going along a reaction path in which the Ir-H bonds become gradually larger and and the H-H distance gradually smaller until the H2 molecule is eliminated from the complex.
Catalytic Deuteration of Benzene
Combinations of oxidative additions and reductive eliminations have many applications in the synthesis of organic molecules using an organometallic reactant.
An example is the deuteration of benzene (Fig. $6$:). Deuterated benzene is an important solvent in NMR spectroscopy. Industrially, the deuteration is done using a dicyclopentadienyltrihydridotantalum(V) catalyst starting out from benzene and D2. The D2 is provided in excess to drive the chemical equilibrium to the right side.
How does this reaction work mechanistically? This 18e Cp2TaH3 is actually a precatalyst that is in chemical equilibrium with H2 and the 16e dicyclopentadienylhydrido tantalum(III) which is the actual catalyst (Fig. $7$:). Elimination of H2 from the Ta(V) catalyst is a reductive elimination (RE) and the re-addition of H2 to the Ta(III) species an oxidative addition (OA). In the presence of benzene, the Ta(III) species can oxidatively add benzene to form an 18e Ta(V) complex. This complex can reductively eliminate H2 to form a 16e Cp2Ta(III)-Ph complex. In the presence of deuterium this complex can add D2 oxidatively to form an 18e Cp2D2Ta(V)-Ph complex. This species can then reductively eliminate a monodeuterated benzene molecule under formation of Cp2Ta(III)-D. In the presence of enough D2 this monodeuterated benzene can be further deuterated in subsequent catalytic cycles.
Nucleophilic Displacement Reactions
Another important reaction is the nucleophilic displacement reaction. In these reactions a complex, typically an anion, acts as a nucleophile. The ligand is added, but no electrons are added to the complex. The reactions can be extremely useful for the synthesis of organic compounds. The utility results mostly from the fact that an organic electrophile is turned into a nucleophile.
For example, tetracarbonyl ferrate (2-), also known as Collman’s reagent reacts with alkyl halides to form alkyl tetracarbonyl ferrate (-) (Fig. $8$:). We can check that the addition of the alkyl group did not add any electrons by counting the electrons before and after the reaction. The Fe(CO)42- anion is an 18e complex. Applying the neutral atom method to the electron counting in [R-Fe(CO)4]- means that Fe contributes 8e, the 1- charge adds 1e, the CO ligands 4x2=8 electrons and the alkyl ligand 1e. This gives 8+1+8+1=18 electrons. Also the oxidation number of Fe has not changed upon the addition of the ligand. It is -2 before and after the reaction.
In the alkyltetracarbonyl ferrate (1-) anion, the group R which was formerly an electrophile in R-X, can now act as a nucleophile due to the fact that the Fe-C bond is polarized toward the C-atom. As a nucleophile it can for instance react with protons to form an alkane. It can also react with oxygen to form a carboxylic acid. In the presence of another alkyl halide R’-X it can form a ketone R-C(O)-R’. It can also insert CO into the Fe-C bonding in a so-called migratory insertion reaction, and the reaction product can react with acid to form an aldehyde. With dihalogens X2 it can form acyl chloride. Collman’s reagent cannot only add alkyl halides by also acyl halides to form acyl tetracarbonyl ferrates (-).
Migratory Insertion and Deinsertion (Elimination) Reactions
Our next reaction is the migratory insertion. In a migratory insertion reaction, there is an anionic ligand X and a neutral unsaturated ligand Y in cis-position, and the two ligands couple to form another anionic ligand XY that is attached to the same metal (Fig. $9$:). This produces a vacancy at the metal, meaning a coordination site at the metal which is not occupied by a ligand. In an insertion reaction it is more common that the anionic ligand moves. The reaction is reversible and there is typically a chemical equilibrium. The reverse reaction is called a deinsertion. To drive the chemical equilibrium to the right side one can add the neutral ligand in excess or add a different neutral ligand. The neutral ligand then adds to the vacancy and prevents the reverse reaction. Typical neutral, unsaturated ligands are CO, olefins, alkynes, carbenes, dioxygen, carbon dioxide, and nitriles. Typical anionic ligands are hydrido, alkyl, aryl, alkoxy, and amido-ligands.
Carbonyl Migratory Insertions
One of the most common insertions are carbonyl insertions into alkyl groups like methyl groups (Fig. $10$:). The carbonyl insertion produces an acyl group. To drive the reaction, the reaction can be carried out in the presence of a neutral ligand L like free CO, which occupies the vacant site.
Olefin Insertion and β-Hydride Elimination
Another very common migratory insertion is the olefin migratory insertion into hydrides (Fig. $11$:). This produces an alkyl group. When the reaction is carried out in the presence of CO which binds to the vacant site, then the reaction is driven to the right side. The reverse of this reaction is the β-hydride elimination. It is called “β”-elimination because of the H-atoms that are attached to the carbon atom in the alkyl chain that is second-closest to the metal are called “β”-hydrogens. The hydrogen atoms that are attached to the carbon which is bound to the carbon bonded to the metal via a metal-carbon bond are called “$\alpha$”-hydrogens.
Instability of Transition Metal Complexes Caused by β-Hydride Elimination
The β-hydride elimination is responsible for the frequent instability of transition metal alkyl complexes that possess β-hydrogen atoms.
For example, the depicted dialkyldiphosphine platinum complex is unstable because it has β-hydrogen atoms (Fig. $12$:). Statistically, the complex can lose a phospine complex at a certain rate due to the lability of the Pt-P bond leading to a complex with a vacant site. This complex can then undergo a β-elimination to form an olefin complex. In this case the olefin is a volatile butene molecule which can detach from the metal and leave the system unless the system is hermetically closed. In contrast, the respective methyl complex is stable because it only has α-hydrogens, but not β-hydrogens.
Carbene Migratory Insertion
Migratory insertions are also possible for carbenes. In this case, a CH2 group bonded to the metal via a M=C double bond plays the role of the neutral, unsaturated ligand. In the shown Re complex the carbene inserts into a Re-CH3 bond to form an ethyl group. This ethyl group is unstable because it can undergo β-hydride elimination to form an alkene complex with a hydride ligand.
Dr. Kai Landskron (Lehigh University). If you like this textbook, please consider to make a donation to support the author's research at Lehigh University: Click Here to Donate. | textbooks/chem/Inorganic_Chemistry/Inorganic_Coordination_Chemistry_(Landskron)/12%3A_Organometallic_Reactions_and_Catalysis/12.01%3A_Reactions_Involving_Gain_or_Loss_of_Ligands.txt |
Organometallic Compounds as Hydrogenation Catalysts
Migratory insertions play an important role in catalysis.
For example a Rh-catalyst called the Wilkinson catalyst is an effective hydrogenation catalyst for olefins. The mechanism of the hydrogenation involves a combination of oxidative additions, olefin migratory insertions, and reductive eliminations (Fig. 12.2.2). Wilkinson’s catalyst is the square planar chloro tris(triphenylphosphine) rhodium(I) complex. This molecule is actually a precatalyst that becomes the actual catalyst when it statistically loses a triphenylphosphine ligand producing chloro bis(triphenylphosphine) rhodium(I). The loss of this ligand is a reversible reaction, and thus the catalyst is in chemical equilibrium with the precatalyst. The actually catalyst is in a second chemical equilibrium with its dimer. The chloro bis(triphenylphosphine) rhodium(I) catalyst can undergo an oxidative addition in the presence of hydrogen to form a trigonal bipyramidal chlorodihydrido bis(triphenylphosphine) (III) rhodium complex. This species is in chemical equilibrium with an octahedral chlorodihydrido tris(triphenyl phosphine) rhodium(III) species that can form due to the presence of free triphenylphosphine ligands in the system. The trigonal bipyramidal species can then add an olefin that binds side-on to the Rh. Because the olefin is in cis-position to the hydride ligand it can undergo an olefin insertion. The Rh-C bond can either form with the first or the second carbon in the carbon chain of the olefin, giving a linear and a branched alkyl complex, respectively. The branched complex can undergo a β-hydride elimination thereby reforming the trigonal bipyramidal Rh-complex, and an olefin. This reaction is a side-reaction because the branched alkyl complex is sterically more crowded than the linear complex. The linear alkyl Rh complex can undergo a reductive elimination to form the linear alkane and the RhCl(PPh3)2 catalyst. This completes the catalytic cycle, and a new cycle can start.
Ziegler-Natta Polymerization
Another example of an organometallic catalytic reaction is the Ziegler-Natta olefin polymerization. This reaction is of high industrial importance for the production of olefins like polyethylene. There are both heterogeneous and homogeneous Ziegler-Natta catalysts. The mechanism for the homogeneous catalysts is generally well understood. Homogeneous catalysts are typically metallocene catalysts.
An example of a zirconium-based catalyst is shown (Fig. 12.2.3). The catalyst is a coordinatively unsaturated complex cation with two cyclopentadienyl rings and a methyl group. The catalyst is formed from its precatalyst, a neutral molecule with an additional chloro ligand. The catalyst oxidatively adds an olefin like an ethylene molecule to the coordinatively unsaturated site. This step is followed an olefin insertion step that produces a propyl group. The migratory insertion leads to the formation of a vacant site, that can be re-oocupied by another ethylene molecule. This molecule can insert into the propyl chain thereby prolonging the propyl chain to a pentyl chain. The olefin insertion step generates another vacant site that can be reoccupied by a new ethylene molecule. Repeating the catalytic cycle many times eventually leads to polyethylene.
Catalytic Olefin Hydroformylation
A further important industrial reaction is the catalytic hydroformylation reaction, also known as oxo-process. It was discovered in 1938 by Otto Roelen at BASF. In the hydroformylation reaction an H atom and a formyl group are added to an alkene to form aldehydes. The reaction can produce both branched an linear aldehydes from terminal alkenes, CO, and H2 using a carbonyl hydrides such as HCo(CO)4 as a catalyst. The reaction is performed at about 100°C at a pressure of up to 100 atm.
Mechanism of the Catalytic Olefin Hydroformylation by HCo(CO)4
How does the hydroformylation work mechanistically?
The mechanism is illustrated for the hydroformylation of propene (Fig. 12.2.5). The actual catalyst HCo(CO)4 is first formed from its precatalyst Co2(CO)8 in the presence of H2 in a dinuclear oxidative addition reaction. The catalyst can undergo a substitution reaction in which a CO ligand is replaced by the olefin that binds side-on to the cobalt. This species can then undergo a migratory olefin insertion reaction. This leads to a mixture of linear and branched alkyl groups attached to the Co. A new CO ligand can add to the vacant site. The alkyl group can then insert into a carbonyl group in another migratory insert step, and the vacant site can be reoccupied by a new CO molecule. Then, H2 is added in an oxidative addition. This is the slowest and rate-limiting step in the catalytic cycle. From the addition product the aldehyde can then be eliminated in a reductive elimination reaction. Addition of CO regenerates the catalyst, and the catalytic cycle can begin again.
Hydrocarbonylations
After the hydroformylation, a number of other hydrocarbonylations were developed, and industrially deployed.
When hydrogen is replaced by H2O hydrocarboxylations of alkenes lead to carboxylic acids (Fig. 12.2.6). With an alcohol instead of H2 hydroalkoxycarbonylcations lead to esters. The employment of amines instead of H2 leads to amides in hydroamidocarbonylation reactions.
The Monsanto Acetic Acid Process
Another carbonylation reaction involving an organometallic catalyst is the Monsanto acetic acid process. It has been introduced by Monsanto in the 1970s for the industrial production of acetic acid from methanol. The reaction involves dual catalysis with HI and [RhI2(CO)2]- as a co-catalysts. How does this reaction work?
In the first step methanol reacts with HI to form methyl iodide. The methyl iodide then reacts with the Rh-catalyst in an oxidative addition reaction in which a methyl and an iodo group are added in trans-fashion to the square-planar Rh-complex to give an octahedral complex. The octahedral complex then undergoes a migratory insertion reaction with CO producing an acyl group and a vacant site. A CO molecule can then add to the vacant site. The acetyl iodide can then be eliminated in a reductive elimination to reform the Rh-catalyst thereby closing the catalytic cycle. The acetyl iodide can then react with methanol to form new methyl iodide and acetic acid. The methyl iodide can start a new catalytic cycle with the Rh-catalyst.
Olefin Metathesis
Olefin metathesis is a reaction which allows to cut and rearrange C=C double bonds in olefins to make new olefins (Fig. 12.2.8). Formally, the carbon-carbon bond of the reactant is cleaved homoleptically and the two carbene fragments are combined in a different way. This reaction is typically an equilibrium reaction, and neither the reactants nor the products are clearly favored. This reaction is catalyzed by molybdenum arylamido carbene complexes or ruthenium carbene complexes.
The former are called Shrock catalysts, and the latter Grubbs catalysts named after their discoverers Richard Shrock and Robert Grubbs who received the Nobel prize for Chemistry in 2005 (Fig. 12.2.9). The Schrock catalysts are more active, but also very sensitive to air and water. The Grubbs catalysts, while less active, are less sensitive to air and water.
Olefin metathesis often allows for simpler preparation of olefins compared to other methods. Olefin metathesis is particularly powerful when one olefin product is gaseous because then it can be quite easily removed from the chemical equilibrium by purging. This drives the chemical equilibrium to the right side. An example is the preparation of 5-decene from 1-hexene. Cleavage of the C=C double bond in the hexene leads to C5 and C1 carbene fragments (Fig. 12.2.10). The two C1 fragments can combine to form ethylene and the two C5 fragments combine to 5-decene. The ethylene is volatile and can be purged from the reaction system thereby driving the chemical reaction to the right side.
The same principles can also be applied to produce polymers from dienes with two terminal C=C double bonds at the chain ends. This is called acylic diene metathesis (ADMET), Fig. 12.2.11. For instance the cleavage of the two terminal double bonds in a diene with seven C atoms leads to C1 and C5 fragments. The C1 fragments can combine to form ethylene, and the C5 fragments can combine to make an unsaturated polymer of the type [CH(CH2)3CH]n. Again, the reaction can be driven to the right side by removing the gaseous ethylene from the reaction mixture through purging.
Another variation of olefin metathesis is ring-opening metathesis polymerization (ROMP). It allows to make polymers from strained cycloolefins, for example norbornene. The reaction driving force is the relief of the strain. Because the strain is removed in the polymer, the chemical equilibrium lies far on the right side. The reaction product in norbornene is a polymer with 5-membered rings that are interconnected by ethylene -CH=CH- units (Fig. 12.2.12).
The opposite of ROMP is ring-closure metathesis (RCM). RCM allows for the preparation of unstrained rings with C=C double bonds from dienes with C=C double bonds that are five or six carbon atoms apart. This distance is suitable to produce unstrained rings. In the shown example a five-membered ring with a C=C double bond is formed from a diene with terminal C=C double bonds that are five atoms apart.
The Mechanism of Olefin Metathesis
What is the mechanism of olefin metathesis?
In the first step, the alkene adds to the the carbene fragment of the catalyst in a 2+2 cycloaddition reaction to produce an unstable intermediate with a highly strained four-membered ring (Fig. 12.2.14). This four membered ring can open to produce the first new alkene product R-CH=CH-R and a metal carbene species. This metal carbene can react with another reactant olefin to form another highly stained 4-ring intermediate via a 2+2 cycloaddition reaction. This ring can then reopen again to produce the second alkene metathesis product, in this case ethylene, and the original catalyst. The regenerated catalyst can then start a new catalytic cycle.
Dr. Kai Landskron (Lehigh University). If you like this textbook, please consider to make a donation to support the author's research at Lehigh University: Click Here to Donate.
Concept Review Questions Chapter 12
Concept Review Questions
Section 1
1) What is an oxidative addition?
2) What is a reductive elimination?
3) What is a cis and a trans addition, respectively?
4) Does the coordination number in oxidative additions always increase? Explain.
5) What is the reverse reaction of an oxidative addition?
6) What is a migratory insertion?
7) Are migratory insertions reversible reactions?
Section 2
1) Why are transition metal alkyl complexes that have alkyl groups longer than 1 carbon atom often unstable?
2) What is meant by beta-hydride elimination?
3) What is alpha-abstraction?
4) What is a nucleophilic displacement reaction?
5) What is meant by an oxidative addition of an intact ligand?
6) Write the catalytic cyclic for the deuteration of benzene with Cp2TaH3 as a catalyst. Name each reaction step.
7) What is Collman’s reagent and how is it being used?
8) What is Wilkinson’s catalyst, and what reactions can it be used for. Write the catalytic cycle.
9) Write done the reaction mechanism for the Ziegler-Natta olefin polymerization.
10) What is carbene migratory insertion?
11) Write down the catalytic cycle for the hydroformylation of olefins.
12) Name three hydrocarbonylation reactions. What are the reactants?
13) Write down the mechanism for the Monsanto acetic acid process.
14) What is olefin metathesis?
15) What is ring-opening olefin polymerization (ROMP)?
16) What is acyclic diene metathesis?
17) Why are olefins with terminal C=C double bonds particularly useful in olefin metathesis?
18) What is ring-closing metathesis?
19) What are the characteristics of Schrock and Grubbs catalysts, respectively?
20) Write down the catalytic cycle for olefin metathesis.
Dr. Kai Landskron (Lehigh University). If you like this textbook, please consider to make a donation to support the author's research at Lehigh University: Click Here to Donate.
Homework Problems Chapter 12
Homework Problems
Section 1
Exercise 1
The coordination compound CH3Co(CO)4 undegoes a migratory insertion. What is the reaction product?
Answer
Section 2
Exercise 1
Identify all oxidative additions in the catalytic cycle of olefin hydroformylation:
Answer
Step D --> E is an oxidative addition
Formation of HCo(CO)4 from Co2(CO)8 is also an oxidative addition.
Exercise 2
1. What are the reaction products of the following olefin metathesis reactions?
a) 2 CH2=CH-Ph -->
b) n CH2=CH-CH=CH2 -->
Answer
a) CH2=CH2 and Ph-CH=CH-Ph
b) CH2=CH2 and (CH=CH)n
Exercise 3
What reaction product would you expect would you expect from the following reaction:
Fe(CO)42- + CH3CH2I -->
Name the reaction.
Answer
Fe(CO)4CH2CH3- + I- Nucleophilic displacement
Exercise 4
Which of the following phosphines would you expect to be most suitable to stabilize metals in low oxidation states:
a) P(CF3)3
b) P(CH3)3
c) PH3
Answer
a) P(CF3)3
Exercise 5
What reaction product would you expect from the following reaction?
Pt(PPh3)4 + 4 NPh3 -->
Answer
There would be no reaction.
Exercise 6
Which of the following complexes would likely be unstable:
a) [PtCl3(C2H5)]2-
b) [PtCl3(CH3)]2-
c) PtCl42-
Answer
a) [PtCl3(C2H5)]2-
Dr. Kai Landskron (Lehigh University). If you like this textbook, please consider to make a donation to support the author's research at Lehigh University: Click Here to Donate. | textbooks/chem/Inorganic_Chemistry/Inorganic_Coordination_Chemistry_(Landskron)/12%3A_Organometallic_Reactions_and_Catalysis/12.02%3A_Organometallic_Catalysts.txt |
Learning Objectives
In this lecture you will learn the following:
• Lewis structure.
• Limitations of Lewis model.
Organometallic Chemistry of Main Group Elements
Our aim in this course is to introduce one of the important disciplines of inorganic chemistry which bridges the organic compounds with main group and transition elements; organometallic chemistry. Organometallic chemistry is defined as the chemistry of chemical compounds containing one or more metal—carbon bonds which are essentially polar (Mδ+—Cδ-) in nature. To begin with it is necessary to understand the bonding concepts that explain the structures of both main group and transition elements. The objectives of first few lectures are to give some insight into the various metal—ligand interactions that would later help in planning the synthesis and also eventually looking into their electronic and kinetic stability. The course is divided into four major sections and the first section deals with the various bonding concepts employed for both main group and transition metal compounds. The second and third sections deal with the organometallic chemistry of main group and transition elements, respectively, while the fourth section would be devoted to applications of organometallic compounds with special emphasis to catalysis.
A set of problems along with solutions are presented at the end and also in Chapter 2. In future, it is planned to provide interactive sections as well.
The pertinent literature and books in this and related areas are listed at the end.
Molecular structure and bonding
The chemical properties of the molecules can be directly correlated to their electronic structures. In this lecture attempts are being made to give an overall view of how bonding concept evolved starting from Lewis approach to the development of molecular orbital theory.
1.1.1: Lewis structures
Lewis proposed that when two atoms come close to each other to establish a bond by sharing an electron pair, a covalent bond will be established. One pair of electrons would give a single bond X—Y; two or three pairs of electrons would leads to the formation of double (X=Y) and triple bonds (X≡Y), respectively. The pairs of valence electrons that are not utilized in bonding are called lone pairs of electrons or simply lone pairs. The lone pair of electrons does not participate in bonding; however, they do influence the shape and the geometry of the molecule and their chemical properties as well.
Lewis introduced octet rule which states that each atom shares its valence electrons with neighboring atoms to have a total of eight (s2p6) electrons in its valence shell to have noble-gas configuration. An exception to this rule is hydrogen as it can have only two valence electrons in its only shell, 1s.
By simply counting the number of valence electrons present on the central atom and its neighbors, Lewis structures can be written in just three easy steps.
1. Consider the valence electrons of all participating atoms; add an electron for each negative charge and subtract one electron for each positive charge.
2. Identify the central atom and write the symbols of the atoms around central atom. In majority of polyatomic molecules, the least electronegative one will be the central atom with an exception of hydrides, for example, H2O, NH3 or H2S.
3. Distribute the electron pairs throughout the molecule to satisfy the octet of all atoms present in the molecule starting from the most electronegative one. Each pair of singly bonded atoms requires one pair of electrons.
4. Each bonding pair should be represented by a single bond and the net charge is assumed to be possessed by the ion (cation or anion) as a whole and not by an individual atom.
For some molecules, Lewis dot structure differs from the experimentally determined structural observations. For example, in acetate ion both the C—O bonds are identical as per X-ray structure determination but the prediction by the Lewis structure is incorrect. The reason is due to resonance.
Limitations of Lewis model
Molecules with odd number of electrons can never satisfy the octet rule.
Example: NO.
Some atoms with fewer valence electrons can never complete octet without formal charges.
A central atom can have more than 8 electrons. Example: SF6
Lewis model does not explain paramagnetic nature of O2.
Geometry and molecular shapes cannot be explained by Lewis model.
Worked examples
Example 1 : ClO2-
Solution :
ClO2- Total number of electrons 7 + 2 x 6 + 1 = 20 = 10 pairs
Identify the central atom and connect the peripheral atoms with a pair of electrons as two dots and count the remaining electron pairs
\(\ce{O:Cl:O}\) 20-4 = 16 electrons left (8 pairs)
Complete the octet of oxygen atoms and count the remaining electron pairs
16-12 = 4 electrons
Complete the octet of chlorine atom,
Since all the electrons are utilized and octet is satisfied, no need of any multiple bonds.
The structure is or or
Example 2 : CO
Solution :
CO (carbon monoxide) Total number of electrons 4 + 6 = 10
Connect the two atoms with a pair of electrons as two dots and count the remaining electron pairs
\(\ce{C:O}\) 10-2 = 8 electrons left
Complete the octet of oxygen atoms and count the remaining electron pairs
8-6 = 2 electrons
Place the remaining electron pair on carbon atom,
All the electrons are utilized and octet is not satisfied for carbon as there is a shortage of four more electrons.
Drag two electron pairs on oxygen atom in between carbon and oxygen to establish two more bonds so that a triple bond exists between C and O which satisfies the octet of both C and O.
The structure is :C:::O: or :C≡O: or |C≡O|
Problems:
Work out Lewis structures for: BF4-, PCl3, PO43-, SO42-, O3, N2 and SO2 | textbooks/chem/Inorganic_Chemistry/Introduction_to_Organometallic_Chemistry_(Ghosh_and_Balakrishna)/01%3A_Introduction/1.01%3A_Bonding_Concepts_in_Main_Group_Chemistry.txt |
Learning Objectives
In this lecture you will learn the following
• Definition of VSEPR.
• Hybridization concept.
• Prediction of shapes and geometries of molecules.
• Bent’s rule and its application in understanding the structural parameters.
The Valence Shell Electron Pair Repulsion Theory (VSEPR)
VSEPR theory is an improved and extension of Lewis model but predicts the shapes of polyatomic molecules. This model was first suggested by Nevil Sidgwick and Herbet Powell in 1940 and later improved by Ronald Gillespie and Ronald Nyholm.
Prediction of molecular shapes and geometries was made easy by this model through the following simple steps.
1. Draw the Lewis structure.
2. Count the total number of bonds and lone pairs around the central atom. (Each single bond would involve one pair of electrons).
3. Arrange the bonding pairs and lone pairs in one of the standard geometries to minimize the electron-electron repulsion.
1. Lone pair electrons stay closer to the nucleus and also they spread out over a larger space than bond pairs and hence large angles between lone pairs.
2. The repulsion follows the order LP—LP > LP—BP > BP—BP.
4. Multiple bonds should be considered as a single bonding region.
Steric numbers:
Another term called steric number is often used in VSEPR theory.
Steric number (SN) = No. of attached atom + No. of lone pairs. Since the lone pair—lone pair repulsions are maximum, the most stable geometry can be obtained by maximizing the distance between steric numbers on the central atom.
Molecular shapes are eventually determined by two parameters: Bond distance, separation between the nuclei of two bonded atoms in a straight line and the bond angle, the angle between any two bonds containing a common atom.
While mentioning the molecular shapes lone pairs may be ignored, however, while defining the geometry both the lone pairs and bond pairs should be considered.
For example: in water molecule the central oxygen atom is in tetrahedral environment with two lone pairs and two O—H bonds (or two bond pairs). The shape of the water molecule is therefore bent (two lone pairs are ignored).
Similarly, in ammonia, the nitrogen atom is in tetrahedral environment with three bonded pairs (three N—H bonds) and one lone pair. The shape of NH3 molecule is pyramidal.
Predicting the molecular geometries
To begin with, draw the Lewis structure.
Count the number of bonding pairs and lone pairs around the central atom.
Arrange the bonding pairs and the lone pairs in one of the standard geometries thereby minimizing electron—electron repulsion.
Multiple bonds count as a single bonding region.
What is Bent’s rule:
More electronegative substituents ‘prefer’ hybrid orbitals having less s-character, and more electropositive substituents ‘prefer’ orbitals having more s-character.
The bond angles in CH4, CF4 and CH2F2 can be explained using Bent’s rule. While a carbon in CH4 and CF4 uses four identical sp³ hybrids for bonding, in CH2F2 the hybrids used are not identical.
The C-F bonds are formed from sp3 + x hybrids, with slightly more p-character and less s-character than an sp³ hybrid, and the hydrogen are bonded by sp3 - x hybrids, with slightly less p-character and slightly more s-character. Increasing the amount of p-character in the C-F bonds decreases the F-C-F bond angle, because for bonding by pure p-orbitals the bond angle would be decreased to 90°.
Molecular shapes determined by VSEPR theory
Molecule
Steric Number
(Number Electron Pairs) (SN)
Geometry
Example
MA2
2
Linear
BeCl2
MA3
3
Trigonal planar
BF3
MA4
4
Tetrahedral
SiF4
MA5
5
Trigonal bipyramidal
PF5
MA6
6
Octahedral
SF6
MA7
7
Pentagonal bipyramidal
IF7
Molecule
SN
Number of lone pairs
Geometry
shape
Example
MA2
2 0
Linear
CO2
MA3
3
3
0
1
Trigonal planar
Trigonal planar
angular
SO3
SO2
MA4
4 0
1
2
Tetrahedral
Tetrahedral
Trigonal pyramidal
Angular
CH4
NH3
H2O
MA5
5 0
1
2
3
Trigonal bipyramidal
Trigonal bipyramidal
Seesaw
T-shaped
linear
AsF5
SF4
ClF3
XeF2
MA6
6 0
1
2
Octahedral
Octahedral
Square pyramidal
Square planar
SF6
BrF5
XeF4
Problems:
1. Draw the structure and depict the geometry around Se atoms in [Se3O6F3]3-, which is a symmetric ionic molecule with cyclic structure, using VSEPR model.
Solution
Apply VSEPR theory to the structure.
Se has six valence electrons. One Se—F and three Se—O (one terminal and two bridging) will add four more electrons to the valence shell of Se, so, 10 e = 5 electron pairs out of which four are bonded pairs and one is a lone pair. A TBP environment is expected.
2. Predict and draw the structure of I3+ using VSEPR model.
Solution
Total number of electron at the central I atom is: 7 + 2 -1 (charge) = 8; 2BP + 2LP, should be tetrahedral and angular shape.
3. Which of H2O and F2O will have the larger X-O-X bond angle?
Solution:
F is more electronegative than H; therefore the space occupied by the O-H bonding pair in the O valence shell will be greater. Hence, H2O will have the larger X-O-X bond angle.
4. Explain why the X-P-X bond angles for the series of POX3 molecules decrease from X = Br (104.1°) to X = Cl (103.3°) to X = F (101.3°)
Solution:
Fluorine is the most electronegative halogen, so it will draw electron density in the P-F bond away from P atom; repulsion of the P-F bonding pairs will be less than the repulsion of P-Cl and P-Br bonding pairs, so the F-P-F bond angle will be the smallest.
5. Show that the following molecules and their corresponding shapes are correct, using VSEPR theory.
BCl3, trigonal planar
[IF5]2-, pentagonal planar
[NH4]+, tetrahedral
SF6; octahedral
XeF4; square planar
AsF5; trigonal bipyramidal
Xe(O)F4, square pyramid
IF7, pentagonal bipyramidal
[H3O]+ tetrahedral
H2Se tetrahedral
6. Work out the geometry of I3- ion using VSEPR model.
Solution | textbooks/chem/Inorganic_Chemistry/Introduction_to_Organometallic_Chemistry_(Ghosh_and_Balakrishna)/01%3A_Introduction/1.02%3A_VSEPR_Theory_and_its_Utility.txt |
Learning Objectives
In this section you will learn the following
• Various synthetic methodologies to make M—C bonds.
• How to choose an appropriate synthetic method.
• Reaction conditions and the role of solvents
An organometallic compound contains one or more metal-carbon bonds.
General Methods of Preparation
Most organometallic compounds can be synthesized by using one of four M-C bond forming reactions of a metal with an organic halide, metal displacement, metathesis and hydrometallation.
a. Reaction with metal and transmetallation
The net reaction of an electropositive metal M and a halogen-substituted hydrocarbon is
$\ce{ 2 M} + \ce{R X} (\text{alkyl or aryl}) \ce{-> MR + MX} \nonumber$
For example
$\ce{ 8 Li + 4CH3Cl -> Li4(CH3)4 + 4LiCl} \nonumber$
$\ce{ Mg + CH3Br -> CH3MgBr (organometal\:halide\:with\:Mg, Al, Zn)} \nonumber$
If, one metal atom takes the place of another, it is called transmetallation
$\ce{ M + M'R -> M' + MR} \nonumber$
$\ce{ 2Ga + 3CH3-Hg-CH3 -> 3Hg + 2Ga(CH3)3} \nonumber$
Transmetallation is favorable when the displacing metal is higher in the electrochemical series than the displaced metal.
b. Metathesis
The metathesis of an organometallic compound MR and a binary halide EX is a widely used synthetic route in organometallic chemistry.
$\ce {MR + EX -> ER + MX} \nonumber$
$\ce {Li4(CH3)4 + SiCl4 -> 4LiCl + Si(CH3)4} \nonumber$
$\ce {Al2(CH3)6 + 2BF3 -> 2AlF3 + 2B(CH3)3} \nonumber$
Metathesis reaction can frequently be predicted from electronegativity or hard and soft acid-base considerations.
Hydrocarbon groups tends to bond to the more electronegative element; the halogen favors the formation of ionic compounds with the more electropositive metal.
In brief, the alkyl and aryl group tends to migrate from the less to the more electronegative element [χ = electronegativity].
When the electronegativities are similar, the correct outcome may be predicted, with care*, by considering the combination of the softer element with organic group and harder element with fluoride or chloride.
*An insoluble product or reactant may change the outcome, e.g.;
$\ce{SnPh4(THF) + HgBr2(THF) -> HgPhBr(s) + PhSnBr (THF)} \nonumber$
HgPhBr turns out to be insoluble in THF
Metathesis reactions involving the same central element are often referred to as redistribution reactions.
$\ce{SiCl4 + SiMe4 -> Me3SiCl + Me2SiCl2 + ...} \nonumber$
$\ce{3GeCl4 + 2AlMe6 -> 3GeMe4 + 4AlCl3} \nonumber$
Al is more electropositive than Ge, this reaction occurs as it is thermodynamically favorable.
c. Hydrometallation
The net outcome of the addition of a metal hydride to an alkene is an alkylmetal compound.
$\ce{EH + H2C=CH2 -> E-CH2-CH3} \nonumber$
The reaction is driven by the high strength of E-C bond relative to that of most E-H bonds, and occurs with a wide variety of compounds that contain E-H bonds.
Hydroboration
Hydrosilylation
Ionic and electron-deficient compounds of Group 1, 2
Organometallic derivatives of all Group 1 metals are known. Amongst, the alkyllithium compounds are most thoroughly studied and useful reagents.
Many of them are commercially available.
MeLi is generally handled in ether solution, but RLi compounds with longer chains are soluble in hydrocarbons.
Commercial preparation:
$\ce{M + RX -> MR (often \: contaminated \: with \: halide)} \nonumber$
The best method would be:
$\ce{HgR2 + 2Li -> 2LiR + Hg} \nonumber$
MeLi exists as a tetrahedral cluster in the solid state and in the solution. Many of its higher homologs exist in solution as hexamers or equilibrium mixture of aggregates ranging up to haxamers.
The larger aggregates can be broken down by Lewis bases, such as, TMEDA.
Common organolithium compounds have one Li per organic group.
Several polylithiated organic molecules containing several lithium atoms per molecule are known.
The simplest example is Li2CH2, which can be prepared by the pyrolysis of MeLi which crystallizes in a distorted antifulorite* structure. However, the finer details of the orientation of the CH2 groups are yet to be established.
*the antifluorite structure is the inverse of the fluorite structure in which the locations ofcations and anions are reversed. Look into the structures of CaF2 (fluorite structure) and K2O (antifluorite structure). An fcc array of cations and all the tetrahedral holes are filled with anions.
Radical anion salts
Sodium naphthalide is an example of an organometallic salt with a delocalized radical anion, C10H8-.
Such compounds are readily prepared by reacting an aromatic compound with an alkali metal in a polar aprotic solvent.
Naphthalene dissolved in THF reacts with Na metal to produce a dark green solution of sodium naphthalide.
$\ce{Na(s) + C10H8(THF) -> Na[C10H8](THF)} \nonumber$
EPR spectra show that the odd electron is delocalized in an antibonding orbital of C10H8.
Formation of radical anion is more favorable when the π of LUMO of the arene is low in energy.
Simple MOT predicts that the energy of LUMO decreases steadily on going from benzene to more extensively conjugated hydrocarbons.
Sodium naphthalide and similar compounds are highly reactive reducing agents.
They are preferred to sodium because unlike sodium, they are readily soluble in ethers.
The resulting homogeneous reaction is generally faster and easier to control than a heterogeneous reaction between one reagent in solution and pieces of sodium metal, which are often coated with unreactive sodium oxide or with insoluble reaction products.
The additional advantage is that by proper choice of the aromatic group the reduction potential of the reagent can be chosen to match the requirements of a particular synthetic task.
Alternative route to delocalized anion is the reductive cleavage of acidic C—H bonds by an alkali metal or alkylmetallic compound.
Example:
Problems:
1. Classify the following reactions into, (i) hydrometallation, metal displacement, metathesis OR transmetallation reactions; (ii) give an example for each case in the form of a balanced chemical equation.
Solution
a. $\ce{M + Mx’R -> M’ + MR} \nonumber$ ….Transmetallation
e.g.: $\ce{2Ga + 3CH3-Hg-CH3 -> 3Hg + 2Ga(CH3)3} \nonumber$
b. $\ce{MR + EX → ER + MX} \nonumber$ ….Metathesis
e.g.: $\ce{Li4(CH3)4 + SiCl4 -> 4LiCl +Si(CH3)4} \nonumber$ or $\ce{Al2(CH3)6 + 2BF3 -> 2AlF3 + 2B(CH3)3} \nonumber$
c. $\ce{EH + H2C=CH2 -> E—CH2—CH3 } \nonumber$ ….Hydrometallation
e.g.:
Problems:
2. For each of the following compounds, indicate those that may serve as
(1) a good carbanion nucleophile reagent,
(2) a mild Lewis acid,
(3) a mild Lewis base at the central atom,
(4) a strong reducing agent. (A compound may have more than one of these properties)
(a) Li4(CH3)4, (b) Zn(CH3)2, (c) (CH3)MgBr, (d) B(CH3)3, (e) Al2(CH3)6, (f) Si(CH3)4, (g) As(CH3)3.
Solution
1. (MeLi)4 - good carbanion nucleophile and strong reducing agent
2. ZnMe2 - reasonable carbanion nucleophile, mild Lewis acid, reducing agent
3. MeMgBr - good carbanion nucleophile
4. BMe3 - mild Lewis acid
5. Al2Me6 - good carbanion nucleophile, strong reducing agent
6. SiMe4 - mild Lewis acid
7. AsMe3 - mild Lewis base | textbooks/chem/Inorganic_Chemistry/Introduction_to_Organometallic_Chemistry_(Ghosh_and_Balakrishna)/02%3A_Organometallic_Chemistry_of_s-_and_p-block_Elements/2.01%3A_General_Methods_of_Preparation.txt |
Learning Objectives
In this lecture you will learn the following
• How to synthesize and handle sodium, lithium compounds.
• Structural features.
Organometallic compounds of alkaline metals
Organic compounds such as terminal alkynes which contain relatively acidic hydrogen atoms form salts with the alkali metals.
$\ce{2Et-C≡CH + 2Na -> 2Na+[Et-C≡C]- + H2} \nonumber$
$\ce{Me-C≡C-H + K[NH2] -> K+[Me-C≡C]- + NH3} \nonumber$
$\ce{C5H6 + Na -> NaCp + 1/2 H2} \nonumber$
NaCp is pyrophoric in air, but air-sensitivity can be lessened by complexing the Na+ with dme.
In the solid state, [Na(dme)][Cp] is polymeric
*Pyrophoric material: is one that burns spontaneously when exposed to air.
Transmetallation:
$\ce{HgMe2 + Na -> 2NaMe + Hg} \nonumber$
Organolithium compounds:
$\ce{nBuCl + 2Li ->[Hydrocarbon\:Solvent] nBuLi + LiCl} \nonumber$
Organolithium compounds are of particular importance among the group 1 organometallics.
Many of them are commercially available as solutions in hydrocarbon solvents.
Solvent choices for reactions involving organometallics of the alkali metals are critical. For example, nBuLi is decomposed by Et2O to give nBuH, C2H4 and LiOEt.
Alkali metal organometallics are extremely reactive and must be handled in air- and moisture-free environments; NaMe, for example, burns explosively in air.
Lithium alkyls are polymeric both in solution and in the solid state.
NMR is very useful in understanding the solution structures; 6Li (I = 1), 7Li (I = ½), 13C (I = ½)
The structures of (tBuLi)4 and (MeLi)4 are similar. nBuLi when mixed with TMEDA, gives a polymeric chain. TMEDA link cubane units together through the formation of Li-N bonds.
Alkyllithium compounds are soluble in organic solvents whereas Na and K salts are insoluble, but are solubilized by the chelating ligand TMEDA. Addition of TMEDA may break down the aggregates of lithium alkyls to give lower nuclearity complexes. E.g. [nBuLi.TMEDA]2
However, detailed studies have revealed that the system is far from simple, and it is possible to isolate crystals of either [nBULi.TMEDA]2 or [(nBuLi)4.TMEDA].
In the case of (MeLi)4, the addition of TMEDA does not lead to cluster breakdown, and the X-ray structure confirms the composition (MeLi)4.2TMEDA, the presence of both tetramers and the amine molecules in the crystal lattice.
N. D. R. Barnett et al. J. Am. Chem. Soc., 1993, 115, 1573.
Lithium alkyls and aryls are very useful reagents in organic synthesis and also in making corresponding carbon compounds of main group elements.
Lithium alkyls are important catalysts in the synthetic rubber industry for the stereospecific polymerization of alkenes.
2.03: Organometallic Compounds of Alkaline Earth Metals (Beryllium and Magn
Objectives
In this lecture you will learn the following
• Organometallic compounds of beryllium and magnesium.
• Structural features of alkyl lithium and beryllium sandwich compounds.
Beryllium
$\ce{HgMe2 + Be -> Me2Be + Hg (at \: 383K)} \nonumber$
$\ce{2PhLi + BeCl2 -> Ph2Be +2LiCl (in\:diethyl\:ether)} \nonumber$
In vapor phase, Me2Be is monomeric with a linear C—Be—C (Be-C = 170 pm).
The solid state structure is polymeric and resembles that of BeCl2.
$\ce{2NaCp + BeCl2 -> Cp2Be + 2NaCl} \nonumber$
The X-ray diffraction at 128 K suggested [(η1-Cp)(η5-Cp)Be].
However, 1H NMR spectrum shows that all protons environments are equivalent even at 163 K.
Also, solid state structure shows the Be atom is disordered over two equivalent sites and NMR data can be interpreted in terms of fluxional process in which the Be atom moves between these two sites.
However, Cp*2Be possesses a sandwich structure with both the rings are coplanar.
Magnesium
Alkyl and aryl magnesium halides (Grignard reagents, RMgX) are extremely well-known on account of their uses in synthetic chemistry.
$\ce{Mg + Rx -> RMgX (in\:diethyl\:ether)} \nonumber$
Transmetallation is useful means of preparing pure Grignard reagents.
$\ce{Mg + RHgBr -> Hg + RMgBr} \nonumber$
$\ce{Mg + R2Hg -> Hg + R2Mg} \nonumber$
Two-coordination at Mg in R2Mg is observed only when the R groups are sufficiently bulky, e.g. Mg{C(SiMe3)3}2.
RMgX are generally solvated and Mg centre is typically tetrahedral.
e.g. EtMgBr.2Et2O; PhMgBr.2Et2O.
Cp2Mg has a staggered sandwich structure.
Solutions of Grignard reagent may contain several species, e.g. RMgX, R2Mg, MgX2, RMg(μ-X)2MgR, which are further complicated by solvation. The position of equilibrium between these species is markedly dependent on concentration, temperature and solvent; strongly donating solvents favour monomeric species in which they coordinate to the metal centre.
Treatment with dioxane results in the precipitation of MgCl2(dioxane) leaving behind pure R2Mg in the solution.
Problems:
1. The compound (Me3Si)2C(MgBr)2.nTHF is monomeric. Suggest a value of ‘n’ and propose a structure for this Grignard reagent.
Solution
2. If a typical Grignard reagent exists as an equilibrium mixture of dialkylmagnesium and magnesium halide, give a method of isolating pure dialkyl magnesium. Your answer should be in the form of balanced chemical equations only.
Solution
$\ce{2RMgX <=> R2Mg + MgX2} \nonumber$
Treatment of equilibrium mixture with dioxane results in the precipitation of, say, MgCl2(dioxane) (if, X = Cl), leaving behind pure R2Mg in the solution.
2.04: Structure and Bonding
Learning Objectives
In this lecture you will learn the following
• Solid state structures of nickel-arsenide, alkyl lithium and alkyl aluminium compounds
tructure and bonding
The slight differences that arise between organometallic compounds and binary hydrogen compounds are mainly due to the tendency of alkyl groups to avoid ionic bonding.
The molecular structures of AlMe3and MeLi differ from AlH3and LiH.
Even the more ionic MeK crystallizes in the nickel-arsenide structure rather than the rock-salt structure adopted by KCl.
Nickel-arsenide structure is typical of soft-cation, soft-anion combinations.
Electron deficient compounds such as AlMe3contain 3c-2e bonds analogous to the B—H—B bridges in diborane.
The Nickel-Arsenide, NiAs, Structure
MeLi in nonpolar solvents consists of tetrahedron of Li atoms with each face bridged by a methyl group. Similar to Al2Me6, the bonding in MeLi consists of a set of localized molecular orbitals. The symmetric combination of three Li 2s orbitals on each face of the Li4 tetrahedron and one sp3 hybrid orbital from CH3 gives an orbital that can accommodate a pair of electron to form a 4c-2e bond.
The lower energy of the C orbital compared with the Li orbitals indicates that the bonding pair of electrons will be associated primarily with the CH3 group, thus supporting the carbanionic character of the molecule. Some analysis has indicated that about 90% ionic character for the Li-CH3 interaction.
The interaction between an sp3 orbital from a methyl group and the three 2s orbitals of the Li atoms in a triangular face of Li4(CH3)4 to form a totally symmetric 4c,2e bonding orbital. The next higher orbital is non-bonding and the uppermost is antibonding.
Me2Be and Me2Mg exist in a polymeric structure with two 3c,2e-bonding CH3 bridges between each metal atom. | textbooks/chem/Inorganic_Chemistry/Introduction_to_Organometallic_Chemistry_(Ghosh_and_Balakrishna)/02%3A_Organometallic_Chemistry_of_s-_and_p-block_Elements/2.02%3A_Organometallic_Compounds_of_Alkali_Metals_%28Sodium_and_Lithium%29.txt |
Learning Objectives
In this section you will learn the following
• Chemical properties of organometallic compounds.
• Reactivity patters and their electronic properties.
• Electronic properties verses chemical reactivity.
• Application of alkyl lithium in organic transformations.
Reactions of organometallic compounds
The reactions of organometallic compounds of electropositive elements are dominated by factors such as the carbanion character of the organic moiety and the availability of a coordination site on the central metal atom. Reaction patterns:
a. Oxidation
• All organometallic compounds are potentially reducing agents.
• Those of electropositive elements are in fact very strong reducing agents (many of them are pyrophoric in nature).
• The strong reducing character also presents a potential explosion hazard if the compounds are mixed with larger amount of oxidzing agents.
Why is it so?
All organometallic compounds of the electropositive metals that have unfilled valence orbitals, or that readily dissociate into fragments with unfilled orbitals are pyrophoric,e.g., $\ce{Li4(CH3)4}$, $\ce{Zn(CH3)2}$, $\ce{B(CH3)3}$ and $\ce{Al2(CH3)6}$
Volatile pyrophoric compounds, such as $\ce{B(CH3)3}$, may be handled in vacuum line and, inert atmosphere techniques are used for less volatile but air-sensitive compounds. Compounds such as $\ce{Si(CH3)4}$ and $\ce{Sn(CH3)4}$ which do not have low-lying empty orbitals, require elevated temperatures to initiate combustion, and can be handled in air.
The combustion of many organometallic compounds takes place by a radical chain mechanism.
b. Nucleophilic character
The partial negative charge of an organic group attached to an electropositive metal makes it a strong nucleophile and Lewis base. This is referred to as its carbanion character even though the compound itself is not ionic. Alkyllithium and alkylaluminium compounds and Grignard reagents are the most common carbanion reagents in laboratory-scale synthetic chemistry. The carbanion character diminishes for the less metallic boron and silicon.
The carbanion character finds many synthetic applications.
Where X = halide, E = B, Si, Ge, Sn, Pb, As and Sb
c. Lewis acidity
Due to the presence of unoccupied orbitals on the metal atom, electron-deficient organometallic compounds are observed to be Lewis acids.
e.g $\ce{B(C6H5)3 + LiC6H5 -> Li[B(C6H5)4]} \nonumber$
This reaction may be viewed as the transfer of the strong base C6H5-from the weak Lewis acid Li+to the stronger acid B(III).
Organometallic species that are bridged by organic groups can also serve as Lewis acids and, in the process, bridge cleavage can take place.
$\ce{Al2(CH3)6 + 2N(C2H5)3 -> 2(CH3)3AlN(C2H5)3} \nonumber$
Electron deficient organometallic species are Lewis acids.
Problems
1. Name each of the following compounds and classify them: (a) SiH(CH3)3, (b) BCl(C6F5)2, (c) Al2Cl2(C6H5)4, (d) Li4(C4H9)4, (e) Rb(CH2H3).
Solution
1. trimethylsilane (tetrahedral monomer, electron-precise);
2. bis(pentafluorophenyl)chloroborane (trigonal monomer, electron-deficient);
3. tetraphenyldichlorodialuminum (two Al-Cl-Al bridges, in this structural form it is electron –precise);
4. butyllithium or more precisely, tetrabutyltetralithium (tetrahedral Li4 array with a phenyl carbon bridging each face, electron deficient);
5. ethylrubidium, salt-like.
2. Sketch the structures of: (a) methyl lithium, (b) trimethyl boron, (c) hexamethyldialuminum, (d) tetramethylsilane, (e) trimethylarsane, and (f) tetraphenylarsonium.
Solution
1. methyl lithium: Li tetrahedron with each face capped by CH3 [see chapter 6]
2. trimethyl boron: planar triangular array of B and C
3. hexamethyldialuminum:four terminal CH3 and two CH3 bridges in a diborane-like structure
4. tetramethylsilane: tetrahedral
5. trimethylarsane: pyramidal
6. tetraphenylarsonium: pseudotetrahedral | textbooks/chem/Inorganic_Chemistry/Introduction_to_Organometallic_Chemistry_(Ghosh_and_Balakrishna)/03%3A_Organometallic_Chemistry_of_p-block_Elements/3.01%3A_Reactions_of_Organometallic_Compounds.txt |
Learning Objectives
In this lecture you will learn the following
• Preparation and reactivity of organoboron and organo aluminium compounds.
• Influence of Lewis acidity on structural features
Organoboron Compounds
BMe3 is colorless, gaseous ( b.p. -22 °C), and is monomeric. It is pyrophoric but not rapidly hydrolyzed by water.
Alkylboranes can be synthesized by metathesis between BX3 and organometallic compounds of metals with low electronegativity, such as RMgX or AlR3.
$\ce{BF3 + 3CH3MgBr -> B(CH3)3 + 3MgBrF (solvent\: used: \: dibutyl\: ether)} \nonumber$
Why dibutyl ether as a solvent: Has much lower vapor pressure than BMe3 and as a result the separation by trap-to-trap distillation on a vacuum line is easy.
Also, there is a very weak association between BMe3 and OBu2(Me3B:OBu2).
Although, trialkyl- and triarylboron compounds are mild Lewis acids, strong carbanion reagents lead to anions of the type [BR4]-.
Example, Na[BPh]4: The bulky anion hydrolyses very slowly in neutral or basic water and is useful for the preparation of large positive cations.
$\ce{Na[BPh]4 + K+ -> K[BPh]4} \nonumber$
K[BPh]4 is insoluble, used for the gravimetric estimation (determination) of potassium, an example of the low solubility of large-cation and large-anion salts in water/
Organohaloboron compounds are more reactive than simple trialkylboron compounds.
Preparation:
$\ce{2 BCl3 + 6AlR3 -> 3R2BCl + 6AlR2Cl (metathesis)} \nonumber$
$\ce{2 BCl3 + BMe3 ->[(diborane)] 3BMeCl2 (redistribution\:reaction)} \nonumber$
Reactions: (Protolysis reactions with ROH, R2NH and other reagents)
$\ce{3BMeCl2 + 2HNR2 -> BMe2(NR2) + [R2NH2]Cl} \nonumber$
$\ce{BMe2Cl + Li(C4H9) -> BMe2(C4H9) + LiCl} \nonumber$
Organoaluminium compounds
With less bulky alkyl groups, dimerization occurs and one of the distinguishing features of alkyl bridge is the small Al-C-Al angle, which is ~ 75°.
The 3c,2e bonds are very weak and tend to dissociate in the pure liquid which increases with increase in the bulkiness of the alkyl group.
$\ce{Al2(CH3)6 <=> 2Al(CH3)3} \nonumber$ K= 1.52 x 10-8
$\ce{Al2(C4H9)6 <=> 2Al(C4H9)3} \nonumber$ K= 2.3 x 10-4
Perpendicular orientation of pheynl groups in Al2Ph6
Triphenylaluminium exists as a dimer with bridging η1-phenyl groups lying in a plane perpendicular to the line joining the two Al atoms.
This structure is favored partly on steric grounds and partly by supplementation of the Al-C-Al bond by electron donation from the phenyl π-orbitals to the Al atoms.
Tendency for bridging: X > Ph > alkyl
3c,2e bonds formed by a symmetric combination of Al and C orbitals
An additional interaction between the pπ orbital on C and an antisymmetric combination of Al orbitals.
Synthesis
Very useful as alkene polymerization catalysts and chemical intermediates.
Expensive carbanion reagents for the replacement of halogens organic groups by metathesis.
Laboratory scale preparations involves: $\ce{2Al + 3Hg(CH3)2 -> Al2(CH3)6 + 3Hg} \nonumber$
Commercial method: $\ce{2Al + CH3Cl -> Al2Cl2(CH3)4} \nonumber$
$\ce{Al2Cl2(CH3)4 + 6Na -> Al2(CH3)6 + 2Al + 6NaCl} \nonumber$
Commercial method for ethylaluminium and higher homologs: $\ce{2Al + 3H2 + 6RHC=CH2 ->[60-110°C][110-200atm] 2Al2(CH2CH2R)6} \nonumber$
The reaction probably proceeds by the formation of a surface Al—H species that adds across the double bond of the alkene in a hydrometallation reaction.
Reactions:
Alkylaluminum compounds are mild Lewis acids and form complexes with ethers, amines and anions. When heated, often β-hydrogen elimination is responsible for the decomposition of ethyl and higher alkylaluminium compounds. E.g. Al(iC4H9)3
Tendency towards bridging structure is: PR2-> X -> H -> Ph-> R-.
Problems:
1. Propose a structure for Al2(Me)4Cl2.
Solution:
Similar to diborane:
2. For these compounds (H3Si)2O and (CH3CH2)2O, which do you expect to have the lower force constant, Si-O-Si bending or C-O-C bending?
Solution:
Lower force constant for Si-O-Si bending.
3. Explain how the difference in reactivity between Al-C and Si-C bonds with O-H groups leads to the choice of different strategies for the synthesis of aluminum and silicon alkoxides.
Solution
$\ce{Al2Me6 + 6MeOH -> 2Al(OMe)3 + 6CH4} \nonumber$
For reaction of Al2Me6 with alcohols, see the text book by Shriver and Atkins.
Tetramethylsilane does not react with methyl alcohol. Therefore, the appropriate reagent is tetrachlorosilane and the reaction is:
$\ce{SiCl3 + 4MeOH -> Si(OMe)4 + 4HCl} \nonumber$
4. Compare formulas of the most stable hydrogen compounds of germanium and arsenic with those of their methyl compounds. Can the differences be explained in terms of the relative electronegativities of C and H?
Solution
GeH4, GeR4; AsH3, AsR3
The stability of hydrides and alkyls are very similar for each element. This may due to similar H and C electronegativity.
5. To buy from a chemical company, the price of trimethylaluminum is higher than that of triethylaluminum. Is it due to the methods of synthesis? Rationalize the price difference.
Solution:
Triethylaluminum can be made in larger quantities by direct reaction of aluminum, hydrogen gas and ethane gas which is a cheaper method.
$\ce{2Al + 3H2 + 6RHC=CH2 ->[60-110°C][110-200atm] 2Al2(CH2CH2R)6} \nonumber$
Preparation of trimethyaluminum involves a more expensive route such as MeCl and aluminum to form Al2Me4Cl2 followed by treatment with sodium metal. The sodium metal and MeCl are not cheap as compared to ethane and hydrogen gases.
$\ce{2Al + CH3Cl -> Al2Cl2(CH3)4} \nonumber$
$\ce{Al2Cl2(CH3)4 + 6Na -> Al2(CH3)6 + 2Al + 6NaCl} \nonumber$ | textbooks/chem/Inorganic_Chemistry/Introduction_to_Organometallic_Chemistry_(Ghosh_and_Balakrishna)/03%3A_Organometallic_Chemistry_of_p-block_Elements/3.02%3A_Organometallic_Compounds_of_Boron_and_Aluminium.txt |
Learning Objectives
In this section you will learn the following
• Chemistry of gallium and indium.
• How to stabilize M—M multiple bonds.
$\ce{3Li4(C2H5)4 + 4GaCl3 -> 2LiCl + 4Ga(C2H5)3} \nonumber$
Trialkylgallium compounds are mild Lewis acids, so the corresponding metathesis reaction in ether produces the complex (C2H5)2OGa(C2H5)3. Similarly excess use of C2H5Li leads to the salt, Li[Ga(C2H5)4].
$\ce{Li4(C2H5)4 + GaCl3 -> 3LiCl + Li[Ga(C2H5)4]} \nonumber$
Alkylindium and alkylthalium compounds may be prepared similar to gallium analogs. InMe3 is monomeric in the gas phase and in the solid the bond lengths indicate that association is very weak. Partial hydrolysis of TlMe3 yields the linear (MeTiMe]+ion, which is isoelectronic and isostructural with HgMe2.
CpIn and CoTl exist as monomers in the gas phase but are associated in solids {Inert-pair effect is displayed for In and Tl}. CpTl is useful as a synthetic reagent in organometallic chemistry because it is not as highly reducing as NaCp.
Species of the type R4E2(single E-E bond) and [R4E2] - (with E-E bond order of 1.5) can be prepared for Ga and In with bulky R groups (R = (Me3Si)2CH, 2,4,6-iPr3C6H2), and reduction of [(2,4,6-iPr3C6H2)4Ga2] to [(2,4,6-iPr3C6H2)4Ga2] - is accompanied by a shortening of the Ga—Ga bond from 252-234 pm.
Using even bulkier substituents, it is possible to prepare gallium(I) compounds, RGa starting from GaI. No structural data are yet available for these monomers
(We are working on it).
Crystallized as dimer but reverts to monomer when dissolved in cyclohexane.
Interest in organometallic comounds of Ga, In and Tl is mainly because of their potential use as precursors to semiconducting materials such as GaAs and InP. Volatile compounds can be used in the growth of thin films by MOCVD (metal organic chemical vapor deposition) or MOVPE (metal organic vapor phase epitaxy) techniques. Precursors include appropriate Lewis base adducts of metal alkyls, e.g. Me3Ga.NMe3 and Me3In.PEt3. Thermal decomposition of gaseous precursors result in semiconductors (III-V semiconductors) which can be deposited in thin films.
$\ce{Me3Ga(g) + AsH3(g) ->[1000-1150K] GaAs(s) + 3CH4(g)} \nonumber$
III-V semiconductors: Derive their name from the old groups 13 and 15, and include AlAs, AlSb, GaP, GaAs, GaSb, InP, InAs and InSb. Off these GaAs is of the greatest commercial interest. Although Si is probably the most important commercial semiconductor, a major advandage of GaAs over Si is that the charge carrier mobility is much greater. This makes GaAs suitable for high-speed electronic devices.
Another important difference is that GaAs exhibits a fuly allowed electronic transition between valence and conduction bands (i.e. it is direct band gap semiconductor) whereas Si is an indirect band gap semiconductor. The consequence of difference is that GaAs (also other III-V types) are more suited than Si for use in optoelectronic devices, since light is emitted more efficiently. The III-Vs have important applications in light-emitting diodes (LEDs).
Problems:
1. Predict the structure of monomeric, Cp3Ga; polymeric Cp3In and CpIn.
Solution:
See the articles Organometallics 1985, 4, 751.
Inorg. Chem. 1972, 11, 2832.
Organometallics 1988, 7, 105.
2. The reaction of [(R3C)4Ga4] ( R = a bulky substituent) (i) with I2 in boiling hexane results in the formation of [(R3C)GaI]2(ii) and [(R3C)GaI2]2(iii). Draw the structure and state the oxidation state for (i) - (iii).
Solution:
3. The I2 oxidation of [(tBu}4In4] leads to the formation of the InII compound [(tBu}4In4I4] in which each indium atom retains a tetrahedral environment. Draw the correct structure.
Solution | textbooks/chem/Inorganic_Chemistry/Introduction_to_Organometallic_Chemistry_(Ghosh_and_Balakrishna)/03%3A_Organometallic_Chemistry_of_p-block_Elements/3.03%3A_Organometallic_Compounds_of_Gallium_and_Indium.txt |
Learning Objectives
In this lecture you will learn the following
• Olefin polymerization.
• Mechanism involved in polymerization process.
Ziegler Natta Polymerization Catalysts
Insertion of aluminum alkyls into olefins was studied by Ziegler. During the systematic investigation of olefin polymerization, Ziegler realized that the most effective catalyst is the combination of TiCl4/AlEt3which can polymerize ethylene at pressure as low as 1 bar. The application of Ziegler method to the polymerization of propylene and its establishment and the investigation of bulk properties was carried out by Natta and hence the methodology is called Ziegler-Natta process.
Important discovery: R3Al + Lewis acids.
In the absence of reaction mechanism with solid proof, it is presumed that the reaction is due to the heterogeneous catalysis in which fibrous TiCl3, alkylated on its surface is considered to be the active catalyst species.
3.05: Organosilicon and Organogermanium Compounds
Learning Objectives
In this lecture you will learn the following
• Organosilicon and organogermanium compounds.
• Compounds with Si=Si and Ge=Ge bonds.
Organosilicon and organogermanium compounds
Organosilicon compounds are extensively studied due to the wide range of commercial applications as water repellents, lubricants, and sealants. Many oxo-bridged organosilicon compounds can be synthesized. e.g. (CH3)3Si—O—Si(CH3)3 which is resistant to moisture and air.
The lone pairs on O are partially delocalized into vacant σ*- orbitals of Si, as a result the directionality of the Si-O bond is reduced making the structure more flexible.
This flexibility permits silicone elastomers to remain rubber-like down to very low temperature.
Delocalization also accounts for low basicity of an O atom attached to silicon as the electrons needed for the O atom to act as a base are partially removed.
The planarity of N(SiH3)3 is also explained by the delocalization of the lone pair on N which makes it very weakly basic.
$\ce{nMeCl + Si/Cu -> Me_{n}SiCl_{4-n}} \nonumber$
$\ce{SiCl4 + 4RLi -> R4Si} \nonumber$
$\ce{SiCl4 + RLi -> RSiCl3} \nonumber$
$\ce{SiCl4 + 2RMgCl -> R2SiCl2 + 2MgCl2} \nonumber$
$\ce{Me2SiCl2 + tBuBi -> tBuMe2SiCl + LiCl}|] Si—C bonds are relatively strong (bond enthalpy is 318 kJ mol-1) and R4Si derivatives possess high thermal stabilities. Et4Si on chlorination gives (ClCH2CH2)4Si, in contrast to the chlorination of R4Ge or R4Sn which yields RnGeCl4-n or RnSnCl4-n. Me2SiCl2 on hydrolysis produce silicones. \[\ce{Me3SiCl + NaCp -> (η^{1}-Cp)SiHMe3} \nonumber$
1-C5Me5) 2SiBr2 on treatment with anthracene/potassium gives Cp*2Si
Solid state structure of Cp*2Si consists of two independent molecules which differ in the relative orientations of the Cp rings.
In one molecule, they are parallel and staggered whereas in the other, they are tilted with an angle of 167°at Si.
The reaction between R2SiCl2 and alkali metal or alkali naphthalides give cyclo-(R2Si)n by loss of Cl-and Si—Si bond formation.
Bulky R groups favour small rings [e.g. (2,6-Me2C6H3)6Si3 and tBu6Si3] while smaller R groups encourage the formation of large rings [Me12Si6, Me14Si7 and Me32Si16]
$\ce[Ph2SiCl2 + Li(SiPh2)5Li -> cyclo-Ph12Si6 + 2LiCl} \nonumber$
Bulky substituents stabilize R2Si=SiR2 compounds. The sterically demanding 2,4,6-iPr3C6H2 provided first example of compound containing conjugated Si=Si bonds.
Has s-cis configuration in both solution and the solid state.
Similar germanium compounds are also known
*The spatial arrangement of two conjugated double bonds about the intervening single bond is described as s- cis if synperiplanar and s-trans if antiperiplanar.
3.06: Organotin and Organolead Compounds
Learning Objectives
In this section you will learn the following
• Organotin and organolead compounds and their preparation.
• Bonding in tin compounds with Sn=Sn double bonds.
• Uses and environmental issues with tin compounds.
• Reactivity of tetraethyl lead.
• Structural features of organolead compounds.
Organotin and organolead compounds
Preparation of Sn(IV) derivatives
$\ce{3SnCl3 + 4R3Al ->[R'2O] 3R4Sn + 4AlCl3} \nonumber$
$\ce{R4Sn + SnCl4 -> R3SnCl + RSnCl3 ->[500K] 2R2SnCl2} \nonumber$
$\ce{SnCl4 + 4RMgBr -> R4Sn + 4MgBrCl} \nonumber$
$\ce{SnCl2 + Ph2Hg -> Ph2SnCl2 + Hg} \nonumber$
Tin(II) organometallics of the type R2Sn, containing Sn-C σ-bonds, are stabilized only if R is sterically demanding.
$\ce{SnCl2 + 2Li[(Me3Si)2CH] -> [(Me3Si)2CH]2Sn} \nonumber$
(monomeric in solution and dimeric in solid state). But the dimer does not possess a planar Sn2R4framework unlike an analogous alkene, and Sn—Sn bond distance (267 pm) is shorter than a normal Sn—Sn single bond (276 pm).
Sn2R4 has a trans bent structure with a weak Sn=Sn double bond
Look into the reactions of R3SnCl with various reagents to form useful tin containing starting materials
The first organotin(II) hydride was reported only in 2000.
Shows dimeric structure in the solid state containing hydride bridges (Sn-Sn = 312 pm).
Commercial uses and environmental problems
Organotin(II) compounds find wide range of applications due to their catalytic and biocidal properties.
• nBu3SnOAc is an effective fungicide and bactericide and also a polymerization catalyst.
• nBu2Sn(OAc)2 is used as a polymerization catalyst and a stabilizer for PVC.
• nBu3SnOSnnBu3 is algicide, fungicide and wood-preserving agent.
• nBu3SnCl is a bactericide and fungicide.
• Ph3SnOH used as an agricultural fungicide for crops such as potato, sugar beet and peanuts.
• The cylic compound (nBu2SnS)3 is used as a stabilizer for PVC.
Tributyltin derivatives have been used as antifouling agents, applied to the underside of ships’ hulls to prevent the build-up of, for example, barnacles.
Global legislation now bans or greatly restricts the use of organotin-based anti-fouling agents on environmental grounds. Environmental risks associated with the uses of organotin compounds as pesticides, fungicides and PVC stabilizers are also a cause for concern.
*A barnacle is a type of arthropod belonging to infraclass Cirripedia in the sub-phylum Crustacea, and is hence related to crabs and lobsters.
Organolead compounds
Tetraethyllead
$\ce{4NaPb + 4EtCl -> Et4Pb + 3Pb + 4NaCl} [at\: 373K\: in\: an\: autoclave] \nonumber$
Laboratory Scale,
$\ce{2PbCl3 + 4RMgBr ->[Et2O][-4MgBrCl] 2(R2Pb) -> R4Pb + Pb} \nonumber$
Thermolysis leads to radical reactions.
$\ce{Et4Pb -> Et3Pb + Et} \nonumber$
$\ce{2Et -> n-C4H10} \nonumber$
$\ce{Et3Pb + Et -> C2H4 + Et3PbH} \nonumber$
$\ce{Et3Pb + Et4Pb -> H2 + Et3Pb + Et3PbCH2CH2} \nonumber$
Tetraalkyl and tetraaryl lead compounds are inert with respect to attack by air and water at room temperature. WHY ????
Me3PbCl consists of linear chain
Solid state structure of Cp2Pb shows polymeric nature, but in the gas phase, discrete Cp2Pb molecules are present which possess the bent structure similar to silicon analogue.
R2Pb=PbR2 are similar to analogues tin compounds
Problems
1. Find out the structures of (Me3SiCH2)3SnF and Me2SnF2
Solution: use VSEPR theory | textbooks/chem/Inorganic_Chemistry/Introduction_to_Organometallic_Chemistry_(Ghosh_and_Balakrishna)/03%3A_Organometallic_Chemistry_of_p-block_Elements/3.04%3A_Zeigler_Natta_Polymerization_Catalysts.txt |
Learning Objectives
In this section you will learn the following
• Organoaresnic and organoantimony compounds.
• Preparation and reactivity of pentavlent As and Sb compounds.
Organic chemistry of non-metal phosphorus, metalloids such as arsine and antimony along with metallic element bismuth is termed as organoelement chemistry. The importance given to organoarsenic compounds earlier due to their medicinal values was waded out after antibiotics were discovered and also their carcinogenic and toxic properties were revealed. Also, the synthetically important organometallic compounds of group 13 and 14 masked the growth of group 15 elements. However, the organoelement compounds of phosphorus, arsenic and antimony find usefulness as ligands in transition metal chemistry due to their σ-donor and π-acceptor abilities which can be readily tuned by simply changing the substituents. These donor properties are very useful in tuning them as ligands to make suitable metal complexes for metal mediated homogeneous catalysis. Although organoelement compounds can be formed in both +3 (trivalent and tricoordinated) and +5(pentavalent and tetra or pentacoordinated) oxidation states, trivalent compounds are important in coordination chemistry.
For organoelement compounds of group 15, the energy of E—C bond decreases in the order, E = P > As > Sb > Bi, and in the same sequence E—C bond polarity increases.
Organometallic compounds of As(V) and Sb(V)
Due to the strong oxidizing nature of pentahalides, the direct alkylation or arylation to generate ER5 is not feasible, but can be prepared in two steps.
A few representative methods of preparation are given below:
$\ce{ Me3As ->[Cl2] Me3AsCl2 ->[MeLi][Et2O] Me5As} \nonumber$
$\ce{Ph3Sb + PhI -> Ph4SbI ->[PhLi][-LI] Ph5Sb} \nonumber$
$\ce{Ph3Bi ->[SO2Cl2][-SO2] Ph3BiCl2 ->[PhMgX][-MgXCl] Ph5Bi} \nonumber$
Structures and properties
Pentaalkyl or pentaaryl derivatives are moderately thermally stable. On heating above 100°C, they form trivalent compounds as shown below:
$\ce{Me5As ->[T>100°] Me3As + CH4 + CH2CH2} \nonumber$
$\ce{Ph5Sb ->[T>200°] Ph3Sb + Ph-Ph} \nonumber$
Reaction with water,
$\ce{Me5As+ H2O -> Me4AsOH + MeH} \nonumber$
Pentavalent compounds readily form “tetrahedral onium” cations and “octahedral and hexacoordinatged ate” anions.
$\ce{Ph5E + BPh3 -> [Ph4E][BPh4]} (E=As, Sb, Bi) \nonumber$
$\ce{Ph5E + LiPh -> Li[EPh6]} \nonumber$
In solid state, Ph5As adopts trigonal bipyramidal geometry, whereas Ph5Sb prefers square based pyramidal geometry although the energy difference between the two is marginal.
The salts of the type [R4E]+ adopt tetrahedral geometry, whereas hexacoordinated anions [R6E]-assume octahedral geometry.
Mixed organo-halo compounds of the type RnEX5-n adopt often dimeric structures due to the presence of lone pairs of electrons on X which can readily coordinate to the second molecule. The following structural types can be anticipated.
The thermal stability of RnEX5-n decreases with decreasing ‘n’. Thermal reactions are essentially the reverse reactions of addition reactions used in the preparation of R5E.
$\ce{R3SbX2 ->[\Delta T] R2SbX + RX} \nonumber$
$\ce{Ph3AsCl2 ->[CO_{2}][100°C] Ph2AsCl + Cl2} \nonumber$
$\ce{Me2AsCl3 ->[50°C] MeAsCl2 + MeCl} \nonumber$
4.02: Organometallic Compounds of As(III) and Sb(III)
Learning Objectives
In this lecture you will learn the following
• Preparation of trivalent compounds.
• Mono and bis derivatives.
• Reaction of organo arsenic and antimony compounds.
• Structural features of organolead compounds.
Direct synthesis
Mono- derivatives
$\ce{2As + 3MeBr ->[\Delta T][Cu] Me2AsBr + MeAsBr2} \nonumber$
$\ce{Me2AsBr + PhLi -> Me2AsPh} \nonumber$
$\ce{MeAsBr2 + 2PhLi -> MeAsPh2} \nonumber$
$\ce{EX3 + 3RMgX -> R3E + 3MgX2} \nonumber$
$\ce{EX3 + 3RLi -> R3E + 3LiX} \nonumber$
Bis derivatives:
In a similar way, a variety of bisphosphines and arsines can be generated.
Reactions of trialkyl derivatives, R3E
The transition metal chemistry of R3E, phosphines, arsines or stibines has been extensively studied because of their distinct donor and acceptor properties. Among them, the phosphines or tertiary phosphines (R3P) are the most valuable ligands in metal mediated homogeneous catalysis. Interestingly, the steric and electronic properties can be readily tuned by changing the substituents on phosphorus atoms. Chapter 16 is fully dedicated to the chemistry of phosphines.
Properties
Trialkyl derivatives are highly air-sensitive liquids with low boiling points and some of them are even pyrophyric. Triphenyl derivatives are solids at room temperature and are moderately stable and oxidizing agents such as KMnO4, H2O2 or TMNO are needed for oxidation to form Ph3E=O.
Cyclic and acyclic derivatives containing E—E bonds
E—E single bonds:
The E—E bond energies suggest that they do not have greater stability and the stability decreases down the group.
The simplest molecules include Ph2P—PPh2, Me2As—AsMe2 prepared by coupling reactions:
$\ce{Me2AsH + Me2AsCl ->[ ][-HCl] Me2As-AsMe2} \nonumber$
$\ce{2Ph2BiCl ->[Na, NH3] Ph2Bi-BiPh2} \nonumber$
The weakness of E—E bonds accounts for many interesting reactions and a few of such reactions are listed below:
Cyclic and polycyclic derivatives can be prepared by employing any of the following methods:
Problems:
1. Confirm that the octahedral structure of [Ph6Bi]- is consistent with VSEPR theory.
Solution:
Octahedral similar to PF6-
5 (Bi valence electrons) + 6 (each Ph ) + 1 (-ve charge) = 12 electrons
i.e. six pairs, octahedral geometry
2. Comment on the stability of BiMe3 and Al2(iBu)6 with respect to their thermal decomposition and give chemical equations for their decomposition.
Solution:
Similar to other heavy p-block elements, Bi—C bonds are weak and readily undergo homolytic cleavage. The resulting methyl radicals will react with other radicals or form ethane.
$\ce{2BiMe3 -> 2Bi + 3CH3-CH3} \nonumber$
The Al2(iBu)6 dimer readily dissociates. At elevated temperature dissociation is followed by β-hydrogen elimination. This type of elimination is common for organometallic compounds that have alkyl groups with β-hydrogens, can form stable M—H bonds, and can provide a coordination site on the central metal.
The decomposition reaction is:
$\ce{Al2(iBu)6 ->[\Delta] 2Al(iBu)3 -> [\Delta \Delta] Al(iBu)2H + (CH3)2C=CH2} \nonumber$
1. Using a suitable Grignard reagent, how would you prepare (i) MeC(Et)(OH)Ph; (ii) AsPh3.
Solution
1. Add a Grignard reagent to a C=O bond, then acidify.
Several possibilities, e.g. $\ce{Me-C(O)-Et + PhMgBr -> Me-C(OMgBr)(Et)(Ph) -> MeC(Et)(OH)Ph \: or \: Me-C(O)-Ph + EtMgBr -> etc} \nonumber$
2. $\ce{AsCl3 + 3PhMgBr -> AsPh3 + 3MgBrCl} \nonumber$ | textbooks/chem/Inorganic_Chemistry/Introduction_to_Organometallic_Chemistry_(Ghosh_and_Balakrishna)/04%3A_Organoelement_Compounds_of_Group_15/4.01%3A_Organometallic_Compounds_of_As%28V%29_and_Sb%28V%29.txt |
Learning Objectives
In this lecture you will learn the following
• Classification of ligands.
• Nature of bonding in phosphines.
• Steric and electronic properties of phosphines.
• Bonding in phosphines and CO.
• Cone angle and its application in catalysis.
Classification of Ligands by donor atoms
Ligand is a molecule or an ion that has at least one electron pair that can be donated. Ligands may also be called Lewis bases; in terms of organic chemistry, they are ‘nucleophiles’.
Metal ions or molecules such as BF3 (with incomplete valence electron shells (electron deficient) are called Lewis acids or electrophiles).
Why do molecules like H2O or NH3 give complexes with ions of both main group and transition metals. E.g [Al(OH2)6]3+ or [Co(NH3)6]3+
Why other molecules such as PF3 or CO give complexes only with transition metals.
Although PF3 or CO give neutral molecules such as Ni(PF3)4 or Ni(CO)4 or Cr(CO)6.
Why do, NH3, amines, oxygen donors, and so on, not give complexes such as Ni(NH3)4.
Classical or simple donor ligands
Act as electron pair donors to acceptor ions or molecules, and form complexes of all types of Lewis acids, metal ions or molecules.
Non-classical ligands, π-bonding or π-acid ligands: Form largely with transition metal atoms.
In this case special interaction occurs between the metals and ligands
These ligands act as both σ-donors and π-acceptors due to the availability of empty orbitals of suitable symmetry, and energies comparable with those of metal t2g (non-bonding) orbitals.
e.g. Consider PR3 and NH3: Both can act as bases toward H+, but P atom differs from N in that PR3 has σ* orbitals of low energy, whereas in N the lowest energy d orbitals or σ* orbitals are far too high on energy to use.
Consider CO that do not have measurable basicity to proton, yet readily reacts with metals like Ni that have high heats of atomization to give compounds like Ni(CO)4.
Ligands may also be classified electronically depending upon how many electrons that they contribute to a central atom. Atoms or groups that can form a single covalent bond are one electron donors.
EXAMPLES: F, SH, CH3 etc.,
Compounds with an electron pair are two-electron donors
EXAMPLE: NH3, H2O, PR3 etc.,
Steric factors in phosphines (Tolman’s cone angle)
Cone angle is very useful in assessing the steric properties of phosphines and their coordination behavior.
The electronic effect of phosphines can be assessed by IR and NMR spectroscopic data especially when carbonyls are co-ligands. In a metal complex containing both phosphines and carbonyl, the ν(CO) frequencies would reveal the σ-donor or π–acceptor abilities of phosphines. If the phosphines employed are strong σ-donors, then more electron density would move from M (t2g orbitals)- π*(CO) and as a result, a lowering in the ν(CO) is observed. In contrast, if a given phosphine is a poor σ-donor but strong π -acceptor, then phosphine(σ*-orbitals) also compete with CO for back bonding which results in less lowering in ν(CO) frequency.
Another important aspect is the steric size of PR3 ligands, unlike in the case of carbonyls, which can be readily tuned by changing R group. This is of great advantage in transition metal chemistry, especially in metal mediated catalysis, where stabilizing the metals in low coordination states is very important besides low oxidation states. This condition can promote oxidative addition at the metal centre which is an important step in homogeneous catalysis. The steric effects of phosphines can be quantified with Tolman’s cone angle.
Cone angle can be defined as a solid angle at metal at a M—P distance of 228 pm which encloses the van der Waal’s surfaces of all ligand atoms or substituents over all rotational orientations. The cone angles for most commonly used phosphines are listed in the following table.
Phosphine
Cone Angle (°)
PH3
87
PF3
104
P(OMe)3
107
PMe3
118
PMe2Ph
122
PEt3
132
PPh3
145
PCy3
170
P(But)3
182
P(mesityl)3
212
Phosphines with different cone angles versus coordination number for group 8 metals:
ML4
ML3
ML2
(Me3P)4Ni
(Me3P)4Pd
(Me3P)4Pt
(Ph3P)3Pt
(tert-Bu3P)2Pt
Tolman Angle and Catalysis
Sterically demanding phosphine ligands can be used to create an empty coordination site (16 VE complexes) which is an important trick to fine tune the catalytic activity of phosphine complexes. | textbooks/chem/Inorganic_Chemistry/Introduction_to_Organometallic_Chemistry_(Ghosh_and_Balakrishna)/04%3A_Organoelement_Compounds_of_Group_15/4.03%3A_Phosphines.txt |
• 5.1: Organometallic Compounds of Zinc and Cadmium
The synthetically important organometallic compounds of group 13 and 14 masked the growth of group 15 elements. However, the organoelement compounds of phosphorus, arsenic and antimony find usefulness as ligands in transition metal chemistry due to their σ-donor and π-acceptor abilities which can be readily tuned by simply changing the substituents.
• 5.2: Organometallic Compounds of Mercury
Dialkylmercury compounds are very versatile starting materials for the synthesis of many organometallic compounds of more elctropositive metals by transmetallation. However, owing to high toxicity of alkylmercury compounds, other synthons are preferred. In striking contrast to the high sensitivity of dimethylzinc to oxygen, dimethylmercury survives exposure to air.
05: Group 12 Elements
Learning Objectives
In this section you will learn the following
• Organometallic compounds of zinc and cadmium.
• Structural features of organozinc compounds
Organometallic compounds of zinc and cadmium
Dialkyl compounds of Zn, Cd and Hg do not associate through alkyl bridges.
Dialkylzinc compounds are only weak Lewis acids, organocadmium compounds are even weaker, and organomercury compounds do not act as Lewis acids except under special circumstances.
The Group 12 metals form linear molecular compounds, such as ZnMe2, CdMe2 and HgMe2, that are not associated in solid, liquid or gaseous state or in hydrocarbon solution.
They form 2c, 2e bonds. Unlike Be and Mg analogs, they do not complete their valence shells by association through alkyl bridges. The bonding in these molecules are similar to d10 metals such as CuI, AgI and AuI with linear geometry ([N≡C-M-C≡N]-, M = Ag or Au). This tendency is sometimes rationalized by invoking pd hybridization in the M+ ion, which leads to orbitals that favor linear attachment of ligands (similar to spd hybridization).
The preference for the linear coordination may be due to the similarity in energy of the outer ns, np and (n-1)d orbitals, which permits the formation of collinear spd hybrids.
The hybridization of s, pz and dz2 with the choice of phases shown here produces a pair of collinear orbitals that can be used to form strong σ-bonds.
Organozinc and organocadmium compounds
Convenient route is metathesis with alkylaluminium or alkyllithium compounds.
With alkyllithium compounds it is the electronegativity which is decisive, whereas between Al and Zn it is hardness considerations correctly predict the formation of softer ZnCH3 and harder AlCl pairs.
$\ce{ZnCl2 + Al2Me6 -> ZnMe2 + Al2Cl2Me4} \nonumber$
Alkylzinc compounds are pyrophoric and readily hydrolyzed, whereas alkylcadmium compounds react more slowly with air. Due to mild Lewis acidity, dialkylzinc and dialkylcadimum compounds form stable complexes with amines, especially with chelating amines.
The Zn—C has greater carbanionic character than the Cd—C bond.
For example, addition of alkylzinc compounds across the carbonyl group of a ketone:
$\ce{ZnMe2 + (CH3)2C=O -> (CH3)2C-O-ZnCH3} \nonumber$
This reaction do not proceed with the less polar alkylcadmium or alkylmercury compounds, but organolithium, organomagnesium and organoaluminium compounds can promote this reaction readily since all of which contain metals with lower electronegativity than zinc.
Interestingly, the cyclodipentadienyl compounds are structurally unusual. CpZnMe is monomeric in the gas phase with a pentahapto Cp group.
In the solid state it is associated in a zig-zag chain, each Cp group being pentahapto with respect to two Zn atoms.
Problems:
1. Do you think that the following reaction proceeds? If so, why and how?
$\ce{ZnCl2 + Al2Me6 -> ZnMe2 + Al2Cl2Me4} \nonumber$
Solution
Al2Me6 being an electron deficient molecule readily exchanges two methyl groups with zinc for two chloride ions. Since chloride ions have sufficient electron in their valence shell act as four electron donor through bridging coordination mode. Al2Cl2Me4 is no longer an electron deficient molecule.
5.02: Organometallic Compounds of Mercury
Learning Objectives
In this section you will learn the following
• Dialkymercury preparation.
• Mercury toxicity.
• Mercury poisoning.
Consider this reaction that proceeds due to both electronegativity and hardness considerations.
$\ce{2RMgX + HgX2 -> HgR2 + MgX2} \nonumber$
Dialkylmercury compounds are very versatile starting materials for the synthesis of many organometallic compounds of more electropositive metals by transmetallation. However, owing to high toxicity of alkylmercury compounds, other synthons are preferred. In striking contrast to the high sensitivity of dimethylzinc to oxygen, dimethylmercury survives exposure to air.
Mercury Toxicity
The toxicity of mercury arises from the very high affinity of the soft Hg atom for sulfhydryl (—SH) groups in enzymes. Simple mercury-sulfur compounds have been studied as potential analogs of natural systems. The Hg atoms are most commonly four-coordinated, as in [Hg2(SMe)6]2-.
Mercury poisoning was a serious concern even from early days. Issac Newton, Alfred Stock worked in the early 20th century. Later in 60s awareness came following the incidence of brain damage and death it caused among the inhabitants in Minamata, Japan. Mercury from a plastic company was allowed to escape into a bay where it found its way into fish that were later eaten. Research has shown that bacteria found in sediments are capable of methylating mercury, and that species such HgMe2 and [HgCH3]+ enter the food chain because they readily penetrate cell walls. The bacteria appear to produce HgMe2 as a means of eliminating toxic mercury ions through their cell walls and into the environment. | textbooks/chem/Inorganic_Chemistry/Introduction_to_Organometallic_Chemistry_(Ghosh_and_Balakrishna)/05%3A_Group_12_Elements/5.01%3A_Organometallic_Compounds_of_Zinc_and_Cadmium.txt |
• 6.1: 18 Valence Electron Rule
Many transition metal organometallic compounds are primarily of interest from the prospectives of chemical catalysis. Unlike the main group organometallic compounds, which use mainly ns and np orbitals in chemical bonding, the transition metal compounds regularly use the (n−1)d, ns and np orbitals for chemical bonding. Partial filling of these orbitals thus render these metal centers both electron donor and electron acceptor abilities.
• 6.2: Synthesis and Stability
Ligands play a vital role in stabilizing transition metal complexes. The stability as well as the reactivity of a metal in its complex form thus depend upon the number and the type of ligands it is bound to. In this regard, the organometallic carbon based ligands come in diverse varieties displaying a wide range of binding modes to a metal. In general, the binding modes of the carbon-derived ligands depend upon the hybridization state of the metal bound carbon atom.
06: General Properties of Transition Metal Organometallic Complexes
Learning Objectives
In this section you will learn the following
• Have an insight about the stability of the transition metal complexes with respect to their total valence electron count.
• Be aware of the transition metal complexes that obey or do not obey the 18 Valence Electron Rule.
• Have an appreciation of the valence electron count in the transition metal organometallic complexes that arise out of the metal-ligand orbital interactions.
The transition metal organometallic compounds exhibit diverse structural variations that manifest in different chemical properties. Many of these transition metal organometallic compounds are primarily of interest from the prospectives of chemical catalysis. Unlike the main group organometallic compounds, which use mainly ns and np orbitals in chemical bonding, the transition metal compounds regularly use the (n−1)d, ns and np orbitals for chemical bonding (Figure \(1\)). Partial filling of these orbitals thus render these metal centers both electron donor and electron acceptor abilities, thus allowing them to participate in σ-donor/π-acceptor synergic interactions with donor-acceptor ligands like carbonyls, carbenes, arenes, isonitriles and etc,.
The 18 Valence Electron (18 VE) Rule or The Inert Gas Rule or The Effective Atomic Number (EAN) Rule: The 18-valence electron (VE) rule states that thermodynamically stable transition metal compounds contain 18 valence electrons comprising of the metal d electrons plus the electrons supplied by the metal bound ligands. The counting of the 18 valence electrons in transition metal complexes may be obtained by following either of the two methods of electron counting, (i). the ionic method and (ii). the neutral method. Please note that a metal-metal bond contributes one electron to the total electron count of the metal atom. A bridging ligand donates one electron towards bridging metal atom.
Example \(1\)
Ferrocene Fe(C5H5)2
Example \(2\)
Mn2(CO)10
Transition metal organometallic compounds mainly belong to any of the three categories.
1. Class I complexes for which the number of valence electrons do not obey the 18 VE rule.
2. Class II complexes for which the number of valence electrons do not exceed 18.
3. Class III complexes for which the valence electrons exactly obey the 18 VE rule.
The guiding principle which governs the classification of transition metal organometallic compounds is based on the premise that the antibonding orbitals should not be occupied; the nonbonding orbitals may be occupied while the bonding orbitals should be occupied.
Class I
In class I complexes, the Δo splitting is small and often applies to 3d metals and σ ligands at lower end of the spectrochemical series. In this case the t2g orbital is nonbonding in nature and may be occupied by 0−6 electrons (Figure \(2\)). The eg* orbital is weakly antibonding and may be occupied by 0−4 electrons. As a consequence, 12−22 valence electron count may be obtained for this class of compounds. Owing to small Δtetr splitting energy, the tetrahedral transition metal complexes also belongs to this class.
Class II
In class II complexes, the Δo splitting is relatively large and is applicable to 4d and 5d transition metals having high oxidation state and for σ ligands in the intermediate and upper range of the spectrochemical series. In this case, the t2g orbital is essentially nonbonding in nature and can be filled by 0−6 electrons (Figure 3). The eg* orbital is strongly antibonding and is not occupied at all. Consequently, the valence shell electron count of these type of complexes would thus be 18 electrons or less.
Class III
In class III complexes, the Δo splitting is the largest and is applicable to good σ donor and π acceptor ligands like CO, PF3, olefins and arenes located at the upper end of the spectrochemical series.
The t2gorbital becomes bonding owing to interactions with ligand orbitals and should be occupied by 6 electrons. The eg* orbital is strongly antibonding and therefore remains unoccupied.
Problems
State the oxidation state of the metal and the total valence electron count of the following species.
1. V(C2O4)33−
Ans: +3 and 14
2. Mn(acac)3
Ans: +3 and 16
3. W(CN)83−
Ans: +5 and 17
4. CpMn(CO)3
Ans: 0 and 18
5. Fe2(CO)9
Ans: 0 and 18
Self Assessment test
State the oxidation state of the metal and the total valence electron count of the following species.
1. TiF62-
Ans: +4 and 12
2. Ni(en)32+
Ans: +2 and 20
3. Cu(NH3)62+
Ans: +2 and 21
4. W(CN)84-
Ans: +4 and 18
5. CH3Co(CO)4
Ans: 0 and 18
Summary
The transition metal complexes may be classified into the following three types.
1. The ones that do not obey the 18 valence electron rule are of class I type
2. The ones that do not exceed the 18 valence electron rule are of class II and
3. The ones that strictly follow the 18 valence electron rule.
Depending upon the interaction of the metal orbitals with the ligand orbitals and also upon the nature of the ligand position in spectrochemical series, the transition metal organometallic compounds can form into any of the three categories. | textbooks/chem/Inorganic_Chemistry/Introduction_to_Organometallic_Chemistry_(Ghosh_and_Balakrishna)/06%3A_General_Properties_of_Transition_Metal_Organometallic_Complexes/6.01%3A_18_Valence_Electron_Rule.txt |
Learning Objectives
In this lecture you will learn the following
• Understand the role lead by ligands in stabilizing organometallic transition metal complexes.
• Know about various synthetic methods available for preparing the organometallic transition metal complexes.
• Understand the various factors like β-elimination and other bimolecular decomposition pathways that contribute to the observed instability of the organometallic transition metal complexes.
• Obtain insight about making stable organometallic transition metal complexes by suppression of the destabilizing factors mentioned.
Ligands play a vital role in stabilizing transition metal complexes. The stability as well as the reactivity of a metal in its complex form thus depend upon the number and the type of ligands it is bound to. In this regard, the organometallic carbon based ligands come in diverse varieties displaying a wide range of binding modes to a metal. In general, the binding modes of the carbon-derived ligands depend upon the hybridization state of the metal bound carbon atom. These ligands can thus bind to a metal in many different ways as depicted below. Lastly, these ligands can either be of (a) purely σ−donor type, or depending upon the capability of the ligand to form the multiple bonds may also be of (b) a σ−donor/π−acceptor type, in which the σ−interaction is supplemented by a varying degree of π−interaction.
Preparation of transition metal-alkyl and transition metal-aryl complexes
The transition metal−alkyl and transition−metal aryl complexes are usually prepared by the following routes discussed below,
a. Metathesis
This involves the reactions of metal halides with organolithium, organomagnesium, organoaluminium, organotin and organozinc reagents.
Of the different organoalkyl compounds listed above, the organolithium and organomagnesium compounds are strongly carbanionic while the remaining main group organometallics like the organoalkyl, organozinc and organotin reagents are relatively less carbanionic in nature. Thus, the main group organometallic reagents have attenuated alkylating power, that can be productively used in partial exchange of halide ligands.
$\ce{TiCl4 ->[Al2Me6] MeTiCl3} \nonumber$
$\ce{NbCl5 ->[ZnMe2] Me2NbCl3} \nonumber$
b. Alkene insertion or Hydrometallation
As the name implies, this category of reaction involves an insertion reaction between metal hydride and alkene as shown below. These type reactions are relevant to certain homogeneous catalytic processes in which insertion of an olefin to M−H bond is often observed.
c. Carbene insertion
This category represents the reaction of metal hydrides with carbenes.
d. Metallate alkylation reaction
This category represents the reaction of carbonylate anions with alkyl halides as shown below.
e. Metallate acylation reaction
This category involves the reaction of carbonylate anions with acyl halides
f. Oxidative addition reaction
Many unsaturated 16 VE transition metal complexes having d8 or d10 configuration undergo oxidative addition reactions with alkyl halides. The oxidative addition reactions proceed with the oxidation state as well as coordination number of the metal increasing by +2.
g. Addition reaction
This category involves the reaction of an activated metal bound olefin complex with a nucleophile as shown below.
Thermodynamic Stability and Kinetic Lability
The transition metal organometallic compounds are often difficult to synthesize under ordinary laboratory conditions and require stringent experimental protocols involving the exclusion of air and moisture for doing so. As a consequence, many homoleptic binary transition metal−alkyl and transition metal−aryl compounds like, Et2Fe or Me2Ni cannot be made under normal laboratory conditions. More interestingly, most of the examples of transition metal−aryl and transition metal−alkyl compounds, known in the literature, invariably contain additional ligands like η5-C5H5, CO, PR3 or halides.
For example,
Transition metal−carbon (TM−C) bond energy values are important for understanding the instability of transition metal organometallic compounds. In general, the TM−C bonds are weaker than the transition metal−main group element (TM−MGE) bonds (MGE = F, O, Cl, and N) and more interestingly so, unlike the TM−MGE bond energies, the TM−C bond energy values increase with increasing atomic number. The steric effects of the ligands also play a crucial role in influencing the TM−C bond energies and thus have to be given due consideration.
Contrary to the popular belief, the difficulty in obtaining transition metal−aryl and transition metal−alkyl complexes does primarily arise from the thermodynamic reasons but rather the kinetic ones. β−elimination is by far the most general decomposition mechanism that contribute to the instability of transition metal organometallic compounds. β−elimination results in the formation of metal hydrides and olefin as shown below.
β−elimination can also be reversible as shown below.
The instability of transition metal organometallic compounds can arise out of kinetic lability like in the case of the β−elimination reactions that trigger decomposition of these complexes. Thus, the suppression of the decomposition reactions provides a viable option for the stabilization of the transition metal organometallic complexes. The β−elimination reactions in transition metal organometallic complexes may be suppressed under any of the following three conditions.
a. Formation of the leaving olefin becomes sterically or energetically unfavorable
In the course of β−elimination, this situation arises when the olefinic bond is formed at a bridgehead carbon atom or when a double bond is formed with the elements of higher periods. For instance, the norbornyl group is less prone to decomposition by β−elimination because that would require the formation of olefinic double bond at a bridgehead carbon atom in the subsequent olefin, i.e. norbornene, and which is energetically unfavorable.
b. Absence of β−hydrogen atom in organic ligands
Transition metal bound ligands that do not possess β−hydrogen cannot decompose by β−elimination pathway and hence such complexes are generally more stable than the ones containing β−hydrogen atoms. For example, the neopentyl complex, Ti[CH2C(CH3)3]4 (m.p 90 °C), and the benzyl complex, Zr(CH2Ph)4 (m.p. 132 °C), exhibit higher thermal stability as both of the neopentyl and benzyl ligands lack β−hydrogens.
c. Central metal atom is coordinatively saturated
Transition metal organometallic complexes in which the central metal atom is coordinatively saturated tend to be more stable due to the lack of coordination space available around the metal center to facilitate β−elimination reaction or other decomposition reactions. Thus, the absence of free coordination sites at the metal is crucial towards enhancing the stability of the transition metal organometallic complexes. For example, Ti(Me)4, which is coordinatively unsaturated can undergo a bimolecular decomposition reaction via a binuclear intermediate (A), is unstable and exhibits a decomposition temperature of –40 °C. On the contrary, Pb(Me)4, that cannot undergo decomposition by such bimolecular pathway, is more stable and distills at 110 °C at 1 bar atmospheric pressure.
The Ti(Me)4 decomposes by dimerization involving the formation of Ti−C (3c−2e) bonds. For Pb(Me)4, such bimolecular decomposition pathway is not feasible, as being a main group element it has higher outer d orbital for extending the coordination number. If the free coordination site of Ti(Me)4 is blocked by another ligand, as in [(bipy)Ti(Me)4], then the thermal stability of the complex, [(bipy)Ti(Me)4], increased significantly. Other bidentate chelating ligands like bis(dimethylphosphano)ethane (dmpe) also serve the same purpose.
Coordinative saturation thus brings in kinetic stabilization in complexes. For example, Ti(Me)4is extremely reactive as it is coordinatively unsaturated, while W(Me)4 is relatively inert for reasons of being sterically shielded and hence, coordinatively saturated. Thus, if all of the above discussed criteria for the suppression of β-elimination are taken care of, then extremely stable organometallic complexes can be obtained like the one shown below.
Problems
1. Arrange the following compounds in the order of their stability.
1. Ti(Et)4
2. Ti(Me)4 and
3. Ti(6-norbornyl)4
Ans: Ti(Et)4 < Ti(Me)4 < Ti(6-norbornyl)4
2. Predict the product of the reaction given below.
$\ce{(BuP)CuCH2CD2C2H5 -> } \nonumber$
Ans: Equi molar amounts of (Bu3P)CuD and CH2=CDC2H5
3. Will the compound β-eliminate,
$\ce{PtH(C≡CH)L2 ->[\Beta -elimination] } \nonumber$
(a). readily, (b). slowly and (c). not at all.
Explain your answer with proper reasoning.
Ans: Not at all as the ß-hydrogens are pointing away from the metal and cannot participate in ß-elimination recation.
Self Assessment test
1. Write the product(s) of the reactions.
Ans:
Summary
Ligands assume a pivotal role in the stabilization of the organometallic transition metal complexes. There are several methods available for the preparation of the organometallic transition metal complexes. The observed instability of the organometallic transition metal complexes can be attributed to two main phenomena namely β-elimination and bimolecular decomposition reaction that severely undermine the instability of these complexes. The suppression of these decomposition pathway thus pave way for obtaining highly stable organometallic transition metal complexes. | textbooks/chem/Inorganic_Chemistry/Introduction_to_Organometallic_Chemistry_(Ghosh_and_Balakrishna)/06%3A_General_Properties_of_Transition_Metal_Organometallic_Complexes/6.02%3A_Synthesis_and_Stability.txt |
• 7.1: Transition Metal Alkyl Complexes
Transition metal σ−bonded organometallic compounds like the metal alkyls, aryls and the hydrides derivatives are by for the most common organometallic species encountered in the world of chemistry. Yet, these compounds remained elusive till as late as the 1960s and the 1970s.
• 7.2: Metal Hydrides
Metal hydrides occupy an important place in transition metal organometallic chemistry as the M−H bonds can undergo insertion reactions with a variety of unsaturated organic substrates yielding numerous organometallic compounds with M−C bonds. Not only the metal hydrides are needed as synthetic reagents for preparing the transition metal organometallic compounds but they also are required for important hydride insertion steps in many catalytic processes.
07: Metal Alkyls and Metal Hydrides
Learning Objectives
In this lecture you will learn the following
• Get a general prospective on the historical background of transition metal organometallic compounds with particular emphasis on metal alkyls.
• Know more about stable metal alkyls.
• Get introduced to transition metal agostic alkyls.
• Develop an understanding of reactions of relevance to metal alkyls like the reductive eliminations, oxidative additions and the halide eliminations.
Transition metal σ−bonded organometallic compounds like the metal alkyls, aryls and the hydrides derivatives are by for the most common organometallic species encountered in the world of chemistry. Yet, these compounds remained elusive till as late as the 1960s and the 1970s.
Historical background
Metal alkyls of the main group elements namely, Li, Mg, Zn, As and Al, have been known for a long time and which over the years have conveniently found applications in organic synthesis whereas development on similar scale and scope in case of the transition metal counterparts were missing till only recently. The origin of the organometallic compounds traces back to 1757, when Cadet prepared a foul smelling compound called cacodyl oxide from As2O3 and CH3COOK, while working in a military pharmacy in Paris. Years later in 1840, R. W. Bunsen gave the formulation of cacodyl oxide as Me2As−O−AsMe2. The next known transition metal organometallic compound happens to be Et2Zn, which was prepared serendipitously in 1848 from the reaction of ethyl iodide (EtI) and Zn with the objective of generating free ethyl radical. Frankland further synthesized alkyl mercury halides like, CH3HgI, from the reaction of methyl iodide (CH3I) and Hg in sunlight. It is important to note that the dialkyl mercury, R2Hg, and the dialkyl zinc, R2Zn, have found applications as alkyl transfer reagents in the synthesis of numerous main group organometallic compounds.
Another notable development of the time was of the preparation of Et4Pb from ethyl iodide (EtI) and Na/Pb alloy by C. J. Lowig and M. E. Schweizer in 1852. They subsequently extended the same method for the preparation of the Et3Sb and Et3Bi compounds. In 1859, aluminumalkyliodides, R2 AlI, were prepared by W. Hallwachs and A Schafarik from alkyl iodide (RI) and Al. The year 1863 saw the preparation of organochlorosilanes, RmSiCl4−m, by C. Friedel and J. M. Crafts while the year 1866 saw the synthesis of halide-free alkyl magnesium compound, Et2Mg, by J. A. Wanklyn from the reaction of Et2Hg and Mg. In 1868, M. P. Schutzenberger reported the first metal−carbonyl complex in the form of [Pt(CO)Cl2]2. In 1890, the first binary metal−carbonyl compound, Ni(CO)4 was reported by L. Mond, who later founded the well−known chemical company called ICI (Imperial Chemical Industries). In 1909, W. J. Pope reported the first σ−organotransition metal compound in the form of (CH3)3PtI. In 1917, the alkyllithium, RLi, compounds were prepared by W. Schlenk by transalkylation reactions. In 1922, T. Midgley and T. A. Boyd reported the utility of Et4Pb as an antiknock agent in gasoline. A. Job and A. Cassal prepared Cr(CO)6 in 1927. In 1930, K. Ziegler showed the utility of organolithium compounds as alkylating agent while in the following year in 1931, W. Heiber prepared Fe(CO)4H2 as the first transition metal−hydride complex. O. Roelen discovered the much renowned hydroformylation reaction in 1938, that went on to become a very successful industrial process worldwide.
The large scale production and the use of silicones were triggered by E. G. Rochow, when he reported the ‘direct synthesis’ from methyl chloride (CH3Cl) and Si using Cu catalyst at 300 °C in 1943. The landmark compound, ferrocene (C5H5)2Fe, known as the first sandwich complex was obtained by P. Pauson and S. A. Miller in 1951. H. Gilman introduced the important utility of organocuprates when he prepared LiCu(CH3)2, in 1952. In the subsequent year 1953, G. Wittig found a new method of synthesizing olefins from phosphonium ylides and carbonyl compounds that fetched him a Nobel prize in 1979. The year 1955 turned out to be a year of path breaking discoveries with E. O. Fischer reporting the rational synthesis of bis(benzene)chromium, (C6H6)2Cr while K. Ziegler and G. Natta announcing the ground breaking polyolefin polymerization process that subsequently gave them the Nobel prizes, E. O. Fischer sharing with G. Wilkinson in 1973 while K. Ziegler and G. Natta shared the same in 1963. In 1956, H. C. Brown reported hydroboration for which he too received the Nobel prize in 1979. In 1963, L. Vaska reported the famous Vaska’s complex, trans−(PPh3)2Ir(CO)Cl, that reversibly binds to molecular oxygen. In 1964, E. O. Fischer reported the first carbene complex, (CO)5WC(OMe)Me. In 1965, G. Wilkinson and R. S. Coffey reported the Wilkinson catalyst, (PPh3)3RhCl, for the hydrogenation of alkenes. In 1973, E. O. Fischer synthesized the first carbyne complex, I(CO)4Cr(CR).
After the early 1970s, there were tremendous outburst in activity, in the area of transition metal organometallic chemistry leading to phenomenal developments having far-reaching consequences in various branches of the main stream and interfacial chemistry. Several Nobel prizes that have been awarded to the area in recent times fully recognized the significance of these efforts with Y. Chauvin, R. R. Schrock, and R. H. Grubbs winning it in 2005 for olefin metathesis and Akira Suzuki, Richard F Heck and E. Negishi receiving the same for the Pd catalyzed C−C cross-coupling reactions in organic synthesis in 2010.
Metal alkyls
In day to day organic synthesis, particularly from the application point of view, the metal alkyls are often perceived as a source of stabilized carbanions for reactions with various electrophiles. The extent of stabilization of alkyl carbanions in metal alkyl complexes depend upon the nature of the metal cations. For example, the alkyls of electropositive metals like that of Group 1 and 2, Al and Zn are regarded as polar organometallics as the alkyl carbanions remain weakly stabilized while retaining strong nucleophilic and basic character of a free anion. These polar alkyls are extremely air and moisture sensitive as in their presence they often get hydrolyzed and oxidized readily. Similar high reactivity was also observed in case of the early transition metal organometallic compounds particularly of Ti and Zr. On the contrary the late transition metal organometallic compounds are much less reactive and stable. For example, the Hg−C bond of (Me−Hg)+ cation is indefinitely stable in aqueous H2SO4 solution in air. Thus, on moving from extremely ionic Na alkyls to highly polar covalent Li and Mg alkyls and to essentially covalent late−transition metal alkyls, a steady decrease in reactivity is observed. This trend can be correlated to the stability of alkyl carbanions that also depended on the nature of hybridization of the carbon center, with sp3 hybridized carbanions being the least stable and hence most reactive, followed by the sp2 carbanions being moderately stable while the sp carbanions being the least reactive and most stable. The trend also correlates well with the respective pKa values observed for CH4 (pKa = ~50), C6H6 (pKa = ~43) and RC≡CH (pKa = ~25).
Stable alkyls
As has been mentioned earlier, that the β−elimination is a crucial destabilizing influence on the transition metal organometallic complexes. Hence, inhibition of this decomposition pathway leads to increased stabilities of organometallic compounds. Thus, many stable alkyl transition metal complexes do not possesses β−hydrogens like, W(Me)6 and Ti(CH2Ph)4. In some cases despite the presence of the β−hydrogens the organometallic complexes are stable as the β−hydrogens are deposed away from the metal center like in, Cr(CHMe2)4, and Cr(CMe3)4. In this category of stable transition metal organometallic compounds also falls the ones that contain β−hydrogens but cannot β−eliminate owing to the formation of a olefinic bond at a bridgehead, which is unfavorable, like in Ti(6−norbornyl)4 and Cr(1−adamantyl)4. Lastly, some 18 VE metal complexes are stable, again despite having β−hydrogens, for reasons of being electronically as well as coordinatively saturated at the metal center owing to attaining the stable 18 electron configuration.
Agostic alkyls
Agostic alkyls are extremely rare but very interesting species that represents a frozen point in a β−elimination pathway that have fallen short of the completion of the decomposition reaction. Thus, these agostic alkyl complexes can be viewed as snap shots of a β−elimination trajectory thereby providing valuable mechanistic understanding of the decomposition reaction. The agostic interaction has characteristic signatures in various spectroscopic techniques as observed from the decreasing JC−Hcoupling constant values in the 1H NMR and the 13C NMR spectra and the lowering of the νCHstretching frequencies in the IR spectroscopy. The agostic alkyl complexes can be definitively proven by X−ray diffraction or neutron diffraction studies. The agostic alkyls thus have activated C−H bonds which are of interest for their utility in chemical catalysis. Quite interestingly, many d0 Ti agostic alkyl complexes do not β−eliminate primarily for the metal center being too electron deficient to donate electron to the σ* C−H orbital as required for the subsequent β−elimination process.
Reductive elimination
Reductive elimination represents a major decomposition pathway of the metal alkyls. Opposite of oxidative addition, the reductive elimination is accompanied by the decrease in the oxidation state and the valence electron count of the metal by two units. The metal alkyl complexes may thus reductively eliminate with an adjacent hydrogen atom to yield an alkane, (R−H) or undergo the same with an adjacent alkyl group to give an even larger alkane (R−R) as shown below.
$\ce{L_{n}MRH -> R-H + L_{n}M} \nonumber$
$\ce{L_{n}MR2 -> R-R + L_{n}M} \nonumber$
The reductive elimination is often facilitated by an electron deficient metal center and by sterically demanding ligand systems. Often d8 metals like Ni(II), Pd(II), and Au(III) and d6 metals in high oxidation state like, Pt(IV), Pd(IV), Ir(III), and Rh(III) exhibit reductive elimination.
Oxidative addition
Unlike the reductive elimination that represents a decomposition pathway of metal alkyls, the oxidative addition reaction represents a useful method for the formation of the metal alkyl complexes. The oxidative addition thus leads to increase in valence electron count and the oxidation state of the metal center by two units. The oxidative addition reactions are often facilitated by low valent electron rich metal centers and by less sterically demanding ligands.
Halide elimination
β−halide elimination is observed for the early transition metals and the f−block elements resulting in the formation of stable alkyl halides. The phenomenon is mostly seen in case of the metal fluorides and arise owing to the very high alkyl−fluoride bond strengths that favor the halide elimination.
Problems
1. Who elucidated the structure of cacodyl oxide?
Ans: R. W. Bunsen in 1840
2. Give the example of the first olefin bound transition metal complex?
Ans: Zeise's salt, Na[PtCl3(C2H4)]
3. Who discovered olefin polymerization?
Ans: K. Ziegler and G. Natta
4. What kind of metal center promotes oxidative addition reactions?
Ans: Electron rich
5. The 18 VE complex would favor/disfavor oxidative addition reactions?
Ans: Disfavor
Self Assessment test
1. O. Roelen discovered which famous reaction?
Ans: Hydroformylation
2. What is the first binary metal−carbonyl complex?
Ans: Ni(CO)4
3. Who discovered the hydroboration reaction?
Ans: H. C. Brown
4. Reductive elimination reaction is favored by what kind of ligands?
Ans: Sterically demanding
5. β−halide elimination is mainly observed for what type of metal halide complexes?
Ans: Metal fluorides
Summary
A broader outlook on metal alkyls is obtained from the study of its historical background thus dispelling many myths about these compounds like them being inherently unstable. It also establishes newer founding principles like these compounds indeed being thermodynamically stable under certain experimental conditions and thus facilitating further attempts to take up the synthesis of these compounds. Another important class of transition metal organometallic compounds are the agostic alkyls, which can be viewed as the ones that have proceeded along but have fallen short of the final sequence of the β−elimination step. While oxidative addition reaction remains a key method for synthesizing metal alkyls, the complementary reaction, i.e, the reductive elimination, represents a decomposition reaction of these compounds. β−halide elimination reactions are observed for early transition metal elements and f−block elements. | textbooks/chem/Inorganic_Chemistry/Introduction_to_Organometallic_Chemistry_(Ghosh_and_Balakrishna)/07%3A_Metal_Alkyls_and_Metal_Hydrides/7.01%3A_Transition_Metal_Alkyl_Complexes.txt |
Learning Objectives
In this lecture you will learn the following
• Know about metal hydrides, their synthesis, characterization and their reactivity.
• Know about σ−complexes and their properties.
• Know about transition metal dihydrogen complexes.
Metal hydrides occupy an important place in transition metal organometallic chemistry as the M−H bonds can undergo insertion reactions with a variety of unsaturated organic substrates yielding numerous organometallic compounds with M−C bonds. Not only the metal hydrides are needed as synthetic reagents for preparing the transition metal organometallic compounds but they also are required for important hydride insertion steps in many catalytic processes. The first transition metal hydride compound was reported by W. Heiber in 1931 when he synthesized Fe(CO)4H2. Though he claimed that the Fe(CO)4H2 contained Fe−H bond, it was not accepted until 1950s, when the concept of normal covalent M−H bond was widely recognized.
The metal hydride moieties are easily detectable in 1H NMR as they appear high field of TMS in the region between 0 to 60 ppm, where no other resonances appear. The hydride moieties usually couple with metal centers possessing nuclear spins. Similarly, the hydride moieties also couple with the adjacent metal bound phosphine ligands, if at all present in the complex, exhibiting characteristic cis (J = 15 − 30 Hz) and trans (J = 90 − 150 Hz) coupling constants. In the IR spectroscopy, the M−H frequencies appear between (1500 − 2200) cm−1 but their intensities are mostly weak. Crystallographic detection of metal hydride moiety is difficult as hydrogen atoms in general are poor scatterer of X−rays. Located adjacent to a metal atom in a M−H bond, the detection of hydrogen atom thus becomes challenging and as a consequence the X−ray crystallographic method systematically underestimates the M−H internuclear distance by ~ 0.1 Å. However, better data could be obtained by performing the X−ray diffraction studies at a low temperature in which the thermal motion of the atoms are significantly reduced. In light of these facts, the neutron diffraction becomes a powerful method for detection of the metal hydride moieties as hydrogen scatters neutrons more effectively and hence the M−H bond distances can be measured more accurately. A limitation of neutron diffraction method is that large sized crystals are required for the study.
Synthesis
Following reactions are employed for synthesizing metal hydrides.
i. Protonation reactions
For this reaction to occur the metal center has to be basic and electron rich.
$\ce{[Fe(CO)4]^{-2} ->[H+] [HFe(CO)4]^{-} ->[H+] H2Fe(CO)4} \nonumber$
ii. From hydride donors
Generally for this method, a main group hydride is reacted with metal halide.
$\ce{WCl6 + LiBEt3H + PR3 -> WH6(PR3)3} \nonumber$
iii. Using dihydrogen (H2) addition
This method involves oxidative addition of H2 and thus requires metal centers that are capable of undergoing the oxidative addition step.
$\ce{WMe6 + PMe2Ph ->[H_{2}] WH6(PMe2Ph)3} \nonumber$
iv. From a ligand
This method takes into account the β−elimination that occur in a variety of metal bound ligand moieties, thereby yielding a M−H bond.
$\ce{RuCl2(PPh3)3 + KOCHMe2 + PPH3 -> RuH2(PPh3)4 + Me2CO + KCl} \nonumber$
Reactions of metal hydrides
Metal hydrides are reactive species kinetically and thus participate in a variety of transformations like the ones discussed below.
i. Deprotonation reactions
The deprotonation reaction can be achieved by a hydride moiety resulting in the formation of H2gas as shown below.
$\ce{WH6(PMe3)3 + NaH -> Na[WH5(PMe3)3] + H2} \nonumber$
ii. Hydride transfer and insertion
In this reaction a hydride transfer from a metal center to formaldehyde resulting in the formation of a metal bound methoxy moiety is observed as shown below.
$\ce{Cp2*ZrH2 + CH2O -> Cp2*Zr(OMe)2} \nonumber$
iii. Hydrogen atom transfer reaction
An example of hydrogen atom transfer reaction is given below.
$\ce{[Co(CN)5H]^{3-} + PhCH=CHCOOH -> [Co(CN)5]^{3-} + PHCH-CH2COOH} \nonumber$
It is interesting to note that the nature of hydrogen atom in a M−H bond can vary from being protic in nature, when bound to electron deficient metal centers as in metal carbonyl compounds, to that of being hydridic in nature, when bound to more electropositive early transition metals. In the latter case, the hydride moieties tend to be basic and exhibit hydride transfer reactions with electrophiles like aldehydes or ketones. Furthermore, the protonation of these basic metal hydrides leads to the elimination of dihydrogen (H2) gas along with the generation of a vacant coordination site at the metal center.
Bridging hydrides
The metal hydrides usually show two modes of binding, namely terminal and bridging. In case of the bridging hydrides, the hydrogen atom can bridge between two or even more metal centers and thus, the bridging hydrides often display bent geometries.
σ−complexes
σ−complexes are rare compounds, in which the σ bonding electrons of a X−H bond further participate in bonding with a metal center (X = H, Si, Sn, B, and P). The σ complexes thus exhibit an askewed binding to a metal center with the hydrogen atom, containing no lone pair, being more close to the metal center and thereby resulting in a side−on structure. Many times if the metal center is electron rich, then further back donation to the σ* orbital of the metal bound X−H moiety may occur resulting in a complete cleavage of the X−H bond.
Metal dihydrogen complexes
The simplest variant of a σ−complex contains a dihydrogen ligand. The first dihydrogen complex was isolated by Kubas, after which many new ones were reported.
Quite expectedly, the dihydrogen moiety bound to a metal in a σ−complex is found to be more acidic (pKa = 0 − 20) when compared to the free dihydrogen molecule (pKa = 35). It is interesting to note that the pKa change associated with the binding of dihydrogen to a metal in a σ−complex relative to that of the free H2 molecule is significantly larger than the change associated with binding of H2O to metal. Owing to this inherent acidity, the deprotonation of the metal bound dihydrogen moiety by a base can thus be appropriately employed for heterolytic activation of the dihydrogen moiety as illustrated below.
The dihydrogen complexes of metals are often referred to as nonclassical hydrides. The electron rich π basic metals are anticipated to split the metal bound dihydrogen moieties resulting in classical dihydride complexes. Along the same line of thinking, the electron deficient and less π basic metal would tend to stabilize a dihydrogen complex. The dihydrogen complexes can also be characterized by the X ray diffraction as well as neutron diffraction methods. In IR spectrum, the metal bound H−H stretch appear in the range (2300 − 2900) cm−1 while in the 1H NMR spectrum the same appear between 0 to −10 ppm as a broad peak. The dihydrogen complexes are often characterized by isotopic labeling studies of metal bound H−D moiety that shows a coupling constant of 20 – 34 Hz as supposed to 43 Hz observed in case of the free H−D molecule.
Problems
1. Predict the product of the reaction.
Ans:
2. Give the oxidation state and total valence electron count of the metal center.
Ans: Oxidation state 0 and 18 VE
3. What kind of metal centers would stabilize metal dihydrogen complexes?
Ans: Electron deficient and less π basic ligands
4. Specify whether the nature of hydrogen moiety in the complex, HCo(CO)4 is acidic or basic?
Ans: Acidic
5. Where do the M−H stretching bands appear in the IR spectrum of metal hydride complexes?
Ans: 1500 to 2200 cm-1
Self Assessment test
1. Predict the product of the reaction.
$\ce{IrCl(Co)(Pph3)2 ->[H_{2}]} \nonumber$
Ans:
$\ce{IrCl(CO)(PPh3)2 ->[H2] IrH2Cl(CO)(PPh3)2} \nonumber$
2. Give the oxidation state and total valence electron count of the metal center.
Ans: Oxidation state +2 and 18 VE
3. What kind of metal centers would stabilize classical dihydride complexes?
Ans: Electron rich and more p basic ligands
4. Specify whether the nature of hydrogen moiety in the complex, IrH5(PCy3)2 is acidic or basic?
Ans: Basic
5. Between X−ray diffraction and neutron diffraction, which is a better method for the characterization of the M−H moiety?
Ans: Neutron diffraction
Summary
Metal hydrides are important compounds in the overall scheme of organometallic chemistry as they are involved in many crucial steps of numerous catalytic reactions. Apart from metal hydrides another important class of compounds are transition metal σ−complexes whose simplest variant are the metal dihydrogen complexes. These σ−complexes and the metal dihydrogen complexes are important for the heterolytic activations of the respective metal bound H−heteroatom and the H−H bonds. | textbooks/chem/Inorganic_Chemistry/Introduction_to_Organometallic_Chemistry_(Ghosh_and_Balakrishna)/07%3A_Metal_Alkyls_and_Metal_Hydrides/7.02%3A_Metal_Hydrides.txt |
Learning Objectives
In this lecture you will learn the following
• The historical background of metal carbonyl complexes.
• The CO ligand and its binding ability to metal.
• Synergism between the ligand to metal forward σ–donation and the metal to ligand backward π–donation observed in a metal-CO interaction.
• The synthesis, characterization and their reactivity of the metal carbonyl compounds.
Metal carbonyls are important class of organometallic compounds that have been studied for a long time. Way back in 1884, Ludwig Mond, upon observing that the nickel valves were being eating away by CO gas in a nickel refining industry, heated nickel powder in a stream of CO gas to synthesize the first known metal carbonyl compound in the form Ni(CO)4. The famous Mond refining process was thus born, grounded on the premise that the volatile Ni(CO)4 compound can be decomposed to pure metal at elevated temperature. Mond subsequently founded the Mond Nickel Company Limited for purifying nickel from its ore using this method.
The carbonyl ligand (CO) distinguishes itself from other ligands in many respects. For example, unlike the alkyl ligands, the carbonyl (CO) ligand is unsaturated thus allowing not only the ligand to σ−donate but also to accept electrons in its π* orbital from dπ metal orbitals and thereby making the CO ligand π−acidic. The other difference lies in the fact that CO is a soft ligand compared to the other common σ−and π−basic ligands like H2O or the alkoxides (RO−), which are considered as hard ligands.
Being π−acidic in nature, CO is a strong field ligand that achieves greater d−orbital splitting through the metal to ligand π−back donation. A metal−CO bonding interaction thus comprises of a CO to metal σ−donation and a metal to CO π−back donation (Figure $1$). Interestingly enough, both the spectroscopic measurements and the theoretical studies suggest that the extent of the metal to CO π−back donation is almost equal to or even greater than the extent of the CO to metal σ−donation in metal carbonyl complexes. This observation is in agreement with the fact that low valent−transition metal centers tend to form metal carbonyl complexes.
In the metal carbonyl complexes, the direct bearing of the π−back donation is observed on the M−C bond distance that becomes shorter as compared to that of a normal M−C single bond distance. For example, the CpMo(CO)3CH3 complex, exhibits two kind of M−C bond distances that comprise of a longer Mo−CH3 distance (2.38 Å) and a much shorter Mo−CO distance (1.99 Å) arising out of a metal to ligand π−back donation. It becomes thus apparent that the metal−CO interaction can be easily characterized using X−ray crystallography. The infrared spectroscopy can also be equally successfully employed in studying the metal−CO interaction. Since the metal to CO π−back bonding involves a π−donation from the metal dπ orbital to a π* orbital of a C−O bond, significant shift of the ν(CO) stretching frequency towards the lower energy is observed in metal carbonyl complexes with respect to that of free CO (2143 cm−1).
Preparation of metal carbonyl complexes
The common methods of the preparation of the metal carbonyl compounds are,
i. Directly using CO
$\ce{Fe ->[CO, \: 200atm, \: 200°C] Fe(CO)5} \nonumber$
The main requirement of this method is that the metal center must be in a reduced low oxidation state in order to facilitate CO binding to the metal center through metal to ligand π−back donation.
ii. Using CO and a reducing agent
$\ce{NiSO4 + CO + S2O4^{2-} -> Ni(CO)4} \nonumber$
This method is commonly called reductive carbonylation and is mainly used for the compounds having higher oxidation state metal centers. The reducing agent first reduces the metal center to a lower oxidation state prior to the binding of CO to form the metal carbonyl compounds.
iii. From carbonyl compounds
This method involves abstraction of CO from organic compounds like the alcohols, aldehydes and CO2.
Reactivities of metal carbonyls
i. Nucleophilic attack on carbon
The reaction usually gives rise to carbene moiety.
ii. Electrophilic attack at oxygen
$\ce{Cl(PR3)4Re-CO + AlMe3 -> Cl(PR3)4Re-CO -> AlMe3} \nonumber$
iii. Migratory insertion reaction
$\ce{MeMn(CO)5 +PMe3 -> (MeCO)Mn(CO)4(PMe3)} \nonumber$
The metal carbonyl displays two kinds of bindings in the form of the terminal and the bridging modes. The infrared spectroscopy can easily distinguish between these two binding modes of the metal carbonyl moiety as the terminal ones show ν(CO) stretching band at ca. 2100-2000 cm−1 while the bridging ones appear in the range 1720−1850 cm−1. The carbonyl moiety can bridge between more than two metal centers (Figure $2$).
Problems
1. How many lone pairs are there in the CO molecule?
Ans: Three (one from carbon and two from oxygen).
2. Despite O being more electronegative than C, the dipole moment of CO is almost zero. Explain.
Ans: Because of the electron donation from oxygen to carbon.
3. What type of metal centers form metal carbonyl complexes?
Ans: Low−valent metal centers.
4. What are the two main modes of binding exhibited by CO ligand?
Ans: Terminal and bridging modes of binding.
Self Assessment test
1. Predict the product of the reaction?
$\ce{Ni + CO_{Excess} ->} \nonumber$
Ans: Three (one from carbon and two from oxygen).
2. Upon binding to a metal center the C−O stretching frequency increases/decreases with regard to that of the free CO?
Ans: Decreases.
3. Explain why do low−valent metal centers stabilize CO binding in metal carbonyl complexes?
Ans: Because metal to ligand π−back donation.
4. Give an example of a good σ−donor and π−donor ligand?
Ans: Alkoxides (RO-).
Summary
CO is a hallmark ligand of organometallic chemistry. The metal carbonyl complexes have been studied for a long time. The CO ligands bind tightly to metal center using a synergistic mechanism that involves σ−donation of the ligand lone pair to metal and followed by the π−back donation from a filled metal d orbital to a vacant σ* orbital of C−O bond of the CO ligand. The metal carbonyl complexes are prepared by several methods. The metal carbonyl complexes are usually stabilized by metal centers in low oxidation states. | textbooks/chem/Inorganic_Chemistry/Introduction_to_Organometallic_Chemistry_(Ghosh_and_Balakrishna)/08%3A_Carbonyls_and_Phosphine_Complexes/8.01%3A_Metal_Carbonyls.txt |
Learning Objectives
In this lecture you will learn the following
• Know about metal phosphine complexes.
• Have an understanding of the steric and electronic properties of the phosphine ligands.
• Obtain a deeper insight about the metal phosphine interactions.
• Be introduced to other π−basic ligands.
Phosphines are one of the few ligands that have been extensively studied over the last few decades to an extent that the systematic fine tuning of the sterics and electronics can now be achieved with certain degree of predictability. Phosphines are better spectator ligands than actor ligands. Tolman carried out pioneering infrared spectroscopy experiments on the PR3Ni(CO)3 complexes looking at the ν(CO)stretching frequencies for obtaining an insight on the donor properties of the PR3 ligands. Thus, a stronger σ−donor phosphine ligand would increase the electron density at the metal center leading to an enhanced metal to ligand π−back bonding and thereby lowering of the ν(CO) stretching frequencies in these complexes. Another important aspect of the phosphine ligand is its size that has significant steric impact on its metal complexes. Thus, unlike CO ligand, which is small and hence many may simultaneously be able to bind to a metal center, the same is not true for the phosphine ligands as only a few can bind to a metal center. The number of phosphine ligands that can bind to a metal center also depends on the size of its R substituents. For example, up to two can bind to a metal center in case of the PCy3 or P(i−Pr)3 ligands, three or four for PPh3, four for Me2PH, and five or six for PMe3. The steric effect of phosphine was quantified by Tolmann and is given by a parameter called Cone Angle that measures the angle at the metal formed by the PR3 ligand binding to a metal (Figure \(1\)).
The Cone Angle criteria has been successfully invoked in rationalizing the properties of a wide range of metal phosphine complexes. One unique feature of the phosphine ligand is that it allows convenient change of electronic effect without undergoing much change in its steric effects. For example, PBu3 and P(OiPr)3 have similar steric effects but vary in their electronic effects. The converse is also true as the steric effect can be easily changed without undergoing much change in the electronic effect. For example, PMe3 and P(o−tolyl)3 have similar electronic effect but differ in their steric effects. Thus, the ability to conveniently modulate the steric and the electronic effects make the phosphine ligands a versatile system for carrying out many organometallic catalysis.
Structure and Bonding
Phosphines are two electron donors that engage a lone pair for binding to metals. These are thus considered as good σ−donors and poor π−acceptors and they belong to the same class with the aryl, dialkylamino and alkoxo ligands. In fact they are more π−acidic than pure σ−donor ligands like NH3and, more interestingly so, their π−acidity can be varied significantly by systematic incorporation of substituents on the P atom. For example, PF3 is more π−acidic than CO. Analogous to what is observed in case of the benchmark π−acidic CO ligand, in which the metal dπ orbital donates electron to a π* orbital of a C−O bond, in the case of the phosphines ligands, such π−back donation occurs from the metal dπ orbital occurs on to a σ* orbital of a P−R bond (Figure \(2\)). In phosphine ligands, with the increase of the electronegativity of R both of the σ and the σ* orbitals of the P−R bond gets stabilized. Consequently, the contribution of the atomic orbital of the P atom to the σ*−orbital of the P−R bond increases, which eventually increases the size of the σ* orbital of the P−R bond. This in turn facilitates better overlap of the σ* orbital of the P−R bond with the metal dπ orbital during the metal to ligand π−back donation in these metal phosphine complexes.
Starting from CO, which is a strong π−acceptor ligand, to moving to the phosphines, which are good σ−donors and poor π−acceptor ligands, to even going further to other extreme to the ligands, which are both good σ−donors as well as π−donors, a rich variety of phosphine ligands thus are available for stabilizing different types of organometallic complexes. In this context the following ligands are discussed below.
π-basic ligands
Alkoxides (RO) and halides like F, Cl and Br belong to a category of π−basic ligands as they engage a second lone pair for π−donation to the metal over and above the first lone pair partaking σ−donation to the metal. Opposite to what is observed in the case of π−acidic ligands, in which the π* ligand orbital stabilizes the dπ metal orbital and thereby affecting a larger ligand field splitting, as consistent with the strong field nature of these ligands (Figure 3), in the case of the π−basic ligands, the second lone pair destabilizes the dπ metal orbitals leading to a smaller ligand field splitting, which is in agreement with the weak field nature of these ligands. The orbitals containing the lone pair of the ligands are usually located on the more electronegative heteroatoms and so they are invariably lower in energy than the metal dπ orbitals. Hence, the destabilization of the metal dπ orbitals occurs due to the repulsion of the filled ligand lone pair orbital with the filled metal dπ orbitals. In case of the situations in which the metal dπ orbitals are vacant, like in d0 systems of Ti4+ ions, the possibility of the destabilization of the metal dπ orbitals do not arise but instead stabilization occurs through the donation of the filled ligand lone pair orbital electrons to the empty metal dπ orbitals as seen in the case of TiF6 and W(OMe)6. Thus, this scenario in π−basic ligands is opposite to that observed in case of the π−acidic ligands, for which the empty π* ligand orbitals are higher in energy than the filled metal dπ orbitals. | textbooks/chem/Inorganic_Chemistry/Introduction_to_Organometallic_Chemistry_(Ghosh_and_Balakrishna)/08%3A_Carbonyls_and_Phosphine_Complexes/8.02%3A_Metal_Phosphines.txt |
Learning Objectives
In this lecture you will learn the following
• The metal alkene complexes.
• The metal−olefin bonding interactions.
• The synthesis and reactivities of the metal−olefin complexes.
• The umpolung reactivities of olefins in the metal alkene complexes.
Though the first metal olefin complex dates back a long time to the beginning of 19th century, its formulation was established only a century later in the 1950s. While reacting $\ce{K2PtCl4}$ with EtOH in 1827, the Danish chemist Zeise synthesized the famous Zeise’s salt $\ce{K[PtCl3(C2H4)]•H2O}$ containing a Pt bound ethylene moiety and which incidentally represented the first metal−olefin complex (Figure $1$).
The metal−olefin bonding interaction is best explained by the Dewar−Chatt model, that takes into account two mutually opposing electron donation involving σ−donation of the olefinic C=C π−electrons to an empty dπ metal orbital followed by π−back donation from a filled metal dπ orbital into the unoccupied C=C π* orbital. Quite understandably so, for the d0 systems, the formations of metal−olefin complexes are not observed. The extent of the C=C forward π-donation to the metal and the subsequent π−back donation from the filled dπ orbital to the olefinic C=C π* orbital have a direct bearing on the C=C bond of the metal bound olefinic moiety in form of bringing about a change in hybridization as well as in the C−C bond distance (Figure $2$).
If the metal to ligand π−back donation component is smaller than the ligand to metal σ−donation, then the lengthening of the C−C bond in the metal bound olefin moiety is observed. This happens primarily because of the fact that the alkene to metal σ−donation removes the C=C π−electrons away from the C−C bond of the olefin moiety and towards the metal center, thus, decreasing its bond order and increasing the C−C bond length. Additionally, as the metal to ligand π−back donation increases, the electron donation of the filled metal dπ orbital on to the π* orbital of the metal bound olefin moiety is enhanced. This results in an increase in the C−C bond length. The lengthening of the C−C bond in metal bound olefin complex can be correlated to the π−basicity of the metal. For example, for a weak π−basic metal, the C−C bond lengthening is anticipated to be small while for a strong π−basic metal, the C−C lengthening would be significant.
Another implication of ligand−metal π−back donation is in the observed change of hybridization at the olefinic C atoms from pure sp2, in complexes with no metal to ligand π−back donation, to sp3, in complexes with significant metal to ligand π−back donation, is observed. The change in hybridization from sp2 to sp3 centers of the olefinic carbon is accompanied by the substituents being slightly bent away from the metal center in the final metalacyclopropane form (Figure $3$). This change in hybridization can be conveniently detected by 1H and 13C NMR spectroscopy. For example, in case of the metalacyclopropane systems, which have strong metal to ligand π−back donation, the vinyl protons appear 5 ppm (in the 1H NMR) and 100 ppm (in the 13C NMR) high field with respect to the respective position of the free ligands.
An interesting fallout of the metal to ligand π−back bonding is the tighter binding of the strained olefins to the metal center as observed in the case of cyclopropene and norbornene. The strong binding of these cyclopropene and norbornene moieties to the metal center arise out of the relief of ring strain upon binding to the metal. Lastly, in the metal−olefin complexes having very little π−back bonding component, the chemical reactivities of the metal bound olefin appear opposite to that of a free olefin. For example, a free olefin is considered electron rich by virtue of the presence of π−electrons in its outermost valence orbital and hence it undergoes an electrophilic attack. However, the metal bound olefin complexes having predominantly σ−donation of the olefinic π−electrons and negligible metal to ligand π−back donation, the olefinic C becomes positively charged and hence undergoes a nuclophilic attack. This nature of reversal of olefin reactivity is called umpolung character.
Synthesis
Metal alkene complexes are synthesized by the following methods.
i. Substitution in low valent metals
$\ce{AgOSO2CF3 + C2H4 -> (C2H4)AgOSO2CF3} \nonumber$
ii. Reduction of high valent metal in presence of an alkene
$\ce{(cod)PtCl2 + C2H4 -> [PtCl3(C2H4)]- + Cl-} \nonumber$
iii. From alkyls and related species
$\ce{Cp2TaCl3 + n-BuMgX -> {Cp2TaBu3}} \nonumber$
$\ce{{Cp2TaBu3} ->[\Beta -elimination][reductive \: elimination] Cp2TaH(1-Butene) + Butene + Butane} \nonumber$
Reaction of alkenes
The metal alkene complexes show the following reactivities.
i. Insertion reaction
These reactions are commonly displayed by alkenes as they insert into metal−X bonds yielding metal alkyls. The reaction occurs readily at room temperature for X = H, whereas for other elements (X = other atoms), such insertions become rare. Also, the strained alkenes and alkynes undergo such insertion readily.
$\ce{PtHCl(PEt3)2 + C2H4 <=> PtElCl(PEt3)2} \nonumber$
ii. Umpolung reactions
Umpolung reactions are observed only for those metal−alkene complexes for which the metal center is a poor π−base and as a result of which the olefin undergoes a nuclophilic attack.
iii. Oxidative addition
Alkenes containing allylic hydrogens undergo oxidative addition to give a allyl hydride complex.
Exercise $1$
Predict the product of the reaction.
$\ce{AuMe(PPh3) + CF2=CF2 -> A -> B} \nonumber$
Answer
A = {(CF2=CF2)AuMe(PPh3)} and
B = Au(CF2-CF2Me)(PPh3)
Exercise $2$
Specify whether the lengthening/shortening of the C−C bond distance in the metal bound olefin moiety is observed as a result of metal to ligand π−back donation?
Answer
Lengthening
Exercise $3$
Draw the structure of Zeise’s salt.
Answer
Exercise $4$
The change in hybridization at the olefinic C from sp2 to sp3 primarily arise due to?
Answer
Metal-ligand π-back donation.
Self Assessment test
1. Predict the product of the reaction.
$\ce{PtCl2- + C2H4 ->} \nonumber$
Ans: [PtCl3(C2H4)]- and Cl-
2. Specify whether the lengthening/shortening of the C−C bond distance in the metal bound olefin moiety is observed as a result of ligand to metal σ− donation?
Ans: Lengthening.
3. Metalacyclopropane intermediate in a metal bound olefin complex is primarily formed due to which kind of interaction?
Ans: Metal−ligand π−back donation
4. The oxidation state of Pt in Zeise’s salt is?
Ans: PtII
Summary
Alkenes are an important class of unsaturated ligands that bind to a metal by σ−donating its C=C π−electrons and also accepts electrons from the metal in its π* orbital of C=C bond. These symbiotic σ−donation and π−back donation in metal bound olefin complexes have a significant impact on their structure and reactivity properties. Quite importantly, the structural manifestations arising out of these forward σ−donation and π−back donation can be characterized by using 1H, 13C NMR and IR spectroscopic methods. | textbooks/chem/Inorganic_Chemistry/Introduction_to_Organometallic_Chemistry_(Ghosh_and_Balakrishna)/09%3A_Complexes_of_bound_Ligands/9.01%3A_Metal_Alkene_Complexes.txt |
Learning Objectives
In this lecture you will learn the following
• The metal−allyl complexes.
• The metal−diene complexes.
• The metal−cyclobutadiene complexes.
• The respective metal−ligand interactions.
The allyl ligand is often referred to as an “actor” ligand rather than a “spectator” ligand. It binds to metals in two ways i.e. in a η1 (monohapto) form and a η3 (trihapto) form (Figure $1$). (i). In its monohapto (η1) form, it behaves as an anionic 1e−donor X type of a ligand analogous to that of a methyl moiety while (ii) in a trihapto (η3) form, it acts as an anionic 3e−donor LX type of a ligand.
Metal−allyl interaction
Of particular interest are the molecular orbitals namely Ψ1, Ψ2 and Ψ3 of the allyl ligand that interact with the metal in a metal allyl complex. The energy of these molecular orbitals increase with the increase in the number of nodes. Of the three, the Ψ1 and Ψ2 orbitals usually engage in ligand to metal σ−donation, with Ψ1 involving in a dative L−type bonding and Ψ2 participating in a covalent X−type bonding with the metal d orbitals (Figure $2$).
Synthesis of the metal allyl complexes
The metal allyl complexes are synthesized by the following methods.
i. From an alkene complex as shown below.
$\ce{Mo(dpe)2(η2 -propene) <=> Mo(dpe)2( η3-allyl)H} \nonumber$
ii. By a nucleophilic attack of an allyl compound as shown below.
iii. By an electrophilic attack of an allyl compound as shown below.
iv. From a diene complex as shown below.
Reactions of metal allyl complexes
The reactivities of the metal allyl complexes toward various species are illustrated below.
i. Reaction with nucleophiles
ii. Reaction with electrophiles
iii. Insertion reaction
iv. Reductive elimination
Diene complexes
1,3−Butadiene is a 4e−donor ligand that binds to a metal in a cisoid conformation. The Dewar−Chatt model, when applied to 1,3−butadiene, predicts that the ligand may bind to metal either as a L22) donor type, similar to that of an alkene, or as an LX22π) donor type, similar to that of a metalacyclopropane form. The L2 binding of 1,3−butadiene is rare, e.g. asin (butadiene)Fe(CO)3, while the LX2 type binding is more common, e.g. as in Hf(PMe3)2Cl2. An implication of the LX2 type binding is in the observed shortening of the C2−C3 (1.40 Å) distance alongside the lengthening of the C1−C2(1.46 Å) and C3−C4 (1.46 Å) distances (Figure $3$).
The molecular orbitals of the 1,3−butadiene ligand comprises of two filled Ψ1 (HOMO−1) and Ψ2(HOMO) orbitals and two empty Ψ3 (LUMO) and Ψ4 (LUMO+1) orbitals. In a metal−butadiene interaction the ligand to metal σ−donation occurs from the filled Ψ2 orbital of the 1,3−butadiene ligand while the metal to ligand π−back donation occurs on to the empty Ψ3 orbital of the 1,3−butadiene ligand (Figure $4$).
Though cisoid binding is often observed in metal butadiene complexes, a few instances of transoidbinding is seen in dinuclear, e.g. as in Os3(CO)10(C4H6), and in mononuclear complexes e.g. as in Cp2Zr(C4H6) (Figure $5$).
Synthesis of metal butadiene complex
Metal butadiene complexes are usually prepared by the same methods used for synthesizing metal alkene complexes. Two noteworthy synthetic routes are shown below.
Metal cyclobutadiene complexes
Cyclobutadiene is an interesting ligand because of the fact that its neutral form, being anti−aromatic (−electrons), is unstable as a free molecule (Figure $6$), but its dianionic form is stable because of being aromatic (−electrons). Consequently, the cyclobutadiene ligand is stabilized by significant metal to ligand π−back donation to the vacant ligand orbitals.
A synthetic route to metal cyclobutadiene complex is shown below.
Problems
1. The hapticities displayed by an allyl moiety in binding to metals are?
Ans: 1 and 3.
2. Identify which molecular orbitals of an allyl moiety engage in σ−interaction with a suitable d orbital of a metal in a η3−metal allyl complex?
Ans: Ψ1 and Ψ2.
3. Predict the product of the reaction.
Ans:
4. Identify which molecular orbitals of a butadiene moiety engage in σ−interaction with a suitable dorbital of a metal in a η4−metal butadiene complex?
Ans: Ψ2.
Self Assessment test
1. Predict the product of the reaction.
$\ce{<(--Ni--)> + CO2 -> } \nonumber$
Ans:
2. Identify which molecular orbitals of a butadiene moiety engage in π−interaction with a suitable dorbital of a metal in a η4−metal allyl complex?
Ans: Ψ3.
3. Mention the type of orientations displayed by butadiene ligands for binding to metal.
Ans: Cisoid (common) and transoid (rare).
4. Comment on the number of π−electrons present in the cyclobutadiene moiety of a metal cyclobutadiene complex.
Ans: 6 π−electrons.
Summary
Allyl, 1,3−butadiene and cyclobutadiene together constitute an important class of σ−donor/π−acceptor ligands that occupy a special place in organometallic chemistry. The complexes of these ligands with metals are important intermediates in many catalytic cycles and hence an understanding of their interaction with metal is of significant importance. In this context, the synthesis, characterization and the reactivities of the organometallic complexes of these ligands are described alongside the respective metal−ligand interactions. | textbooks/chem/Inorganic_Chemistry/Introduction_to_Organometallic_Chemistry_(Ghosh_and_Balakrishna)/09%3A_Complexes_of_bound_Ligands/9.02%3A_Metal_Allyl_and_Diene_Complexes.txt |
Learning Objectives
In this lecture you will learn the following
• The cyclopentadienyl ligands.
• The synthesis and reactivity of metal−cyclopentadienyl complexes.
• The metal−cyclopentadienyl interaction.
Cyclopentadienyl moiety acts as an important “spectator” ligand and is quite ubiquitous in organometallic chemistry. It remains inert to most nucleophiles and electrophiles and solely engages in stabilizing organometallic complexes. The cyclopentadienyl ligands form a wide array of organometallic compounds exhibiting different formulations that begin with the so-called “piano stool” CpMLn (n = 2,3 or 4) type ones and extends to the most commonly observed “metallocene” Cp2M type ones to even go beyond further to the “bent metallocene” Cp2MXn (n = 1,2 or 3) type ones. In the “piano stool” CpMLn structure, the cyclopentadienyl (Cp) ligand is regarded as the “seat” of the piano stool while the remaining L ligands are referred to as the “legs” of the piano stool. Though the cyclopentadienyl ligand often binds to metal in a η5 (pentahapto) fashion, e. g. as in ferrocene, the other form of binding to metal at lower hapticities, like that of the η3 (trihapto) binding e. g. as in (η5−Cp)(η3−Cp)W(CO)2 and that of the η1 (monohapto) binding e. g. as in (η5−Cp)(η1−Cp)Fe(CO)2, are also seen on certain rare occasions.
The binding modes of the cyclopentadienyl ligand in metal complexes can be ascertained to a certain degree by 1H NMR in the diamagnetic metal complexes, in which the Cp−protons appear as a singlet between 5.5−3.5 ppm while the β and γ hydrogens come at 7−5 ppm.
Cyclopentadienyl−metal interaction
The frontier molecular orbital of the cyclopentadienyl ligand contains 5 orbitals (Ψ1−Ψ5) residing in three energy levels (Figure 9.3.1 ). The lowest energy orbital Ψ1 does not contain any node and is represented by an a1 state, followed by a doubly degenerate e1 states that comprise of the Ψ2 and Ψ3orbitals, which precede another doubly degenerate e2 states consisting of Ψ4 and Ψ5 orbitals.
The above frontier molecular orbital diagram becomes more intriguing on moving over to the metallocenes that contain two such cyclopentadienyl ligands. Specifically, in the Cp2M system, (e. g. ferrocene) each of these above five molecular orbital of the two cyclopentadienyl ligands combines to give ten ligand molecular orbitals in three energy levels (Figure 9.3.2 ). Of these, the orbitals that subsequently interact with the metal orbitals to generate the overall molecular orbital correlation diagram for the Cp2M type of complexes are shown below (Figure 9.3.3 ).
Generic metallocene Cp2M type complexes are formed for many from across the 1st row transition metal series along Sc to Zn. The number of unpaired electrons thus correlates with the number unpaired electrons present in the valence orbital of the metal (Figure 4). Of the complexes of the 1st row transition metal series, the manganocene exists in two distinct forms, one in a high-spin form with five unpaired electrons, e.g. as in Cp2Mn and the other in a low-spin form with one unpaired electron, e.g. as in Cp*2Mn owing to the higher ligand field strength of the Cp* ligand. Cobaltocene, Cp2Co, has 19 valence electrons (VE) and thus gets easily oxidized to the diamagnetic 18 VE valence electron species, Cp2Co+. Of these metallocenes, the much-renowned ferrocene, Cp2Fe is a diamagnetic 18 VE complex, whose molecular orbital diagram is shown above (Figure 9.3.3 ).
Bent metallocenes
Bent metallocenes are Cp2MXn type complexes formed of group 4 and the heavier elements of groups 5−7. In these complexes the frontier doubly degenerate e2g orbitals of Cp2M fragment interacts with the filled lone pair orbitals of the ligand (Figure 9.3.5 ).
Synthesis of cyclopentadienyl-metal complexes
The metal−cyclopentadienyl complexes are synthesized by the following methods.
i. from Cp−
ii. from Cp+
iii. from hydrocarbon
Reactivity of cyclopentadienyl-metal complexes
The reactivity of cyclopentadienyl−metal complexes of the type Cp2M is shown for a representative nickellocene complex.
i. reaction with NO
ii. reaction with PR3
iii. reaction with CO
iv. reaction with H+
Problems
1. Comment on the p−acceptor property of the cyclopentadienyl ligand.
Ans: The ligand being anionic shows very little π-acceptor properties.
2. Give the total valence electron count at the metal in a nickellocene complex.
Ans: 20 electrons.
3. Explain why the metal center in cobalticene gets easily oxidized.
Ans: 19 electrons cobalticene gets easily oxidized to 18 electron Cp2Co+.
4. Specify the number unpaired electrons present in chromocene.
Ans: 2
Self Assessment test
1. Specify the number of unpaired electron present in vanadocene.
Ans: The ligand being anionic shows very little π-acceptor properties.
2. What different hapticities are exhibited by cyclopentadienyl ligand?
Ans: 1, 3, and 5.
3. Specify the hapticities of the cyclopentadienyl ligands in Cp2W(CO)2.
Ans: 5 and 3.
4. Specify the hapticity of the cyclopentadienyl ligands in CpRh(CO)2(PMe3).
Ans: 3.
Summary
Cyclopentadienyl moiety is almost synonymous with the transition metal organometallic complexes as the ligand played a pivotal role at the early developmental stages of the field of organometallic chemistry in the 1960s and 1970s. An important quality of the cyclopentadienyl ligand is that it behaves as an extremely good “spectator” ligand being inert to nucleophiles and electrophiles and displays uncanny ability towards stabilizing metal complexes of elements from across the different parts of the periodic table. Cyclopentadienyl moiety thus forms several types of complexes of different formulations like that of the “piano stool” CpMLn (n = 2,3 or 4) types, the metallocene Cp2M types and the bent metallocene Cp2MXn (n = 1,2 or 3) types. Cyclopentadienyl metal complexes make valuable catalysts for many chemical transformations of interest to academia and industries alike. The cyclopentadienyl moiety participates in a complex interaction with the metal involving ligand frontier molecular orbitals and the metal valence orbitals. Cyclopentadienyl metal complexes can be accessed by many methods. | textbooks/chem/Inorganic_Chemistry/Introduction_to_Organometallic_Chemistry_(Ghosh_and_Balakrishna)/09%3A_Complexes_of_bound_Ligands/9.03%3A_Metal_Cyclopentadienyl_Complexes.txt |
Learning Objectives
In this lecture you will learn the following
• The oxidative addition reactions.
• The reductive elimination reactions.
• Various mechanistic pathways prevalent for these reactions.
Oxidative addition (OA) is a process that adds two anionic ligands e. g. A and B, that originally are a part of a A-B molecule, like in H2 or Me−I, on to a metal center and is of significant importance from the perspective of both synthesis and catalysis. The exact reverse of the same process, in which the two ligands, A and B, are eliminated from the metal center forming back the A−B molecule, is called the reductive elimination (RA). As A and B are anionic X type ligands, the oxidative addition is accompanied by an increase in the coordination number, valence electron count as well as in the formal oxidation state of the metal center by two units. The oxidative addition step may proceed by a variety of pathways. It requires the metal center to be both coordinatively unsaturated and electron deficient.
Oxidative addition transfers a single mononuclear metal center having 16 VE to a 18 VE species upon oxidative addition. Another frequently observed pathway is that a 18 VE complex looses a ligand to become a 16 VE species which then undergoes an oxidative addition. Apart from above two types, another possible pathway for oxidative addition proceeds as a binuclear oxidative addition in which each of the two metal centers undergo change in oxidation state, electron count and coordination number by one unit instead of two. This type of a binuclear oxidative addition is observed for a 17 VE metal complex or for a binuclear 18 VE metal complex having a metal−metal bond and, for which the metal has a stable oxidation state at a higher positive oxidation state by one unit.
It is interesting to note that in the oxidative addition the breakage of A−B σ−bond occurs as a result of a net transfer of electrons from the metal center to a σ*−orbital of the A−B bond, thus resulting in the formation of the two new M−A and M−B bonds. The oxidative addition is facilitated by electron rich metal centers having low oxidation state whereas the reductive elimination is facilitated by metal centers in higher oxidation state.
Table $1$. Common types of oxidative addition reactions.
Abbreviations: Lin. = linear, Tet. = tetrahedral, Oct. = octahedral, Sq. Pl. = square planar, TBP = trigonal bipyramidal, Sq. Pyr. = square pyramidal: 7-c, 8-c = 7- and 8-coordinate.
In principle, the oxidative addition is the reverse of reductive elimination, but in practice one may dominate over the other. Thus, the favorability of one over the other is depends on the position of equilibrium, which is further dependent on the stability of the two oxidation states of the metal and on the difference of bond strengths of A−B versus that of the M−A and M−B bonds. For example, metal hydride complexes frequently undergo reductive elimination to give alkanes but rarely an alkane undergoes oxidative addition to give an alkyl hydride complex. Along the same line, alkyl halides frequently undergo oxidative addition to a metal giving metal−alkyl halide complexes but these complexes rarely reductively eliminate to give back alkyl halides. Usually the oxidative addition is more common for 3rd row transition metals because they tend to possess stronger metal ligand bond strengths. The oxidative addition is also favored by strong donor ligands, as they stabilize the higher oxidation state of the metal. The oxidative addition reaction can expand beyond transition metals as observed in the case of the Grignard reagents as well as for some main group elements.
Oxidative addition may proceed by several pathways as discussed below.
Concerted oxidative addition pathway
Oxidative addition may proceed by a concerted 3−centered associative mechanism involving the incoming ligand with the metal center. Specifically, the addition proceeds by the formation of a σ−complex upon binding of an incoming ligand say, H2, followed by the cleavage of the H−H bond as a result of the back donation of electrons from the metal to the σ*−orbital of the H−H bond. Such type of addition is common for the H−H, C−H and Si−H bonds. As expected these proceed by two steps (i) the formation of a σ−complex and (ii) the oxidation step. For example, the oxidative addition of H2 to Vaska’s complex (PMe3)2Ir(CO)Cl proceeds by this pathways.
SN2 pathway
This pathway of oxidative addition is operational for the polarized AB type of ligand substrates like the alkyl, acyl, allyl and benzyl halides. In this mechanism, the LnM fragment directly donates electrons to the σ*−orbital of the A−B bond by attacking the least electronegative atom, say A, of the AB molecule and concurrently initiating the elimination of the most electronegative atom of the AB molecule in its anionic form, B. These reactions proceed via a polar transition state that is accompanied by an inversion of the stereochemistry at the atom of attack by the metal center and are usually accelerated in polar solvents.
Radical pathway
This type of oxidative addition proceeds via a by radical pathway that generally are vulnerable to the presence of impurities. The radical processes can be of non−chain and chain types. In a non−chain type of mechanism, the metal (M) transfer one electron to the σ*−orbital of the RX bond resulting in the formation of a radical cation M+• and a radical anion RX−•. The generation of the two radical fragments occurs by the way of the elimination of the anion X from the radical anion RX−• leaving behind the radical R while the subsequent reaction of X anion with the radical cation M+• generates the other radical MX in the course of the reaction. Such type of non−chain type of oxidative addition is observed for the addition of the alkyl halide to Pt(PPh3)3 complexes.
$\ce{PtL3 ->[fast] PtL2} \nonumber$
$\ce{PtL2 + RX -> ^{•}PtL2 + ^{•-}RX ->[slow] ^{•}PtXL2 + ^{•}R} \nonumber$
$\ce{^{•}PtXL2 + ^{•}R ->[fast] RPtXL2} \nonumber$
The other type in this category is the chain radical type reaction that is usually observed for the oxidative addition of EtBr and PhCH2Br to the (PMe3)2Ir(CO)Cl complex. For this process a radical initiator is required and the reaction proceeds along a series of known steps common to a radical process.
$\ce{R^{•} + IR^{I}Cl(CO)L2 -> RIr^{II•}Cl(CO)L2} \nonumber$
$\ce{RIr^{II•}Cl(CO)L2 + RX -> RXIr^{III}Cl(CO)L2 + R^{•}} \nonumber$
$\ce{2R^{•} -> R2} \nonumber$
Ionic pathway
This is kind of pathway for the oxidative addition reaction is common to the addition of hydrogen halides (HX) in its dissociated H+ and Xforms. The ionic pathways are usually of the following two types (i) the ones in which the starting metal complex adds to H+ prior to the addition of the halide Xand (ii) the other type, in which the halide anion Xadds to the starting metal complex first, and then the addition of proton H+ occurs on the metal complex.
Reductive Elimination
The reductive eliminations are reverse of the oxidative addition reactions and are accompanied by the reduction of the formal oxidation state of the metal and the coordination numbers by two units. The reductive eliminations are commonly observed for d8 systems, like the Ni(II), Pd(II) and Au(III) ions and the d6 systems, like the Pt(IV), Pd(IV), Ir(III) and Rh(III) ions. The reaction may proceed by the elimination of several groups.
$\ce{L_{n}MRH -> L_{n}M + R-H} \nonumber$
$\ce{L_{n}MR2 -> L_{n}M + R-R} \nonumber$
$\ce{L_{n}MH(COR) -> L_{n}M + RCHO} \nonumber$
$\ce{L_{n}MR(COR) -> L_{n}M + R2CO} \nonumber$
$\ce{L_{n}MR(SiR3) -> L_{n}M + R-SiR3} \nonumber$
Binuclear Reductive Elimination
Similar to what has been observed in the case of binuclear oxidative addition, the binuclear reductive elimination is also observed in some instances. As expected, the oxidation state and the coordination number decrease by one unit in the binuclear reductive elimination pathway.
$\ce{2MeCH=CHCu(PBu3) ->[heat] MeCH=CHCH=CHMe} \nonumber$
$\ce{ArCOMn(CO)5 + HMn(CO)5 -> ArCHO + Mn2(CO)10} \nonumber$
Problems
1. What kind of metal centers favor oxidative addition?
Ans: Electron rich low valent metal centers.
2. Complete the sentence correctly.
(a) Reductive elimination is frequently observed in coordinatively saturated/unsaturated metal complexes.
(b) Reductive elimination is accompanied by increase/decrease in the oxidation state of the metal.
(c) Oxidative addition is accompanied by increase/decrease in the coordination number of the metal
Ans:
(a) Saturated.
(b) Decrease in the oxidation state by two units.
(c) Increase in the coordination number by two units
3. State the various mechanistic pathways involved in oxidative addition reactions.
Ans: Concerted oxidative addition, SN2 mechanism, radical and ionic mechanism.
4. Complete the reaction.
Ans:
Self Assessment test
1. What kind of metal centers favor reductive elimination?
Ans: Electron deficient high valent metal centers
2. Complete the sentence correctly.
(a) Oxidative addition is frequently observed in coordinatively saturated/unsaturated metal complexes.
(b) Oxidative addition is accompanied by increase/decrease in the oxidation state of the metal.
(c) Reductive elimination is accompanied by increase/decrease in the coordination number of the metal.
Ans:
(a) Unsaturated.
(b) Increase in the oxidation state by two units.
(c) Decrease in the coordination numbers by two units.
3. How does the geometry of the square planar complexes change upon oxidative addition reactions?
Ans: Square planar to octahedral.
4. Complete the reaction.
Ans:
Summary
The oxidative addition and the reductive elimination reactions are like the observe and reverse of the same coin. The oxidative addition is generally observed for metal centers with low oxidation state and is usually accompanied by the increase in the oxidation state, the valence electron count and the coordination number of the metal by two units. Being opposite, the reductive elimination is seen in the case of the metal centers with higher oxidation state and is accompanied by the decrease in the oxidation state, the valence electron count and the coordination number of the metal by two units. The oxidative addition may proceed by a variety of pathways that involve concerted, ionic and the radical based mechanisms. More interestingly, the oxidative addition and reductive elimination reactions are not solely restricted to the mononuclear metal complexes but can also be observed for the binuclear complexes. | textbooks/chem/Inorganic_Chemistry/Introduction_to_Organometallic_Chemistry_(Ghosh_and_Balakrishna)/10%3A_Reaction_Mechanisms/10.01%3A_Oxidative_Addition_and_Reductive_Elimination.txt |
Learning Objectives
In this lecture you will learn the following
• The insertion and elimination reactions.
• The various mechanistic pathways by which these reactions proceed.
• Their relevance in some important catalytic cycles.
Unlike what we have learned about the oxidative addition and the reductive elimination reactions, that facilitate the addition or removal of 1−electron and 2−electron ligands on to a metal center, the insertion and the elimination reactions perform the subsequent transformation of these ligands from within the same coordination sphere of a metal. Thus, in an insertion reaction a metal bound 2 electron A=B type of a ligand can insert on to a M−X bond resulting in a new metal bound 1−electron ligand like, M−A−B−X, which is formed as a result of the formations of the M−A and B−X bonds. The insertion reaction thus leads to the generation of one vacant site created at the initial metal bound A=B site. Thus a primary requirement for reverse elimination reaction to occur is the presence of a cis vacant site.
Insertions are of two types, 1,1−insertion and 1,2−insertion. In 1,1−insertion both the metal M and the ligand X of the M−X bond end up on the same atom like in the M−A(X)−B moiety formed after the insertion of the A=B molecule in the M−X bond, whereas in the 1,2−insertion, these end up on the adjacent atoms like in the M−A−B−X moiety formed after the insertion of the A=B molecule in the M−X bond. The type of insertion depends on the type of the ligand undergoing the insertion like η1−ligand showing 1,1−insertion and η2−ligands showing 1,2−insertion. For example, the CO ligand exclusively undergoes 1,1−insertion while the C2H4 ligand undergoes 1,2−insertion. The SO2 ligand remains the only exception as it can bind by both η1−(by S−donor site) and η2−(by S− and O−donor sites) modes and thus shows both type of insertions.
Though the insertion and the elimination reactions are mutually reversible, owing to the thermodynamical reasons one is favored over the other. For example, SO2 is known to insert into the M−R bond with no report of its elimination is known of it, while for N2 ligand, no report of its insertion is known but it’s elimination from a M−N=N−R bond is known.
$\ce{M-R + SO2 -> M-SO2R} \nonumber$
$\ce{M-N=N-Ar -> M-Ar = N2} \nonumber$
The CO ligand inserts readily into a metal−alkyl bond. Sterically demanding substituents (R) is found to accelerate the reaction as a bulky R group in an acyl moiety in the final M−CO−R bond is far removed from the metal center than that in the starting M−R bond.
Isonitriles readily insert into M−R and M−H bonds giving η2-bound iminoacyls.
Olefins usually inserts across a M−H bond and such insertions are of relevance to the commercially important olefin polymerization process. In certain cases the 1,2−insertions of olefins give species exhibiting agostic insertions.
ß-Elimination
β−elimination is just a reverse of 1,2−insertion and is a major cause of decomposition of metal alkyl bond having a b−hydrogen atom. A pre-requisite for the β−elimination reaction to occur is the presence of an adjacent vacant site next to the metal alkyl bond undergoing the β−elimination. The β−elimination step results in the formation of a metal hydride species that also contain a metal bound olefin moiety.
α-elimination
In absence of a β−hydrogen, a metal bound alkyl moiety may undergo the cleavage of a C−H bond at the α, γ and δ positions. For example, a methyl moiety may α−eliminate to give a metal bound methylene hydride moiety.
Problems
1. Give an example of a ligand that undergo 1,1−insertion.
Ans: CO
2. Give an example of a ligand that undergo 1,2−insertion.
Ans: C2H4
3. Give an example of a ligand that undergo both the 1,1−insertion and the 1,2−insertion reactions.
Ans: C2H4
Self Assessment test
1. On what type of bonds does CO insertion usually occur?
Ans: Metal−alkyl (M−R) bonds.
2. On what type of bonds does isonitrile insertion usually occur?
Ans: Metal-alkyl (M-R) and the metal-hydride (M-H) bonds.
3. On what type of bonds does C2H4 insertion usually occur?
Ans: Metal-hydride (M-H) bonds.
4. State an important requirement for the occurrence of an elimination reaction.
Ans: The presence of an adjacent vacant site.
Summary
Insertion and elimination reactions are important sequences that carryout transformation of the metal bound ligands to the corresponding product from within the same coordination sphere of the metal center and thus together represent key steps of an overall catalytic cycle. The insertions are highly ligand dependent and may proceed by 1,1−insertion and 1,2−insertion mechanisms. Elimination reactions are just reverse of the insertion reactions and they too may proceed by several pathways. The two pathways that are commonly observed are the β−elimination and the α−eliminations, even though other type of elimination pathways exists. | textbooks/chem/Inorganic_Chemistry/Introduction_to_Organometallic_Chemistry_(Ghosh_and_Balakrishna)/10%3A_Reaction_Mechanisms/10.02%3A_Insertion_and_Elimination_Reactions.txt |
Learning Objectives
In this lecture you will learn the following
• Ligand activation by metal that leads to a direct external attack at the ligand.
• Nucleophilic addition and nucleophilic abstraction reactions.
• Electrophilic addition and electrophilic abstraction reactions.
The nucleophilic and electrophilic substitution and abstraction reactions can be viewed as ways of activation of substrates to allow an external reagent to directly attack the metal activated ligand without requiring prior binding of the external reagent to the metal. The attacking reagent may be a nucleophile or an electrophile. The nucleophilic attack of the external reagent is favored if the LnM fragment is a poor π−base and a good σ−acid i.e., when the complex is cationic and/or when the other metal bound ligands are electron withdrawing such that the ligand getting activated gets depleted of electron density and can undergo an external attack by a nucleophile Nu, like LiMe or OH. The attack of the nucleophiles may result in the formation of a bond between the nucleophiles and the activated unsaturated substrate, in which case it is called nucleophilic addition, or may result in an abstraction of a part or the whole of the activated ligand, in which case it is called the nucleophilic abstraction. The nucleophilic addition and the abstraction reactions are discussed below.
Nucleophilic addition
An example of a nucleophilic addition reaction is shown below.
Carbon monoxide (CO) as a ligand can undergo nucleophilic attack when bound to a metal center of poor π−basicity, as the carbon center of the CO ligand is electron deficient owing to the ligand to metal σ−donation not being fully compensated by the metal to ligand π−back donation. Thus, activated CO ligand undergoes nucleophilic attack by the lithium reagent to give an anionic acyl ligand, which upon alkylation generates the famous Fischer carbene complex.
Nucleophilic abstraction
An example of a nucleophilic abstraction reaction is shown below.
Electrophilic addition
Similar to the nucleophilic addition and abstraction reactions, the electrophilic counterparts of these reactions also exist. An electrophilic attack is favored if the LnM fragment is a good π−base and a poor σ−acid i.e., when the complex is anionic with the metal center at low−oxidation state and/or when the other metal bound ligands are electron donating such that the ligand getting activated becomes electron rich from the π−back donation of the metal center and thus can undergo an external attack by an electrophile E+ like H+ and CH3I. The attack of the electrophiles may result in the formation of a bond between the electrophile and the activated unsaturated substrate, in which case it is called electrophilic addition, or may result in an abstraction of a part or the whole of the activated ligand, in which case it is called the electrophilic abstraction.
Electrophilic abstraction
An example of an electrophilic abstraction reaction is shown below.
Alkyl abstractions are often achieved by Hg2+ that can proceed in two ways, (i) by an attack at the α−carbon of a metal alkyl bond leading to an inversion of configuration at the alkyl carbon and (ii) by an attack at the metal center leading to retention of configuration at the alkyl carbon. The inversion of configuration proceeds by the following pathway.
The retention of configuration proceeds by the following pathway. | textbooks/chem/Inorganic_Chemistry/Introduction_to_Organometallic_Chemistry_(Ghosh_and_Balakrishna)/10%3A_Reaction_Mechanisms/10.03%3A_Nucleophilic_and_Electrophilic_Addition_and_Abstraction.txt |
Learning Objectives
In this lecture you will learn the following
• The application of organometallic complexes in homogeneous catalysis.
• Alkene isomerization.
• Alkene and the arene hydrogenations.
• Transfer hydrogenation.
One of the most important exploits of the organometallic chemistry is its application in the area of homogeneous catalysis. The field has now expanded its territory to accommodate in equal measures many large-scale industrial processes as well as numerous small scale reactions of the day-to-day organic synthesis. A few representative examples of organometallic catalysis are outlined below.
Alkene Isomerization
Alkene isomerization is a transformation that involve a shift of a double bond to an adjacent position followed by 1,3−migration of a H atom. The isomerization reaction is transition metal catalyzed.
The alkene isomerization reaction may proceed by two pathways, (i) one through a η1−alkyl intermediate and (ii) the other through η3−allyl intermediate. In the η1−alkyl pathway, an alkene first binds to a metal at a vacant site next to M−H bond and then subsequently undergoing an insertion into the M−H bond thus creating back the vacant site. The resultant species then undergoes a H atom transfer from the alkyl moiety to give the isomerized olefin along with the regeneration of the M−H species.
The η3−allyl mechanism requires the presence of two vacant sites. This mechanism goes through a η3−allyl intermediate formed by a C−H activation at the allylic position of the olefin formed after binding to the metal and alongside leads to the formation of a M−H bond. Subsequent H transfer from the metal back to the η3−allyl moiety leads to the alkene isomerized product.
Alkene Hydrogenation
The transition metal catalyzed alkene hydrogenation reactions are of significant industrial and academic interest. These reactions involve the H2 addition on a C=C bond of olefins to give alkenes. The alkene hydrogenation may proceed by three different pathways namely the (i) oxidative addition (ii) heterolytic activation and (iii) the homolytic activation of the H2 molecule.
The oxidative addition pathway is commonly observed for the Wilkinson’s catalyst (PPh3)3RhCl and is the most studied among all of the three pathways that exist. The catalytic cycle initiates with the oxidative addition of H2 followed by alkene coordination. The resultant species subsequently get converted to the hydrogenated product.
The second pathway proceeds by the heterolytic activation of the H2 molecule and requires the presence of a base like NEt3, which facilitates the heterolytic cleavage by abstracting a proton from the H2 molecule and leaving behind a hydride H ion that participates in the hydrogenation reaction. This type of mechanism is usually followed by the (PPh3)3RuCl2 type of complexes.
$\ce{X-Y + M-Z -> M-X + Y-Z} \nonumber$
Homolytic cleavage of H2 is the third pathway for the alkene hydrogenation. It is the rarest of all the three methods and proceeds mainly in a binuclear pathway. Paramagnetic cobalt based Co(CN)53− type catalysts carries out alkene hydrogenation by this pathway via the formation of the HCo(CN)53−species.
Arene hydrogenations
Examples of homogeneous catalysts for arene hydrogenation are rare though it is routinely achieved using catalysts like Rh/C under the heterogeneous conditions. A representative example of a homogeneous catalyst of this class is (η3−allyl)Co[P(OMe)3]3 that carry out the deuteration of benzene to give the all-cis-C6H6D6 compound.
Transfer hydrogenation
This is a new kind of a hydrogenation reaction in which the source of the hydrogen is not the H2molecule but an easily oxidizable substrate like isopropyl alcohol. The method is particularly useful for the reduction of ketones and imines but not very effective for the olefins.
$\ce{Me2CHOH + RCH=CH2 -> Me2C=O + RCH2CH3} \nonumber$
Summary
The applications of organometallic compounds in homogeneous catalysis have transcend the boundaries of industry to meet the day-to-day synthesis in laboratory scale reactions. The alkene isomerization is one such application of homogeneous catalysis by the transition metal organometallic complexes. The hydrogenation reactions of alkene, arene, ketone and imine substrates are achieved by several types of the transition metal organometallic catalysts. They also proceed by different mechanisms involving oxidative addition, heterolytic and homolytic cleavages of the H−H bond. The transfer hydrogenation reaction uses easily oxidizable substrates like i−PrOH instead of H2 as the hydrogenation source.
11.02: Homogeneous Catalysis - II
Learning Objectives
In this lecture you will learn the following
• The hydroformylation reaction and its mechanism.
• The C−C cross-coupling reactions and their mechanisms.
It is truly an exciting time for the field of organometallic chemistry as its potentials in homogeneous catalysis are being realized in an unprecedented manner. The growth in the field organometallic chemistry has been rightly acknowledged by the award of three Nobel prizes in over a decade in the areas of asymmetric hydrogenation (Nyori and Knowles in 2001), olefin metathesis (Grubbs, Schrock and Chauvin in 2006) and palladium mediated C−C cross coupling reactions (Suzuki, Negishi and Heck, 2010). A few representative examples of such landmark discoveries of homogeneous catalysis by organometallic compounds are discussed below.
Hydroformylation reaction
Hydroformylation, popularly known as the "oxo" process, is a Co or Rh catalyzed reaction of olefins with CO and H2 to produce the value-added aldehydes.
The reaction, discovered by Otto Roelen in 1938, soon assumed an enormous proportion both in terms of the scope and scale of its application in the global production of aldehydes. The metal hydride complexes namely, the rhodium based HRh(CO)(PPh3)3 and the cobalt based HCo(CO)4 complexes, catalyzed the hydroformylation reaction as shown below.
C−C cross-coupling reactions
The palladium catalyzed cross-coupling reactions are a class of highly successful reactions with applications in the organic synthesis to have emerged recently. The reactions carry out a coupling of the aryl, vinyl or alkyl halide substrates with different organometallic nucleophiles and as such encompasses a family of C−C cross-coupling reactions that are dependent on the nature of nucleophiles like that of the B based ones in the Suzuki-Miyuara coupling, the Sn based ones in the Stille coupling, the Si based ones in the Hiyama coupling, the Zn based ones in the Negishi coupling and the Mg based ones in the Kumada coupling reactions (Figure \(1\)).
An unique feature of these reactions is the exclusive formation of the cross-coupled product without the accompaniment of any homo-coupled product. Another interesting feature of these coupling reactions is that they proceed via a common mechanism involving three steps that include the oxidative addition, the transmetallation and the reductive elimination reactions (Figures \(2\) and \(3\)).
Summary
Organometallic complexes play a pivotal role in several successful homogeneous catalysis reactions like that of the hydroformylation and the C−C cross-coupling reactions. These reactions are important because of the fact that both of the hydroformylation and the C−C cross-coupling reactions give more value added products compared to the starting reactants. The palladium catalyzed C−C cross-coupling reactions are a class of highly successful reactions that have permanently impacted the area of organic synthesis in a profound way to an extent that the 2010 Nobel prize has been conferred on one of these reactions thereby recognizing the importance of the C−C cross-coupling recations. | textbooks/chem/Inorganic_Chemistry/Introduction_to_Organometallic_Chemistry_(Ghosh_and_Balakrishna)/11%3A_Applications/11.01%3A_Homogeneous_Catalysis_-_I.txt |
Learning Objectives
In this lecture you will learn the following
• The characterization techniques of organometallic compounds.
• The NMR analysis of these compounds.
• The IR analysis of these compounds.
• The X-ray single crystal diffraction studies of these compounds.
The characterization of an organometallic complex involves obtaining a complete understanding of the same right from its identification to the assessment of its purity content, to even elucidation of its stereochemical features. Detailed structural understanding of the organometallic compounds is critical for obtaining an insight on its properties and which is achieved based on the structure-property paradigm.
Synthesis and isolation
Synthesis and isolation are two very important experimental protocols in the overall scheme of things of organometallic chemistry and thus these needs to be performed carefully. The isolation of the organometallic compounds is essential for their characterization and reactivity studies. Fortunately, many of the methods of organic chemistry can be used in organometallic chemistry as the organometallic compounds are mostly nonvolatile crystalline solids at room temperature and atmospheric pressure though a few examples of these compounds are known to exist in the liquid [(CH3C5H4Mn(CO)3] and even in the vapor [Ni(CO)4] states. The organometallic compounds are comparatively more sensitive to aerial oxygen and moisture, and because of which the manipulation of these compounds requires stringent experimental skills to constantly provide them with anaerobic environment for their protection. All of these necessities led to the development of the so-called special Schlenk techniques, requiring special glasswares and which in conjunction with a high vacuum line and a dry box allow the lab bench-top manipulation of these compounds. Successful isolation of organometallic compounds naturally points to the need for various spectroscopic techniques for their characterizations and some of the important ones are discussed below.
1H NMR spectroscopy
The 1H NMR spectroscopy is among the extensively used techniques for the characterization of organometallic compounds. Of particular interest is the application of 1H NMR spectroscopy in the characterization of the metal hydride complexes, for which the metal hydride moiety appear at a distinct chemical shift range between 0 ppm to −40 ppm to the high field of tetramethyl silane (TMS). This upfield shift of the metal hydride moiety is attributed to a shielding by metal d−electrons and the extent of the upfield shift increases with higher the dn configuration. Chemical shifts, peak intensities as well as coupling constants from the through-bond couplings between adjacent nuclei like that of the observation of JP-H, if a phosphorous nucleus is present within the coupling range of a proton nucleus, are often used for the analysis of these compounds. The 1H NMR spectroscopy is often successfully employed in studying more complex issues like fluxionality and diastereotopy in organometallic molecules (Figure $1$).
The paramagnetic organometallic complexes show a large range of chemical shifts, for example, (η6−C6H6)2V exhibits proton resonances that extend even up to 290 ppm.
13C NMR spectroscopy
Although the natural abundance of NMR active 13C (I = ½) nuclei is only 1 %, it is possible to obtain a proton decoupled 13C{1H} NMR spectra for most of the organometallic complexes. In addition, the off−resonance 1H decoupled 13C experiments yield 1JC-H coupling constants, which contain vital structural information, and hence are very critical to the 13C NMR spectral analysis. For example, the 1JC-H coupling constants directly correlate with the hybridization of the C−H bonds with sp center exhibiting a 1JC-H coupling constant of ~250 Hz, a sp2 center of 160 Hz and a sp3 center of 125 Hz. Similar to what is seen in 1H NMR, a phosphorous−carbon coupling is also observed in a 13C NMR spectrum with the trans coupling (~100 Hz) being larger than the cis coupling (~10 Hz).
31P NMR spectroscopy
The 31P NMR spectroscopy, which in conjunction with 1H and 13C NMR spectroscopies, is a useful technique in studying the phosphine containing organometallic complexes. The 31P NMR experiments are routinely run under 1H decoupled conditions for simplification of the spectral features that allow convenience in spectral analysis. Thus, for this very reason, many mechanistic studies on catalytic cycle are conveniently undertaken by 31P NMR spectroscopy whenever applicable.
IR spectroscopy
Qualitative to semi-quantitative analysis of organometallic compounds using IR spectroscopy are performed whenever possible. In general the signature stretching vibrations for chemical bonds are more conveniently looked at in these studies. The frequency ($\nu$) of a stretching vibration of a covalent bond is directly proportional to the strength of the bond, usually given by the force constant (k) and inversely proportional to the reduced mass of the system, which relates to the masses of the individual atoms.
$\nu = \frac{1}{2\pi C} \:\sqrt {\frac{k}{m_{r}}} \nonumber$
$m_{r} = \frac{m_{1}m_{2}}{m_{1} + m_{2}} \nonumber$
The organometallic compounds containing carbonyl groups are regularly studied using IR spectroscopy, and in which the $\ce{CO}$ peaks appear in the range between 2100−1700 cm-1 as distinctly intense peaks.
Crystallography
The solid state structure elucidation using single crystal diffraction studies are extremely useful techniques for the characterization of the organometallic compounds and for which the X-ray diffraction and neutron diffraction studies are often undertaken. As these methods give a three dimensional structural rendition at a molecular level, they are of significant importance among the various available characterization methods. The X-ray diffraction technique is founded on Bragg’s law that explains the diffraction pattern arising out of a repetitive arrangement of the atoms located at the crystal lattices.
$2d \sin \theta =n \lambda \nonumber$
A major limitation of the X-ray diffraction is that the technique is not sensitive enough to detect the hydrogen atoms, which appear as weak peaks as opposed to intense peaks arising out of the more electron rich metal atoms, and hence are not very useful for metal hydride compounds. Neutron diffraction studies can detect hydrogens more accurately and thus are good for the analysis of the metal hydride complexes.
Summary
Along with the synthesis, the isolation and the characterization protocols are also integral part of the experimental organometallic chemistry. Because of their air and moisture sensitivities, specialized experimental techniques that succeed in performing the synthesis, isolation and storage of these compounds in an air and moisture-free environment are often used. The organometallic compounds are characterized by various spectroscopic techniques including the 1H NMR, 13C NMR and IR spectroscopies and the X-ray and the neutron diffraction studies. | textbooks/chem/Inorganic_Chemistry/Introduction_to_Organometallic_Chemistry_(Ghosh_and_Balakrishna)/12%3A_Physical_Methods_in_Organometallic_Chemistry/12.01%3A_Characterization_of_Organometallic_Complexes.txt |
• 13.1: Metal-Carbenes
The organometallic compounds containing metal−ligand multiple bonds of the types, M=X and M≡X (X = C, N, O) are of current interest as they are valuable intermediates in many important catalytic cycles. In this regard, considerable attention has been paid towards developing an understanding of the metal−ligand multiply bonded systems like that of the metal carbene LnM=CR2 type complexes and of the metal carbyne LnM≡CR type complexes.
• 13.2: Metal-Carbynes
The metal−ligand multiply bonded systems even extended beyond the doubly bonded Fischer and the Schrock carbenes to the triply bonded LnM≡CR type Fischer carbyne and the Schrock carbyne complexes. Similar to carbene that exists in a singlet and a triplet spin state, the carbyne also exists in two other spin states i.e in a doublet and a quartet form.
13: Multiply-Bonded Ligands
Learning Objectives
In this lecture you will learn the following
• The metal−ligand multiple bonding and their relevance in.
• The Fischer type carbene complexes.
• The Schrock type carbene complexes.
The organometallic compounds containing metal−ligand multiple bonds of the types, M=X and M≡X (X = C, N, O) are of current interest as they are valuable intermediates in many important catalytic cycles. In this regard, considerable attention has been paid towards developing an understanding of the metal−ligand multiply bonded systems like that of the metal carbene LnM=CR2 type complexes and of the metal carbyne LnM≡CR type complexes. A detailed account of the metal−carbene complexes is presented in this chapter.
Metal carbene complexes
Carbenes are highly reactive hexavalent species that exist in two spin states, i.e. (i) in a singlet form (), in which two electrons are paired up and (ii) in a triplet form (), in which the two electrons remain unpaired. Of the two, the singlet form is the more reactive one. The instability of carbene accounts for its unique reactivity like that of the insertion reaction, which has aroused significant interest in recent years. The singlet carbene and the triplet carbene bind differently to metals, with the singlet one yielding Fischer type carbene complexes while the triplet one yielding Schrock type carbene complexes (Figure $1$).
The LnM=CR2 type Fischer carbene complexes comprise of two dative covalent interactions that include (i) a LnM←CR2 type ligand to metal σ−donation and (ii) a LnM→CR2 type metal to ligand π−back donation. The Fischer type carbene complexes are usually formed with metal centers at a low oxidation state. These are also commonly observed for the more electron rich late−transition metals that participate in the LnM→CR2 type metal to ligand π−back donation. Another characteristic of the Fischer type carbene complex is the presence of the heteroatom substituents like R = OMe or NMe2 on the carbene CR2 moiety which makes the carbene carbon significantly cationic (δ+) to facilitate the LnM→CR2 type metal to ligand π−back donation.
Similarly, the LnM=CR2 type Schrock carbene complexes comprise of two covalent interactions that involve one electron donation towards the σ−bond from each of the metal LnM and the carbene CR2fragments. Schrock carbene complexes are thus formed with the metal centers having high oxidation state and are usually observed for electron deficient early−transition metals (Figure $2$).
Carbene complexes can be prepared by the following methods.
i. by the reaction with electrophiles
ii. by H/H+ abstraction reactions as shown below
iii. from low−valent metal complexes
$\ce{L_{n}M + CH2N2 -> L_{n}MCH2 + N2} \nonumber$
Because of the electronically different metal−ligand interaction that exist between the LnM and the carbene CR2 moiety, the reactivity of Fischer and Schrock carbene complexes are completely different. For example, the Fischer type carbene complexes undergo attack by nucleophiles at its carbene−C center.
The Schrock type carbene complexes on the other hand undergo attack by electrophiles at its carbene−C center.
Summary
The metal−ligand multiple bonding is of significant interest as many of the compounds containing such bonds are important intermediates in various catalytic cycles. The metal−ligand doubly bonded carbene systems can exist in two varieties like the Fischer type and the Schrock type carbene complexes. Due to their different electronic structures, the reactivities of these Fischer type and the Schrock type carbene complexes differ significantly, with the former undergoing nucleophilic attack while the later undergo electrophilic attack at their respective carbene−C centers. | textbooks/chem/Inorganic_Chemistry/Introduction_to_Organometallic_Chemistry_(Ghosh_and_Balakrishna)/13%3A_Multiply-Bonded_Ligands/13.01%3A_Metal-Carbenes.txt |
Learning Objectives
In this lecture you will learn the following
• The metal−ligand multiple bonding and their relevance in.
• The Fischer type carbyne complexes.
• The Schrock type carbyne complexes.
The metal−ligand multiply bonded systems even extended beyond the doubly bonded Fischer and the Schrock carbenes to the triply bonded LnM≡CR type Fischer carbyne and the Schrock carbyne complexes. Similar to carbene that exists in a singlet and a triplet spin state, the carbyne also exists in two other spin states i.e in a doublet and a quartet form.
Upon binding to the metal in its doublet spin state as in the Fischer carbene system, the carbyne moiety donates two electrons via its sp hybridized lone pair containing orbital to an empty metal d orbital to yield a LnM←CR type ligand to metal dative bond. It also makes a covalent π−bond through one of its singly occupied pz orbital with one of the metal d orbitals. The carbyne−metal interaction consist of two ligand to metal interactions namely a dative one and a covalent one that together makes the carbyne moiety a LX type of a ligand. In addition to these two types of ligand to metal bonding interactions, there remains an empty py orbital on the carbyne−C atom that can accommodate electron donation from a filled metal d orbital to give a metal to ligand π−back bonding interaction (Figure $1$).
Analogously, in the quartet carbyne spin state in the Schrock carbyne systems three covalent bonds occur between the singly occupied sp, pyand pz orbitals of carbyne−C moiety with the respective singly occupied metal d orbitals (Figure $1$).
Similar to what has been observed earlier in the case of the Fischer carbenes and Schrock carbenes, the Fischer carbyne complexes are formed with metal centers in lower oxidation states for e.g. as in Br(CO)4W≡CMe, while the Schrock carbyne complexes are formed with metals in higher oxidation state, e.g. as in (t−BuO)3W≡Ct−Bu.
Carbyne complexes can be prepared by the following methods.
i. The Fischer carbyne complexes can be prepared by the electrophilic abstraction of a methoxy group from a methoxy methyl substituted Fischer carbene complex.
$\ce{L(CO)4M=C(OMe)Me + 2BX3 -> [L(CO)4M≡CMe]+BX_{4}^{-} + BX2(OMe) -> X(CO)4M≡CMe} \nonumber$
ii. Schrock carbynes can be prepared by the deprotonation of a α−CH bond of a metal−carbene complex.
$\ce{CPCl2Ta=CHR ->[(i)PMe3][(ii)Ph3P=CH2] Cp(PMe3)ClTa≡CR} \nonumber$
iii. by an α−elimination reaction on a metal−carbene complex
$\ce{Cp*Br2Ta=CHt-Bu ->[(i)dmpe][(ii)Na/Hg] Cp*(dmpe)HTa≡Ct-Bu} \nonumber$
iv. by metathesis reaction
$\ce{(t-BuO)3W≡W(Ot-Bu)3 + t-BuC≡Ct-Bu -> 2(t-BuO)3W≡Ct-Bu} \nonumber$
The reactivities of Fischer and the Schrock carbynes mirror that of the Fischer and Schrock carbenes. For example, the Fischer carbyne undergo nucleophilic attack at the carbyne−C atom while the Schrock carbyne undergo electrophilic attack at the carbyne−C atom.
Summary
The theme of metal−ligand multiple bonding extends beyond the doubly bonded Fischer and the Schrock carbene systems to even triply bonded Fischer and the Schrock carbyne systems. The carbyne moieties in these Fischer and the Schrock carbyne systems respectively exist in a doublet and a quartet spin state. The carbyne complexes are generally prepared from the respective carbene analogues by the abstraction of alkoxy (OR), proton (H+), hydride (H) moieties, the α−elimination reactions and the metathesis reactions. The reactivity of the Fischer and the Schrock carbyne complexes parallel the corresponding Fischer and the Schrock carbene counterparts with regard to their reactivities toward electrophiles and nucleophiles. | textbooks/chem/Inorganic_Chemistry/Introduction_to_Organometallic_Chemistry_(Ghosh_and_Balakrishna)/13%3A_Multiply-Bonded_Ligands/13.02%3A_Metal-Carbynes.txt |
Learning Objectives
In this lecture you will learn the following
• The alkene metathesis reactions and their different variants.
• The application of metal carbenes in alkene metathesis reactions.
• The functional group tolerance, air and moisture sensitivity and high efficiency as important catalyst attributes for the alkene metathesis reactions.
The application of organometallic chemistry in homogenous catalysis is progressively increasing with the fast pace of discovery of new catalysts in the area. The benefits of organometallic catalysis have now percolated to all facets of the chemical world that span from the confines of the industry to the day-to-day small scale use in organic synthesis in academic laboratories. Quite a few of these applications of organometallic complexes in homogeneous catalysis have made a permanent imprint on the ever going developmental process that is constantly transforming our day-to-day life. An example of such a success story is of alkene metathesis, which is described in this chapter.
Alkene metathesis
Alkene metathesis reactions are gaining wide popularity in synthesizing unsaturated olefinic compounds as well as the unsaturated polymeric counterparts. Central to this catalysis is a metal carbene intermediate that reacts with olefins to give different olefinic compounds or even the unsaturated olefinic polymers depending upon the reaction conditions of the metathesis reaction. Metathesis is an unusual transformation in which a C=C is broken and also formed during catalysis to generate new unsaturated olefins.
$\ce{RCH=CHR + R'CH=CHR' <=> 2RCH=CHR'} \nonumber$
Though a large variety of metal−carbene catalysts have been developed for the metathesis reaction, only a few have been found to be functional group tolerant. Thus a critical step in broadening the utility of metathesis reaction has been in developing catalysts that are functional group tolerant. In this regard, the early-transition metal based carbene catalysts like that of the Ti based ones are highly oxophilic and hence are intolerant to the functional groups. On the other hand, the more electron-rich Mo and W based catalysts are of intermediate character. Finally, the late-transition metal based Ru catalysts are found to be exceptionally tolerant toward functional groups but all the while exhibiting high reactivity toward olefinic bonds. In this context notable are the Grubb’s Ru catalyst, which is easy to handle, and the Schrock’s Mo catalyst, which display high activity.
The metathesis reaction as such stands for a family of related reactions all of which involve a “cutting and stitching” of olefinic bonds leading to different unsaturated products. When two different olefin substrates are used, the reaction is called the “cross metathesis” owing to the fact that the olefinic ends are exchanged.
$\ce{RCH=CHR + R'CH=CHR' <=> 2RCH=CHR'} \nonumber$
The metathesis reactions can even extend further to the conjugated dienes that can undergo Ring Closing Metathesis (RCM) in systems where the ring strain is not too high in the final product. The reverse of Ring Closing Metathesis (RCM) is called the Ring Opening Metathesis (ROM), and which is usually favored in the presence of large excess of C2H4.
The variants of metathesis often used in producing polymers are, (i) the Acyclic Diene Metathesis (ADMET) and (ii) the Ring Opening Metathesis Polymerization (ROMP), in which the relief of ring-strain of cycloalkenes drives the polymerization reaction forward. Both of these reactions, produce long chain polymers in a living fashion and as a result of which these reactions are useful for producing block copolymers −(AAABBBB) n−.
Though several possibilities have been debated for the mechanism of the metathesis reaction, the one proceeding via a metalacyclobutane intermediate has gained credence.
Several important industrial applications have emerged out of the metathesis reaction like that of the commercial synthesis of the housefly pheromone.
$\ce{Me(CH2)7CH=CH2 + Me(CH2)12CH=CH2 -> Me(CH2)7CH=C(CH2)12Me [housefly \: pheromone] + C2H4 + \: other \: products} \nonumber$
Similarly, the polycyclopentadiene polymer, which is formed from the Ring Opening Metathesis Polymerization (ROMP) of dicyclopentadiene substrate, is used for bullet proof related applications because of its exceptional strength owing to its cross-linked nature.
Summary
Alkene metathesis represents a distinct class of related chemical reactions that involve the “cutting and stitching” of olefinic bonds to give unsaturated organic products. Depending upon the nature of the product formed, different type of alkene metathesis reactions exist like the alkene metathesis, cross-metathesis, Ring Closing Metathesis (RCM), Ring Opening Metathesis (ROM), Acyclic Diene Metathesis (ADMET), and the Ring Opening Metathesis Polymerization (ROMP). A commonality that runs through all of these different varieties of the metathesis reaction is its mechanism that involves a catalytically active metal−carbene species. The mechanism is said to be proceed via a 4−membered metalacyclobutane intermediate. The alkene metathesis has found important applications in organic synthesis as well as in the chemical industry. | textbooks/chem/Inorganic_Chemistry/Introduction_to_Organometallic_Chemistry_(Ghosh_and_Balakrishna)/14%3A_Metathesis/14.01%3A_Catalytic_Applications_of_Organometallic_Compounds-_Alkene_Metathesis.txt |
Introduction to Solid State Chemistry is a first-year single-semester college course on the principles of chemistry. This unique and popular course satisfies MIT’s general chemistry degree requirement, with an emphasis on solid-state materials and their application to engineering systems.
01: Lectures
ATOMS
The familiar model of an atom is that of a small nucleus composed of protons and neutrons surrounded by rapidly moving electrons. Typically, the atomic diameter is on the order of $10^{-10} \mathrm{~m}$ while that of the nucleus is on the order of $10^{-15} \mathrm{~m}$. Protons and neutrons have about the same mass (1.00728 and 1.00867 amu respectively) and each is about 1800 times as heavy as an electron. A neutron is electrically neutral, but a proton has a positive charge $\left(+1.6 \times 10^{-19}\right.$ coulomb* ) which is exactly the opposite of the negative charge of an electron. In a neutral atom, the number of electrons around the nucleus equals the number of protons in the nucleus.
The number of protons in the nucleus (the "atomic number", Z) characterizes a chemical element. Atoms of a given element all have the same number of protons, yet may have different masses. The atomic mass number of an atom, $A$, is given by $A=Z+N$, where $N$ is the number of neutrons in the nucleus. Since an element is characterized solely by $Z$, it follows that atoms of a given chemical element may have a varying number of neutrons. Subspecies of chemical elements with the same $Z$ but differing $\mathrm{N}$ and $\mathrm{A}$ are called isotopes. The atomic weight of an element is the weighted average of the atomic masses of the various naturally occurring isotopes of the element, and the atomic weight scale is based on a value of exactly 12, after the carbon isotope that has an atomic mass number of 12.
*
Generally adopted International Unit System.
NUCLEI
The nucleus of an atom weighs less than the sum of the weights of its isolated component particles. The difference between the actual mass and that of the components is called the mass defect. The mass defect, $\Delta \mathrm{m}$, is related to the binding energy within the nucleus, $\Delta \mathrm{E}$ (in Joules), through Einstein's equation:
$\Delta \mathrm{E}=\Delta \mathrm{m} \mathrm{c}^{2}$
where $\mathrm{c}$ is the velocity of light (in $\mathrm{m} \cdot \mathrm{s}^{-1}$ ) and $\Delta \mathrm{m}$ is the mass defect (in $\mathrm{kg}$ ). The nuclear forces which bind protons and neutrons together are strong, and the binding energy per nuclear particle (nucleon) amounts to about $1.4 \times 10^{-12}$ Joules. The greatest nuclear binding energy is found in nuclei of medium atomic number (such as $\mathrm{Fe}$ ) where $\mathrm{N}$ is approximately equal to $Z$. For nuclei of larger atomic number, such as uranium, $N$ is about equal to $1.5 \mathrm{Z}$, and the binding energy per nucleon is less. As a consequence of this decreased nuclear stability, some isotopes (of uranium, for example) are unstable. That is, if the uranium isotope, ${ }_{92} U^{235}(Z=92, A=235)$, is bombarded with neutrons, the following reaction can take place:
$\ce{_{92}^{235}U + ^1n -> _{39}^{94}Y + _{53}^{140}I + 2^1n}$
Here the reaction products are smaller nuclei of increased stability. (In the above convention, subscripts indicate the atomic number and the superscripts the mass number.) Notice that one incident neutron generates fission products including two neutrons - the basis for chain reactions in nuclear reactors and nuclear explosions.
Heavy nuclei (even light nuclei) which have an unfavorable ratio in the number of protons and neutrons can spontaneously decay by the emission of $\alpha$ particles (helium ions) or $\beta$ particles (electrons). These nuclei are referred to as radioactive. The rate at which the decay of such unstable nuclei takes place varies greatly and is indicated by the half-life of the material. In one half-life period, half of the unstable nuclei will have emitted radiation and thus will have changed their character (atomic number). In two half-life periods, only $1 / 4$ of the nuclei will have survived. In three half-lives, only $1 / 8$ of the original nuclei remain, etc. For example, the half-life of gamma-emitting "radio" cobalt, ${ }_{27} \mathrm{Co}^{60}$ (used for $\mathrm{X}$-ray therapy), is 5.3 years, whereas that of radioactive ${ }_{6} \mathrm{C}^{14}$ is 5700 years. (Much more radiation is emitted per second by a given number of $\mathrm{Co}^{60}$ atoms than by the same number of $C^{14}$ atoms.)
EXTRA-NUCLEAR ELECTRONS IN ATOMS
The first atomic theory in quantitative agreement with some experimentally determined facts was proposed in 1913 by Niels Bohr. He postulated (for atomic hydrogen) that:
• The (extra-nuclear) electron can assume only distinct (quantized) energy levels or states.
• In such energy levels, the electrons in motion will not radiate (loose) energy; on changing energy levels, radiation equivalent to the energy difference between the levels is involved.
$\mathrm{E}_{\mathrm{rad}}=\mathrm{E}_{2}-\mathrm{E}_{1}=hv$
• The stable states of the atom involve motion of the electrons in circular orbits.
• The angular momentum of electrons in orbit (mvr) is an integral number of $\mathrm{h} / 2 \pi$ units:
$\mathrm{mvr}=\mathrm{n} \dfrac{\mathrm{h}}{2 \pi}$
• Newtonian mechanics applies to orbiting electrons.
A pictoral view of the Bohr atomic model visualizes electrons orbiting at a velocity ( $\mathrm{v}$ ) in well-defined spherical orbitals of radius ( $r$ ) around the nucleus. The angular momentum of the orbiting electrons can, as stated above, only assume certain values of $n \times h / 2 \pi$ where $n$ (called the principal quantum number) can assume any positive integer value, i.e.1, 2, 3, 4, etc. to infinity (fig. 1).
Hydrogen Atom
Assuming coulombic interaction and the applicability of Newtonian mechanics, the following values may be directly obtained for the hydrogen atom:
• The radius of an electron orbit (spherical) in hydrogen is given by:
\begin{aligned}
&r_{n}=\dfrac{n^{2} h^{2} \varepsilon_{0}}{\pi m^{2}} \quad(n=1,2,3,4, \text { etc. }) \
&r_{n}=n^{2} \times \text { constant }=n^{2}\left(0.529 \times 10^{-10}\right) m
\end{aligned}
It can be seen that the smallest electron orbit in hydrogen - the stable orbit - is given for $n=1$ at $r_{0}=0.529 \AA$. It is also seen that $r$ increases with $n^{2}\left(r=n^{2} r_{0}\right)$.
• The energy of an orbiting electron characterized by the principal quantum number (n) is given as:
$E_{n}=-\dfrac{1}{n^{2}} \times \dfrac{m e^{4}}{8 h^{2} \varepsilon_{0}^{2}}=-\dfrac{1}{n^{2}} \times$ constant [Joule]
This relationship indicates that electron orbits with increasing principal quantum number $(n)$ assume decreasing negative values $\left(1 / n^{2}\right)$ with the limiting value $(n \rightarrow \infty$ ) being zero. For convenience, it is frequently customary to express electronic energies in terms of wave numbers $(\bar{v})$ with the units of $\left[\mathrm{m}^{-1}\right]$. This conversion may readily be made since $E=h v$. Thus:
$\bar{v}=\dfrac{1}{\lambda}=-\dfrac{1}{n^{2}} \times \dfrac{m e^{4}}{8 h^{3} c \varepsilon_{0}^{2}}\left[m^{-1}\right]$
Upon substituting numerical values for all the constants, we obtain:
$\bar{v}=-\dfrac{10973500}{n^{2}}=-\dfrac{1}{n^{2}} \times 10973500\left[\mathrm{~m}^{-1}\right]$
(where $\mathrm{n}$ is again the principal quantum number which may assume the values 1,2,3,4, etc. The value of 10973500 is referred to as the Rydberg constant $\left(\mathrm{R}\right.$; also $\mathrm{R}_{\mathrm{H}}$ and $R_{\infty}$ ) to honor the man who obtained its value from spectroscopic studies prior to the establishment of the Bohr model.)
The energy changes associated with electronic transitions and the accompanying emission of radiation (or absorption of radiation) may simply be calculated as follows:
$\text { Emission }=\Delta \mathrm{E}=\left(\mathrm{E}_{\mathrm{n}_{1}}-\mathrm{E}_{\mathrm{n}_{2}}\right)=h v$
Here $n_{2}$ is the principal quantum number of the outer orbit and $n_{1}$ is that of the inner orbit. In terms of $\bar{v}$, the wave number, we obtain for emission of radiation:
$\bar{v}=\left[\left[-\dfrac{10973500}{n_{1}^{2}}\right]-\left[-\dfrac{10973500}{n_{2}^{2}}\right]\right]$
or
\begin{aligned}
&\Delta \mathrm{E}=10973500\left[\frac{1}{\mathrm{n}_{2}^{2}}-\frac{1}{\mathrm{n}_{1}^{2}}\right] \times \mathrm{hc} \
&\Delta \mathrm{E}=\mathrm{R}\left(\frac{1}{\mathrm{n}_{2}^{2}}-\frac{1}{\mathrm{n}_{1}^{2}}\right] \mathrm{xcc} \
&\Delta \mathrm{E}=2.1798 \times 10^{-18}\left[\frac{1}{\mathrm{n}_{2}^{2}}-\frac{1}{\mathrm{n}_{1}^{2}}\right] \mathrm{J} \
&\Delta \mathrm{E}=\mathrm{E}_{\mathrm{n}_{2}}-\mathrm{E}_{\mathrm{n}_{1}} \
&\text{Absorption} =\Delta \mathrm{E}=\mathrm{E}_{\mathrm{n}_{2}}-\mathrm{E}_{\mathrm{n}_{1}} \
&\bar{v}=R\left[\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right]
\end{aligned}
Electronic transitions (fig. 2) are conveniently represented in so-called term schemes.
Multi-Electron Systems
The limitations of the Bohr atom became apparent soon after its establishment. For example, the uranium atom would require 92 electronic orbits and it would have to assume a size which is incompatible with the experimental facts. The first modifications of the Bohr model were made by Sommerfeld, who introduced additional quantum conditions: Taking into account that individual sharp spectral lines split into several lines (of slightly differing $\lambda$ ) if light emission takes place in a magnetic field (Zeeman effect), the following picture emerged from the Bohr atom: electronic spherical orbits (shells)
may have suborbitals (orbits) which are elliptical of varying eccentricity - controlled by an additional quantum number, the orbital quantum number (I). The eccentric orbitals may assume only particular spatial orientations controlled by the magnetic number $(\mathrm{m})$, in an external magnetic field. The observed magnetic behavior of the elements required an additional quantization (Uhlenbeck) - the introduction of the spin quantum number (s).
According to the Bohr-Sommerfeld quantum theory, electrons move about the nucleus of an atom in well-defined orbits, each of which is characterized by four numbers, called quantum numbers. An electron moving in an orbit close to the nucleus has a larger (negative) energy than one in an outer orbit. Energy is therefore necessary to move an electron from an inner to an outer orbit (with smaller negative energy). Conversely, energy is released (as radiation) during the transition of an electron from an outer to an inner orbit. The (electromagnetic) radiation released consists of discrete entities or quanta, which are called photons. The energy of emitted photons is given by the difference in energy between the two orbits involved and is related to the frequency, $v$, or the wavelength, $\lambda$, of the electromagnetic radiation by:
$\mathrm{E}_{\text {photon }}=\mathrm{E}_{2}-\mathrm{E}_{1}=\mathrm{hv}=\mathrm{h}(\mathrm{c} / \lambda) \quad(\mathrm{c}=\lambda v)$
where $h$ is called the Planck's constant, $E_{2}$ and $E_{1}$ are the energies of the outer and inner orbits, respectively, and $c$ is the velocity of light. Through (spectroscopic) observations of the frequencies or wavelengths of radiation emitted by an excited atom (atom with an excited electron), the differences of electron energy levels can be determined. Figure 1 shows such energy levels obtained from measurements of the wavelengths of radiation emitted by atomic hydrogen excited electrically in a gas discharge tube.
To specify the spatial location and energies of electrons in an atom requires the use of four quantum numbers. The "principal quantum number", $(n)$, determines the size of the shell in which a particular electron orbits around the nucleus; it also largely determines its energy. The value of $n$ is restricted to integral values: $n=1,2,3$ and so forth. $A$ value of $n=1$ signifies that the electron exists in the lowest energy state and its orbit is in the innermost allowed shell, as close to the nucleus as possible. Successively higher energy states are represented by $n=2,3$, etc. It should thus be clear that the principal quantum number of the outermost orbiting electron determines, in principle, the size of a given atom. For convenience, letters are frequently used to characterize the electronic shells of $n=1,2,3,4$, etc. In letter notation:
\begin{aligned}
&\mathrm{n}=1=\mathrm{K} \text { shell } \
&\mathrm{n}=2=\mathrm{L} \text { shell } \
&\mathrm{n}=3=\mathrm{M} \text { shell } \
&\text { etc. }
\end{aligned}
The second quantum number, ($I$), is referred to as the "orbital" or "azimuthal quantum number". It specifies the angular momentum of orbiting electrons and, to a minor extent, their energy. "I" can only assume integral values ranging from $I=0$ to $I=(n-1)$ (where $n$ is the principal quantum number). Accordingly, for $n=1$ (the $K$ shell), I can only assume a value of 0. For $n=2$, either $I=0$ or $I=1$ is possible, etc. For convenience again, letters are used to specify the electronic state (orbit) in a given shell (characterized by $\mathrm{n}$ ) corresponding to the second quantum number:
\begin{aligned}
&I=0=\mathrm{s} \text { (orbit) } \
&I=1=\mathrm{p} \text { (orbit) } \
&I=2=\mathrm{d} \text { (orbit) } \
&I=3=\mathrm{f} \text { (orbit) }
\end{aligned}
An electron with $I=0$ is in an s-quantum state (in short, it is an s-electron), one with $I=1$ is in a p-state, etc. Thus, an electron with a principal quantum number $n=3$ and an orbital quantum number $I=1$ is called a $3 p$ electron (the electron in the $M$ shell is in a p orbital).
The third quantum number, $(\mathrm{m})$, called the "magnetic quantum number", controls the number of allowed spatial orientations ("degeneracy") of each orbit characterized by $I$ in a given shell [characterized by $(\mathrm{n})$ ]. (Degenerate states are of identical energy. The "degeneracy" disappears in the presence of a magnetic field where different spatial orientation of orbits assume different energy values.) The total number of allowed orbital orientations for any orbital characterized by $l$ is $(2 I+1)$, corresponding to $m$ values of $I$, $(I-1),(I-2), \ldots, 0,-1,-2, \ldots,-l$. An electronic state with $I=0$ necessarily has only $\mathrm{m}=0$ and thus has no directional orientation in space; it is a spherical orbit. On the other hand, a p-state $(I=1)$ allows $m$ values of $-1,0,+1$. The resulting three possible orientations are perpendicular to one another (as shown in fig. 4).
The first three quantum numbers $(\mathrm{n}, /$ and $\mathrm{m})$ define atomic electron orbitals. They are related respectively to the size, shape, and spatial orientation of the orbital. The fourth quantum number, s, (called the "electron spin quantum number"), can be interpreted as determining for an orbiting electron the direction of electronic spin around its own axis. " $s$ " can assume values of $+1 / 2$ (parallel spin) and $-1 / 2$ (anti-parallel spin). For notational purposes, a positive s is conveniently represented as $\uparrow$ and a negative $s$ as $\downarrow$.
We may now summarize our findings about the electronic states in atoms. Any electron in an atom is defined by four quantum numbers, viz.:
• The principal quantum number (n), which may have any positive integral value except zero. (In practice, because of the instability of heavy nuclei, atoms containing electrons with $\mathrm{n}$ values greater than 7 are unknown.)
• The orbital or azimuthal quantum number ($I$), which is an integer and may have any positive value less than $n$, including zero.
$0 \leq I<n$
• The magnetic quantum number $(\mathrm{m})$, which is also an integer and may have any positive or negative value equal or less than I, including zero.
$-l \leq m \leq+l$
• The spin quantum number (s) which may only assume one of two values, namely $+1 / 2$ and $-1 / 2$.
$s=\pm 1 / 2$
An important law enables us to make use of these quantum rules for the characterization of electronic states in multi-electron systems: the Pauli Exclusion Principle. It states that in any atom no two electrons may have the same four quantum numbers. From this principle it follows that each electronic orbital can accommodate at most two electrons differing by their spin quantum number which will be $+1 / 2$ for one electron and $-1 / 2$ for the other electron. (The Pauli principle is based on the fact that the separate existence of any electron depends upon its non-destruction by interference, i.e. on its wave nature.)
A further useful fact that generally simplifies understanding electronic structures is that (with exceptions stated shortly) the quantum states for electrons follow the rule (Aufbau Principle) that the lowest $m$, I and $n$ numbers, consistent with Pauli's exclusion principle, are selected first by electrons in multi-electron atoms. (For the spin quantum numbers, the $+1 / 2$ value is given priority over the $-1 / 2$ value.) Applicability of this rule is restricted to systems in which the orbitals, defined by a selected set of quantum numbers with lowest possible numerical value, correspond to orbitals of lowest energy since in all instances the lowest energy levels are filled first. The build-up of the electronic states of an atom is obtained by placement of the electrons first in the orbitals of lowest energy (the aufbau principle or "construction principle").
Wave mechanics (introduced by Schrödinger), unlike the quantum theory based on a planetary model, asserts that an electron in an atom cannot be considered as a particle having an orbit with a definite radius. Instead, there is a probability of an electron being at certain spatial positions. Hence, the location of an electron is best described in terms of its probability density distribution, which is sometimes called an electron cloud. The spatial symmetry of the probability distribution depends upon the electronic state. The electron cloud is spherically symmetric for s-electrons, but more complicated for electrons in a p-state. Examples of these distributions are shown in fig. 3 for $1 \mathrm{~s}$ and $2 p$ electrons.
ENERGY LEVELS AND THE AUFBAU PRINCIPLE
Consider the atom of an element containing one extra-nuclear electron: hydrogen; for electro-neutrality the charge on the nucleus must be $+1$. This orbiting electron (in the ground state - the lowest possible, most stable state) will have the lowest available quantum numbers in $n, I$ and $m$. That is, $n=1$; hence $I=0$ and $m=0$; also, $s=+1 / 2$.
In the two-electron atom helium, one electron will have the same quantum numbers as the electron in hydrogen and the other electron will have quantum numbers $n=1, I=0$, $m=0$ and $s=-1 / 2$. The element of atomic number 3 (lithium) will have two electrons with the same quantum numbers as the helium electrons, plus one electron with the quantum numbers $n=2, I=0, m=0, s=+1 / 2$. [Note that $n$ must be 2 for the third electron since states such as $(n=1, I=1)$ or $(n=1, I=0, m=1)$ or $(n=1, I=0, m=-1)$ are not allowed. Similarly, the quantum numbers for the additional electron in beryllium (atomic number 4) are $n=2, I=0, m=0, s=-1 / 2$. The next electron for boron (atomic number 5 ) has the values $n=2, I=1, m=-1, s=+1 / 2$, and so on.
In applying the aufbau principle to the orbital filling with increasing atomic number we have thus far inferred the tendency of the energies of electrons to follow in the same order as the principal quantum numbers; that is, successive electron shells are filled with increasing $\mathrm{Z}$. This concept does not hold in all instances. For example, electrons with the quantum numbers $n=3, I=2$ (and various $m$ values) are of higher energy than those with $n=4, I=0$. Other inversions occur with $n=4$ and higher. These apparent irregularities in the aufbau principle, the result of an energetic overlap of orbitals in successive shells, lead to partial shell fillings and the appearance of groups of so-called transition elements. The groups are characterized (with minor exceptions) by identical outermost electron shell configurations and therefore do not exhibit, with increasing atomic number, the (expected) change in properties observed on regular shell filling for elements 2-18. (In the transition elements, with increasing atomic number the electrons are accommodated in lower lying shells which remained empty because of the above mentioned energy overlap of their orbitals.) The energy levels of the various orbitals in shells with increasing $\mathrm{n}$ are schematically indicated in fig. 4.
Another factor in the application of the aufbau principle is Hund's Rule which states that in atoms the electrons tend at first to fill up given orbitals ( $m$ levels) singly (with unpaired spins and spin quantum number equal to $+1 / 2$ ). Only after all $m$ levels associated with a particular I value in a given shell have been used for single-electron occupation does doubling of electrons into $m$ levels occur. Thus the extra electron for carbon (atomic number 6 ) has the quantum numbers $n=2, I=1, m=0, s=+1 / 2$ rather than $n=2, I=1, m=-1, s=-1 / 2$. Both the effects of energy inversion and Hund's rule are very apparent when considering the electronic configuration of the elements $Z=18, 19,20$ and $21$.
Relative energies of the orbitals in neutral, many-electron atoms. Electrons will always assume lowest available energy states. Accordingly, with increasing atomic numbers, the 4s states will be filled prior to the 3d states, for example.
The s, p, d, and f orbital sets:
$\begin{array}{clcc} \begin{array}{l} \text { Types of } \ \text { orbitals } \end{array} & \text { Orbital quantum numbers } & \begin{array}{c} \text { Total number of } \ \text { orbitals in set } \end{array} & \begin{array}{c} \text { Total number electrons } \ \text { accommodated. } \end{array} \ \hline \mathrm{s} & \mathrm{I}=0 ; \mathrm{m}=0 & 1 & 2 \ \mathrm{p} & \mathrm{l}=1 ; \mathrm{m}=1,0,-1 & 3 & 6 \ \mathrm{~d} & \mathrm{I}=2 ; \mathrm{m}=2,1,0,-1,-2 & 5 & 10 \ \mathrm{f} & \mathrm{l}=3 ; \mathrm{m}=3,2,1,0,-1,-2,-3 & 7 & 14 \end{array}$
Figure 4
The presently used notation is not the most convenient way of designating the energy levels (and wave functions) occupied by the electrons in atoms. The more informative way is a code written as follows: The principal quantum number is given first, followed by the conventional letter which designates the azimuthal (orbital) quantum number and, in a superscript, the number of electrons with that azimuthal quantum number. The code for the electronic ground state in hydrogen, therefore, is $1 \mathrm{~s}^{1}$, for helium $1 \mathrm{~s}^{2}$, lithium $1 s^{2} 2 s^{1}$, beryllium $1 s^{2} 2 s^{2}$, boron $1 s^{2} 2 s^{2} 2 p^{1}$, etc. (See fig. 5 for example.) You will note that the magnetic quantum number has not been specified, but this can be done when required by writing $p_{x}$ when $m=-1, p_{y}$ when (m=+1\) and $p_{z}$ when $m=0$. Similar codes exist for the $d$ and $f$ levels. Usually such subscripts are not specified and in that case a consideration of Hund's rule will not be required.
An important fact to remember is that the number of electrons in any $s, p, d$ or $f$ level is limited: There are, as you know, a maximum of two electrons in any level with spin numbers $+1 / 2$ and $-1 / 2$. For the $p$ levels, / is one and, therefore, three $m$ values $(-1,0$, $+1$ ) are permitted, each with two electrons of spin number $+1 / 2$ and $-1 / 2$, resulting in a maximum of six $p$ electrons in any $p$ orbital system. The reader should confirm that ten electrons are the maximum in a d system of orbitals and 14 in an $\mathrm{f}$ system of orbitals.
IONIZATION POTENTIAL
Electrons may be removed from isolated atoms by bombardment with other electrons and by heat, for examples. The work (energy) required to remove the most weakly bound (outermost) electron from an isolated atom is known as the "ionization energy". This energy is sometimes listed in units of Joules. More often ionization energies are given in terms of the "ionization potential". The ionization potential is the potential (V) that will accelerate an electron at rest so that it acquires a kinetic energy, sufficient to extract the outermost (most loosely bound) electron from an atom. This potential is 13.595 Volt (as listed in the PT) for the ionization of a hydrogen atom. You can also say the electron is bound to the proton by an (negative) energy equivalent to the energy of an electron accelerated by a potential of $13.595$ Volt. [If you make a dimensional analysis, you will recognize that charge (e) times potential (V) has the dimensions of energy $\left(\mathrm{kg}^{2} \mathrm{~m}^{2} \cdot \mathrm{sec}^{-2}\right)$ and 1 electron Volt $(1 \mathrm{eV})$ corresponds to $1.6 \times 10^{-19}$ Joules.] (Note that the first ionization energies are given as first ionization potentials $(\mathrm{V})$ in the Periodic Table of the Elements; see also the 3.091 courseware menu.)
It is significant that inert or noble gases have some of the highest ionization potentials. This reflects the fact that these elements have just enough electrons to completely fill a shell or a subshell which form stable configurations. On the other hand, the ionization potentials of alkali metal atoms (Li, Ni, etc.) are low - the lowest, that of $\mathrm{Cs}$, is only $3.89 \mathrm{eV}$. The reason for this is that alkali metal atoms have one outer s-electron beyond the stable electronic structure of an inert gas atom. Consequently this single electron in
the outermost new shell can be removed relatively easily. (The second and successive ionization potentials of such atoms are increasingly greater.)
From a chemical and physical viewpoint, it is found that whether an atom is neutral or ionized as well as isolated or combined leads to distinctly different characteristics. It should be recognized that the notation used previously for neutral atoms can also be employed for electronic structures of ions. The electronic configurations of an iron atom and two iron ions are shown below.
$\begin{array}{lll}\text { Fe: } \quad 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{6} 4 s^{2} & \text { (neutral atom) } \ \mathrm{Fe}^{+2}: 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{6} & \text { (ferrous ion) } \ \mathrm{Fe}^{+3}: 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{5} & \text { (ferric ion) }\end{array}$
ELECTRON AFFINITY
As we shall see later, the tendency of some atoms to accept, as well as to lose, electrons is important in determining how atoms combine with each other. Whereas removal of an electron always requires the expenditure of energy, acceptance of one extra electron by an atom generally is accompanied by the release of energy. The amount of energy released on acquisition of an electron is called the electron affinity (or "first electron affinity"). The elements on the left of the periodic table (metals) and the inert gases on the outer right have low electron affinities, whereas the non-metals have higher ones. The particularly high electron affinities of halogen atoms $(\mathrm{F}, \mathrm{Cl}$, etc.) may be attributed to the fact that these elements all lack one electron which would provide the stable electronic structure of an inert gas. By acquiring an additional electron, they increase their stability since the additional electron goes into the orbital lacking one electron and thus results in a stable, $s^{2} p^{6}$, octet configuration. (See fig. 5.)
ATOMIC SIZE
The size, or volume, of an isolated atom is difficult to define explicitly since, in the electron cloud model of the atom, the probability density distribution theoretically reaches zero only at infinity. Nevertheless, the electron density falls off so rapidly at a short distance from the nucleus that some approximation of size can be made. In the hydrogen atom, for example, the electron density is very nearly zero at a distance of $1.2 \AA\left(1.2 \times 10^{-10} \mathrm{~m}\right)$ from the nucleus. The problem of defining atomic size is simplified in molecules and solids in that rather precise dimensions can be determined from interatomic distances which can be measured by diffraction techniques. Thus, in the $\mathrm{H}_{2}$ molecule the atoms are only $0.72 \AA$ apart, as determined from the distance between the nuclei. In this case the radius of the hydrogen atoms is taken to be $0.37 \AA$, even though the size of the $\mathrm{H}_{2}$ molecule is considerably more than four times this value.
It should be apparent that the atomic radius depends upon whether an atom is isolated or combined with other atoms. The radius of an isolated atom is called the van der Waals radius, that of a bound atom in a molecule is the covalent radius, and that of a bound atom in a metal is the metallic radius. Van der Waals, covalent and metallic radii for some elements are listed in some periodic tables of the elements. The radii of positive ions (cations) and those of negative ions (anions) differ from the van der Waals radii. (See the 3.091 courseware menu on Athena.)
SAMPLE PROBLEM
For hydrogen, calculate the radius ( $r$ ) and energy level $\left(E_{T}\right)$ of the electron in the lowest energy state (ground state).
Solution
The size of the first allowed orbit can be calculated by requiring that the centrifugal force of the orbiting electron be balanced by the coulombic attraction to the nucleus.
$F_{\text {centr }}=\dfrac{m v^{2}}{r} \quad F_{a t t}=\dfrac{e^{2}}{4 \pi \varepsilon_{0} r^{2}} \quad \dfrac{m v^{2}}{r}=\dfrac{e^{2}}{4 \pi \varepsilon_{0} r^{2}}$
According to Bohr, the angular momentum of the electron (mvr) is quantized in units of $\mathrm{h} / 2 \pi$:
$m v r=\dfrac{\mathrm{nh}}{2 \pi} \quad \mathrm{n}=1,2,3, \ldots$
With these quantum conditions, $r$ for the first allowed orbit becomes:
\begin{aligned}
\frac{m v^{2}}{r} &=\frac{e^{2}}{4 \pi \varepsilon_{0} r^{2}} \
m^{2} v^{2} r^{2} &=\frac{m e^{2} r}{4 \pi \varepsilon_{0}} \
\frac{n^{2} h^{2}}{4 \pi^{2}} &=\frac{m e^{2} r}{4 \pi \varepsilon_{o}} \
r &=\frac{n^{2} h^{2} \varepsilon_{0}}{\pi m e^{2}}=n^{2} x \text { constant }
\end{aligned}
For $n=1:$
$r_{0}=0.529 \times 10^{-10} \mathrm{~m}=0.529 \AA$
The kinetic energy of the electron is $1 / 2 \mathrm{mv}^{2}$. The potential energy as a function of distance from the nucleus is given by Coulomb's law as:
$-\dfrac{e^{2}}{4 \pi \varepsilon_{0} r}$
The total energy is the sum of the kinetic and potential energies.
\begin{aligned}
&\mathrm{E}_{\mathrm{T}}=\mathrm{E}_{\mathrm{K}}+\mathrm{E}_{\mathrm{P}} \
&\mathrm{E}_{\mathrm{T}}=\frac{1}{2} m \mathrm{v}^{2}-\frac{\mathrm{e}^{2}}{4 \pi \varepsilon_{0} r}
\end{aligned}
The balance of attractive and repulsive forces requires that:
$\begin{array}{lll} F_{\text {centr }}=F_{\text {att }} & \text { Since: } & \dfrac{m v^{2}}{r}=\dfrac{e^{2}}{4 \pi \varepsilon_{0} r^{2}} \ \dfrac{m v^{2}}{r}=\dfrac{e^{2}}{4 \pi \varepsilon_{0} r^{2}} & \text { Then: } & \dfrac{m v^{2}}{2}=\dfrac{e^{2}}{8 \pi \varepsilon_{0} r} \end{array}$
Therefore:
$\mathrm{E}_{\mathrm{T}}=\dfrac{\mathrm{e}^{2}}{8 \pi \varepsilon_{0} \mathrm{r}}-\dfrac{\mathrm{e}^{2}}{4 \pi \varepsilon_{0} \mathrm{r}}=-\dfrac{\mathrm{e}^{2}}{8 \pi \varepsilon_{0} \mathrm{r}}$
Substitution for $r$ now gives:
$E_{T}=-\dfrac{1}{n^{2}} \times \dfrac{m e^{4}}{8 h^{2} \varepsilon_{0}^{2}}$
For $n=1$:
\begin{aligned}
&\mathrm{E}_{\mathrm{T}}=-2.179 \times 10^{-18} \mathrm{~J} \
&\mathrm{E}_{\mathrm{T}}=-13.6 \mathrm{eV}
\end{aligned}
DEFINITIONS
ATOM: The basic constituent of matter, consisting of a nucleus surrounded by electrons.
ATOMIC MASS NUMBER (A): The combined number of protons and neutrons in a specific nucleus.
ATOMIC NUMBER (Z): The number of protons in a specific nucleus, characteristic of a chemical element.
ATOMIC WEIGHT:
(1) The weighted mass of the naturally occurring atoms which constitute an element, relative to the value of exactly 12 for $C^{12}$.
(2) The mass, in grams, of one mole of a chemical element.
ELECTRON: A negatively charged subatomic particle.
ELECTRON AFFINITY: The amount of energy released when an electron is added to the lowest possible energy level of a neutral atom.
ELECTRON DENSITY DISTRIBUTION/PROBABILITY DENSITY DISTRIBUTION: The spatial distribution of an electron in an atom, depending upon its set of quantum numbers.
ELECTRONIC CONFIGURATION: The shorthand notation used to designate the occupancy of energy levels in an individual atom.
ENERGY LEVEL: The discrete energy state of an electron in an atom, depending upon its set of quantum numbers.
FIRST IONIZATION POTENTIAL: The work which must be expended to remove an electron from a neutral atom in its ground state.
GROUND STATE: The lowest energy state of an atom. All electrons occupy energy levels sequentially from the lowest level.
HALF–LIFE: The time required for half of the atoms of an unstable, radioactive isotope to decay or for a reaction to go to 50% completion.
ION: A charged atom having either an excess or a deficiency of electrons relative to its nuclear charge.
ISOTOPE: The form of an element having the same atomic number but different atomic mass number (or atomic weight) than other forms of the element. As a rule, isotopes of an element exhibit virtually identical chemical behavior, but may exhibit quite different nuclear and physical behaviors.
NEUTRON: A neutral subatomic particle having approximately the same mass as a proton.
NUCLEUS: An extremely dense, small portion of an atom, with a radius of about $10^{–5}$ that of an atom and containing in excess of 99% of the atomic mass.
ORBITAL: A pictorial designation pertaining to the existence of an electron within a subshell, depending upon the magnetic quantum number, m. Each orbital may contain a maximum of two electrons having opposite spins- i.e., electron spin quantum numbers of $+1/2$ or $–1/2$.
PAULI EXCLUSION PRINCIPLE: The statement that each electron in an atom must have a specific set of unique quantum numbers.
PERIODIC TABLE: A chart arraying the chemical elements in order of increasing atomic number and in groups having similar chemical behavior and similar outer electronic configurations.
PHOTON: A quantum, or minimum unit, of electro magnetic energy. The energy is equal to Planck’s constant, h, times the frequency, $ν$, of the radiation.
PROTON: A positively charged subatomic particle whose charge is exactly opposite that of an electron and whose mass is about 1800 times that of an electron.
QUANTUM NUMBERS: A series of discrete numbers which catalog the state of an electron and which can be derived from wave mechanics.
SHELL: A pictorial designation pertaining to electrons having the same principal quantum number, n, often indicative of the overall energy level of the electron.
SUBSHELL: A pictorial designation pertaining to the state of electrons within a shell, depending upon the second quantum number, $l$, and indicative of the spatial distribution of the electron. | textbooks/chem/Inorganic_Chemistry/Introduction_to_Solid_State_Chemistry/01%3A_Lectures/1.01%3A_Atomic_and_Electronic_Structure.txt |
The electronic configurations of the elements, as specified in the previous chapter, apply in principle only to isolated atoms - atoms separated by distances over which no mutual interactions of their electronic orbitals can occur (infinite distance). This condition is never met in condensed phases (i.e., liquids and solids); it is only encountered in high vacua where atoms move over long distances without mutual interaction. Under normal conditions, particularly in the mentioned condensed phases, atoms are separated over distances controlled, in essence, by the dimension of their respective outermost occupied electronic orbitals. Whenever the outer electron shells of two or more atoms come in contact with each other (overlap to any extent), the potential for interaction (reaction) exists. Spontaneous reaction (consisting of a rearrangement of the electronic orbitals and/or actual transfer of electrons from one atom to another) will take place whenever such a rearrangement results in a lower energy configuration. This means that the driving force for mutual interaction and rearrangement of electronic configurations is in most instances (but not always) manifested through the release of heat (a form of energy) to the environment. [Typical case: $\mathrm{Mg}+1 / 2 \mathrm{O}_{2} \rightarrow$ Reaction Product $(\mathrm{MgO})+($ light + heat) i.e. energy.] Since the same amount of energy must be supplied to the system if the original state - i.e., separation of the species – is to be reestablished, the reaction partners are “bonded together” by comparable energies. The “strength” of the bonds can obviously vary from system to system with the nature of the electronic rearrangement.
Even the inert gases, particularly the heavier ones like xenon, are capable of forming associations with other atoms. Sometimes we find that two atoms assume a more stable state by sharing electrons; at other times, an atom may transfer electrons to another atom in order to achieve a greater stability. In still other instances, the rearrangement may simply be an orbital distortion or an internal charge redistribution. In either event the mutual benefit that accrues is the formation of what is commonly called a chemical bond. Through these bonds atoms combine with each other to form very different kinds of particles referred to as molecules and ions.
NATURE OF CHEMICAL BONDS
In order for a chemical bond to be formed between two atoms, there must be a net decrease in the energy of the system (the two atoms): the ions or molecules produced by electronic rearrangements must be in a lower energy state than the atoms were prior to interaction, prior to bond formation.
Since atoms of each of the elements have different electronic structures, the variety of possible chemical bonds (differing from each other in at least some small way) is considerable and is even further increased by the effects of neighboring atoms on the bond under consideration. The modes of bond formation can be categorized into two basic types, each representing a type of bonding. The bonding types are called electrovalent (or ionic) bonding and covalent bonding. Electrovalent bonding arises from complete transfer of one or more electrons from one atom to another; covalent bonding arises from the sharing of two or more electrons between atoms. Since these models represent the limiting cases, we can anticipate that most real bonds will fall between these two extremes. (Two additional types of bonding, metallic bonding and Van der Waals bonding, will be discussed later.)
Before discussing these models in detail it is appropriate to consider the relationships between the electronic structures of atoms and their chemical reactivity. The inert gases (Group VIII) are the most stable elements with regard to bond formation, i.e. toward electronic rearrangements. It is therefore useful to examine the reasons for their stability. Inert gases all have electronic structures consisting of filled subshells. For all but helium the outer (or valence) shell contains eight electrons, with filled $s$ and $p$ sublevels $\left(n s^{2} p^{6}\right)$. The electronic structure of helium is $1 s^{2}$, which is equivalent to the structure of the other inert gases since there is no $1 \mathrm{p}$ sublevel. Inert gases have high ionization energies because each electron in the sublevel of highest energy is poorly screened from the nucleus by other electrons in its same sublevel. Each electron "sees" relatively high positive charge on the nucleus, and a large amount of energy is required to remove it from the atom. Inert gases have very low electron affinity because any added electron must enter a significantly higher energy level. We find, therefore, that the electronic structures of inert gases are particularly resistant to changes by either loss or gain of electrons and, further, that atoms of other elements with fewer or more electrons than inert gas configuration tend to gain or lose electrons, respectively, to achieve such inert gas structure.
ELECTROVALENT (IONIC) BONDING
An electrovalent bond is formed by the transfer of one or more electrons from one atom to another. Consider first atoms that have electronic structures differing from an inert gas structure by only a few, (1, 2 or 3) electrons. These include the representative elements of Groups I, II and III in the Periodic Table, which have respectively 1, 2 and 3 electrons more than a neighboring inert gas, and the representative elements of Groups V, VI and VII, which have respectively 3, 2 and 1 electrons less than a neighboring inert gas (fig. 1).
Fig.1 Stable and Unstable Valence Shell Configurations.
The elements of Groups I, II and III can form the electronic structure of an inert gas by losing their outer 1, 2 and 3 (valence) electrons. (The resulting species are positively charged ions.)
In a similar electron transfer which, however, involves the acquisition of electrons in the outer valence levels, elements of Groups V, VI and VII form an inert gas electronic structure (by formation of negatively charged ions).
It is through the electron transfer between an electron-losing element and an electron-gaining element that compounds are formed which involve electrostatic attraction (electrovalent bonds) of oppositely charged species called ions.
(In the notation [Na•], the dot indicates the outermost electron which is in excess of the rare gas configuration. It is referred to as a valence electron.)
Elements immediately following the inert gases (in the horizontal columns of the Periodic Table) lose electrons, and those immediately preceding the inert gases gain electrons on interaction. The resulting compounds are called electrovalent; the valence number (charge on the ion) of a particular element when it forms an electrovalent compound is given by the number of electrons lost or gained in changing from the atomic to the ionic state.
The stoichiometric formula of an electrovalent compound reflects the ratio (usually very simple) of positive to negative ions that gives a neutral aggregate. Hence, the ions $\mathrm{Na}^{+}$ and $\mathrm{F}^{-}$ form a compound whose formula is $\mathrm{NaF}$ because these ions are singly charged and are present in the compound in a one-to-one ratio. Magnesium nitride, composed of $\mathrm{Mg}^{2+}$ and $\mathrm{N}^{3-}$, has the formula $\mathrm{Mg}_{3} \mathrm{~N}_{2}$ because this composition represents electroneutrality.
The electrovalent bond is the result of electrostatic attraction between ions of opposite charge. This attractive force accounts for the stability of these compounds, typified by $\mathrm{NaF}, \mathrm{LiCl}, \mathrm{CaO}$, and $\mathrm{KCl}$. The ions individually possess the electronic structures of neighboring inert gases; their residual charge arises from an imbalance in the number of electrons and protons in their structures. Isolated ions and simple isolated pairs of ions, as represented by the formula $\mathrm{NaCl}$, exist only in the gaseous state. Their electrostatic forces are active in all directions; they attract oppositely charged species and thus can form regular arrays, resulting in ordered lattice structures, i.e. the solid state (fig. 2). Even in the liquid state and in solutions (where disruptive thermal forces reach values close to that of the attractive electrostatic bonding forces) attraction between ions and with other species remains effective.
Energetics of Ionic Bonding
Ionic bonding is the simplest type of chemical bonding to visualize, since it is totally (or almost totally) electrostatic in nature. The principle of the energetics of ionic bond formation is realized when considering the formation of $\mathrm{NaCl}$ (our common salt) from its constituents $\mathrm{Na}$ (metal) and $\mathrm{Cl}_{2}$ (chlorine gas). Formally, this reaction is:
\begin{aligned}
&\mathrm{Na}(\mathrm{s})+1 / 2 \mathrm{Cl}_{2}(\mathrm{~g}) \rightarrow \mathrm{NaCl}(\mathrm{s}) \
&\Delta \mathrm{H}=-414 \mathrm{~kJ} / \mathrm{mol}
\end{aligned}
The equation as written indicates that 1 mole sodium reacts with $1 / 2$ mole chlorine $\left(\mathrm{Cl}_{2}\right)$ under formation of 1 mole (ionically bonded) sodium chloride; this reaction is accompanied by the release $(-)$ of $414 \mathrm{~kJ}$ of energy $(\Delta \mathrm{H})$, referred to as the heat of reaction. From earlier considerations it is clear that electronic rearrangement (reaction or bond formation) takes place because the resulting solid product $(\mathrm{NaCl})$ is at a lower energy state than the sum total of the energies of the original components.
The energetics associated with ionic bond formation may be determined quantitatively by considering the energy changes associated with the individual steps leading from the starting materials to the final product (Haber-Born cycle).
The bond formation in NaCl may be formally presented as an electron-transfer reaction:
The reactions involved in this process which result in the formation of 1 mole of solid salt are:
(1) Ionization of $\mathrm{Na}$:
$\mathrm{Na}$ (gas) $\rightarrow \mathrm{Na}^{+}+1 \mathrm{e}$ (E.I. $=+497 \mathrm{~kJ} / \mathrm{mole})$
The energy change associated with this step, energy of ionization (E.I.), is
$+497 \mathrm{~kJ} /$ mole.
(2) Acquisition of one electron by $\mathrm{Cl}$:
$\mathrm{Cl}$ (gas) $+1 \mathrm{e} \rightarrow \mathrm{Cl}^{-}$ (E.A. $=-364 \mathrm{~kJ} / \mathrm{mole}$)
The energy change associated with this step, electron affinity (E.A.), is $-364 \mathrm{~kJ} / \mathrm{mole}$. The minus sign reflects an energy release, a lowering of the energy state associated with the achievement of stable rare gas configuration by chlorine.
So far, the energy balance appears positive ( $\Delta \mathrm{E}=+133 \mathrm{~kJ}$ ); this means the reaction is not favored since the final products are at a higher energy state than the starting products. However, there are additional steps involved since:
(3) Vaporization of $\mathrm{Na}$:
$\mathrm{Na}$ (metal) $\rightarrow \mathrm{Na}$ (gas) $\left(\Delta \mathrm{H}_{\mathrm{V}}=+109 \mathrm{~kJ} /\right.$ mole)
The energy required to transform $\mathrm{Na}$ (metal) into $\mathrm{Na}$ (gas), the latent heat of vaporization $\left(\Delta \mathrm{H}_{\mathrm{v}}\right)$, is $109 \mathrm{~kJ} /$ mole (now reaction appears even less favorable).
(4) Dissociation of $\mathrm{Cl}_{2}$:
$\mathrm{Cl}_{2} \rightarrow 2 \mathrm{Cl} \quad(\text { E.D. }=+242 \mathrm{~kJ} \text { ) }$
The energy associated with breaking up the (stable) chlorine molecule into two reactive chlorine atoms, dissociation energy (E.D.), is $+242 \mathrm{~kJ} / \mathrm{mole}$. (Since the formation of one mole $\mathrm{NaCl}$ involves only 1 mole $\mathrm{Cl}$, and since 2 moles are formed from one mole $\mathrm{Cl}_{2}$, the energy required in this step is $1 / 2 \mathrm{E} . D$., or $121 \mathrm{~kJ}$.)
The total energy change associated with reactions (1) through (4), $\Delta \mathrm{H}_{\mathrm{T}}=+364 \mathrm{~kJ}$ This indicates that one or more basic reactions which lead to an overall decrease in energy are still unconsidered.
(5) Unconsidered as yet is the coulombic attraction of the reaction products which are of opposite charge and an energy term associated with the formation of the "solid state" product, $\mathrm{NaCl}$. The two ionic species, originally at zero energy (infinite distance of separation), attract each other with an accompanying energy change (decrease).
The energetics associated with the approach of the reaction partners (fig. 3) is best considered by fixing the position of one, for example $\mathrm{Na}^{+}$, and letting the other $\left(\mathrm{Cl}^{-}\right)$ approach: as $\mathrm{Cl}^{-}$ approaches $\mathrm{Na}^{+}$, E decreases according to Coulomb's law with $\mathrm{e}^{2} / 4 \pi \varepsilon_{0} r$. With the approach of the two oppositely charged ions, the outermost electronic shells will come into contact and a repulsive force will become active as the shells interpenetrate. The repulsive force $\left(E_{\text {rep }}\right)$ increases with $+b / r^{12}$ and thus is only active in the immediate vicinity of the sodium ion, but at that stage increases rapidly. The two ions at close proximity are under the influence of both attractive and repulsive forces and will assume a distance of separation at which the two forces balance each other - a distance which is referred to as the equilibrium separation ($r_o$) and which corresponds to the energy minimum for the NaCl molecules:
$E_{\text {coul }}=-\dfrac{e^{2}}{4 \pi \varepsilon_{0} r_{0}}+\dfrac{b}{r_{0}^{12}}$
where $e=$ electronic charge, $\varepsilon_{0}=$ permittivity of free space $\left(8.85 \times 10^{-12} \mathrm{~F} / \mathrm{m}\right)$ and $(\mathrm{b})$ is a constant. ( $r_{0}$ for gaseous molecules may be obtained from physical measurements.)
If the presently considered coulombic (attractive) energy term is taken into consideration as reaction (5), it will decrease the overall positive energy change, $\Delta \mathrm{H}_{\bar{T}}$; it will, however, not make it change sign from (+) to (-). Thus, $\mathrm{NaCl}$ molecules in gaseous form are not the final reaction product; nor would reaction occur if $\Delta \mathrm{H}_{\mathrm{T}}$ remained positive.
Considering an ionically bonded, gaseous $\mathrm{Na}^{+} \mathrm{Cl}^{-}$ molecule, it is clear that its electrostatic forces $(+)$ and $(-)$ are not saturated - they remain active in all possible directions. This means that $\mathrm{Cl}^{-}$ will attract $\mathrm{Na}^{+}$ ions from other directions as will the $\mathrm{Na}^{+}$ atoms attract additional $\mathrm{Cl}^{-}$ ions. The result of these attractive forces in all directions is the formation of a "giant size" ionic body - a "solid" body of macroscopic dimensions. From the preceding it is recognized that the minimum energy configuration is given by a body in which $\mathrm{Na}^{+}$ and $\mathrm{Cl}^{-}$ are arranged with extreme periodicity and order since any ion located outside of its "equilibrium position" will be in a "higher energy configuration"; such an ordered body is referred to as a crystalline solid, or frequently just called a solid (fig. 2).
The total energy change associated with the formation of one mole of crystalline ionic solid from its ionic constituents is given as:
$\mathrm{E}_{\text {cryst }}=-\dfrac{M \mathrm{~N}_{\mathrm{A}} \mathrm{e}^{2}\left(\mathrm{Q}_{1} \mathrm{Q}_{2}\right)}{4 \pi \varepsilon_{0} r_{o}}\left(1-\dfrac{1}{\mathrm{n}}\right)$
where $\mathrm{M}=1.747$ ("Madelung" constant for $\mathrm{NaCl}$, reflecting multiple interactions for the particular geometric arrangement of ions in the solid), $\mathrm{N}_{\mathrm{A}}=$ Avogadro's number, $\mathrm{Q}=$ number of charges per ion ( 1 for $\mathrm{Na}^{+}$ and $\left.\mathrm{Cl}^{-}\right), \mathrm{r}_{0}=$ equilibrium distance of separation of ions, and $n=$ repulsive exponent $(n=12$ for $\mathrm{NaCl})$.
[The relationship for the crystal energy $\left(\Delta \mathrm{E}_{\text {cryst }}\right)$ is readily obtained from equilibrium energy considerations:
$\mathrm{E}_{\text {coul }}=-\dfrac{\mathrm{e}^{2}}{4 \pi \varepsilon_{0} r}+\dfrac{\mathrm{b}}{4 \pi \varepsilon_{0} r^{n}}$
At equilibrium distance of ion separation $\left(r_{0}\right),(\mathrm{dE} / \mathrm{dr})=0$. Thus:
$\left(\dfrac{d E}{d r}\right)_{r_{0}}=\dfrac{e^{2}}{4 \pi \varepsilon_{0} r_{o}^{2}}-\dfrac{n b}{4 \pi \varepsilon_{0} r_{o}^{n+1}}=0$
and
$\mathrm{b}=\dfrac{\mathrm{e}^{2} \mathrm{r}_{\mathrm{o}}^{\mathrm{n}-1}}{\mathrm{n}}$
and
\begin{aligned}
&E_{o(\text { Coul })}=-\frac{e^{2}}{4 \pi \varepsilon_{o} r_{o}}+\frac{e^{2}}{n 4 \pi \varepsilon_{o} r_{o}} \
&E_{o(\text { Coul })}=-\frac{e^{2}}{4 \pi \varepsilon_{o} r_{o}}\left(1-\frac{1}{n}\right)
\end{aligned}
For a "molar" crystal with ionic charges $\mathrm{Q}_{1}$ and $\mathrm{Q}_{2}$, the molar $\Delta \mathrm{E}_{\mathrm{o}(\text { Coul) }}$ is thus given as:
$\Delta \mathrm{E}_{\text {cryst }}=-\dfrac{\mathrm{MN}_{\mathrm{A}} \mathrm{Q}_{1} \mathrm{Q}_{2} \mathrm{e}^{2}}{4 \pi \varepsilon_{\mathrm{o}} \mathrm{r}_{\mathrm{o}}}\left(1-\dfrac{1}{\mathrm{n}}\right)$
Here $M$ and $N_{A}$ stand, as indicated above, for the conventional terms, the Madelung constant and Avogadro's number.]
For the present system ( $\mathrm{NaCl})$ :
$\Delta \mathrm{E}_{\text {cryst }}=-777 \mathrm{~kJ} /$ mole.
Now considering reactions (1) through (5):
$\Delta \mathrm{H}_{\text {Total }}=-414 \mathrm{~kJ} /$ mole
which is identical with the value experimentally determined and given for the reaction:
\begin{aligned}
&\mathrm{Na}(\text { metal })+\mathrm{Cl}_{2} \text { (gas) } \rightarrow \mathrm{NaCl} \text { (solid) } \
&\Delta \mathrm{H}_{\text {Reaction }}=-414 \mathrm{~kJ} / \text { mole }
\end{aligned}
In the Haber-Born cycle, the reaction energy $(\Delta \mathrm{H})$ associated with the formation of $\mathrm{NaCl}$ from $\mathrm{Na}+\mathrm{Cl}_{2}$ may be summarized as:
$\Delta \mathrm{H}=\text { E.I. }+\text { E.A. }+\Delta \mathrm{H}_{\mathrm{V}}+1 / 2 \text { E.D. }+\Delta \mathrm{H}_{\text {cryst }}$
$\Delta \mathrm{H}$, the heat of reaction, may thus be obtained from the energetics of the steps leading to the end product. In most instances, however, the reaction energy $(\Delta \mathrm{H})$ is determined experimentally in a calorimeter and the Haber-Born cycle is used to obtain the value of $\Delta \mathrm{E}_{\text {cryst }}$ or $\mathrm{E} . \mathrm{A} .$, which are both extremely hard to come by.
Conclusions: A primary drive for atomic interactions leading to "bonding" is the achievement of valence shell octets which exhibit a high degree of stability. If atoms on the left and right side of the Periodic Table interact, i.e. atoms with a large difference in electron affinity ( $\triangle \mathrm{E} . \mathrm{A} .)$, stabilization is achieved by electron octet formation through charge transfer. The reaction products exhibit opposite charges ( $\oplus$ cations, $\ominus$ anions) and are subject to Coulombic attraction - ionic bonds are formed. Since the electrostatic forces are non-directional and non-saturated, energy minimization will result in the formation of macroscopic bodies that are highly ordered on the atomic scale, crystalline ionic solids. Ionic solids have mostly predictable, basic properties:
• atomic arrangements are a function of the ion size ratio, the charge ratio of ions, and their electronic structure;
• electrical and thermal conductivities are expected to be low because the high stability of the octets formed results in bound electrons which do not contribute to conduction (there is a large "energy gap" to be crossed for electrons to move into a higher energy state);
• optically, ionic solids are mostly transparent, or translucent, reflecting octet stability of the electrons and macro- or micro-crystallinity;
• melting points are high, increasing with the electronic charges on the cations and anions;
• ionic solids are normally hard and brittle.
COVALENT BONDING
Mechanical Concepts and Conclusions
In 1924 L. DeBroglie advanced the hypothesis that all matter in motion possesses wave properties and can be attributed a particle wavelength
$\lambda_{P}=\mathrm{h} / \mathrm{mv}$
where $\mathrm{h}$ is the Planck constant, $\mathrm{m}$ is the mass of the matter and $v$ is its velocity.
The credibility of this hypothesis was established by Davisson and Germer in 1927 when they demonstrated that electrons (like electromagnetic radiation) are diffracted by crystal lattices. An important consequence of the dual nature of matter (it exhibits both particle and wave properties) is the uncertainty principle established in 1927 by W. Heisenberg. It states that it is impossible to simultaneously know with certainty both the momentum and position of a moving particle:
$\left(\Delta p_{x}\right)(\Delta x) \gtrsim h$
We can paraphrase the uncertainty principle in the following manner: If the energy of a particle is known (measured) with high precision, its location is associated with a high degree of uncertainty.
If electrons occupied simple orbits (as postulated by Bohr-Sommerfeld), their momentum and position could be determined exactly at any moment - in violation of the uncertainty principle. According to Heisenberg, if the energy of an electron is specified with precision (sharpness of spectral lines), its location can only be specified in terms of the probability of finding this electron in a certain location (volume element). These arguments give rise to the concepts of probability density and electron cloud which are inherent to the wave-mechanical electron concept emanating from the solutions of Schrödinger’s wave equation which relates the energy of an orbiting electron to its wave properties. When solving exactly the Schrödinger wave equation for an electron in a hydrogen atom, a quantization results according to which electrons can only assume certain energies which are in quantitative agreement with those obtained from the Bohr theory.
In comparison to the Bohr theory, significant differences are observed for the orbital quantization (orbital quantum number $I$ ) which specifies the orbital shape. For $n=1$, $I=0$ (1s orbital), wave mechanics predicts a spherical electron density distribution with a maximum probability density $\left(\psi^{2}\right)$ at a distance of $0.529 \AA\left(\mathrm{a}_{0}\right)$ from the nucleus. For $I=1$ (p orbitals), however, it is found that the orbitals (three) form lobes aligned with rectangular coordinates (see LN-1, fig. 3). For $I=2$ (d orbitals), five complex orbital configurations are obtained.
Previous considerations suggest that all elements attempt to assume a stable octet configuration with eight electrons in the valence shell. (For hydrogen, with only one shell occupied, the stable configuration consists of two electrons in the $\mathrm{K}$ shell - which is the maximum number of electrons that can be accommodated.) Stable octet formation is possible through electron transfer (and ionic bond formation) when, for example, elements in columns IA, IIA and IIIA react with elements in columns VA, VIA and VIIA, respectively. In these instances, the reaction partners exhibit rather pronounced differences in electron affinity and upon reaction one assumes octet configuration by losing one or more electrons while the other does so by acquiring the missing number of electrons.
This mechanism is clearly not possible if $\mathrm{H}$ reacts with $\mathrm{H}$ to form an $\mathrm{H}_{2}$ molecule where two atoms are bonded together. The same argument holds for the formation of $\mathrm{N}_{2}, \mathrm{Cl}_{2}$ and $\mathrm{O}_{2}$ molecules. Inert gas configuration (octet configuration) in such elements is achieved by a mechanism called orbital sharing and the resulting bond is called covalent, or electron-pair bond.
A covalent bond is somewhat more difficult to visualize than an ionic or electrovalent bond because it involves the sharing of a pair of electrons between atoms. The stability of this bond can be attributed to the complex mutual attraction of two positively charged nuclei by the shared pair of electrons. In principle, the bond can be understood if it is recognized that both electrons in the bonding orbital spend more time between the two nuclei than around them and thus must exercise attractive forces which constitute the bond. In this arrangement it is clear that each electron, regardless of its source, exerts an attractive force on each of the “bonded” nuclei. The pair of electrons in a covalent bond is unique to the extent that the Pauli exclusion principle precludes the presence of additional electrons in the same orbital. Furthermore, the pairing phenomenon neutralizes the separate electronic spins of the single electrons, and the resulting electron pair with its zero spin momentum interacts less strongly with its surroundings than do two independent electrons. [Covalent bonds are conveniently symbolized through the dot notation, introduced first by Lewis (fig. 4).]
LEWIS NOTATION:
In LEWIS notation the covalent molecular bond is indicated as a BAR or as two DOTS (standing for the paired electrons)
Formal valence shell octet stabilization can be achieved by electron sharing, H whereby one electron from each reaction partner share - spin paired - the molecular bonding orbital.
Quantum mechanics makes it possible to rigorously describe these bonds for very simple cases such as the hydrogen molecule, which is composed of two protons and two electrons. It can thus be shown that the potential energy for the system reaches a minimum for a certain equilibrium distance between the nuclei, with increased electron density between the nuclei. At shorter distances between the nuclei repulsive forces are found to increase very rapidly.
In the hydrogen molecule $\left(\mathrm{H}_{2}\right)$, the two hydrogen atoms are effectively linked together by one molecular electron orbital, termed a $\sigma$ orbital, which comprises both atoms and contains two electrons. (fig. 5) Each of these two electrons can be considered to have originated from one of the two atoms - they were originally both $1 \mathrm{~s}$ electrons with the same spin value $(s=+1 / 2)$. In the molecular orbit comprising both atoms, the spins of
the two electrons must align anti-parallel (opposite spin). This spin-pairing process results in a considerable release of energy and thus contributes significantly to the stability (strength) of the covalent bond formed. It is interesting that according to Newtonian mechanics, no stable configuration can arise from the placement of two electrons into the same region (orbit) - wave mechanics (see below), however, predicts increased stability from such configurations. (Similar spin-pairing occurs in the filling of atomic orbitals.) In the molecular case, spin-pairing has the consequence that both the probability distributions $\left(\psi^{2}\right)$ and spatial distribution of electrons are such that maximum overlap of orbitals of combining atoms occurs.
The bond formed in hydrogen is a single bond, containing two electrons with paired spins. This formulation predicts that there will be a fixed internuclear separation which, for the hydrogen molecule, has the value of $0.74 \AA$. [ $\sigma$ bonds are formed not only by s orbital overlap, but also by p-s and p-p orbital overlap as well as by the overlap of $s$ or $p$ orbitals with hybridized orbitals; to be discussed below (see fig. 5).]
It is important to recognize that octet stabilization by electron orbital sharing results in bond properties which differ fundamentally from those encountered in ionic (electrovalent) bonding: with the formation of the covalent bond between the hydrogen atoms $\left(\mathrm{H}_{2}\right.$ molecule formation) the bond forming capabilities of the two hydrogen atoms are saturated; the final product is a distinct molecule $\left(\mathrm{H}_{2}\right)$ rather than a giant-sized solid body which is obtained as the final product with ionic bond formation. Covalent solid bodies, however, also do exist: they are formed if the elements involved have the capability of forming more than one bond. For example, carbon will form four covalent bonds in tetrahedral configuration - the result is diamond, a covalently bonded, three dimensional network.
Diatomic Molecules Involving Dissimilar Atoms
Consider the compound hydrogen chloride (in the gaseous, liquid or solid state). This compound is not ionic (because the energy state on complete ionization would be higher. Instead, bonding between hydrogen and chlorine atoms is accomplished by the sharing of electrons in a molecular orbital, thus forming a single covalent bond (fig. 6).
The electrons involved are the 1 s electron of the hydrogen atom and the unpaired $3 p_{z}$ electron of the chlorine atom. After spin-pairing, a typical $\sigma$ bond results in which each atom attains a noble gas structure: in $\mathrm{HCl}$ the hydrogen atom is "involved" with two electrons, as in helium, and the chlorine atom with 18 electrons, as in argon. The other (inner) electrons of chlorine do not participate in the bonding; they are termed non-bonding.
According to the above considerations, the bonding of $\mathrm{HCl}$ may be taken as very similar to that of $\mathrm{H}_{2}$. However, in the hydrogen molecule the electrons participating in the $\sigma$ bond formation do not favor the proximity of either of the hydrogen nuclei - instead, they are equally shared. On the other hand, a preference for one nucleus (chlorine) is shown by the electrons of the $\sigma$ bond in hydrogen chloride. This is because electrons, for energetic reasons, favor the environment of the more electronegative atom; in hydrogen chloride this is the chlorine atom and, consequently, the $\sigma$ bond electrons spend more time in the vicinity of the chlorine atom. This situation results in the chlorine end of the molecule being fractionally negatively charged $\left(\delta^{-}\right)$ and the hydrogen end being fractionally positively charged $\left(\delta^{+}\right)$. We denote such an internal charge redistribution by the symbolism $\mathrm{H}^{\delta+}-\mathrm{Cl}^{\delta-}$, where the $\delta$ sign indicates a partial electronic charge; the bond is said to be polar. (The hydrogen molecule does not a priori exhibit such an "asymmetric charge distribution"; it may, however, acquire a temporary polarization.) Molecules with asymmetric electronic charge distribution have a permanent dipole moment, the value of which is given by the product of the fractional charge ( $\delta^{+}$ must be equal to $\delta^{-}$) and the distance of charge separation $(\mathrm{L})$. Although dipole moments can be measured, their values do not allow us to directly calculate the polarity of bonds between atoms since the detailed internal charge distribution is unknown. The direction of electron drift and, to some extent, the magnitude of its effect can be estimated from the magnitude of the difference in electron affinity between the reaction partners. Because of a limited data base on electron affinity, L. Pauling introduced a related term, the relative electronegativity (see below). From the above considerations it should be clear that no sharp dividing line exists between ionic and covalent bonding. We might consider a completely ionic bond to result in cases where the electron drift is such that one atom (the cation) becomes entirely deficient in one or more electrons, and the other atom (the anion) becomes correspondingly electron-rich, the bonding electrons being entirely under the influence of the latter. Hydrogen halides are $\sigma$ bonded and have dipole moments; the bonds are referred to as polar covalency. The homonuclear diatomic molecules formed by the halogens, viz. $\mathrm{F}_{2}, \mathrm{Cl}_{2}, \mathrm{Br}_{2}$ and $\mathrm{I}_{2}$, are all $\sigma$-bonded systems involving spin-pairing of the various $p_{z}$ electrons $\left(2 p_{z}, 3 p_{z}\right.$, $4 p_{z}$ and $5 p_{z}$ respectively). They have no permanent dipole moments.
Energetics of Covalent Bonding
Pauling extensively treated the energetics of polar covalencies, encountered in all heteronuclear systems (such as $\mathrm{H}-\mathrm{Cl}$ ) which have permanent dipole moments $\left(\mathrm{H}^{\delta+}-\mathrm{Cl}^{\delta-}\right)$. His approach visualizes the bonding to consist of two components - a pure covalency and an ionic bonding component with attraction resulting from the interaction of the fractional charges on the nuclei involved.
Pauling determines the basic covalent bonding component, for example in the formation of $\mathrm{HCl}$, from the experimentally obtained bond energies associated with the molecular species of the components, $\mathrm{H}_{2}$ and $\mathrm{Cl}_{2}$. For this purpose he made the basic assumption that the covalent bond component between the dissimilar atoms is given by the geometric mean of the pure covalent bond energies associated with these molecular species. Thus:
Pure Covalent Bond Energy of $\mathrm{HCl}$:
\begin{aligned}
\mathrm{BE}_{\mathrm{HCl}} &=\sqrt{\mathrm{BE}_{\mathrm{H}_{2}} \times \mathrm{BE}_{\mathrm{Cl}_{2}}} \
& \mathrm{H}_{2}: \mathrm{H}-\mathrm{H} \text { (Bond Energy }=\mathrm{BE}_{\mathrm{H}_{2}}=430 \mathrm{~kJ} / \mathrm{mole} \text { ) } \
&\mathrm{Cl}_{2}: \mathrm{Cl}-\mathrm{Cl} \text { (Bond Energy }=\mathrm{BE}_{\mathrm{Cl}}=238 \mathrm{~kJ} / \mathrm{mole} \text { ) } \
&\mathrm{BE}_{\mathrm{HCl}}=\sqrt{430 \times 238}=320 \mathrm{~kJ} / \text { mole }
\end{aligned}
The ionic contribution $(\Delta)$ to the polar covalent bonding is then obtained from the difference between the experimentally determined bond energy (for $\mathrm{HCl}$, for example), which obviously must contain both components, and the calculated pure covalent bond energy:
$\left.\Delta=\left[\mathrm{BE}_{\mathrm{HCl}} \text { (experimental }\right)\right]-\left[\mathrm{BE}_{\mathrm{HCl}} \text { (theoretical - covalent) }\right]$
$\text{(The experimentally determined} \mathrm{BE}_{\mathrm{HCl}}=426 \mathrm{~kJ} / \mathrm{mole}.)$
$\Delta=426-320=106 \mathrm{~kJ} / \text { mole }$
In connection with the presently discussed work, Linus Pauling established the now generally used electronegativity scale, or, better, the scale of relative electronegativities. This scale, included in the Periodic Table of the Elements, is extremely helpful since values of the electron affinity are known thus far for only very few elements.
The electronegativity $(\mathrm{x})$ scale lists the relative tendency of the neutral elements to attract an additional electron. (The values of $x$ are conventionally listed in electron Volts.) The scale listed was obtained by arbitrarily fixing the value of $x_{H}=2.2$. Pauling obtained the values for the other elements by relating differences in electronegativity of reaction partners to the fractional ionic character (ionic bonding component) of the bond established between them:
\begin{aligned}
&\text { experimental } \mathrm{BE}_{\mathrm{AB}}=\sqrt{\mathrm{BE}_{\mathrm{AA}} \times \mathrm{BE}_{\mathrm{BB}}}+\mathrm{k}\left(\mathrm{x}_{\mathrm{A}}-\mathrm{x}_{\mathrm{B}}\right)^{2} \
&\text { or } \
&\Delta=96.3\left(\mathrm{x}_{\mathrm{A}}-\mathrm{x}_{\mathrm{B}}\right)^{2} \mathrm{~kJ}
\end{aligned}
According to Pauling, the bonding character between two different elements may be defined as:
ionic bonding for: $\Delta x>1.7$; covalent bonding for: $\Delta x<1.7$
The fractional ionicity of polar covalent bonding as listed in the P/T is obtained by the relationship:
$\%$ ionic bonding $=\left(1-\mathrm{e}^{-0.25\left(\mathrm{x}_{\mathrm{A}}-\mathrm{x}_{\mathrm{B}}\right)^{2}}\right) \times 100$
Bonding In Polyatomic Molecules
In the formation of covalent bonds between atoms in polyatomic molecules, the conditions that atomic orbitals distort so that maximum overlap may be achieved when bonding occurs produces more extensive changes of orbital geometry than is the case with diatomic molecules. It will be remembered that the changes which result in a $\sigma$ bond formation in diatomic molecules are a distortion of the s orbitals or $p$ lobes so that strongest bond formation results from the maximum overlap between the two joined nuclei. For polyatomic molecules extensive alterations in the spatial disposition of atomic orbitals do occur, and very often the unique orbital geometry of the original atomic orbitals is completely lost (fig. 7). It is sometimes convenient to view these alterations as occurring in each atom prior to bonding by a process involving the mixing, or hybridization, of atomic orbitals.
Bonding Involving Carbon
In methane $\left(\mathrm{CH}_{4}\right)$ the four outer, or valence, electrons of carbon are shared with the electrons of hydrogen; there is spin-pairing (resulting in bond formation) between each individual hydrogen electron and one of the carbon valence electrons. The noble gas structure is thus attained by each nucleus: the carbon nucleus "sees" eight outer electrons and each $\mathrm{H}$ nucleus "sees" two electrons. Consider now the orbitals which are involved in more detail. Each hydrogen has one spherical valence orbital (the 1s orbital) containing one electron, and covalent bond formation results from its distortion, overlap with the valence orbitals of carbon and spin-pairing.
According to the Aufbau principle and Hund's rule, in its ground state carbon has two electrons in the filled $\mathrm{K}$ shell and four electrons in the $\mathrm{L}$ shell: two 2 s electrons and two $2 p$ electrons in singly occupied orbitals which are capable of covalent bond formation. This configuration provides, in principle, only two orbitals $\left(2 p_{x}, 2 p_{y}\right)$ for covalent bond formation. However, with two covalencies carbon will not yield the desirable octet configuration. Such a configuration can be obtained if one of the two 2s electrons is "promoted" into the empty $2 p_{z}$ orbital since this process results in four singly occupied orbitals - all of which, being singly occupied, are capable of bond formation. The dissimilar singly occupied orbitals can assume (and therefore will assume) upon bond formation a lower energy configuration involving a process called hybridization whereby the four orbitals (of two different types) hybridize into four identical orbitals of maximum equal spacing from each other (fig. 8). Thus the hybridized orbitals ($sp^3$ hybrids) are lobes emanating from the carbon atom into the corners of a tetrahedron, forming bond angles of $109^{\circ} 28^{\prime} . \mathrm{sp}^{3}$ hybridization is characteristic for carbon; however, other types of hybridization, $\mathrm{sp}^{2}$ and $\mathrm{sp}$, are encountered in other elements as well. Boron, for example, will tend to promote one of its two 2s electrons into a $2 p$ state and, by hybridization, form three equivalent $s p^{3}$ orbitals which assume planar orientation with band angles of $120^{\circ}$. Beryllium forms sp hybrid orbitals of linear orientation (the bond angle is $\left.180^{\circ}\right)$. All hybrid orbitals are capable of $\sigma$ bond formation.
The great diversity of carbon compounds (it forms more compounds than all the rest of the elements in the Periodic Table) can be attributed to the hybridization capability of the carbon orbitals $\left(\mathrm{sp}^{3}, \mathrm{sp}^{2}\right.$ and $\left.\mathrm{sp}\right)$. As a consequence, carbon forms not only axisymmetric $\sigma$ bonds (through axial orbital overlap), but also $\pi$ bonds (through lateral orbital overlap). In the compound ethane $\left(\mathrm{H}_{2} \mathrm{C}=\mathrm{CH}_{2}\right)$, the interactive carbon atoms undergo $\mathrm{sp}^{2}$ hybridization for a $\sigma$ bond by overlap of two $\mathrm{sp}^{2}$ hybrid orbitals and form, in addition, a $\pi$ bond by lateral overlap of the remaining non-hybridized $p$ orbitals. (Double bonds involve one $\sigma$ and one $\pi$ bond.) In acetylene $(\mathrm{HC} \equiv \mathrm{CH})$, we find sp hybridization and, by axial overlap, $\sigma$ bond formation as well as lateral overlap of the remaining non-hybridized $p_{x}$ and $p_{y}$ orbitals which form two $\pi$ bonds (fig. 9).
METALLIC BONDING
The stability of both covalent and metallic bonds may be regarded as arising from the potential energy lowering experienced by valence electrons under the influence of more than one nucleus. In metals, where the valence electrons are not as tightly bound to their ion cores, we cannot expect the formation of strong electron-pair bonds. The bond energies of known diatomic molecules of metallic elements are, in fact, smaller than those consisting of nonmetallic elements: $104 \mathrm{~kJ} /$ mole for $\mathrm{Hg}_{2}$. Only diatomic molecules of the semi-metals have relatively high binding energies $\left(385 \mathrm{~kJ} / \mathrm{mole}\right.$ for $\mathrm{As}_{2}$, $293 \mathrm{~kJ} / \mathrm{mole}$ for $\mathrm{Sb}_{2}$ and $163 \mathrm{~kJ} /$ mole for $\mathrm{Bi}_{2}$ ). These values reflect multiple bonding. Much greater stability is possible in larger aggregates of atoms such as bulk metals.
The known properties of metals, such as low electrical resistance and malleability, support the conceptual view that the valence electrons in metals never remain near any
particular atom very long, but drift in a random manner through the lattice of ion cores. We may therefore visualize metals as a lattice of ion cores being held together by a gas of free electrons.
(Bonding in metallic systems is discussed in more detail in LN-3.)
SECONDARY (van der WAALS) BONDING
Primary bonding (ionic, covalent and metallic) is strong and the energies involved range from about 100 to 1000 kJ/mole. In contrast, secondary bonding is weak, involving energies ranging from about 0.1 to 10 kJ/mole. While this type of bonding, also referred to as “residual”, is weak, it is essential in the functioning of our environment. Coke would likely be gaseous and not a bubbly, refreshing brew were it not for secondary bonding, nor would catalytic converters function.
The energy difference between the liquid and vapor states of a given system is given by the heat of vaporization, i.e. the heat required to convert a given liquid into a vapor (normally) at the boiling point temperature at 1 atm pressure. The energy difference is due to intermolecular attraction between molecules at close distance of separation. This phenomenon of attraction through secondary bonding can best be considered between a single pair of molecules, but recognizing that the forces are of longer range. Four types of intermolecular forces can be identified.
1. Dipole-Dipole Interaction: Molecules with permanent dipoles (such as water, alcohol and other organic compounds with functional groups) exert a net attractive force on each other as a result of varying degrees of alignment of oppositely charged portions of the molecules (fig. 10). For two polar molecules with a dipole moment of $(\mu)$ separated by a distance of (r), the energy of attraction can be quantified as:
where $\mu$ is the dipole moment, $r$ is the distance of approach of the oppositely charged molecular portions, $k$ is the Boltzmann constant (see $P / T)$ and $T$ is the absolute temperature in $\mathrm{K}$.
The molar energies of attraction associated with dipole-dipole interaction range from 0 to about $10 \mathrm{~kJ} /$ mole. These forces are primarily responsible for the liquid state (at room temperature) of most polar organic molecules; they are a contributing factor for $\mathrm{H}_{2} \mathrm{O}$ to be liquid at room temperature, and are responsible for alcohol being a liquid.
2. Dipole-Induced Dipole Interaction: A dipole in one molecule can interact with and polarize the electrons of a neighboring non-polar molecule, thus generating an induced dipole which will experience an attractive force with the polarizing, polar dipole (fig. 11). P. Debye showed that in a molecule with a "polarizability" of $(\alpha)$ the attractive potential arising from dipole-induced dipole interaction is given as:
$\mathrm{E}_{\text {Dipole-Induceddipole }}=-\dfrac{2 \alpha \mu^{2}}{\mathrm{Dr}^{6}}$
Induced dipole interaction is important in aqueous solutions and very effective during adsorption of inert molecules on active solid substrates.
3. London Dispersion Forces: It is a well-known fact that all substances, including rare gases and hydrogen, assume liquid state at finite temperature, an indication of the existence of attractive interatomic and intermolecular forces, even in the absence of permanent dipole systems. The origin of this force has been proposed by F. London in 1930. Accordingly, orbiting electrons will at any instance generate a "temporary dipole", the configuration of which changes as the electrons move. Since all atoms of a given system similarly experience temporary instantaneous dipole moments, their effect is expected to be cancelled because of the statistically random orientation of dipoles. It is evident that, should the dipoles be synchronized in a given assembly of atoms, then a net attractive force would result (fig. 12). But since such an attractive force constitutes a lowering of the energy of a given system, synchronization can and will take place because all systems will attempt to assume minimum energy configuration. The London dispersion force can be formulated as:
$\mathrm{E}_{\text {London }}=-\mathrm{K} \dfrac{\alpha^{2}}{\mathrm{r}^{6}}$
The attractive London forces are small, as manifested by the very low boiling points of the smaller rare gases, of hydrogen and nitrogen.
4. Hydrogen Bonding: The short and long range dipole interactions calculated from molecular dipole moments are inadequate in explaining a multitude of phenomena in organic as well as some inorganic systems. L. Pauling studied such
phenomena and concluded the existence of highly specific attractive interaction between hydrogen that is acidic (carries a fractional positive charge) and the elements O, F, N and, to a lesser extent, S in both organic and inorganic molecules. This interaction, which can as yet not be formulated, is referred to as hydrogen bonding; its magnitude, ranging up to 40 kJ/mole, is significantly larger than that of any other secondary bonding type. Hydrogen bonding is considered instrumental in controlling most properties of water, is a key element in the structure of nucleic acid and thought to be an essential component in memory functions of the human brain. | textbooks/chem/Inorganic_Chemistry/Introduction_to_Solid_State_Chemistry/01%3A_Lectures/1.02%3A_Chemical_Bonding.txt |
INTRODUCTION
Characteristic properties of metals include: (1) electrical conductivity, (2) opaqueness and (3) malleability. A very simple model in which the metallic crystal is viewed as a lattice of positive ions surrounded by a “gas” of free electrons provides a crude understanding of the first and third properties. If the crystal has a gas of free electrons, it is easy to see why the application of an electric field will result in the motion of these electrons and thus for the high electrical conductivity. This model also allows one to explain the malleability of metals, as shown in fig. 1b. When a metallic crystal is subject to forces that displace one plane of atoms with respect to another, the environment of the charged species is left unchanged. In contrast, the displacement of neighboring planes in an ionic crystal as a result of a distorting force will lead to cleavage, largely because of changes in the interaction of the charged species (Figure $\PageIndex{1, left}$).
The free electron gas theory or Drude–Lorentz (D-L) theory also explains in principle the nature of the attractive bonding forces which hold the metallic ions (the metal) together: the crystal is held together by electrostatic forces of attraction between the positively charged metal ions and the non-localized, negatively charged electrons - the electron gas. The theory in its original form assumes that the classical kinetic theory of gases is applicable to the electron gas ; mutual repulsion between electrons was ignored and the electrons were expected to have velocities which are temperature dependent according to a Maxwell-Boltzmann distribution law.
While the Drude-Lorentz theory of metallic bonding was considered a useful model, several shortcomings soon became apparent. The most notable failure consisted of the unexplainable discrepancy between the observed and predicted specific heats of metals (energy in the form of heat, required to increase the temperature of $1 \mathrm{~g}$ of a given metal by $1^{\circ} \mathrm{C}$ ). The D-L theory predicted much larger specific heats than are observed (because the Maxwell-Boltzmann energy distribution has no restrictions as to the number of species allowed to have exactly the same energy). [If there are restrictions as to the number of electrons with identical energy (Pauli exclusion principle!), we have to apply a different form of statistics (Fermi-Dirac statistics).]
ENERGY BAND THEORY BASED ON FORMATION OF MOLECULAR ORBITALS
A different model, and one that is more closely related to the models of the chemical bond discussed earlier, is the band model. This model, proposed by Bloch before the development of the molecular-orbital approach to chemical bonding, is actually a molecular-orbital model of metallic crystals. The orbitals characteristic of the whole crystal are obtained as linear combinations of the atomic orbitals of the individual atoms.
From the large electrical conductivity of metals, it appears that at least some of the electrons can move freely through the bulk of the metal. Since even lithium with only one valence electron has eight nearest neighbors (crystallizes in a body-centered cubic lattice), it is clear that the atoms can not be bonded to each other by localized electron-pair bonds (for in that case the lithium atom, which altogether has only three electrons, would have to supply eight or even fourteen valence electrons to establish bonding with the nearest and next-to-nearest neighbor atoms). The fact that in a metal lattice one atom can simultaneously interact with a large number of others can be explained by considering that metals constitute an extreme example of delocalized bonding.
Conditions in metals are not as easily assessed as in organic hydrocarbon chains because metal lattices are three-dimensional structures. However, several important results from the theory of metallic bonding can be understood on a highly simplified model in which the three-dimensional lattice is replaced by a one-dimensional system. Consider the formation of a linear array of lithium atoms from individual lithium atoms:
$\mathrm{Li} \rightarrow \mathrm{Li}-\mathrm{Li} \rightarrow \mathrm{Li}-\mathrm{Li}-\mathrm{Li} \rightarrow \mathrm{Li}-\mathrm{Li}-\mathrm{Li}-\mathrm{Li} \rightarrow \ldots$
The first stage is the formation of a lithium molecule, $\mathrm{Li}_{2}$, which is comparable to the hydrogen molecule, $\mathrm{H}_{2}$. The two lithium atoms are bound together by a pair of valence electrons; each lithium atom supplies its 2s-electron which, through orbital overlap, forms a covalent molecular bond (fig. 2). The molecule formed occurs in lithium vapor and differs from the $\mathrm{H}_{2}$ molecule only by its greater interatomic distance ( $\mathrm{Li}-\mathrm{Li}= 2.67 \times 10^{-10} \mathrm{~m}, \mathrm{H}-\mathrm{H}=0.75 \times 10^{-10} \mathrm{~m}$ ) and smaller energy of formation. The larger separation and the consequent reduction in bond energy can be attributed to the size of the $\mathrm{Li}^{+}$ atomic core.
In the determination of molecular orbital states by linear combination of atomic orbitals (LCAO) we obtain more than one solution from the Schrödinger equation. We have so far concentrated mostly on the low energy states, leading to $\sigma$ or $\pi$ bonds. We have largely ignored the so-called antibonding states which in covalent molecules have energies higher than the original energy states of the isolated reaction partners (presently Li) and, therefore, remain unoccupied (or empty) in two-atomic molecules. In any multi-atom system (such as a piece of metal), the total number of molecular orbital states (electronic states) is always the same as the total number of original atomic states (conservation of electronic states). However, not all of these states are necessarily occupied by electrons.
Consider the hypothetical linear molecule $\mathrm{Li}_{3}$. Since the valence electron cloud is spherical, the central lithium atom can not give preference to either of its neighbors. The three atomic valence electron clouds overlap to form one continuous distribution and two others with nodes (antibonding states), i.e., three molecular orbitals in all (Fig. 3).
As the length of the chain is increased, the number of electronic states into which the atomic $2 s$ state splits also increases, the number of states always equaling the number of atoms. The same occurs when lithium chains are placed side-by-side or stacked on top of each other, so that finally the space lattice of the lithium crystal is obtained. It is of great significance that these electronic states have energies which are bounded by an upper and lower limiting value (see fig. 4). Within these limits the states form an energy band of closely spaced values (one gram of lithium contains nearly $10^{23}$ atoms). Similarly, energy bands can also result from overlapping $p$ and $d$ orbitals. The electronic states (orbitals) within an energy band are filled progressively by pairs of electrons in the same way that the orbitals of an atom were filled in accordance with the Pauli principle. This means that for lithium the electronic states of the 2s band will be exactly half-filled.
It is of interest to consider why lithium atoms or $\mathrm{Li}_{2}$ molecules combine to form a metal lattice. In the lithium lattice the smallest distance between neighboring atoms is $3.03 \times 10^{-10} \mathrm{~m}$, which is larger than in the $\mathrm{Li}_{2}$ molecule. This reflects the fact that bonds between pairs of atoms in the metal are weaker than they are in the molecule. Nevertheless, the metallic form of lithium is more stable than the molecular form because in the metal one atom has many more neighbors than in the $\mathrm{Li}_{2}$ molecule. As a result, the binding energy per gram atom of lithium (i.e., per $6.92 \mathrm{~g}$ of lithium) is $163 \mathrm{~kJ}$ for the metal lattice, but only $56 \mathrm{~kJ}$ for one mole of molecule.
The possibility of hybridization (first advanced by L. Pauling to explain metallic bonding) is also a likely factor for the formation of metallic bonds. Thus strong bonds can be formed when the valence electron clouds become concentrated along the direction in which the bonding partners are situated. According to Pauling the situation can be described by mesomeric limiting formulae, just as in the theory of unsaturated carbon compounds. One reason given for hydrogen not forming a metal is its inability to hybridize.
The appearance of bands of allowed electron energies relates to the overlap of electron $\psi$ functions. The width of each energy band is a function of the crystal structure because it determines the number of nearest neighbors in the crystal. Different metal atoms with the same valence, particularly elements in the same group of the periodic table, can often replace each other in arbitrary proportions without altering either the lattice type or the structure of the energy bands. This explains why such metals tend to form a complete series of solid solutions. Metallic alloys consist of such solid solutions or of heterogeneous mixtures of such solutions. Within certain limits, even metal atoms of different valence can be interchanged in a lattice.
Structure of Metals
According to the above considerations the band structure of Li metal can be represented as shown in fig. 5. According to previous reasoning, the $2 \mathrm{~s}$ band has $\mathrm{N}$ states ( $\mathrm{N}=$ number of atoms) and accommodates $n 2 s$ electrons (where $n$ is the number of electrons per atom in the $2 \mathrm{~s}$ state times N). Thus, this band has only half of the states filled since each state can accommodate two electrons of opposite spin (Pauli exclusion principle). In accordance with the Aufbau principle, the lowest energy states of the band are filled first and the upper states remain empty - but can readily be occupied by electrons upon thermal excitation or the application of an electric field. Since the width of the energy band is of the order of a few volts, spacings of states within the band are of the order of $\sim 10^{-20} \mathrm{eV}\left(1 \mathrm{eV}=1.6 \times 10^{-19} \mathrm{~J}\right)$, electrons can readily acquire the energy necessary to move into excited states, be accelerated, and move through the metal as conducting electrons. Partly filled bands thus constitute conduction bands.
Fig. 5 Schematic energy band configuration for Li .
The conduction mechanism in Mg, for example, appears complicated by the fact that each $3 s$ state of the valence shell in the atoms is doubly occupied $\left(3 s^{2}\right)$. Thus the $3 s$ band must be filled completely and no electronic conduction would in principle be expected. Electronic conduction, however, is observed because of a partial overlap of the $3 s$ and the empty $3 p$ bands. With this overlap, electrons can be activated into empty $3 p$ states and exhibit conduction, as in the partly filled $s$ band in Li.
Structure of Insulators and Semiconductors (Molecular Crystals)
The same band model can also account for the lack of conductivity in covalent crystals, such as diamond. In the diamond crystal, with a two-electron bond between every atom and each of its neighbors, the bonding molecular orbitals form a band at much lower energies than the energy of the atomic $\mathrm{sp}^{3}$ hybrid orbitals, and the antibonding orbitals form a band at energies much higher than the energy of the atomic orbitals. Also, since each atom contributes four orbitals and four electrons, there are just enough electrons to fill the bonding orbitals (corresponding to an electron pair bond joining every atom to each of its neighbors) and the only vacant orbitals are those in the high energy band of antibonding orbitals (see fig. 6).
Fig. 6 Band structure of insulators and semiconductors (molecular crystals); the conditions depicted reflect a molar crystal of carbon (diamond).
Both insulators and semiconductors have the same basic band structure $-$ the primary difference is the width of the forbidden energy gap $\left(\mathrm{E}_{\mathrm{g}}\right)$ between the valence and the conduction band.
Insulators, generically, are materials with very high resistivity (see Table I), comprising glasses, polymers, refractories, composites, liquids and gases. In the present context, an insulator is a molecular crystal, such as diamond (C) or sapphire, with a band gap $\left(E_{g}\right)$ in excess of $4 \mathrm{eV}$ (arbitrary value). Generally such materials will not conduct electricity since their valence band is filled and the energy required to transfer electrons from the valence band to the empty conduction band is far in excess of both the thermal energies at room temperature and the energy provided by radiation of the visible spectrum ( $2 \mathrm{eV})$. Therefore, insulators (in single crystal form) are normally transparent (colorless); however, if light is excessively or totally scattered at internal heterogeneities (such as grain boundaries), they may be translucent and even opaque. It should also be recognized that impurities $\left(\mathrm{Cr}^{3+}\right.$ in $\left.\mathrm{Al}_{2} \mathrm{O}_{3}\right)$ or particular point defects (color centers) may impart a color to the transparent insulator crystals. The color arises because of partial absorption of white light and selective transmission of the other portions of the visible spectrum.
Semiconductors: The conventional semiconductors, silicon (Si) and germanium (Ge), have a band gap $\left(\mathrm{E}_{\mathrm{g}}\right)$ of $1.1$ and $0.7 \mathrm{eV}$ respectively and therefore absorb visible radiation; they are opaque (fig. 7). Considering the statistical nature of the thermal energy distribution in the solid matrix (Maxwell-Boltzmann), a significant number of electrons in the valence band will, at room temperature, acquire sufficient energy to cross the existing energy gap and thus provide for semiconductivity. The conductivity will therefore increase with temperature, contrary to metallic systems, until electron scattering effects, due to increased lattice vibrations (which decrease the mobility of electrons), begin to dominate.
The value of semiconductors for solid state device fabrication lies in the fact that the number and type of conducting electric charge carriers [electrons are $n$-type (negative), holes are p-type (positive)] can be controlled through incorporation of appropriate dopant elements. Thus the substitutional incorporation of Group V elements (Sb, As, P) provides for shallow donor levels in the band gap at about $0.01 \mathrm{eV}$ from the conduction band. The substitutional incorporation of Group III elements (B, Al) generates acceptor levels in the band gap at about $0.01 \mathrm{eV}$ from the valence band. The two types of impurities are almost completely ionized at room temperature and give rise to extrinsic n-type and p-type conductivity – the basis for the formation of junction devices such as diodes and transistors (fig. 8).
Fig. 7 Optical behavior of insulators and semiconductors
Of increasing importance are compound III-V (adamantine) semiconductors, such as GaAs, InSb, InP and GaP (compounds of Group III and Group V elements). Together these compounds provide eight valence electrons and, by $\mathrm{sp}^{3}$ hybridization, are able to form a diamond-like, covalent crystal structure with semiconductor properties. These
Fig. 8 Extrinsic p- and n-type semiconductors.
compounds (GaAs, for example) exhibit electron mobilities which are higher than those of silicon and, therefore, are of considerable interest for advanced device technology.
Table I. ELECTRICAL RESISTIVITIES OF METALS AND NONMETALS AT $20^{\circ} C^{*}$
*From American Institute of Physics Handbook, Dwight E. Gray, ed. McGraw-Hill, New York (1963), pp. 4–90; 9–38.
**Note the different units in the two columns
APPENDIX
Approximate Energy Values for Some Events and Reactions
$\begin{array}{ll} \text { Solar Energy Emission } & \sim 10^{30} \mathrm{~kJ} / \text { day } \ \text { Convertible Solar Energy on Earth } & \sim 10^{18} \mathrm{~kJ} / \text { year } \ \text { Energy Consumption on Earth } & \sim 8 \times 10^{17} \mathrm{~kJ} / \text { year } \ \text { Super-Nova Explosion } & \sim 10^{39} \mathrm{~kJ} \ 100 \text { Mega-Ton H-Bomb Explosion } & \sim 10^{15} \mathrm{~kJ} \ \text { Atlas Blast-Off } & \sim 10^{8} \mathrm{~kJ} \ \text { Lethal X-Ray Dose } & \sim 1 \mathrm{~kJ} \ \text { 25ф Piece Falling from Pocket } & \sim 10^{-4} \mathrm{~kJ} \ \text { Bee's Wing Beat } & \sim 10^{-6} \mathrm{~kJ} \ \text { Moonlight/Face/Second } & \sim 10^{-8} \mathrm{~kJ} \end{array}$ | textbooks/chem/Inorganic_Chemistry/Introduction_to_Solid_State_Chemistry/01%3A_Lectures/1.03%3A_Bonding_in_Metals_Semiconductors_and_Insulators__Band_Structure.txt |
In an assembly of atoms or molecules a solid phase is formed whenever the interatomic (intermolecular) attractive forces significantly exceed the disruptive thermal forces and thus restrict the mobility of atoms, forcing them into more-or-less fixed positions. From energy considerations it is evident (as discussed in LN-2) that in such solids the atoms or molecules will always attempt to assume highly ordered structures which are characterized by symmetry. Depending on the nature of the acting interatomic forces, all solids may be subdivided into:
1. Ionic solids (NaCl)
2. Covalent solids (Diamond)
3. Metallic solids (Fe, Ni, etc.)
4. Van der Waals solids (ice, solid He)
Solids as we encounter them in nature may or may not reflect the internally ordered arrangement in their appearance. We find, for example, well-formed quartz crystals, garnets, diamonds and snowflakes which are all characterized by flat bounding planes which intersect at characteristic angles. On the other hand, we also observe rounded stones and man-made cast solid objects with no external evidence of internal order (fig. 1).
To understand the external appearance of the solid state it is necessary to consider the formation of solids from different phases. Solids, for example, are formed upon cooling of liquids (melts) - by freezing or solidification; this solidification process normally proceeds in total confinement and the resulting “cast” structure will have an external appearance which reflects in detail the confining geometry (and not the internal order). Moreover, depending on solidification conditions, the solid body may be either a single crystal or polycrystalline. Polycrystalline solids (in excess of 95% of the solid state encountered) may be thought of as an assembly of microscopic single crystals with random orientation held together like a maze structure by the interwoven irregular shapes of the individual crystals.
A typical example of an "unconfined" phase transformation is the formation of snowflakes where the external boundaries of the solid have assumed crystalline appearance, reflecting in detail the internally ordered molecular $\left(\mathrm{H}_{2} \mathrm{O}\right)$ arrangement. Another unconfined formation of a solid is precipitation from solution (sugar crystals, $\mathrm{CuSO}_{4}$ and the like). Similarly, the formation of crystals from the vapor phase leads to bodies which externally reflect elements of internal order.
CRYSTAL STRUCTURE
From the earlier discussion it should be apparent that, when strong interatomic forces exist, atoms tend to pack closely together - the closeness of packing being particularly pronounced in the solid state. In this case atoms can be regarded as hard spheres and the problem of close packing can be treated as one in which the whole assembly has a tendency toward efficient packing. A little thought or a few simple experiments with ping-pong balls quickly convince us that regular arrangements of the spheres generally lead to more compact assemblies than irregular arrangements (fig. 2). The same principle applies to arrangement of atoms in the solid state. Where strong attractive forces are exerted we find that the atoms or molecules concerned arrange themselves in a regular three-dimensional pattern. It is this regularity which is the basis of crystallinity in materials: i.e., a crystal structure is nothing more than an orderly array of atoms or molecules. This definition of a crystal is distinct from the popular concept based on observation of external symmetry of crystals, often seen during the study of elementary chemistry, in which some crystals appear cubic, others needle-shaped and so on. The regular external shape is obtained only when the conditions of crystallization are favorable to development of flat, geometric faces. In most instances, particularly with metals, these conditions are absent, and the crystals have irregular surfaces even though the internal arrangement is perfectly geometric.
Atomic arrays in crystals are conveniently described with respect to a three-dimensional net of straight lines. Consider a lattice of lines, as in fig. 3, dividing space into equal-sized prisms which stand side-by-side with all faces in contact, thereby filling all space and leaving no voids. The intersections of these lines are points of a space lattice, i.e., a geometrical abstraction which is useful as a reference in describing and correlating symmetry of actual crystals. These lattice points are of fundamental importance in describing crystals for they may be the positions occupied by individual atoms in crystals or they may be points about which several atoms are clustered. Since prisms of many different shapes can be drawn through the points of a space lattice to partition it into cells, the manner in which the network of reference lines is drawn is arbitrary. It is not necessary that the lines be drawn so that atoms lie only at corners of unit prisms. In fact, it is more convenient to describe some crystals with respect to prisms in which atoms lie at prism centers or at the centers of prism faces as well as at prism corners.
An important characteristic of a space lattice is that every point has identical surroundings: the grouping of lattice points about any given point is identical to the grouping about any other point in the lattice. In other words, if we could move about in the lattice, we would not be able to distinguish one point from another because rows and planes near each point would be identical. If we were to wander among the atoms of a solid metal or chemical compound, we would find the view from any lattice point exactly the same as that from any other.
There are fourteen space lattices (fig. 4). That is, no more than fourteen ways can be found in which points can be arranged in space so that each point has identical surroundings. Of course, there are many more than fourteen ways in which atoms can be arranged in actual crystals; thus there are a great number of crystal structures. Too often the term “lattice” is loosely used as a synonym for “structure”, an incorrect practice which is frequently confusing. The distinction can be clearly seen if we remember that a space lattice is an array of points in space. It is a geometrical abstraction which is useful only as a reference in describing and correlating symmetry of actual crystals. A crystal structure, however, is the arrangement of atoms or molecules which actually exists in a crystal. It is a dynamic, rather than a static, arrangement and is subject to many imperfections. Although any crystal structure has an inherent symmetry which corresponds to one of the fourteen space lattices, one, two or several atoms or molecules in the crystal structure may be associated with each point of the space lattice. This symmetry can be maintained with an infinite number of different actual arrangements of atoms - making possible an endless number of crystal structures.
To specify a given arrangement of points in a space lattice, it is customary to identify a unit cell with a set of coordinate axes, chosen to have an origin at one of the lattice points (fig. 5). In a cubic lattice, for example, we choose three axes of equal length that are mutually perpendicular and form three edges of a cube. Each space lattice has some convenient set of axes, but they are not necessarily equal in length or orthogonal. Seven different systems of axes are used in crystallography, each possessing certain characteristics as to equality of angles and equality of lengths. These seven crystal systems are tabulated in Table I (to be considered in conjunction with fig. 4).
Table I. The Seven Crystal Systems
$\begin{array}{lll} \hline \text { System } & \text { Parameters } & \text { Interaxial Angles } \ \hline & a \neq b \neq c & \alpha \neq \beta \neq \gamma \ \text { Triclinic } & a \neq b \neq c & \alpha=\gamma=90^{\circ} \neq \beta \ \text { Monoclinic } & a \neq b \neq c & \alpha=\beta=\gamma \ \text { Orthorhombic } & a=b \neq c & \alpha=\beta=\gamma \ \text { Tetragonal } & a=b=c & \alpha=\beta=\gamma=90^{\circ} \ \text { Cubic } & a=b \neq c & \alpha=\beta=90^{\circ}, \gamma=120^{\circ} \ \text { Hexagonal } & a=b=c & \alpha=\beta=\gamma \neq 90^{\circ} \ \text { Rhombohedral } & a=c h \ \hline \end{array}$
The network of lines through the points of a space lattice (fig. 3) divides it into unit cells (see also fig. 4). Each unit cell in a space lattice is identical in size, shape and orientation to every other unit cell. It is the building block from which the crystal is constructed by repetition in three dimensions. The unit cells of the fourteen space lattices are shown in fig. 4. All crystal structures are based on these fourteen arrangements.
The body-centered cubic, face-centered cubic and hexagonal lattices are common and of prime importance in metals. Some of the metals associated with nuclear applications, such as uranium and plutonium, have crystal structures which are more complicated than these three relatively simple types. In general, crystalline ceramics also are more complex.
CELLS VS PRIMITIVE CELLS
In the literature we often find reference to unit cells and to primitive cells. The primitive cell may be defined as a geometrical shape which, when repeated indefinitely in three dimensions, will fill all space and is the equivalent of one atom. The unit cell differs from the primitive cell in that it is not restricted to being the equivalent of one atom. In some cases the two coincide. For instance, in fig. 4 all fourteen space lattices are shown by their unit cells. Of these fourteen, only seven (which are those?) are also primitive cells.
Primitive cells are drawn with lattice points at all corners, and each primitive cell contains the equivalent of one atom. For instance, a simple cubic unit cell has an atom at each corner. However, at any of these given corners, this atom must be shared with seven other identical cubes which fill the volume surrounding this point. Thus there is effectively only 1/8 of the atom which can be assigned to that particular unit cell. Since there are eight corners in a cube, there is the equivalent of one atom, and thus the primitive cell and unit cell coincide.
Continuing, consider the body-centered cubic (BCC) lattice. In this case there is one atom at the center of the cube and one atom contributed by the eight corners. This cell, then, has two atoms and, to avoid confusion, should be termed a unit cell. In the face-centered cubic lattice there are six face atoms, but each face atom is shared by two cells. Consequently, each face contributes 1/2 an atom. The faces thus contribute three atoms and the corners one, for a total of four atoms in the unit cell. The face-centered cubic structure (FCC) can also be considered as four interpenetrating simple cubic cells.
In the study of crystals the primitive cell has limited use because the unit cell more clearly demonstrates the symmetrical features of a lattice. In other words, the unit cell can usually be visualized readily whereas the primitive cell cannot. For example, the cubic nature of the face-centered cubic lattice is immediately apparent in the unit cell, but it is not nearly so obvious in the rhombohedral primitive cell.
PACKING OF ATOMS
A crystal structure is a regular array of atoms arranged on one of the fourteen space lattices. The least complicated crystal structures are those having a single atom at each lattice point. Polonium has the simplest structure, being simple cubic. In normal metals, the atoms (or positive ions) are held together by a cloud of free electrons so that each atom tends to be attracted equally and indiscriminately to all its geometrically nearest neighbors by the free electrons passing between them. This condition fosters the formation of closely packed structures of the types which can be demonstrated by efficiently packing uniformly-sized spheres into a given volume.
You can gain a better grasp of packing by conducting an experiment as follows. Assume we have a quantity of small spheres which we are required to efficiently pack into a box (a ‘two-dimensional’ approach can be made using a handful of pennies). After some shuffling, it is obvious that the closest possible packing is obtained when the spheres are in contact and their centers occupy positions which correspond to the apices of equilateral triangles (fig. 6). It is also evident that there are two sets of triangles – one set with vertices pointing away from the observer (points up) and the other set with vertices pointing toward the observer (points down).
When a second layer is added, there is closest-packing if the spheres in this new layer rest in the hollows formed by the spheres of the first layer. The centers of the spheres in the second layer will lie above the centers of the points-up triangles or above the centers of the points-down triangles, but not both simultaneously. Which set is used is immaterial. For our discussion, however, assume the second layer is centered on the points-up triangles.
When we start adding a third layer, the spheres will again rest in the hollows formed by the spheres in the second layer. And again we have the option of placing the third layer on the points-up or on the points-down triangles. If we center the third layer on the points-down triangles, we find the third layer is directly above the first layer. If additional layers are added using an alternate stacking sequence (i.e., alternately centering the layers on the points-up and the points-down triangles), the sequence can be written as ABABABAB. . . This arrangement of spheres, translated to an arrangement of atoms, is the hexagonal close-packed (HCP) structure - very important, but not discussed in detail here.
Many elements having covalent bonding form arrangements in which the coordination number is (8–N), where N is the number of valence electrons. What, then, is the coordination number of the HCP structure just demonstrated? The geometry of the structure shows that any one atom has twelve equidistant neighbors. It is apparent that if, in any layer, a given sphere is placed in the adjoining layer, it fits in the hollow formed by three spheres and consequently is tangent to three spheres in the adjoining layer. Thus any given atom in the HCP structure is tangent to twelve other atoms – six in its own layer and three each in two adjoining layers.
When the third layer of spheres was added in the above discussion, we assumed this layer was centered on the points-down triangles of the second layer. What happens if the centers of the points-up triangles of the second layer are used instead? The distribution of spheres in the third layer is the same as in the first two layers, but does not lie directly above either of these two layers. If a fourth layer is added, centered on the points-up triangles of the third layer, we find the fourth layer is directly above the first layer and duplicates it completely. The stacking sequence for this structure can be written as ABCABCABC. . . This arrangement has the same density of packing and the same coordination number as the HCP structure. However, it is the face-centered cubic (FCC) structure.
The HCP and FCC crystal structures have the same density of packing and the same coordination number. Therefore we might expect the behavior of the two HCP and FCC structures to be very much alike with regard to physical and mechanical properties. This, however, in most instances is not the case.
The maximum density of packing is found only in the HCP and FCC crystal structures. Why the metallic bond does not always produce one or the other of these two densest arrangements of atoms is as yet subject of intensive studies. The BCC unit cell contains two atoms, and the coordination number is eight. There is partial compensation for this in the fact that there are six next-nearest neighbors at distances only slightly greater than that of the eight nearest neighbors. Some characteristics of cubic structures are given in Table II.
An interesting example of the type of crystal structure obtained in covalently bonded elements which obey the (8–N) coordination number rule (where N is the number of valence electrons) is the diamond structure. This structure is found in carbon, germanium, silicon, and tin at low temperatures and in certain compounds. Each atom has four nearest neighbors – a configuration that is variously called diamond cubic, body-centered tetrahedral or tetrahedral cubic.
TABLE II. Characteristics of Cubic Lattices
$\begin{array}{lcccc} &\underline{ \text { Simple } }& \underline{\text { Body-Centered } }& \underline{\text { Face-Centered }} \ \text { Unit Cell Volume } & \mathrm{a}^{3} & \mathrm{a}^{3} & \mathrm{a}^{3} \ \text { Lattice Points Per Cell }& 1 & 2 & 4 \ \text { Nearest Neighbor Distance } & a & \frac{a \sqrt{3}}{2} & \frac{a}{\sqrt{2}} \ \text { Number of Nearest Neighbors } & 6 & 8 & 12 \ \text { Second Nearest Neighbor Distance } & a \sqrt{2} & a & a \ \text { Number of Second Neighbors } & 12 & 6 & 6 \ \end{array}$
Many compounds crystallize in variations of cubic forms. Rock salt (NaCl), for example, is typical of many oxides, fluorides, chlorides, hydrides and carbides. It is sometimes considered as simple cubic with alternate Na and Cl atoms on the cube corners. Actually the structure is two interpenetrating FCC lattices – one of Na and the other of Cl – and the corner of one is located at point 1/2, 0, 0 of the other.
Many other oxides, fluorides and some intermetallic compounds have the fluorite $\left(\mathrm{CaF}_{2}\right)$ structure. This is $\mathrm{FCC}$ with $\mathrm{Ca}$ at the cube corners and face centers and $\mathrm{F}$ at all quarter points along the cube diagonals.
LATTICE PLANES AND DIRECTIONS
It is desirable to have a system of notation for planes within a crystal or space lattice such that the system specifies orientation without giving position in space. Miller indices are used for this purpose. These indices are based on the intercepts of a plane with the three crystal axes - i.e., the three edges of the unit cell. The intercepts are measured in terms of the edge lengths or dimensions of the unit cell which are the unit distances from the origin along the three axes. For instance, the plane that cuts the $x$-axis at a distance from the origin equal to one-half the $x$-dimension of the cell is said to have an $x$-intercept equal to $1 / 2$, and if it cuts the $y$-axis at $1 / 2$ the $y$-dimension of the cell, the $y$-intercept is $1 / 2$, regardless of the relative magnitudes of the $x$ - and $y$-dimensions. If a plane is parallel to an axis, it intercepts the axis at infinity.
To determine Miller indices (hkl) of a plane, we take the following steps:
1. Find the intercepts on the three axes in multiples or fractions of the edge lengths along each axis.
2. Determine the reciprocals of these numbers.
3. Reduce the reciprocals to the three smallest integers having the same ratio as the reciprocals.
4. Enclose these three integral numbers in parentheses, e.g., (hkl).
A cube has six equivalent faces. If we have a definite orientation and wish to discuss one specific plane of these six, it is possible to specify this plane by using the proper Miller indices. Parentheses are used around the Miller indices to signify a specific plane. On the other hand, it is often advantageous to talk about planes of a “form” – i.e., a family of equivalent planes such as the six faces of a cube. To do this it is customary to use the Miller indices, but to enclose them in curly brackets (braces). Thus the set of cube faces can be represented as {100} in which
$\{100\}=(100)+(010)+(001)+(\overline{100})+(0 \overline{1} 0)+(00 \overline{1})$
This notation thus provides a shorthand scheme to avoid writing the indices for all six cube faces.
The utility of the scheme is even more evident in the case of the (110) planes – i.e., the dodecahedral planes (in a cubic system), where
\begin{aligned}
\{110\}=&(110)+(101)+(011)+(\overline{1} 10)+(\overline{1} 01)+(0 \overline{1} 1) \
&+(1 \overline{1} 0)+(10 \overline{1})+(01 \overline{1})+(\overline{110})+(\overline{1} 0 \overline{1})+(0 \overline{11})
\end{aligned}
The equivalent form for the orthorhombic system is
\begin{aligned}
&\{110\}=(110)+(\overline{110})+(1 \overline{1} 0)+(\overline{110}) \
&\{101\}=(101)+(\overline{1} 01)+(10 \overline{1})+(\overline{1} 0 \overline{1}) \
&\{011\}=(011)+(0 \overline{1} 1)+(01 \overline{1})+(0 \overline{11})
\end{aligned}
The octahedral planes for the cube are
\begin{aligned}
\{111\}=&(111)+(\overline{1111})+(1 \overline{1} 1)+(11 \overline{1}) \
&+(\overline{111})+(\overline{111 \overline{1}})+(1 \overline{11})+(\overline{111})
\end{aligned}
Direction indices are defined in a different manner. A line is constructed through the origin of the crystal axis in the direction under consideration and the coordinates of a point on the line are determined in multiples of lattice parameters of the unit cell. The indices of the direction are taken as the smallest integers proportional to these coordinates and are closed in square brackets. For example, suppose the coordinates are $x=3 a, y=b$ and $z=c / 2$, then the smallest integers proportional to these three numbers are 6, 2 and 1 and the line has a [621] direction. As further examples, the $x$-axis has direction indices [100], the $y$-axis [010] and the $z$-axis [001]. A face diagonal of the $x y$ face of the unit cell has direction indices [110], and a body diagonal of the cell has direction indices [111]. Negative indices occur if any of the coordinates are negative. For example, the $-y$ axis has indices [010]. A full set of equivalent directions, i.e., directions of a form, are indicated by carets: $\langle u v w\rangle$. | textbooks/chem/Inorganic_Chemistry/Introduction_to_Solid_State_Chemistry/01%3A_Lectures/1.04%3A_The_Nature_of_Crystalline_Solids.txt |
HISTORICAL INTRODUCTION
X-rays were discovered during the summer of 1895 by Wilhelm Röntgen at the University of Würtzburg (Germany). Röntgen was interested in the cathode rays (beams of electrons) developed in discharge tubes, but it is not clear exactly which aspects of cathode rays he intended to study. By chance he noticed that a fluorescent screen $\left(\mathrm{ZnS}+\mathrm{Mn}^{++}\right)$ lying on a table some distance from the discharge tube emitted a flash of light each time an electrical discharge was passed through the tube. Realizing that he had come upon something completely new, he devoted his energies to investigating the properties of the unknown ray " $X$ " which produced this effect. The announcement of this discovery appeared in December 1895 as a concise ten page publication.
The announcement of the discovery of X-rays was received with great interest by the public. Röntgen himself prepared the first photographs of the bones in a living hand, and use of the radiation was quickly adopted in medicine. In the succeeding fifteen years, however, very few fundamental insights were gained into the nature of X-radiation. There was some indication that the rays were waves, but the evidence was not clear-cut and could be interpreted in several ways. Then, at the University of Munich in 1912, Max von Laue performed one of the most significant experiments of modern physics. At his suggestion, Paul Knipping (who had just completed a doctoral thesis with Röntgen) and Walter Friedrich (a newly appointed assistant to Sommerfeld) directed a beam of X-rays at a crystal of copper sulfate and attempted to record the scattered beams on a photographic plate. The first experiment was unsuccessful. The result of a second experiment was successful. They observed the presence of spots produced by diffracted X-ray beams grouped around a larger central spot where the incident X-ray beam struck the film. This experiment demonstrated conclusively that X-radiation consisted of waves and, further, that the crystals were composed of atoms arranged on a space lattice.
ORIGIN OF X-RAY SPECTRA
The interpretation of X–ray spectra according to the Bohr theory (LN-1) of electronic levels was first (and correctly) proposed by W. Kossel in 1920: the electrons in an atom are arranged in shells (K, L, M, N, corresponding to n = 1, 2, 3, 4, ..., etc.). Theory predicts that the energy differences between successive shells increase with decreasing n and that the electron transition from n = 2 to n = 1 results in the emission of very energetic (short wavelength) radiation (fig. 1), while outer shell transitions (say, from $n=5$ to $n=4$ ) yield low energy radiation (long wavelength). For hydrogen, you recall, the wave number of the emitted radiation associated with a particular electron transition is given by the Rydberg equation:
$\bar{v}=\left(\dfrac{1}{n_i^2}-\dfrac{1}{n_f^2}\right) R$
For "hydrogen-like" atoms with the atomic number $Z$ (containing one electron only) the corresponding Rydberg equation becomes:
$\bar{v}=\left(\dfrac{1}{n_i^2}-\dfrac{1}{n_f^2}\right) R Z^2$
From this relationship it is apparent that the energy difference associated with electron transitions increases strongly with the atomic number and that the wavelength of radiation emitted during such transitions moves with increasing $Z$ from the $10^{-7} \mathrm{~m}$ range to the $10^{-10} \mathrm{~m}$ range (radiation now defined as $\mathrm{X}$-rays).
To bring about such inner shell transitions requires the generation of an electron vacancy: an electron must be removed, for example from the $\mathrm{K}$ shell $(\mathrm{n}=1)$, of an atom. Such a vacancy is conveniently produced in an X-ray tube by an electron beam (generated by a heated filament which is made a cathode) impinging, after being subjected to an accelerating potential of several $\mathrm{kV}$, into a target material made anode (fig. 2). The impinging electrons will transfer part of their energy to electrons of the target material and result in electronic excitation. If the energy of the arriving electrons is high enough, some may knock out a $\mathrm{K}$ shell electron in the target and thus generate a vacancy. [It should be clear that a $\mathrm{K} \rightarrow \mathrm{L}$ excitation cannot take place since the $\mathrm{L}$ shell is filled: excitation must involve $(n=1) \rightarrow(n=\infty)$.] When such a vacancy is generated, it can readily be filled by an electron from the $L$ shell or the $M$ shell of the same atom. These internal electron transitions give rise to the emission of "characteristic" X-radiation which, because of its short wavelength, has extremely high "penetrating" power.
Since an electron beam is used to generate $X$-rays, the $X$-ray tube has to be evacuated: to dissipate the energy flux arriving at the target, the anode support (onto which the target is mounted) is water-cooled.
Under standard operating conditions, the characteristic radiation emitted by the target comprises two sharp lines, referred to as $\mathrm{K}_\alpha$ and $\mathrm{K}_\beta$ lines (fig. 3). They are associated, respectively, with electron transitions from $n=2$ to $n=1$ and from $n=3$ to $n=1$.
Emitted X-ray spectra were extensively studied by H.G.J. Moseley who established the relationship between the wavelength of characteristic radiation and the atomic number $\mathrm{Z}$ of the radiation emitting target material (fig. 4). Experimentally he found that the $\mathrm{K}_\alpha$ lines for various target materials (elements) exhibit the relationship:
$\lambda K_\alpha \propto \dfrac{1}{Z^2} \quad\left(\bar{v} K_\alpha \propto Z^2\right)$
Moseley’s empirical relationship (which reflects a behavior in agreement with the Rydberg equation) can be quantified. While the energy levels associated with outer electron transitions are significantly affected by the “screening” effect of inner electrons (which is variable and cannot as yet be determined from first principles), the conditions associated with X-ray generation are simple. Very generally, the screening effect of the innermost electrons on the nuclear charge is accounted for in an effective nuclear charge $(Z-\sigma)$ and the Rydberg equation assumes the form:
$\bar{v}=\left(\dfrac{1}{n_i^2}-\dfrac{1}{n_f^2}\right) R(Z-\sigma)^2$
where $\sigma=$ screening effect
Considering the transition $n_2 \rightarrow n_1$, screening of the full nuclear charge is only provided by the one electron remaining in the K shell. Thus it is possible to use the ν modified Rydberg equation, taking $\sigma = 1$. Accordingly, we have:
$\bar{v} K_\alpha=\left(\dfrac{1}{2^2}-\dfrac{1}{1^2}\right) R(Z-1)^2=-\dfrac{3}{4} R(Z-1)^2$
where: $\mathrm{R}=$ Rydberg constant and $\mathrm{Z}=$ atomic number of the target material. [The minus sign (-) only reflects radiative energy given off by the system.]
Similarly, for the characteristic $L_\alpha$ series of spectral lines ( $n=3$ to $n=2$ ) we find, after removal of one $\mathrm{L}$ electron, that the screening of the electrons in the $\mathrm{K}$ shell and the remaining electrons in the $L$ shell reduces the nuclear charge by 7.4 (empirical value).
$\overline{\mathrm{v}} \mathrm{L}_\alpha=\left(\dfrac{1}{3^2}-\dfrac{1}{2^2}\right) \mathrm{R}(\mathrm{Z}-7.4)^2=-\dfrac{5}{36} \mathrm{R}(\mathrm{Z}-7.4)^2$
A second look at the X-ray spectrum of a Mo target, obtained with an electron accelerating potential of $35 \mathrm{kV}$ (fig. 5), shows that the characteristic radiation $\left(\mathrm{K}_\alpha, \mathrm{K}_\beta\right)$ appears superimposed on a continuous spectrum (continuously varying $\lambda$ ) of lower and varying intensity. This continuous spectrum is referred to as bremsstrahlung (braking radiation) and has the following origin. Electrons, impinging on the target material, may lose their energy by transferring it to orbiting electrons, as discussed above; in many instances, however, the electrons may come into the proximity of the force fields of target nuclei and, in doing so, will be "slowed down" or decelerated to a varying degree,
ranging from imperceptible deceleration to total arrest. The energy lost in this slowing down process is emitted in the form of radiation (braking radiation, or bremsstrahlung). This energy conversion, as indicated, can range from partial to complete (fig. 6). The
incident electrons have an energy of $\mathrm{e} \cdot \mathrm{V}$ (electronic charge times accelerating Voltage) in the form of kinetic energy $\left(\mathrm{mv}^2 / 2\right)$, and their total energy conversion gives rise to a Shortest Wavelength $(S W L)$ - the cut-off of the continuous spectrum for decreasing values of $\lambda$ (fig. 5). Analytically, we have:
$\mathrm{eV}=\mathrm{h} v_{\max }=\mathrm{h} \dfrac{\mathrm{c}}{\lambda_{\mathrm{SWL}}} \quad ; \quad \lambda_{\mathrm{SWL}}=\frac{\mathrm{hc}}{\mathrm{eV}}$
From this relationship it is evident that the cut-off of the continuous spectrum toward decreasing $\lambda^{\prime}$'s $\left(\lambda_{\mathrm{SWL}}\right)$ is controlled by the accelerating potential (fig. 5).
“FINE STRUCTURE” OF CHARACTERISTIC X-RAYS
It is customary to consider the characteristic $\mathrm{X}$-ray spectral lines as discrete lines $\left(\mathrm{K}_\alpha\right.$, $K_\beta, L_\alpha, L_\beta$, etc.). In reality, they are not discrete since the electron shells involved in the associated electron transitions have energy sublevels (s, p, d orbitals). These sublevels give rise to a "fine structure" insofar as the $\mathrm{K}_\alpha$ lines are doublets composed of $\mathrm{K}_{\alpha 1}$ and $\mathrm{K}_{\alpha 2}$ lines. Similarly, $L_\alpha, L_\beta$, etc., exhibit a fine structure.
These considerations suggest that X-ray spectra contain information concerning the energetics of electronic states. Obviously, analysis of X-rays emitted from a target of unknown composition can be used for a quantitative chemical analysis. [This approach is taken routinely in advanced scanning electron microscopy (SEM) where X-rays, generated by the focused electron-beam, are analyzed in an appropriate spectrometer.]
In fundamental studies it is also of interest to analyze soft (long $\lambda$ ) $X$-ray spectra. For example, take the generation of $\mathrm{X}$-rays in sodium ( $\mathrm{Na})$. By generating an electron vacancy in the $\mathrm{K}$ shell, a series of $\mathrm{K}_\alpha$ and $\mathrm{K}_\beta$ lines will result. The cascading electron generates vacancies in the $2 p$ level, which in turn can be filled by electrons entering from the 3s level (generation of "soft" X-rays). If the X-rays are generated in a Na vapor, the $3 s \rightarrow 2 p$ transition will yield a sharp line; on the other hand, if $X$-rays are analyzed in sodium metal, the same transition results in the emission of a continuous broad band, about $30 \AA$ in width. This finding confirms the existence of an energy band (discussed earlier).
An analysis of the width and intensity distribution of the X-ray band provides experimental data concerning the energy band width and the energy state density distribution within the energy band (fig. 7).
X-RAYS FOR STRUCTURAL ANALYSIS
The extensive use of $X$-rays for the analysis of atomic structural arrangements is based on the fact that waves undergo a phenomenon called diffraction when interacting with systems (diffracting centers) which are spaced at distances of the same order of magnitude as the wavelength of the particular radiation considered. X-ray diffraction in crystalline solids takes place because the atomic spacings are in the $10^{-10} \mathrm{~m}$ range, as are the wavelengths of $\mathrm{X}$-rays.
DIFFRACTION AND BRAGG’S LAW
The atomic structure of crystalline solids is commonly determined using one of several different X-ray diffraction techniques. Complementary structure information can also be obtained through electron and neutron diffraction. In all instances, the radiation used must have wavelengths in the range of $0.1$ to $10 \AA$ because the resolution (or smallest object separation distance) to which any radiation can yield useful information is about equal to the wavelength of the radiation, and the average distance between adjacent atoms in solids is about $10^{-10} \mathrm{~m}(1 \AA)$. Since there is no convenient way to focus $\mathrm{X}$-rays with lenses and to magnify images, we do not attempt to look directly at atoms. Rather, we consider the interference effects of $X$-rays when scattered by the atoms, comprising a crystal lattice. This is analogous to studying the structure of an optical diffraction grating by examining the interference pattern produced when we shine visible light on the grating. (The spacing of lines on a grating is about 0.5 to 1 µm and the wavelength of visible radiation ranges from 0.4 to 0.8 µm.) In the optical grating the ruled lines act as scattering centers, whereas in a crystal it is the atoms (more correctly, the electrons about the atom) which scatter the incident radiation.
The geometrical conditions which must be satisfied for diffraction to occur in a crystal were first established by Bragg. He considered a monochromatic (single wavelength) beam of X-rays with coherent radiation (X-rays of common wave front) to be incident on a crystal, as shown in fig. 8. Moreover, he established that the atoms which
constitute the actual scattering centers can be represented by sets of parallel planes (in which the atoms are located) which act as mirrors and "reflect" the X-rays. In cubic systems the spacing of these planes, $\mathrm{d}_{(\mathrm{hkl})}$ (see $\left.\mathrm{LN}-4\right)$, is related to the lattice constant
(a):
$\mathrm{d}_{(\mathrm{hkl})}=\dfrac{\mathrm{a}}{\sqrt{\mathrm{h}^2+\mathrm{k}^2+\mathrm{I}^2}} \tag{1}$
For constructive interference of the scattered X-rays (the appearance of a diffraction peak) it is required that the beams, scattered on successive planes, be "in phase" (have again a common wave front) after they leave the surface of the crystal. In terms of the beams labeled 1 and 2 in fig. 8 this requires that the distance $\overline{\mathrm{AB}}+\overline{\mathrm{BC}}$ be equal to an integral number of wavelengths $(\lambda)$ of the indicent radiation. Accordingly:
$\overline{\mathrm{AB}}+\overline{\mathrm{BC}}=\mathrm{n} \lambda \quad(\mathrm{n}=1,2,3, \ldots)$
Since $\overline{\mathrm{AB}}=\overline{\mathrm{BC}}$ and $\sin \theta=\frac{\overline{A B}}{d_{(h k)}}\left[\overline{A B}=d_{(h k)} \sin \theta\right]:$
$\mathrm{n} \lambda=2 \mathrm{~d}_{(\mathrm{hkl})} \sin \theta \tag{2}$
This relation is referred to as Bragg's Law and describes the angular position of the diffracted beam in terms of $\lambda$ and $d_{(h k l)}$. In most instances of interest we deal with first order diffraction ($n=1)$ and, accordingly, Bragg's law is:
$\lambda=2 d_{(h k l)} \sin \theta$
[We are able to make $n=1$ because we can always interpret a diffraction peak for $n=2,3, \ldots$ as diffraction from (nh $n k n l)$ planes - i.e., from planes with one-nth the interplanar spacing of $\left.d_{(\mathrm{hkl})} \cdot\right]$.
If we consider fig. 8 as representative for a "diffractometer" set-up (fig. 11), we have a collimated beam of X-rays impinging on a (100) set of planes and at $2 \theta$ to the incident beam a detector which registers the intensity of radiation. For a glancing incident beam (small $\theta)$ the detector will register only background radiation. As $\theta$ increases to a value for which $2 d \sin \theta=\lambda$, the detector will register high intensity radiation - we have a diffraction peak. From the above it is evident that the diffraction angle $(\theta)$ increases as the interplanar spacing, $\mathrm{d}_{(\mathrm{hkl})}$, decreases.
The diffraction experiment as presently considered is intended to provide quantitative information on the volume (the lattice constant a) and shape characteristics (SC, BCC, FCC) of the unit cell. The intensity of diffraction peaks depends on the phase relationships between the radiation scattered by all the atoms in the unit cell. As a result, it happens quite often that the intensity of a particular peak, whose presence is predicted by Bragg's law, is zero. (This is because Bragg's law deals not with atom positions, but only with the size and shape of the unit cell.) For example, consider the intensity of the (100) diffraction peak of a crystal which has a BCC unit cell. The phase relationships show that the X-rays scattered at the top and bottom faces of the unit cell, (100) planes, interfere constructively, but are $180^{\circ}$ out of phase with the X-rays scattered by the atom at the center of the unit cell. The resultant intensity is therefore zero. The rules which govern the presence of particular diffraction peaks in the different cubic Bravais lattices (SC, BCC and FCC) are given in Table I.
TABLE I. Selection Rules for Diffraction Peaks in Cubic Systems
$\begin{array}{lcc} \underline{\text { Bravais Lattice } }& \underline{\text { Reflections Present }} & \underline{\text { Reflections Absent }} \ \text { Simple Cubic } & \text { All } & \text { None } \ \ \text { Body-Centered Cubic } & (\mathrm{h}+\mathrm{k}+\mathrm{l})=\text { even } & (\mathrm{h}+\mathrm{k}+\mathrm{l})=\text { odd } \ \ \text { Face-Centered Cubic } & \begin{array}{l} \mathrm{h}, \mathrm{k}, \mathrm{l} \text { unmixed } \ \text { (either all odd } \ \text { or all even) } \end{array} & \mathrm{h}, \mathrm{k}, \mathrm{l} \text { mixed } \ & \begin{array}{c} \end{array} \end{array}$
The rules given are strictly true only for unit cells where a single atom is associated with each lattice point. (Unit cells with more than one atom per lattice point may have their atoms arranged in positions such that reflections cancel. For example, diamond has an FCC Bravais lattice with two atoms per lattice point. All reflections present in diamond have unmixed indices, but reflections such as {200}, {222} and {420} are missing. The fact that all reflections present have unmixed indices indicates that the Bravais lattice is FCC – the extra missing reflections give additional information as to the exact atom arrangement.)
A hypothetical diffraction experiment: A material is known to be of simple cubic structure; determine a, the lattice constant, by X-ray diffraction. In theory, the question may be answered by placing the crystal into a diffractometer, rotating it into all possible positions relative to the incident $X$-ray beam and recording all diffracting $2 \theta$ values. From the above we know that the smallest observed $\theta$ value must correspond to diffraction on $\{100\}$ planes and also that $d_{(100)}=a$. We may now use Bragg's equation to determine a, the lattice constant:
\begin{aligned}
&\lambda=2 d \sin \theta=2 a \sin \theta \
&a=\frac{\lambda}{2 \sin \theta}
\end{aligned}
There are two simplifying assumptions in this problem: (1) we know the system is SC and (2) we are able, through rotation, to bring all planes present into diffraction conditions.
EXPERIMENTAL APPROACHES TO X-RAY DIFFRACTION
In the context of this course we are interested in making use of X-ray diffraction for the purpose of (a) identifying (cubic) crystal systems, (b) determining the lattice constant, a, and (c) identifying particular planes or meaningful orientations. The possible approaches can, in principle, be identified through an examination of Bragg's law. The Bragg condition for particular $\mathrm{d}_{(\mathrm{hkl})}$ values can be satisfied by adjusting either one of two experimental variables: (a) $\lambda$, the wavelength of the X-ray beam used, or (b) $\theta$, the orientation of the crystal planes relative to the incident X-rays.
(a) Fixed $\theta$. Variable $\lambda$: One means of satisfying Bragg's law is to irradiate a stationary single crystal ( $\theta$ fixed for all planes within the crystal) with an X-ray beam of "white" radiation, which contains the characteristic and continuous spectrum produced by an X-ray tube. (For $\lambda$ variable we have the simultaneous exposure of a crystal to a range of $\lambda$ values). Each set of planes will reflect (diffract) the particular $\lambda$ which satisfies the Bragg condition for the fixed $\theta$. The diffracted beams may conveniently be recorded with a Polaroid camera or, alternately, with an electronic imaging device. It is possible to analyze either the transmitted or the back-reflected X-rays. This experimental procedure is referred to as the Laue technique (fig. 9); it is mostly conducted in the back-reflection mode. Note that the approach taken makes it possible to determine the values of $\theta$ for each reflection, but not the corresponding $\lambda$. Therefore, the technique cannot be used, for example, to determine lattice constants. However, it is very valuable if particular planes or crystal orientations are to be identified.
(b)Fixed $\lambda$ (Monochromatic X-Rays), Variable $\theta$: The basic prerequisite for this approach is the availability of a monochromatic $X$-radiation of known wavelength $(\lambda)$. Such radiation can be conveniently obtained by using a crystal (i.e., its diffracting property) as a filter or monochromator (fig. 10). Filter action is achieved by positioning the crystal in such a way that the unfiltered radiation emitted by the X-ray tube becomes incident at an angle, $\theta$, on a set of low index planes which satisfy Bragg's law for the highest intensity radiation $\left(K_\alpha\right)$ emitted. The condition of a fixed $\lambda$ and variable $\theta$ is experimentally used in two techniques. Using a diffractometer (fig. 11), we place a sample (ground to a powder) into the center of a rotating stage and expose it to a monochromatic X-ray beam. The sample is rotated into diffraction condition and the diffraction angle determined.
In the Debye-Scherrer method (fig.12) the sample is ground to a powder and placed (in an ampoule) into the center of a Debye-Scherrer camera. Exposed to monochromatic X-rays, in this way a large number of diffracted cone-shaped beams are generated such that the semiangles of the cones measure $2\theta$, or twice the Bragg angle for the particular diffracting crystallographic planes. The reason diffracted beams are cone-shaped is that the planes in question (within the multitude of randomly oriented grains) give rise to diffraction for any orientation around the incident beam as long as the incident beam forms the appropriate Bragg angle with these planes – thus there is a rotational symmetry of the diffracted beams about the direction of the incident beam. Those planes with the largest interplanar spacing have the smallest Bragg angle, $\theta$.
In a Debye-Scherrer arrangement, after exposing a powder of a crystalline material to monochromatic X-rays, the developed film strip will exhibit diffraction patterns such as indicated in fig. 12. Each diffraction peak (dark line) on the film strip corresponds to constructive interference at planes of a particular interplanar spacing $\left[\mathrm{d}_{(\mathrm{hkl})}\right]$. The problem now consists of "indexing" the individual lines - i.e., determining the Miller indices (hkl) for the diffraction lines:
\begin{aligned}
\text { Bragg: } \quad \lambda &=2 \mathrm{~d}_{(\mathrm{hkl})} \sin \theta & ; & \mathrm{d}_{(\mathrm{hkl})} &=\frac{\mathrm{a}}{\sqrt{\mathrm{h}^2+\mathrm{k}^2+\mathrm{l}^2}} \
\lambda^2 &=4 \mathrm{~d}_{(\mathrm{hkl})}^2 \sin ^2 \theta & ; & \mathrm{d}_{(\mathrm{hkl})}^2 &=\frac{\mathrm{a}^2}{\left(\mathrm{~h}^2+\mathrm{k}^2+\mathrm{l}^2\right)}
\end{aligned}
Substitution and rearrangement of above yields:
$\dfrac{\sin ^2 \theta}{\left(h^2+k^2+\left.\right|^2\right)}=\dfrac{\lambda^2}{4 a^2}=\text { const. }$
Accordingly, we find that for all lines ($\theta$ values) of a given pattern, the relationship
$\dfrac{\sin ^2 \theta_1}{\left(h^2+k^2+\left.\right|^2\right)_1}=\dfrac{\sin ^2 \theta_2}{\left(h^2+k^2+\mathrm{I}^2\right)_2}=\dfrac{\sin ^2 \theta_3}{\left(h^2+k^2+\mathrm{I}^2\right)_3}=\text { const. }$
holds. Since the sum $\left(h^2+k^2+l^2\right)$ is always integral and $\lambda^2 / 4 a^2$ is a constant, the problem of indexing the pattern of a cubic system is one of finding a set of integers $\left(h^2+k^2+l^2\right)$ which will yield a constant quotient when divided one by one into the observed $\sin ^2 \theta$ values. (Certain integers such as $7,15,23$, etc. are impossible because they cannot be formed by the sum of three squared integers.)
Indexing in step-by-step sequence is thus performed as follows: $\theta$ values of the lines are obtained from the geometric relationship of the unrolled film strip. Between the exit hole of the $X$-ray beam $\left(2 \theta=0^{\circ}\right)$ and the entrance hole $\left(2 \theta=180^{\circ}\right)$ the angular relationship is linear (fig. 12). The increasing $\theta$ values for successive lines are indexed $\theta_1, \theta_2, \theta_3$, etc., and $\sin ^2 \theta$ is determined for each. If the system is simple cubic we know that all planes present will lead to diffraction and the successive lines (increasing $\theta$ ) result from diffraction on planes with decreasing interplanar spacing: (100), (110), (111), (200), (210), (211), (220), etc. From equation (3) above we recognize:
$\dfrac{\sin ^2 \theta_1}{1}=\dfrac{\sin ^2 \theta_2}{2}=\dfrac{\sin ^2 \theta_3}{3}=\dfrac{\sin ^2 \theta_4}{4}=\dfrac{\sin ^2 \theta_5}{5}=\text { const. }$
If the system is $\mathrm{BCC}$, however, we know from the selection rules that only planes for which $(h+k+l)=$ even will reflect. Thus:
$\dfrac{\sin ^2 \theta_1}{2}=\dfrac{\sin ^2 \theta_2}{4}=\dfrac{\sin ^2 \theta_3}{6}=\dfrac{\sin ^2 \theta_4}{8} \text { etc. }=\text { const. }$
[SC can be differentiated from BCC through the fact that no sum of three squared integers can yield 7, but 14 can be obtained from planes (321)].
For FCC systems, the selection rules indicate reflections on planes with unmixed h,k,l indices:
$\dfrac{\sin ^2 \theta_1}{3}=\dfrac{\sin ^2 \theta_2}{4}=\dfrac{\sin ^2 \theta_3}{8}==\text { const. }$
After proper indexing, the constant is obtained:
$\dfrac{\sin ^2 \theta}{\left(h^2+k^2+L^2\right)}=\text { const. }$
and the particular Bravais lattice is identified. The lattice constant of the unit cell is subsequently obtained, knowing the wavelength of the incident radiation:
\begin{aligned}
&\dfrac{\sin ^2 \theta}{\left(\mathrm{h}^2+\mathrm{k}^2+\mathrm{I}^2\right)}=\mathrm{const} .=\dfrac{\lambda^2}{4 \mathrm{a}^2} \
&a^2=\dfrac{\lambda^2}{4 \sin ^2 \theta}\left(h^2+k^2+1^2\right) \
&a=\dfrac{\lambda}{2 \sin \theta} \sqrt{\left(h^2+k^2+1^2\right)}
\end{aligned} | textbooks/chem/Inorganic_Chemistry/Introduction_to_Solid_State_Chemistry/01%3A_Lectures/1.05%3A_X-rays_and_X-ray_Diffraction.txt |
INTRODUCTION
Real crystals are never perfect: they always contain a considerable density of defects and imperfections that affect their physical, chemical, mechanical and electronic properties. The existence of defects also plays an important role in various technological processes and phenomena such as annealing, precipitation, diffusion, sintering, oxidation and others. It should be noted that defects do not necessarily have adverse effects on the properties of materials. There are many situations in which a judicious control of the types and amounts of imperfections can bring about specific characteristics desired in a system. This can be achieved by proper processing techniques. In fact, “defect engineering” is emerging as an important activity.
All defects and imperfections can be conveniently considered under four main divisions: point defects, line defects or dislocations, planar defects or interfacial or grain boundary defects, and volume defects. We can also add here macroscopic or bulk defects such as pores, cracks and foreign inclusions that are introduced during production and processing of the solid state. Point defects are inherent to the equilibrium state and thus determined by temperature, pressure and composition of a given system. The presence and concentration of other defects, however, depend on the way the solid was originally formed and subsequently processed.
Briefly consider the effects of imperfections or crystal defects on a few important properties of solids. The electrical behavior of semiconductors, for example, is largely controlled by crystal imperfections. The conductivity of silicon can thus be altered in type (n or p) and by over eight orders of magnitude through the addition of minute amounts of electrically active dopant elements. In this case, each atom of dopant, substitutionally incorporated, represents a point defect in the silicon lattice. The fact that such small amounts of impurity atoms can significantly alter the electrical properties of semiconductors is responsible for the development of the transistor and has opened up the entire field of solid state device technology. Practically none of the semiconducting properties that led to these engineering accomplishments are found in a “perfect” crystal. They are properties peculiar to the defective solid state.
The existence of dislocations (line defects) in crystals provides a mechanism by which permanent change of shape or mechanical deformation can occur. A crystalline solid free of dislocations is brittle and practically useless as an engineering material. While the existence of dislocations in crystals insures ductility (ability to deform), the theoretical strength of crystalline solids is drastically reduced by their presence.
We should recognize that dislocations play a central role in the determination of such important properties as strength and ductility. In fact, virtually all mechanical properties of crystalline solids are to a significant extent controlled by the behavior of line imperfections.
The ability of a ferromagnetic material (such as iron, nickel or iron oxide) to be magnetized and demagnetized depends in large part on the presence of two-dimensional imperfections known as Bloch walls. These interfaces are boundaries between two regions of the crystal which have a different magnetic state. As magnetization occurs, these defects migrate and by their motion provide the material with a net magnetic moment. Without the existence of Bloch walls all ferromagnetic materials would be permanent magnets. In fact, electromagnets would not exist if it were not for this type of defect.
The presence of surface defects such as cracks causes brittle materials like glass to break at small applied stresses. This fact is familiar to anyone who has broken a glass tube by first filing a small notch (or crack) into the surface. Removal of cracks from the surface of glass either by etching in hydrofluoric acid or by flame polishing almost always raises the fracture strength. For example, glass in the absence of any surface cracks has a fracture strength of $\sim 10^{10} \mathrm{Newton} / \mathrm{m}^2$ (as opposed to real glass which has a fracture strength of $\sim 0^7$ Newton $/ \mathrm{m}^2$ ).
POINT DEFECTS
Formation of Point Defects
An incontrovertible law of nature states: “Nothing is perfect”. This law applies to humans as well as to the inorganic world of crystalline solids and can be formulated as the 2nd law of thermodynamics:
$\mathrm{F}=\mathrm{H}-\mathrm{TS} \tag{1}$
where $\mathrm{F}$ is the free energy of a given system, $\mathrm{H}$ is the heat content or enthalpy and TS is the entropy, or disorder, term. If a reaction takes place at a temperature $T$, we find the change in $\mathrm{F}(\Delta \mathrm{F})$ related to a change in $\mathrm{H}(\Delta \mathrm{H})$, the heat content, and possibly also a change in TS $(T \Delta S)$. Such is the case when defects are formed in a perfect solid: The energy distribution in a solid (Maxwell-Boltzmann) suggests that a number of individual atoms may acquire enough thermal energy to be displaced from the equilibrium lattice site into an interstitial position. This process of point defect formation requires energy and leads to lattice strain which constitutes, as discussed earlier, an increase in the heat content of the system ($\Delta H$ is positive and increases linearly with the number of defects formed). The departure from perfection by the generation of defects leads to disorder ($\Delta S$ is positive). The magnitude of disorder generated ($\Delta S$) is very large during the initial step from perfection to slight disarray, but the increase in disorder (with a given number of defects generated) decreases as the overall disorder increases.
Correspondingly the term $T \Delta S$ drops rapidly at the beginning and then flattens out. The net result (fig. 1), free energy, exhibits a minimum for a certain number of defects in the
solid [equilibrium defect density = f (temperature)]; the $F_{\text{minimum}}$ suggests also that the transition from perfection to equilibrium defect structure is spontaneous: it occurs naturally!
While the detailed mechanisms for the formation of atomic vacancies in solids are still the subject of extensive research, the associated equilibrium energetics are clear: calculations of the thermal energy of atoms in a lattice show that the average vibrational energy of lattice atoms is much less than $1 \mathrm{eV}$ (the approximate energy change associated with vacancy formation, i.e., the least amount of energy required to form a vacancy) at room temperature. Therefore a lattice atom will only acquire the energy $\Delta \mathrm{H}_{\mathrm{d}}$, the energy required to form the defect, upon the occurrence of a large energy fluctuation. Since the relative probability of an atom having an energy $\Delta \mathrm{H}_d$ or more in excess of the ground state energy is $e^{-\Delta H_d / k T}$, the probability that an atomic site is vacant varies in the same way. In a (molar) crystal containing $\mathrm{N}$ atomic sites, the number $\mathrm{n}_{\mathrm{d}}$ of vacant sites is, therefore,
$\mathrm{n}_{\mathrm{d}}=\mathrm{ANe}^{-\Delta \mathrm{H}_{\mathrm{d}} / \mathrm{kT}} \tag{2}$
where
• $\mathrm{n}_{\mathrm{d}}$ is the number of defects (in equilibrium at $\mathrm{T}$ )
• $\mathrm{N}$ is the total number of atomic sites per mole
• $\Delta \mathrm{H}_{\mathrm{d}}$ is the energy necessary to form the defect
• $\mathrm{T}$ is the absolute temperature $(\mathrm{K})$
• $\mathrm{k}$ is the Boltzmann constant
• $\mathrm{A}$ is a proportionality constant
Point Defects in “Pure” Metallic Systems
Point defects in “pure” crystalline metals are defects of atomic dimensions, such as impurity atoms, the absence of a matrix atom and/or the presence of a matrix atom in the wrong place. Some of these point defects are shown in fig. 2. An impurity atom that occupies a normal lattice site is called a substitutional impurity atom and an impurity atom found in the interstice between matrix atoms is called an interstitial impurity atom. Whether a foreign atom will occupy a substitutional
or interstitial site depends largely on the size of the atom relative to the size of the site. Small atoms are usually interstitial impurities, while larger atoms are usually substitutional impurities.
A vacancy is an atom site, normally occupied in the perfect crystal, from which an atom is missing. Often the term “vacancy” is used to denote a so-called Schottky defect, which is formed when an atom or an ion leaves a normal lattice site and repositions itself in a lattice site on the surface of the crystal. This may be the result of atomic rearrangement in an existing crystal at a high temperature when atomic mobility is high because of increased thermal vibrations. A vacancy may also originate in the process of crystallization as a result of local disturbances during the growth of new atomic planes on the crystal surface. Vacancies are point defects of a size nearly equal to the size of the original (occupied) site; the energy of the formation of a vacancy is relatively low - usually less than 1 eV.
The number of vacancies at equilibrium at each temperature in a crystal can be determined from eq. (2), in which $\Delta \mathrm{H}_{\mathrm{d}}$ is the energy necessary to take an atom from a regular site of the crystal and place it on the surface for a Schottky-type defect. When a solid is heated a new higher equilibrium concentration of vacancies is established, usually first at crystal surfaces and then in the vicinity of dislocations and grain boundaries which provide sites for the atoms which have left their normal lattice site. Vacancies gradually spread throughout the crystal (from the surfaces into the bulk). On cooling the vacancy concentration is lowered by "diffusion of vacancies" to grain boundaries or dislocations, which act as sinks. In both cases, the new equilibrium vacancy concentration is established only after a finite amount of time. The rate at which vacancies move from point to point in the lattice decreases exponentially with decreasing temperature. Thus, on very rapid cooling (quenching) from a high temperature near the melting point most of the vacancies do not have time to diffuse to sinks and are said to be "frozen in". This gives a considerably greater ("non-equilibrium") concentration of vacancies in quenched specimens than that indicated by the thermal equilibrium value.
The concentration of vacant lattice sites in pure materials is very small at low temperatures - about one vacancy every $10^8$ atom sites - and increases with increasing temperature to about one vacancy every $10^3$ sites at the melting temperature. Vacancies are important because they control the rate of matrix (or substitutional) atom diffusion - i.e., atoms are able to move around in a crystalline solid primarily because of the presence of vacancies. (The mechanism by which they move is the same as that associated with moving a car in a filled parking lot to the exit). This is shown schematically in fig. 3. Self-interstitials are generally not encountered in close-packed
metallic systems, but may be introduced by irradiation. For example, high-energy neutrons from atomic fission can knock metal atoms from their regular sites into interstitial sites, creating vacancy-interstitial pairs.
Point Defects in Ionic Solids
Point defects in ionic structures differ from those found in pure elements because of the charge neutrality requirement. For example, in a pure monovalent ionic material a cation vacancy must have associated with it either a cation interstitial or an anion vacancy to maintain charge neutrality. Similar requirements hold for anion vacancies. A vacancy pair defect (migration of a cation and an anion to the surface) is usually called a Schottky imperfection, and a vacancy-interstitial pair defect is referred to as a Frenkel imperfection (an anion or cation has left its lattice position, which becomes a vacancy, and has moved to an interstitial position). These two types of imperfections are shown in fig. 4. Self-interstitials are much more common in ionic structures than in pure elements because many ionic compounds have relatively large interstitial sites available. That is, there are often interstitial sites in the unit cell that have nearly the
same surroundings as normal atom sites. (For example, in BeO the Be atoms fill only one-half the available tetrahedral sites, leaving four possible cation interstitial sites per unit cell. Thus a Be atom could go from a regular lattice site to an almost equivalent interstitial site with little distortion of the lattice.)
Foreign atoms in ionic crystals produce defects that also must maintain charge neutrality. For example, in NaCl a monovalent cation, such as lithium, may simply replace one of the sodium ions as a substitutional impurity. But a divalent cation, such as calcium, replacing a sodium ion must be accompanied by either a cation vacancy or an anion interstitial if charge neutrality is to be maintained. Correspondingly, monovalent impurity cations in a divalent structure (e.g., Na in MgO) must be accompanied by an appropriate number of cation interstitials or anion vacancies.
Point Defects in Covalently Bonded Solids
Substitutional impurities in covalently bonded materials can create a unique imperfection in the electronic structure if the impurity atom is from a group in the periodic table other than the matrix atoms. For example, you already considered Group V and Group III elements in a Group IV matrix, such as As or B in Si.
When foreign atoms are incorporated into a crystal structure, whether in substitutional or interstitial sites, we say that the resulting phase is a solid solution of the matrix material (solvent) and the foreign atoms (solute). The term “solid solution”, however, is not restricted to the low solute contents of doped semiconductor systems; there are many solid solutions, such as metallic alloys, that comprise a wide composition range.
LINE DEFECTS
Line imperfections, or dislocations, in crystalline solids are defects that produce lattice distortions centered about a line. A dislocation is simply the edge of an extra inserted fractional plane of atoms (fig. 5). Normally the symbol $\perp$ is used to represent a positive
dislocation (extra fractional plane) and $\top$ is used to represent a negative dislocation (missing fractional plane).
The importance of dislocations is readily demonstrated in the deformation of crystalline materials. The plane in which a dislocation moves through the lattice is called a slip plane. With an applied shear stress the dislocation moves, atomic row by atomic row, and one part of the crystal is displaced relative to the other. When the dislocation has passed through the crystal, the portion of the crystal above the slip plane has shifted one atomic distance relative to the portion below the slip plane. In other words, the motion of the dislocation has caused the crystal to change its shape - to be permanently deformed (fig. 6).
Please note: on either side of the dislocation the crystal lattice is essentially perfect, but in the immediate vicinity of the dislocation the lattice is severely distorted. For a positive edge dislocation, the presence of the extra half plane causes the atoms above the slip plane to be put in compression, while those below the slip plane are put in tension. Consequently, the edge dislocation will have a stress field around it that is compressive above the slip plane and tensile below the slip plane.
Plastic Deformation By Slip:
When single crystals of metal (or semiconductor) are pulled in tension, they will begin to deform (elongate) plastically at relatively low stress levels, and “blocks” of the crystals slide over one another because of dislocation motion. Simultaneously, so-called slip lines appear on their surface. It is found that deformation by slip occurs most easily on planes with high atomic density and with large interplanar spacing, while the direction of slip is in all instances an atomically “close-packed direction”. For FCC structures we therefore observe as the primary slip system {111} planes in direction, while in BCC structures the primary slip occurs on {110} planes in directions. (It should be noted that an alternate deformation mechanism is “deformation twinning”, presently not to be considered.)
Dislocation Climb:
Climb is the name given to the motion of dislocations when the extra “half” plane is extended farther into a crystal or partially withdrawn from it. Clearly, the climb process is not a motion of the plane, but rather its growth or shrinking as a result of the addition of atoms or “vacancies” respectively from the environment of the dislocation (fig. 7).
Multiplication of Dislocations:
Since during slip each dislocation leaves the matrix, macroscopic deformation could not take place given normal dislocation densities in the range of $10^6-10^8 / \mathrm{cm}^3$. Examination of the deformed crystals indicates that multiplication of dislocations takes place during deformation. While there are a multitude of multiplication mechanisms, the one most extensively studied is the Frank-Read Source (not to be discussed in detail).
Dislocation Interactions:
The relative ease with which dislocations move across a solid matrix can be attributed to the severe displacements of atoms in the core of dislocations. If these local stresses are reduced, the mobility of dislocations - and thus the ease of slip - is reduced. It is found that impurities in the vicinity of dislocation cores tend to reduce the local distortion energy of the dislocations and thus stabilize the system against slip. In many systems impurities are intentionally added (e.g., solid solution hardening) to increase the strength of materials. Similarly, micro-precipitates tend to impede dislocation motion (e.g., precipitation hardening).
INTERFACIAL IMPERFECTIONS
The several different types of interfacial, or planar imperfections, in solids can be grouped into the following categories:
1. Interfaces between solids and gases, which are called free surfaces;
2. Interfaces between regions where there is a change in the electronic structure, but no change in the periodicity of atom arrangement, known as domain boundaries;
3. Interfaces between two crystals or grains of the same phase where there is an orientation difference in the atom arrangement across the interface; these interfaces are called grain boundaries;
4. Interfaces between different phases, called phase boundaries, where there is generally a change of chemical composition and atom arrangement across the interface.
Grain boundaries are peculiar to crystalline solids, while free surfaces, domain boundaries and phase boundaries are found in both crystalline and amorphous solids.
Free Surfaces
Because of their finite size, all solid materials have free surfaces. The arrangement of atoms at a free surface differs slightly from the interior structure because the surface atoms do not have neighboring atoms on one side. Usually the atoms near the surface have the same crystal structure but a slightly larger lattice parameter than the interior atoms.
Perhaps the most important aspect of free surfaces is the surface energy $(\gamma)$ associated with surfaces of any solid. The source of this surface energy may be seen by considering the surroundings of atoms on the surface and in the interior of a solid. To bring an atom from the interior to the surface, we must either break or distort some bonds - thereby increasing the energy. The surface energy is defined as the increase in energy per unit area of new surface formed. In crystalline solids, the surface energy depends on the crystallographic orientation of the surface - those surfaces that are planes of densest atomic packing are also the planes of lowest surface energy. This is because atoms on these surfaces have fewer of their bonds broken or, equivalently, have a larger number of nearest neighbors within the plane of the surface. Typical values of surface energies of solids range from about $10^{-1}$ to $1 \mathrm{~J} / \mathrm{m}^2$. Generally, the stronger the bonding in the crystal, the higher the surface energy.
Surface energies can be reduced by the adsorption of foreign atoms or molecules from the surrounding atmosphere. For example, in mica the surface energy of freshly cleaved material in a vacuum is much higher than the surface energy of the same surface cleaved in air. In this instance, oxygen is adsorbed from the air to partially satisfy the broken bonds at the surface. Impurity atom adsorption makes it almost impossible to maintain atomically clean surfaces. As a result, surface properties such as electron emission, rates of evaporation and rates of chemical reactions are extremely dependent on the presence of any adsorbed impurities. These properties will be different if the measurements are made under conditions giving different surface adsorption.
Grain Boundaries
Grain boundaries separate regions of different crystallographic orientation. The simplest form of a grain boundary is an interface composed of a parallel array of edge dislocations. This particular type of boundary is called a tilt boundary because the misorientation is in the form of a simple tilt about an axis, parallel to the dislocations. Tilt boundaries are referred to as low-angle boundaries because the angle of misorientation is generally less than $10^{\circ}$.
When a grain boundary has a misorientation greater than $10^{\circ}$ or $15^{\circ}$, it is no longer practical to think of the boundary as being made up of dislocations because the spacing of the dislocations would be so small that they would lose their individual identity. The grain boundary represents a region a few atomic diameters wide where there is a transition in atomic periodicity between adjacent crystals or grains.
Grain boundaries have an interfacial energy because of the disruption in atomic periodicity in the vicinity of the boundary and the broken bonds that exist across the interface. The interfacial energy of grain boundaries is generally less than that of a free surface because the atoms in a grain boundary are surrounded on all sides by other atoms and have only a few broken or distorted bonds.
Solids with grain boundaries are referred to as polycrystalline, since the structure is composed of many crystals - each with a different crystallographic orientation. In the case of iron the grain boundary structure can be revealed by preferential chemical attack (etching) at the grain boundaries, while the grain structure in polyethylene is revealed by the use of polarized light. The grain structure is usually specified by giving average grain diameter or by using a scheme developed by the American Society for Testing and Materials (ASTM). In the ASTM procedure the grain size is specified by a “grain size number” (n) where
$N=2^{n-1}$
with $\mathrm{N}$ equal to the number of grains per square inch when the sample is viewed at 100X magnification. For example, at a magnification of $X=100$, a material with grain size number 8 will show 128 grains per inch ${ }^2$ - this material in effect has (at $X=1$ ) $1.28 \times 10^6$ grains per square inch. If the grains are approximately square in cross section, this corresponds to an average grain dimension of $8.8 \times 10^{-4} \mathrm{in}^*$.
In polycrystalline samples the individual grains usually have a random crystallographic orientation with respect to one another, and the grain structure is referred to as randomly oriented. In some instances, however, the grains all have the same orientation to within a few degrees. In this instance the material is said to have a preferred orientation or texture.
Phase Boundaries
A phase is defined as a homogeneous, physically distinct and mechanically separable portion of the material with a given chemical composition and structure. Phases may be substitutional or interstitial solid solutions, ordered alloys or compounds, amorphous substances or even pure elements; a crystalline phase in the solid state may be either polycrystalline or exist as a single crystal.
Solids composed of more than one element may - and often do - consist of a number of phases. For example, a dentist's drill, something painfully familiar to all of us, consists of a mixture of small single crystals of tungsten carbide surrounded by a matrix of cobalt. Here the cobalt forms a continuous phase. Polyphase materials such as the dentist's drill are generally referred to as composite materials. Composite materials have great importance in the engineering world because they have many attractive properties that set them apart from single-phase materials. For example, the dentist’s drill has good abrasive characteristics (due to the hard carbide particles) and good toughness and impact resistance (due to the continuous cobalt matrix). Neither the tungsten carbide nor the cobalt has both abrasion resistance and impact resistance, yet the proper combination of the two phases yields a composite structure with the desired properties.
*
ASTM has as yet not issued specifications in SI units!
The nature of the interface separating various phases is very much like a grain boundary. Boundaries between two phases of different chemical composition and different crystal structure are similar to grain boundaries, while boundaries between different phases with similar crystal structures and crystallographic orientations may be analogous to low-angle grain boundaries in both energy and structure.
The concept of a solid consisting of a continuous phase and a discontinuous phase (or phases) leads to a simple classification of the various types of composite materials. Table 1 gives this classification, which is based on the structure (whether amorphous or crystalline) of the continuous and discontinuous phases.
TABLE 1
Classification of Composite or Multiphase Materials
$\begin{array}{lll} \text { Continuous Phase } & \begin{array}{l} \text { Discontinuous Phase } \ \text { (or Phases) } \end{array} & \text { Examples } \ \hline \text { Crystalline } & \text { Crystalline } & \begin{array}{l} \text { All metallic systems such as cast } \ \text { iron, steel, soft solder, etc.; most } \ \text { natural rocks such as granite and } \ \text { marble. } \end{array} \ \ \text { Crystalline } & \text { Amorphous } & \text { None of practical significance. } \ \ \text { Amorphous } & \text { Crystalline } & \begin{array}{l} \text { Most man-made ceramics such } \ \text { as building bricks and electrical } \ \text { insulator porcelain, concrete, } \ \text { partially crystalline polymers, } \ \text { some polymer-crystalline particle } \ \text { composites. } \end{array} \ \ \text { Amorphous } & \text { Amorphous } & \begin{array}{l} \text { Fiberglass, asphalt, wood, } \ \text { hydrated cement, other gels. } \end{array} \end{array}$ | textbooks/chem/Inorganic_Chemistry/Introduction_to_Solid_State_Chemistry/01%3A_Lectures/1.06%3A_The_Imperfect_Solid_State.txt |
INTRODUCTION
When a liquid phase is cooled to below its freezing temperature, it usually transforms into a crystalline solid, i.e. it crystallizes. Some liquids, because of complex molecular configuration or slow molecular transport, do not “crystallize” (assume an ordered configuration) upon being cooled to low temperatures, but instead form a rigid disordered network, known as glass, which is very similar in structure to that of the liquid. Most metals and alkali halides crystallize easily upon cooling through the freezing temperature because the structural rearrangement from the liquid to the crystalline state is simple and bonding is nondirectional. At temperatures just above the freezing temperature, most metals and molten salts have fluidities approximately equivalent to that of water at room temperature. Thus, the required structural rearrangements can take place relatively easily.
In contrast, many inorganic silicates form glasses upon cooling because the fluidity at and even above the freezing temperature is very low. This glass formation is related to the high silicon-oxygen single-bond energies and the directional bonding requirements imposed by $\mathrm{sp}^3$ hybridization of silicon. The disordered liquid cannot flow easily and thus cannot undergo the rearrangements required for crystallization. Moreover, the fluidity decreases very rapidly as the temperature is lowered. [For large organic and polymeric liquids, crystallization is difficult because of their chain lengths. In polymer systems the intermolecular bonding (Van der Waals) is weak and expected to permit individual molecules to readily slide past one another. Thus, it is the "difficult" structural rearrangement required for crystallization that induces glass formation in these systems.]
The fluidity of liquids (the reciprocal of the viscosity) is a measure of their ability to flow. In turn, the viscosity ($\eta$) (fig. 1) is a measure of their resistance to flow.
Viscosity is formulated as the ratio of an applied shear stress to the resultant shear-strain rate - its dimensions are Newton seconds per $\text{meter}^2$. Liquid flow, involving the motion of molecules past one another, requires the breaking and making of new bonds. Thus the fluidity, like chemical reactions and diffusion in solids (to be discussed later) is expected to exhibit an exponential temperature dependence and may be modelled as:
$\dfrac{1}{\eta} \propto e^{-}\left(\dfrac{\text { Bond Energy }}{k T}\right)$
It is noteworthy that inorganic glasses occur in nature as do organic glasses (such as the sap from trees) whereas metallic glasses do not occur naturally and have been manufactured only within the last two decades. The major obstacle to achieving metallic glasses was our inability to cool metallic melts to room temperature at rates high enough so that ordering (and thus crystallization) is prevented.
The glass transition point, $T_g$, (temperature at which a supercooled liquid becomes a glass) is for glasses what the melting point, $T_m$, is for crystalline solids. Characteristic for almost all glass-forming materials is that at $T_g$ a change in the temperature dependence of the density $(\Delta \mathrm{V} / \Delta \mathrm{T})$ takes place [where the viscosity is in the range $10^4$ to $10^6 \mathrm{Ns} / \mathrm{m}^2$ (the viscosity of $\mathrm{H}_2 \mathrm{O}$ is $1.5 \times 10^{-3} \mathrm{Ns} / \mathrm{m}^2$, or about nine orders of magnitude less)]. At higher temperatures (lower viscosities) the structural units are able to reorganize themselves quickly as a quasi-equilibrium liquid. At temperatures below $T_g$ reorganization among the structural units virtually ceases and the resulting rigid material is referred to as a glass. The experimental value obtained for $T_g$ depends on the time scale of the experiment. On slow cooling, for example, the structural units have more time to reorganize and, thus, $T_g$ is lower (fig. 2).
METALLIC GLASSES
The existence of metallic glasses shows the generality of the glass-forming process. In addition, the physical picture in this case is especially simple. Even with the highest presently attainable cooling rates (up to $10^{7 \circ} \mathrm{C} / \mathrm{s}$ ), no "pure" metal has yet been obtained in a glassy state. However, if the liquid is an alloy of two (or more) metals whose atoms differ greatly in size, the crystallization process is more difficult and can be prevented in many instances by extremely rapid cooling of the melt $\left(10^{6 \circ} \mathrm{C} / \mathrm{sec}\right)$. The two experimental techniques for rapid cooling are (1) propelling a liquid drop at high velocity and "splatting" it against a cooled copper surface, (2) flattening a liquid drop between a piston and an anvil or between two rolls and (3) "spinning" the melt by transferring it in a continuous mode onto a rapidly rotating, cooled wheel (fig. 3). X-ray patterns of the solidified alloys are similar to those of liquid metals - this and other evidence suggest that they are "amorphous", that their structure is that of a glass.
Another method for preparing metallic glasses is by vacuum deposition onto a substrate cooled to approximately 80K. Again, pure metals crystallize under these conditions, but suitable alloys form glasses, some of which are stable when brought to room temperature. An example is the silver-copper system in which the two solid metals have only about eight percent solubility in each other. If a melt containing 50 wt.$\%$ silver and 50 wt.$\%$ copper is slowly cooled, two crystalline phases are produced. When this same alloy is obtained by vacuum deposition, a metastable metallic glass is formed. Initially the glass has metallic electrical conductivity comparable to that of the liquid alloy. When the glass is heated for $1 / 2$ hour at $380 \mathrm{~K}$, it converts to a crystalline, metastable face-centered cubic solid solution. Only on heating at about $500 \mathrm{~K}$ does the equilibrium, two-phase structure form. Many pure nonmetallic elements, however, such as S, Se, Ge and Si, form a stable glass under suitable conditions.
THE GLASSY STATE
The properties and chemical composition of the glass used for optical purposes differ greatly from those of the glass used for baking dishes. Likewise, the glass used for window panes differs from the glass used for artwork or optical purposes. However, (most) inorganic glasses have one thing in common: they are ceramic materials (compounds that contain metallic and non-metallic elements, most often oxygen) with the amorphous structure of a frozen liquid.
Quartz glass is not crystalline at room temperature because the rearrangement of interatomic bonds between $\mathrm{Si}$ and $\mathrm{O}$ required for the establishment of order is impeded: the viscosity of glass is so high at room temperature that millions of years would be required for it to crystallize (see above). Hence, glass can be viewed as a perfectly rigid "liquid", being exceptional in that its extraordinarily high viscosity slows down its flow properties even on an expanded time scale.
Upon cooling of liquid “glass” we observe that $\Delta V/\Delta T$ (the contraction with lowering of T) changes. This may be seen by following the three basic steps in the production of glass: (1) the melting of quartz sand (minute crystals of silica), (2) the shaping of the glass while in a viscous state and (3) the controlled cooling of the shaped article. The $\mathrm{SiO}_4$ tetrahedra in the sand (crystals) are arranged in an ordered pattern (fig. 4), but in the molten state the crystal matrix breaks up into strings and rings of tetrahedra in irregular patterns (fig. 5). Because the temperature is very high, the groupings continually break up and reform, making the mass fluid.
Upon cooling of molten $\mathrm{SiO}_2$ the tetrahedra form larger groupings and the glass becomes more viscous, permitting it to be shaped by blowing, rolling, pressing, etc. With continuing cooling, larger groupings of tetrahedra form as their kinetic energy decreases. These two phenomena are responsible for the increase in viscosity and the shrinkage that takes place during the cooling period. Beyond the glass transformation temperature changes in configuration of the tetrahedra virtually cease because the glass, to all intents and purposes, has lost its fluidity.
OXIDE GLASSES
The bonding forces responsible for the formation of network structure in crystalline $\mathrm{SiO}_2$ produce similar networks in glasses of other oxides such as $\mathrm{B}_2 \mathrm{O}_3$ and $\mathrm{P}_2 \mathrm{O}_5$ which are thus also considered as primary network formers (the present discussion will be limited to silica glasses). If an oxide, such as $\mathrm{Na}_2 \mathrm{O}$, is added to silica glass, a bond in the network is broken and the relatively mobile sodium ion becomes a part of the structure (fig. 6). Examples of such network modifiers are $\mathrm{N}_2 \mathrm{O}, \mathrm{K}_2 \mathrm{O}, \mathrm{Li}_2 \mathrm{O}, \mathrm{CaO}, \mathrm{MgO}$ and $\mathrm{PbO}$.
With increase in the amount of modifier, the average number of oxygen-silicon bonds forming bridges between silicon between silicon atoms decreases as follows:
O/Si Ratio in Glass Bridging-O-Si Bonds/Si atom
2.0 4 (fig. 7)
2.5 3
3.0 2
Two bridging bonds per silicon atom will correspond, for example, to a linear chain structure, and any further increase in modifier would reduce the length of the chain. The principal effect of a modifier is to lower the melting and working temperature by decreasing the viscosity. An excess of modifier can make the structural units in the melt sufficiently simple and mobile that crystallization occurs in preference to the formation of a glass.
The compositions and properties of typical glasses are listed in Table 1. Soda-lime glass not only has the advantage of being made from inexpensive raw materials, but also the practical convenience of low working temperatures (fig. 8). Most glass-forming operations (pressing, drawing, etc.) begin at the temperature corresponding to the working point, but can continue while the glass cools.
ANNEALING OF GLASS
In order to relieve the internal stresses that may develop during glass formation, all commercial glassware is annealed (slowly cooled) immediately after shaping. This slow cooling operation is carried out in industry by placing the shaped objects on a conveyor belt which transports them first through a reheating zone and then through sections having successively lower temperatures in a furnace called the lehr.
TABLE 1. Typical Oxide Glasses
The transformation of a liquid to a glass actually takes place over a range of temperatures at which the glass shrinks (thermal contraction). The glass does not undergo a change from disordered liquid structure to crystal structure as iron does at its transformation temperature; in glass, the transformation temperature is the temperature below which viscosity prevents any further configurational changes. Any contraction beyond the transformation temperature range is due only to the lower kinetic energy of the groupings.
The transformation temperature of a given “glass composition” depends on its constituents and upon the rate of cooling. Slow cooling results in a lower transformation range because the tetrahedra will have more time to rearrange (to some degree). This results in tighter packing of tetrahedra as the mass reaches its transformation range. When the glass reaches room temperature, its volume will be smaller when cooled slowly than glass melt which has been cooled rapidly. Hence, slower cooling from the melt results in a denser glass. (See fig. 2.)
Glass is thermodynamically unstable at room temperature because configuration changes (which require eons of years to come about!) would result in a lower free energy configuration. However, the changes are so slow that they may be considered negligible.
PROPERTIES OF GLASSES
The strong covalent bonds that exist between the atoms of inorganic glasses plus the lack of crystal structure give glass unusual characteristics. Thus, quartz glass is:
1. (a) Chemically stable;
2. (b) A poor conductor of heat and electricity;
3. (c) Inherently transparent;
4. (d) Inherently strong.
Chemical Stability
All glasses are immune to oxidation (decay) because their atoms are fully oxidized. Glass composed of 100 percent silica tetrahedra is extraordinarily inert. It resists the action of most acids (the most notable exception is hydrofluoric acid, utilized to etch glass), but it is attacked by strong alkalis. (The bottles containing liquid alkalis on a chemist’s shelf have a clouded surface.)
Conductivity
The ability of a material to conduct electricity is dependent upon the presence of electrons in the conduction band, separated in glasses by as much as 10eV from the valence band. Because the electrons in glass are tightly bonded, glass is a very poor conductor of electricity - in fact, the large energy gap makes glass an excellent insulator. Glass is also a poor conductor of heat. Thus, hot common glass may crack when cold water is poured onto it because the surface exposed to the cold water will shrink while the dimensions of the interior remain unaffected.
Strength
Glass is harder than many types of steel and is also very elastic, as evidenced by the speed and accuracy with which glass marbles rebound when they meet.
Glass has "no" crystal structure, and hence the phenomenon of slip cannot take place. This, together with the strong bonding between atoms, gives glass a very high compressive strength and a theoretical tensile strength of about $10^7 \mathrm{kN} / \mathrm{m}^2$ (significantly higher than that of steel). (Glass fibers with an actual tensile strength of $4 \times 10^7 \mathrm{kN} / \mathrm{m}^2$ have been produced.) Since glass has a "liquid structure" it may be considered "saturated with dislocations". [If a piece of metal has a very high dislocation density (eg., due to cold working) the dislocations interfere with each other's movements. Therefore, with increasing dislocation density a piece of metal becomes harder and stronger, but also more brittle.] Since the molecular structures in glass are unable to move, the presence of minute cracks or imperfections in glass permits stress concentrations to localize and exceed the bond strength between atoms - common glass will crack. Thus, in actual practice, the strength of glass is, by a factor of 100 to 1000, less than the theoretical strength, and glass is brittle. For example, a freshly made electric light bulb may not crack when initially dropped, but its surface becomes damaged and, after it rebounds and strikes the floor a few more times, it breaks.
The inherent high strength of fibrous glass is utilized in the fiberglass sections employed for boats and automobile bodies. Before being pressed into shape fiberglass is mixed with a synthetic resin which serves to protect the fibers from scratching. Since the fibers thus retain their high strength, fiberglass sections are very strong for their weight.
STRENGTHENED GLASS
The glass scientist has produced objects made of glass which can be struck with a hammer or dropped from a tall building without breaking. Two techniques are employed in strengthening glass - one physical and one chemical. Both techniques are based on the fact that, when there is the slightest imperfection (minute scratch) on the surface, glass remains extraordinarily strong in compression but becomes weak in tension. Thus the strengthening treatment consists of prestressing a glass object by inducing compressive strains in its exterior and thereby enabling it to counteract any tensile stresses which develop under tension.
In physical prestressing, the shaped object is heated to just below its softening point (surface flows are annealed out) and its surface is then chilled by means of a blast of air or an oil bath. Under these conditions the exterior of the glass cools and contracts immediately, but since glass is a poor conductor of heat the interior will not contract - the glass will not crack because the interior remains plastic. When the interior starts to cool, it cannot contract because the exterior has already set, but, in attempting to contract, the interior continues to draw the exterior together. A built-in compressive stress then develops in the outer layers of the glass. When a tensile stress is applied to this glass, it is counteracted by these compressive stresses. Thus the prestressed glass will not shatter until the surface compression is exceeded.
As long as its surface remains intact, prestressed glass continues to be strong and shock-resistant. However, if a deep scratch develops on the surface, the tensile stresses set up in the interior by the surface compression are released and the glass will shatter into thousands of tiny particles. Therefore prestressed articles must be cut to exact size and all holes drilled prior to prestressing. Among the many uses for this type of glass are plate-glass doors, side windows of automobiles and portholes of ships.
Ion exchange is a chemical technique for prestressing glass objects. The first step is to place the shaped object into a molten salt bath containing potassium ions, which replace the sodium ions on the surface of the glass. The potassium ions are larger than the sodium ions and, in accommodating them, the surface of the glass becomes more crowded - thus inducing compressive strains on the exterior. Glass of this type, characterized by very high strength, is useful for a great variety of applications. Its flexural strength can be as high as $10^6 \mathrm{kN} / \mathrm{m}^2$ compared with $10^5 \mathrm{kN} / \mathrm{m}^2$ for untreated glass.
DEVITRIFICATION OF GLASS
Under certain conditions glass will become “contaminated” with crystalline particles. A glass in this condition is termed devitrified, which is simply another way of stating that it has partially or completely crystallized. Devitrified glass, unless of the very special variety that is deliberately produced, is undesirable since the crystalline areas form large crystals and are extremely weak and brittle, as well as being translucent only. Crystalline segregations in glass are known as stones
RECRYSTALLIZED GLASS (GLASS CERAMICS)
Recrystallized glass, also known as polycrystalline glass, is commonly produced by adding nucleating agents to the glass batch. Subsequently the glass can be formed into a desired shape by any of the conventional glass forming processes and then be heat-treated to promote recrystallization. Recrystallized glass possesses increased impact strength, hardness and thermal shock resistance compared with conventional non-crystalline glasses. One commonplace application of a recrystallized glass is in the manufacture of the so-called refrigerator-to-oven cooking dishes.
COLORED GLASSES
We recognize that glass is normally colorless due to the fact that all electrons are “tightly bonded” and no electronic excitations in the energy range of the visible light spectrum are possible. Glass, however, may be made to absorb selectively in the visible spectrum (thus becoming colored) by any one of three processes:
(1) Addition of ions of transition metals. Such ions provide electronic excitation possibilities to visible light. Typical ions added are:
\begin{aligned}
&\mathrm{Cr}^{++} \quad &&\text{blue} \
&\mathrm{Cr}^{+++} \quad &&\text{green} \
&\mathrm{Co}^{++} \quad &&\text{pink} \
&\mathrm{Mn}^{++} \quad &&\text{orange} \
&\mathrm{Fe}^{++} \quad &&\text{blue-green}
\end{aligned}
(2) Addition of colloidal particles ( $\phi$ $40$ to $2000 \AA$ ). These small particles, while not impeding light transmission (because of their small size) exhibit selective absorption and complementary reflection. Au particles (at $10^{-4} \mathrm{~g} / \mathrm{cm}^3$ ) give colors which vary with particle size:
\begin{aligned}
&\mathrm{Au} \quad &&\phi 4-10 \mathrm{~nm} \quad &\text{pink} \
&\mathrm{Au} \quad &&\phi 10-75 \mathrm{~nm} \quad &\text{ruby} \
&\mathrm{Au} \quad &&\phi 75-110 \mathrm{~nm}\quad &\text{green} \
&\mathrm{Au} \quad &&\phi 110-170 \mathrm{~nm} \quad &\text{brown}
\end{aligned}
(3) Addition of colored crystals. Some glasses are colored by means of crystals dispersed throughout. An example is the scarlet glass made by the early Egyptians (brought about by the addition of red copper oxide). Another example is red color from the addition of red $\mathrm{Pb}_2 \mathrm{CrO}_6$ whereas $\mathrm{Cr}_2 \mathrm{O}_3$ yields green color.
GLASS FIBER
Glass fiber, or fiberglass* as it is commonly known, is glass in fiber form. It is made by one of several processes, each of which involves the drawing out of the filaments from glass in the viscous state. Modern technical developments date from World War I when Germany had to find a substitute for asbestos as insulating material.
The process whereby continuous filament glass fiber is made is particularly interesting. The glass batch is melted in the glass tank furnace and the molten glass is molded into ordinary sized marbles. These marbles are fed into an electric furnace which has a platinum-alloy bushing containing many tiny holes. When the glass marbles reach the holes, the glass flows through them and is drawn down vertically, forming individual filaments. These filaments can then be twisted into a “yarn” and spun on normal textile spindles as required. About one third of an ounce of glass produces about one hundred miles of filament at a rate of about 6,000 feet per minute.
Glass wool, on the other hand, is produced by a different process. The most common technique, known as the Crown process, produces a thick matte of short glass fibers which are held together by a polymer binder. In the Crown method, a thick stream of molten glass is poured into a rapidly rotating steel dish that has tiny vents around its periphery. The glass is forced through these vents by centrifugal force, forming relatively short fibers with diameters of about 0.0007 cm. The matte is then passed through curing ovens where the binder sets and is then cut into sizes suitable for insulation.
*
Fiberglass is really a registered trade name.
SHEET AND PLATE GLASS PRODUCTION
The main types of flat glass produced are drawn sheet and float glass. The processes involved are highly automated.
Flat drawn glass is used for windows and other applications where accuracy of thickness and a high surface finish are not vital. In this process the molten glass is drawn up vertically from the end of a large rectangular tank furnace, the thickness of the drawn sheet being controlled principally by the size of the slot in the steel form held just above the surface of the molten glass. A steel starting dummy is used to commence the drawing action, but this is cracked off once it passes through. Once started, this process is continuous, the cooled and solidified sheet merely being cracked off in suitable lengths once it leaves the rolls.
The float process is the most recent advance in the production of plate glass with a high surface finish. In the float process, flat, fire-finished, stress-free glass is formed by preventing contact between the glass sheet and anything solid. The glass leaves the glass-tank in a continuous molten strip and is floated immediately onto the surface of a bath of molten tin. The tin bath is surrounded by a non-oxidizing atmosphere and, while in this atmosphere, the glass is heated sufficiently to prevent the development of internal stresses. As the glass sheet leaves the float bath chamber it enters an annealing furnace. Glass produced by the float process is free from distortion and has fire-finished surfaces that are smoother and flatter than those of normal polished plate glass. | textbooks/chem/Inorganic_Chemistry/Introduction_to_Solid_State_Chemistry/01%3A_Lectures/1.07%3A_Glasses.txt |
1. INTRODUCTION
We can readily understand that chemical reactions normally are preceded by collisions of atoms, ions, or molecules. However, the rate of such collisions in solids, liquids and gases is so great that all reactions would be very rapid were it only necessary for collisions to occur. “Chemical” reactions will not proceed more rapidly than molecular collisions allow, but many reactions proceed much more slowly. It is thus apparent that not every molecular collision leads to reaction. We have also seen from earlier examples that the driving force for physical as well as chemical reactions is the (free) energy change - which must be negative for reactions to occur spontaneously. However, while providing a “go/no-go” answer, this criterion alone cannot give us any information concerning the rate at which reactions occur, nor can it tell us which factors influence the reaction velocity.
Rates at which reactions occur vary considerably. For example, the nuclear reaction:
\begin{aligned}
&\mathrm{U}_{92}^{238} \rightarrow \mathrm{Th}_{90}^{234}+\alpha \quad &&\text{is } 50 \% \text{ completed } \left(\tau_{1 / 2}\right) \text{ after } 5 \times 10^9 \text{ years}. \ \
&\tau_{1 / 2}=5 \times 10^9 \text{years}
\end{aligned}
On the other hand, the chemical reaction:
\begin{aligned}
\mathrm{SO}_4^{-}+\mathrm{H}^{+} &\rightarrow \mathrm{HSO}_4^{-} \quad &&\text{is } 50\% \text{ completed after about } 10^{-4} \mathrm{~s}. \
\tau_{1 / 2}&=3 \times 10^{-4} \quad &&\text{seconds}
\end{aligned}
Reaction kinetics (rate theory) deals to a large extent with the factors which influence the reaction velocity. Take corrosion (rusting of iron), for example. We all know that it requires air and water to provoke rusting and we also know, much to our sorrow, that rusting proceeds much more rapidly near the ocean where salt is present. We also know that the rate of rusting depends strongly on the composition of iron (pure Fe and steel corrode much less rapidly than cast iron, for example). It is primarily kinetic studies which lead to the elucidation of chemical reactions and, in the case of corrosion, to the development of more corrosion resistant materials.
The principal experimental approach to the study of the reaction process involves the measurement of the rate at which a reaction proceeds and the determination of the dependence of this reaction rate on the concentrations of the reacting species and on the temperature. These factors are grouped together in the term reaction kinetics and the results for a given reaction are formulated in a rate equation which is of the general form:
Rate = k(T) x function of concentration of reactants
The quantity k(T) is called the rate constant and is a function only of the temperature if the term involving the reactant concentrations correctly expresses the rate dependence on concentration. Thus the experimental information on the reaction process is summarized in the rate equation by the nature of the concentration function and by the value and temperature dependence of the rate constant.
2. EXPERIMENTAL METHODS IN REACTION RATE STUDIES
Since we know that chemical reactions are temperature dependent, kinetic investigations will require rigorous temperature control (thermostats). Furthermore, we have to be able to observe and investigate concentration changes of reactions and products. For this purpose it frequently is customary to interrupt a reaction (for example, by quenching - the abrupt lowering of temperature) and to make a chemical analysis. In other instances, particularly for very fast reactions, a direct measurement of concentration changes is physically impossible. In such cases it is necessary to quantitatively follow reactions indirectly, through the accompanying changes in specific physical properties such as:
(1) electrical conductivity
(2) optical absorption
(3) refractive index
(4) volume
(5) dielectric constant
as well as by other means. In the last twenty years, for example, the use of isotopes (as “tracer” elements) has become a valuable tool for the study of reaction kinetics (in slow reactions).
3. CONCENTRATION DEPENDENCE OF REACTION RATES
Normally experimental data of kinetic investigations are records of concentrations of reactants and/or products as a function of time for constant temperatures (taken at various temperatures).
Theoretical expressions for reaction rates (involving concentration changes) are differential equations of the general form:
$\dfrac{\mathrm{dc}}{\mathrm{dt}}=\mathrm{f}\left(\mathrm{c}_1^{\mathrm{m}}, \mathrm{c}_2^{\mathrm{n}}, \mathrm{c}_3^{\mathrm{o}} \ldots\right)$
where (c) are concentration terms which have exponents that depend on details of the reaction.
If we want to compare the theory with the experiment, it is therefore necessary to either integrate the theoretical laws or to differentiate experimental concentration vs time data.
The rate laws are of importance since they provide analytical expressions for the course of individual reactions and enable us to calculate expected yields and optimum conditions for “economic” processes.
In most instances the differential rate equation is integrated before it is applied to the experimental data. Only infrequently are slopes of concentration vs time curves taken to determine dc/dt directly (fig. 1).
Let us look at a simple reaction, the decay of $\mathrm{H}_2 \mathrm{O}_2$ to water and oxygen:
$2 \mathrm{H}_2 \mathrm{O}_2 \rightarrow 2 \mathrm{H}_2 \mathrm{O}+\mathrm{O}_2$
Experimentally we obtain a curve (such as fig. 2) by plotting the concentration of remaining $\mathrm{H}_2 \mathrm{O}_2$ in moles/liter, normally written [$\mathrm{H}_2 \mathrm{O}_2$], as a function of time. The rate is then given by the time differential:
Rate $\left(\dfrac{\text { moles }}{\text { liter } \cdot \mathrm{s}}\right)=\dfrac{-\mathrm{d}\left[\mathrm{H}_2 \mathrm{O}_2\right]}{\mathrm{dt}}$
If we now determined and plotted the variation of the reaction rate, $\left[-\mathrm{d}\left[\mathrm{H}_2 \mathrm{O}_2\right] /(\mathrm{d}\right.$ time $\left.)\right]$, with concentration, $\left[\mathrm{H}_2 \mathrm{O}_2\right]$, we would find a straight line (fig. 3). Thus:
$-\dfrac{\mathrm{d}\left[\mathrm{H}_2 \mathrm{O}_2\right]}{\mathrm{dt}}=\mathrm{k}\left[\mathrm{H}_2 \mathrm{O}_2\right]$ or generally, $-\dfrac{\mathrm{dc}}{\mathrm{dt}}=\mathrm{kc}$
Upon integration:
$\int \dfrac{d c}{c}=-k \int d t$
we have:
$\ln \mathrm{c}=-\mathrm{kt}+\text { const. }$
Taking $\mathrm{c}=\mathrm{c}_{\mathrm{o}}$ for $\mathrm{t}=0$, we get $\left(\right.$ const $\left.=\ln \mathrm{c}_{\mathrm{o}}\right)$ and:
$\ln \mathrm{c}=-\mathrm{kt}+\ln \mathrm{c}_{\mathrm{o}}$
or:
$\mathrm{c}=\mathrm{c}_{\mathrm{o}} \mathrm{e}^{-\mathrm{kt}} \quad \text { and } \quad \mathrm{k}=\dfrac{1}{\mathrm{t}} \times \ln \dfrac{\mathrm{c}_{\mathrm{o}}}{\mathrm{c}}$
Alternately:
$k=\dfrac{2.3}{t} \times \log \dfrac{c_0}{c} \quad \text { and } \quad k=\dfrac{2.3}{t_2-t_1} \times \log \dfrac{c_1}{c_2}$
From the above we see that for decay of $\mathrm{H}_2 \mathrm{O}_2$ a plot of $(\ln \mathrm{c})$ vs. $t$ results in a straight line (fig. 4). This behavior is characteristic of first order reactions in which the concentration exponent $(\mathrm{n})$ is "one":
$-\dfrac{\mathrm{dc}}{\mathrm{dt}}=\mathrm{kc}$
First Order Reactions: Radioactive Decay
From the relationship:
$\mathrm{k}=\dfrac{2.3}{\mathrm{t}_2-\mathrm{t}_1} \times \log \dfrac{\mathrm{c}_1}{\mathrm{c}_2}$
which relates the rate constant $(k)$ to the concentration change $(\Delta c)$ for the time interval $(\Delta t)$, we can show that (for first order reactions) the time required to complete a reaction to $1 / 2$ or $1 / 4$ or any fraction of the initial concentration is independent of the initial concentration of reactant present (fig. 5).
When discussing radioactive decay it is customary to call the “rate constant” (k) the “decay constant”. It is furthermore customary to consider the time it takes to decrease the number of originally present species (normally called reactants) to 0.5 (by 50%). We then talk about the half-life. Accordingly, the above equation can be reformulated as:
$\mathrm{k}=\dfrac{2.3}{\mathrm{t}_{1 / 2}} \times \log \dfrac{\mathrm{c}_1}{\dfrac{\mathrm{c}_1}{2}}=\dfrac{2.3}{\mathrm{t}_{1 / 2}} \times \log 2$
or:
$k=\dfrac{0.693}{t_{1 / 2}}$
This means that from the half-life $\left(\mathrm{t}_{1 / 2}\right)$ we may obtain the decay constant (rate constant in general). Vice versa, knowing the rate constant, we can calculate the time it takes to complete $50 \%$ of the reaction (or decay).
4. REACTION ORDER
Reactions of first order, such as the decay of $\mathrm{H}_2 \mathrm{O}_2$ or radioactive decay, do not require molecular or atomic collisions - in principle they reflect inherent instability. A multitude of chemical reactions do involve collisions, however. For example, take the decay of HI:
$2 \mathrm{HI} \rightarrow \mathrm{H}_2+\mathrm{I}_2$
The rate law for this reaction reflects the requirement of a collision in the concentration exponent:
$-\dfrac{\mathrm{d}[\mathrm{HI}]}{\mathrm{dt}}=\mathrm{k}[\mathrm{HI}] \times[\mathrm{HI}]=k[H \mathrm{H}]^2$
The reaction is referred to as a second order reaction which can readily be identified since upon integration the rate law yields:
$\dfrac{1}{[\mathrm{HI}]}=\mathrm{kt}+\text { const }$
or, more generally:
$\dfrac{1}{\mathrm{c}}=\mathrm{kt}+\text { const }$
A plot of $1 / c$ vs $t$ will yield a straight line.
It is interesting to note that a determination of the reaction order from experimental data supplies insight to the details of how molecules and atoms react with each other. Even though rate data involve measurements of gross numbers of molecules, their interpretation (through the reaction order) permits us to formulate the probable step (or steps) which individual molecules undergo.
5. TEMPERATURE DEPENDENCE OF REACTION RATES
Although a chemical reaction may be "thermodynamically" favored (which means the free energy of the system will be lowered as a result of the reaction), reaction may nevertheless not take place. Thus, $\mathrm{H}_2$ and $\mathrm{O}_2$ gases can exist in intimate contact over a considerable temperature range before reaction takes place. Nevertheless, a lighted match or a platinum powder catalyst can nucleate an immediate explosive reaction. Another example with which we are familiar is glass: some glasses remain "supercooled liquids" for thousands of years at room temperature unless reheated to some definite temperature for at least some minimum period of time. Many different factors may influence the rate of a reaction - for example:
(1) Existing interatomic (or intermolecular) bonds must be broken.
(2) Atoms must be moved to and away from the reaction site.
(3) A new boundary is required wherever a new "phase" is to be nucleated.
All three steps cited require work or a supply of energy. Furthermore, all three steps are temperature dependent. At the beginning of this chapter we have stated:
Rate $=k(T) \times$ function of concentration of reactants
In other words: the reaction rate is proportional to the concentration of reactants. The proportionality constant $k$ is a function of temperature $k(T)$. The temperature dependence of $k$ (which represents the temperature dependence of the rate of a given reaction) can be studied by performing experiments (with constant concentrations) at different temperatures. From such experiments it can be seen that the value of the rate constant ($k$) is much greater at higher temperatures - the reactions proceed faster
In 1899 Arrhenius showed that the rate constant of reactions increased in an exponential manner with T. By an empirical procedure he found that a plot of log k versus 1/T gives a linear relation.
Remember: In all kinetic and thermodynamic calculations it is mandatory to use the "absolute temperature scale" in Kelvin (K) which is given by:
$\mathrm{K}={ }^{\circ} \mathrm{C}+273.16$
Thus OK $=-273.16 \mathrm{C}$, which corresponds to the thermodynamic absolute zero temperature. Linear plots of $\log \mathrm{k}$ vs $1 / \mathrm{T}$ imply the relation:
$\ln k \propto \dfrac{1}{T}$
$k \propto e^{\text {const/T }}$
In view of later deductions this empirical relation can be conveniently written as:
$\mathrm{k}=\mathrm{A} \mathrm{e}^{-\mathrm{E} / \mathrm{RT}}$
where $A$ is a proportionality constant sometimes called the reaction constant, $R$ is the gas constant and equals $8.31 \mathrm{~J} /$ mole $\mathrm{K}, \mathrm{E}$ is the activation energy in Joules/mole and e is the base of the natural (Naperian) system of logarithms. Since $\ln x=2.3 \log _{10} x$, we can rewrite the above equation:
$\ln \mathrm{k}=\ln \mathrm{A}-\dfrac{\mathrm{E}}{\mathrm{RT}}$
or
$\log k=\log A-\dfrac{E}{2.3 \times 8.3 \times \mathrm{T}}$
If the above equation is obeyed, a plot of log k versus 1/T (fig. 6) will be a straight line from which the characteristic activation energy can be determined since the slope is –E/19.15. The reaction constant (log A) may in turn be evaluated by extrapolating the rate line to the point at which 1/T equals zero.
It is important to realize that many reactions involve a succession of steps, in which case the rate is controlled by the slowest step. The corresponding values of E and A must be determined by experimental methods.
6. THE ARRHENIUS THEORY
Arrhenius developed a primarily qualitative theory for molecular reactions which led to empirical expressions for the rate constant. Later theories (which are beyond the scope of 3.091) elaborate and make these original ideas more quantitative.
The very simple (reversible) vapor-phase reaction of hydrogen and iodine to give hydrogen iodide can be used to illustrate the ideas of Arrhenius. The reaction
$\mathrm{H}_2+\mathrm{I}_2 \rightarrow 2 \mathrm{HI}$
apparently proceeds by a one-step (one collision) four-center process such that the path of the reaction can be depicted as follows:
For more complicated systems, even when a mechanism has been postulated, it is not so easy to see how the electrons and atoms move around as the reaction proceeds. However, even Arrhenius recognized that any reaction process can proceed first by means of the formation of some “high-energy species” (which we now call the “activated-complex”) and secondly by the breakdown of this complex into products.
If the activated-complex is assumed to have an energy, $\mathrm{E}_{\mathrm{a}}$, greater than the reactants, then, in analogy to earlier considerations, the number of activated-complex molecules compared with the number of reactant molecules can be written in terms of the Boltzmann distribution as:
$\dfrac{\text { [activated }-\text { complex molecules] }}{[\text { reactant] }}=e^{-E_a / R T}$
The rate of reaction thus becomes proportional to the concentration of activated-complex molecules:
$\text{Rate} \propto (\text{activated-complex molecules})$
or
$\text { Rate }=A \times e^{-E_a / R T} \times \text { reactants }$
Making
$k=A e^{-E_a / R T}$
we find
$\text { Rate }=\mathrm{k} \text { [reactants] }$
Moreover, the theory says that the empirical constant, $\mathrm{E}_{\mathrm{a}}$, is to be interpreted as the energy of the activated-complex compared with that of the reactant molecules.
The idea of an activated-complex can be presented on a plot (fig. 7) of the energy of
the system as ordinate versus the reaction coordinate as abscissa. The reaction coordinate is not any single internuclear distance, but rather depends on all the internuclear distances that change as the reactant molecules are converted into product molecules. In general it is impossible, and for the present purpose unnecessary, to give a quantitative description of the reaction coordinate. It consists of the transformation from reactants to products.
The Arrhenius theory leads to a considerable improvement in our understanding of the reaction process. It is, however, still a very qualitative theory in that it does not show how the pre-exponential factor A depends on the molecular properties of the reaction system, nor does it attempt to predict the value of $\mathrm{E}_{\mathrm{a}}$.
7. THE ACTIVATION ENERGY
The basic distribution of molecular (or atomic) energies at two different temperatures is given in fig. 8. It can be shown that $e^{-E_a / R T}$ is the fraction of molecules or atoms having
an energy of $\mathrm{E}_{\mathrm{a}}$ or greater. The Arrhenius equation holds only if the interacting species have between them at least the certain critical energy $\mathrm{E}_{\mathrm{a}}$. Since the fraction having this energy or greater is $e^{-E_a / R T}$, the reaction rate is proportional to this quantity.
As the value of $\mathrm{E}_{\mathrm{a}}$ increases, the energy requirement increases and it becomes more difficult for the molecules to acquire this energy. In contrast, $e^{-E_a / R T}$ increases rapidly with increasing temperature (T).
The change of a reaction rate with increasing temperature is usually much greater than expected from the corresponding increase in the average velocity of molecules and atoms. The average velocity of molecules and atoms is proportional to the square root of the absolute temperature. Thus, if the temperature is raised $10^{\circ}$ from $298^{\circ}$ to $308^{\circ}$, the average velocity increase, $(308 / 298)^{1 / 2}$, is but $2 \%$ whereas the rate of reaction increases by about $100 \%$. From this we must conclude that the reaction rate is controlled not only by the number of collisions, but also by the activation energy.
We now realize that in order for a reaction to proceed we have to supply activation energy. We also know that for given conditions of concentration and T the rate of any reaction is inversely proportional to the energy of activation.
Since, particularly in industry, time is a major factor which can make a process economically feasible or unfeasible, considerable efforts have been put into accelerating reactions by means of catalysts. In catalytic reactions the “catalyst”, which effectively lowers the required activation energy (fig. 9), is characterized by the following criteria:
(1) unchanged chemically at the end of a reaction;
(2) required in small amounts only;
(3) catalytic action is frequently proportional to its surface area;
(4) catalysis can be selective: if different reactions are possible, catalysis can enhance the rate of either one without affecting the alternate reaction.
For example:
\begin{aligned}
\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH} \rightarrow & \mathrm{H}_2 \mathrm{C}=\mathrm{CH}_2+\mathrm{H}_2 \mathrm{O} & \text { on } \gamma \text { Alumina } \
& \text { (important in polymer chemistry) } \
&\left(\gamma \text { Alumina: a cubic form of } \mathrm{Al}_2 \mathrm{O}_3\right) \
\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH} \rightarrow & \mathrm{CH}_3-\mathrm{C}_H^{\prime \mathrm{O}}+\mathrm{H}_2 \quad &\text { on Copper }
\end{aligned}
While the exact action of catalysis is still quite unclear, we do have some concrete information. The most frequently encountered case, heterogeneous catalysis (the presence of certain solids which increase gas reactions) is attributed to activation by adsorption. From infrared studies we know that upon adsorption of compounds the bonds within the compound are weakened and reactions can subsequently occur with decreased activation energy.
Catalysis has certainly had considerable impact on our daily life. Until 1940 gasoline was exclusively made from crude oil. Also, possibly more important, nitrates and ammonia came primarily from Chile (Chile saltpeter). In the Fischer-Tropsch process, coal and steam are converted into gasoline hydrocarbons with the aid of $\mathrm{Ni}-\mathrm{Co}$ catalysts. The Haber-Bosch process, $\mathrm{N}_2+3 \mathrm{H}_2 \rightarrow 2 \mathrm{NH}_3$ (ammonia) $+99 \mathrm{~kJ}$, is likely the most important catalytic reaction practiced: From the "Le Chatelier" principle (if in a reaction the number of molecules present decreases, the reaction can be accelerated by applying increased pressures) we know that high pressures will favor this synthesis, but we also know that high temperature will favor the reverse reaction, i.e., decay of ammonia; at high temperatures complex molecular structures tend to decay to more elemental, basic species. Catalysis $\left(\mathrm{Fe}, \mathrm{Al}_2 \mathrm{O}_3+\mathrm{K}_2 \mathrm{O}\right.$ in solid form) makes it possible at $400^{\circ} \mathrm{C}$ and at about 600 atm to convert about $60 \%$ of the gas mixture $\mathrm{N}_2+3 \mathrm{H}_2$ to $\mathrm{NH}_3$. Nitrates (important as fertilizers and explosives) are predominantly produced by catalytic oxidation.
\begin{aligned}
&4 \mathrm{NH}_3+5 \mathrm{O}_2 \rightarrow 4 \mathrm{NO}+6 \mathrm{H}_2 \mathrm{O} \quad \text { over Pt at } 900^{\circ} \mathrm{C} \
&2 \mathrm{NO}+\mathrm{O}_2 \rightarrow 2 \mathrm{NO}_2 \
&3 \mathrm{NO}_2+\mathrm{H}_2 \mathrm{O} \rightarrow 2 \mathrm{HNO}_3 \text { (nitric acid) }+\mathrm{NO}
\end{aligned}
Other examples are:
* $\mathrm{SO}_2+1 / 2 \mathrm{O}_2 \rightarrow \mathrm{SO}_3$ over $\mathrm{V}_2 \mathrm{O}_5$
* Synthetic rubber from butadiene and styrene
* Methanol from $\mathrm{CO}+2 \mathrm{H}_2 \rightarrow \mathrm{CH}_3 \mathrm{OH}$ over $\mathrm{ZnO}+\mathrm{Cr}_2 \mathrm{O}_3$
* Formaldehyde from over $\mathrm{Cu}$ (important for plastics)
* In nature we have a large number of catalysts in the form of enzymes. | textbooks/chem/Inorganic_Chemistry/Introduction_to_Solid_State_Chemistry/01%3A_Lectures/1.08%3A_Theory_of_Reaction_Rates.txt |
1. DIFFUSION
At any temperature different from absolute zero all atoms, irrespective of their state of aggregation (gaseous, liquid or solid), are constantly in motion. Since the movement of particles is associated with collisions, the path of a single particle is a zigzag one. However, an aggregation of “diffusing” particles has an observable drift from places of higher to places of lower concentration (fig. 1). For this reason diffusion is known as a transport phenomenon.
In each diffusion reaction (heat flow, for example, is also a diffusion process), the flux (of matter, heat, electricity, etc.) follows the general relation:
Flux = (conductivity) x (driving force)
In the case of atomic or molecular diffusion, the “conductivity” is referred to as the diffusivity or the diffusion constant, and is represented by the symbol D. We realize from the above considerations that this diffusion constant (D) reflects the mobility of the diffusing species in the given environment and accordingly assumes larger values in gases, smaller ones in liquids, and extremely small ones in solids.
The “driving force” for many types of diffusion is the existence of a concentration gradient. The term “gradient” describes the variation of a given property as a function of distance in the x-direction. If a material exhibits a linear variation of concentration with distance in the x-direction, we speak of a constant concentration gradient in the x-direction. The gradient itself is the rate of change of the concentration with distance (dc/dx), which is the same as the slope of a graph of concentration vs. position ($\Delta c/\Delta x)$ (see fig. 2).
Steady State and Nonsteady Diffusion
Diffusion processes may be divided into two types: (a) steady state and (b) nonsteady state. Steady state diffusion takes place at a constant rate - that is, once the process starts the number of atoms (or moles) crossing a given interface (the flux) is constant with time. This means that throughout the system dc/dx = constant and dc/dt = 0.
Nonsteady state diffusion is a time dependent process in which the rate of diffusion is a function of time. Thus dc/dx varies with time and dc/dt ≠ 0. Both types of diffusion are described quantitatively by Fick’s laws of diffusion. The first law concerns both steady state and nonsteady state diffusion, while the second law deals only with nonsteady state diffusion.
2. STEADY STATE DIFFUSION (FICK’S FIRST LAW)
On the basis of the above considerations, Fick’s First Law may be formulated as:
$\mathrm{J}=-\mathrm{D}\left(\dfrac{\mathrm{dc}}{\mathrm{dx}}\right)$ In words: The diffusive flux is proportional to the existing concentration gradient.
The negative sign in this relationship indicates that particle flow occurs in a "down" gradient direction, i.e. from regions of higher to regions of lower concentration. The flux $\mathrm{J}$ can be given in units of atoms $/ \mathrm{cm}^2 \mathrm{~s}$, moles/ $\mathrm{cm}^2 \mathrm{~s}$, or equivalents. Correspondingly, the diffusivity (D) will assume the dimensions $\mathrm{cm}^2 / \mathrm{s}$, as can be seen from a dimensional analysis:
$\mathrm{J}\left(\dfrac{\text { moles }}{\mathrm{cm}^2 \mathrm{~s}}\right)=-\mathrm{D}\left(\dfrac{\mathrm{dc}}{\mathrm{dx}}\right)\left(\dfrac{\text { moles } \cdot \mathrm{cm}^{-3}}{\mathrm{~cm}}\right)$
Thus: $D=\mathrm{cm}^2 / \mathrm{s}$
Like chemical reactions, diffusion is a thermally activated process and the temperature dependence of diffusion appears in the diffusivity as an “Arrhenius-type” equation:
$D=D_0 e^{-E_a / R T}$
where $D_o$ (the equivalent of $A$ in the previously discussed temperature dependence of the rate constant) includes such factors as the jump distance, the vibrational frequency of the diffusing species and so on. Selected values of $D, D_0$ and $E_a$ are given in Table 1 (a) and (b).
TABLE 1
(a) Selected Values of Diffusion Constants (D)
\begin{aligned}
&\begin{array}{llrl}
\underline{\text { Diffusing Substance }} & \underline{\text { Solvent }} & \underline{T\left({ }^{\circ} \mathrm{C}\right)} & \underline{D\left(\mathrm{~cm}^2 \cdot \mathrm{s}^{-1}\right)} \
\mathrm{Au} & \mathrm{Cu} & 400 & 5 \times 10^{-13} \
\mathrm{Cu} \text { (Self-Diffusing) } & (\mathrm{Cu}) & 650 & 3.2 \times 10^{-12} \
\mathrm{C} & \mathrm{Fe} \text { (FCC) } & 950 & 10^{-7} \
\text { Methanol } & \mathrm{H}_2 \mathrm{O} & 18 & 1.4 \times 10^{-5} \
\mathrm{O}_2 & \text { Air } & 0 & 0.178 \
\mathrm{H}_2 & \text { Air } & 0 & 0.611
\end{array}\
\end{aligned}
(b) Selected Values of $D_o$ and $E_a$ for Diffusion Systems
Solute Solvent (host structure) $\underline{D_o, \mathrm{cm}^2 \mathrm{s}}$ $\underline{E_a}$, kJoules/mole
1. Carbon fcc iron 0.2100 142
2. Carbon bcc iron 0.0079 76
3. Iron fcc iron 0.5800 185
4. Iron bcc iron 5.8000 251
5. Nickel fcc iron 0.5000 276
6. Manganese fcc iron 0.3500 283
7. Zinc copper 0.0330 159
8. Copper aluminum 2.0000 142
9. Copper copper 11.0000 240
10. Silver silver 0.7200 188
A typical application of Fick's first law: Determine the rate at which helium (He), held at 5 atm and $200^{\circ} \mathrm{C}$ in a Pyrex glass bulb of $50 \mathrm{~cm}$ diameter and a wall thickness ( $\mathrm{x}$ ) of $0.1 \mathrm{~cm}$, diffuses through the Pyrex to the outside. Assume that the pressure outside the tube at all times remains negligible (see fig. 3). (For the diffusion of gases it is
customary, although not necessary, to replace the diffusion constant $D$ with the Permeation constant $\mathrm{K}$, normally given in units of $\mathrm{cm}^2 / \mathrm{s} \cdot$ atm. Using the gas laws, $\mathrm{K}$ is readily converted to $\mathrm{D}$ if so desired.)
In the present system
$\mathrm{K}=1 \times 10^{-9} \mathrm{~cm}^2 / \mathrm{s} \cdot \mathrm{atm}$
We can now set up the diffusion equation:
J=-\mathrm{K}\left(\frac{\mathrm{dP}}{\mathrm{dx}}\right) \quad \begin{aligned} &\text { (and operate with pressures instead } \ &\text { of concentrations) } \end{aligned}
We may now formally separate the variables and integrate:
\begin{aligned}
J \mathrm{dx} &=-\mathrm{KdP} \
\int_{\mathrm{x}=0}^{\mathrm{x}=0.1} \mathrm{Jdx} &=-\int_{\mathrm{p}_2=5}^{p_1=0} \mathrm{KdP} \
\mathrm{J} &=\mathrm{K} \frac{5.0}{0.1}
\end{aligned}
We can forego the integration since $(\mathrm{dP} / \mathrm{dx})=(\Delta \mathrm{P} / \Delta \mathrm{x})$ and we may immediately write:
\begin{aligned}
\mathrm{J}(\text { total }) &=-\mathrm{K} \frac{\Delta \mathrm{P}}{\Delta \mathrm{x}} \times \mathrm{A}=5 \times 10^{-8} \times 7.9 \times 10^3 \
\mathrm{~J} &=3.9 \times 10^{-4} ? ?
\end{aligned}
The units of the flux may be obtained from a dimensional analysis:
$\mathrm{J}=-\mathrm{K} \times \dfrac{\Delta \mathrm{P}}{\Delta \mathrm{x}} \times \mathrm{A}=\dfrac{\mathrm{cm}^2}{\mathrm{~s} \cdot \mathrm{atm}} \dfrac{\mathrm{atm}}{\mathrm{cm}} \dfrac{\mathrm{cm}^2}{1}=\dfrac{\mathrm{cm}^3}{\mathrm{~s}}$
The total flux is $3.9 \times 10^{-4} \mathrm{~cm}^3 / \mathrm{s}$ (with the gas volume given for $0^{\circ} \mathrm{C}$ and $1 \mathrm{~atm}$ ). If the total gas flow by diffusion were to be determined for a specified time interval, the volume would be multiplied by the indicated time.
3. NONSTEADY STATE DIFFUSION (FICK’S SECOND LAW)
The quantitative treatment of nonsteady state diffusion processes is formulated as a partial differential equation. It is beyond the scope of 3.091 to treat the equations in detail but we can consider the second law qualitatively and examine some relevant solutions quantitatively.
The difference between steady state and nonsteady state diffusion conditions can readily be visualized (fig. 4). In the first case we have, for example, the diffusion of gas
from an infinite volume ($\mathrm{P}_1$ const) through a membrane into an infinite volume ($\mathrm{P}_2$ const). The pressure gradient across the membrane remains constant as does the diffusive flux. In the second case we deal with diffusion from a finite volume through a membrane into a finite volume. The pressures in the reservoirs involved change with time as does, consequently, the pressure gradient across the membrane.
(You are not required to be familiar with the following derivation of the Second Fick’s Law, but you must know its final form.)
Consider a volume element (between x and x+dx of unit cross sectional area) of a membrane separating two finite volumes involved in a diffusion system (fig. 5). The flux of a given material into the volume element minus the flux out of the volume element equals the rate of accumulation of the material into this volume element:
$J_x-J_{x+d x}=\dfrac{\partial \bar{c}}{\partial t} d x$
[ $\bar{c}$ is the average concentration in the volume element and $\bar{c} d x$ is the total amount of the diffusing material in the element at time (t).]
Using a Taylor series we can expand $J_{x+d x}$ about $x$ and obtain:
$J_{x+d x}=J_x+\dfrac{\partial J_x}{\partial x} d x+\dfrac{\partial^2 J_x}{\partial x^2} \dfrac{d x^2}{2}+\ldots$
Accordingly, as $\mathrm{dx} \rightarrow 0$:
$\dfrac{\partial}{\partial x}\left(D \dfrac{\partial c}{d x}\right)=\dfrac{\partial c}{\partial t}$
and if $D$ does not vary with $x$ (which is normally the case) we have the formulation of Fick's Second Law:
$\dfrac{\partial \mathrm{C}}{\partial \mathrm{t}}=\mathrm{D} \dfrac{\partial^2 \mathrm{C}}{\partial \mathrm{x}^2} \quad \text { (Fick's Second Law) }$
In physical terms this relationship states that the rate of compositional change is proportional to the “rate of change” of the concentration gradient rather than to the concentration gradient itself.
The solutions to Fick’s second law depend on the boundary conditions imposed by the particular problem of interest. As an example, let us consider the following problem (encountered in many solid state processes):
A frequently encountered situation is the diffusion of a component 2 into an infinite region of a material 1 (fig. 6) [planar diffusion of doping elements into semiconductors
for the generation of junction devices (p-n junctions, junction transistors)]. The boundary conditions are: the concentration of component (2) at the surface of the solid phase $(x=0)$ remains constant at $c_2$ and the concentration of component (2) in the solid prior to diffusion is uniformly $c_{2^{\prime}}$ (a). Under these boundary conditions the solution to Fick's second law assumes the form:
$\dfrac{c_2-c}{c_2-c_{2^{\prime}}}=\operatorname{erf}\left(\dfrac{x}{2 \sqrt{D t}}\right)$
If no component (2) is originally in the solid matrix (1) (b), the above solution is simpler:
\begin{aligned}
\dfrac{c_2-c}{c_2} &=\operatorname{erf}\left(\dfrac{x}{2 \sqrt{D t}}\right) \
1-\dfrac{c}{c_2} &=\operatorname{erf}\left(\dfrac{x}{2 \sqrt{D t}}\right) \
\dfrac{c}{c_2} &=1-\operatorname{erf}\left(\dfrac{x}{2 \sqrt{D t}}\right) \
\dfrac{c}{c_2} &=\operatorname{erfc}\left(\dfrac{x}{2 \sqrt{D t}}\right) \
c &=c_2 \text { erfc }\left(\dfrac{x}{2 \sqrt{D t}}\right)
\end{aligned}
An analysis shows that the last form of the solution to Fick's law relates the concentration (c) at any position ( $\mathrm{x}$ ) (depth of penetration into the solid matrix) and time (t) to the surface concentration $\left(c_2\right)$ and the diffusion constant $(D)$. The terms erf and erfc stand for error function and complementary error function respectively - it is the Gaussian error function as tabulated (like trigonometric and exponential functions) in mathematical tables. Its limiting values are:
\begin{aligned}
&\text{erf}(0) &&= 0 \
&\text{erf}(\infty) &&= 1 \quad &&\text{And for the complementary error function:} \
&\text{erf}(-\infty) &&=-1 \quad &&\text{erfc = (1-erf)}
\end{aligned}
Another look at the above solution to the diffusion equation shows that the concentration (c) of component (2) in the solid is expressed in terms of the error function of the argument $x / 2 \sqrt{D t}$. To determine at what depth a particular concentration $\left(\mathrm{c}^*\right)$ of (2) will appear, we substitute this concentration for (c) and obtain:
$\dfrac{c^*}{c_2}=\operatorname{erfc}\left(\dfrac{x}{2 \sqrt{D t}}\right)=K$
Since the error function is a constant, its argument must also be a constant:
$\dfrac{x}{2 \sqrt{D t}}=K^{\prime}$
Therefore, under the given boundary conditions:
\begin{aligned}
&x=K^{\prime} 2 \sqrt{D t} \
&x=K^{\prime} \cdot \sqrt{D t} \
&x \propto \sqrt{t}
\end{aligned}
The depth of penetration of a specified concentration is found to be proportional to the square root of the time of diffusion.
4. SELF-DIFFUSION
As previously indicated, the thermal motion of atoms in a lattice is a random process and as such will lead to local displacements of individual atoms. This random movement of atoms within a lattice (self-diffusion), which is not associated with any existing concentration gradients, can be readily demonstrated with the aid of “radioactive elements”. For example, nickel appears in nature in the form of several “stable isotopes”:
$\mathrm{Ni}_{28}^{58}, \mathrm{Ni}_{28}^{60}, \mathrm{Ni}_{28}^{61}, \mathrm{Ni}_{28}^{62}$ and $\mathrm{Ni}_{28}^{64}$
If $\mathrm{Ni}_{28}^{58}$ is irradiated with neutrons in a nuclear reactor, it will capture a neutron and become $\mathrm{Ni}_{28}^{59}$ which is radioactive (a radio-isotope).
$\mathrm{Ni}_{28}^{58}+\mathrm{n} \rightarrow \mathrm{Ni}_{28}^{59} \rightarrow \mathrm{Co}_{27}^{59}+\gamma+\beta$
Nickel 59 is characterized by its instability which leads to the emission of $\beta$ and $\gamma$ radiation, with a half-life of $8 \times 10^4$ years. Since this radiation can be measured by appropriate radiation detectors, it is possible to use $\mathrm{Ni}_{59}$ as a "tracer element" for studies of self-diffusion.
The radioactive nickel (which is identical to ordinary nickel with the exception of its radioactive properties) is electroplated onto normal nickel. This specimen is subsequently placed into a furnace and heated up to close to its melting point for an extended period of time. After removing the specimen, it is sectioned into slices parallel to the surface which contained the radioactive tracer element. With the aid of a radiation detector it can now be shown that the $\mathrm{Ni}_{59}$, which originally was only at the surface, has diffused into the bulk material while simultaneously some bulk nickel has counter-diffused in the other direction. If this sample is heat-treated for a much longer time, sectioning and counting will reveal a completely uniform distribution of the radio-tracer element. It can thus be shown that self-diffusion does occur in solids, and quantitative measurements with tracer elements even permit the determination of self-diffusion coefficients.
5. DIFFUSION MECHANISMS
The diffusion process in interstitial solid solutions, like that of carbon in iron, can readily be understood as a result of considerable differences in atomic diameters. However, the fact that $\mathrm{Au}$ diffuses faster in $\mathrm{Pb}$ than $\mathrm{NaCl}$ diffuses in water at $15^{\circ} \mathrm{C}$ cannot be readily explained. The magnitude of the observed activation energy indicates that a mechanism whereby atoms simply change places with each other has to be excluded. More reasonable mechanisms were suggested by Frenkel and Schottky. They proposed the existence of point defects (vacancies) in crystals which provide a mechanism by which atoms can move (diffuse) within a crystal. The concentration of such vacancies, as you recall, can be calculated from simple statistical calculations.
In most solids we are not dealing with single crystals but rather with polycrystalline materials which contain a large number of grain boundaries (internal surfaces). As expected, the rate of diffusion along grain boundaries is much higher than that for volume diffusion ( $D_{\text {volume }}<D_{\text {g-boundary }}$ ). Finally, surface diffusion, which takes place on all external surfaces, is even higher $\left(D_{\text {volume }}<D_{\text {g-boundary }}<D_{\text {surface }}\right)$. The respective activation energies for diffusion are:
$E_a$ surface $<E_a$ grain boundary $<E_a$ volume
Diffusion in Non-Metals
In non-metallic systems diffusion takes place by the same mechanisms as in metallic systems. Oxygen, for example, diffuses through many oxides by vacancy migration. In crystalline oxides and in silicate glasses as well, it is found that oxygen diffuses much more rapidly than the metallic ion. In glasses containing alkali atoms $\left(\mathrm{Na}^{+}, \mathrm{K}^{+}\right)$, the respective rates of diffusion are:
$D_{\text {alkali }}>D_{\text {oxygen }}>D_{\text {silicon }}$
corresponding to differences in bonding strengths. In polymer materials diffusion requires the motion of large molecules since intramolecular bonding is much stronger than intermolecular bonding. This fact explains that the diffusion rates in such materials are relatively small.
Gaseous Diffusion in Solids
Some gases, like hydrogen and helium, diffuse through some metals with ease even at room temperature. Helium, for example, will diffuse through quartz and steel and limits the ultimate vacuum obtainable in ultra-high vacuum systems. Hydrogen similarly diffuses readily through $\mathrm{Ni}$ at elevated temperatures. $\mathrm{H}_2$ also diffuses at high rates through palladium - a phenomenon which is used extensively for hydrogen purification since that material is impervious to other gases.
TABLE 2: The Error Function
$z$ $erf(z)$ $z$ $erf(z)$
0 0 0.85 0.7707
0.025 0.0282 0.90 0.7970
0.05 0.0564 0.95 0.8209
0.10 0.1125 1.0 0.8427
0.15 0.1680 1.1 0.8802
0.20 0.2227 1.2 0.9103
0.25 0.2763 1.3 0.9340
0.30 0.3286 1.4 0.9523
0.35 0.3794 1.5 0.9661
0.40 0.4234 1.6 0.9763
0.45 0.4755 1.7 0.9838
0.50 0.5205 1.8 0.9891
0.55 0.5633 1.9 0.9928
0.60 0.6039 2.0 0.9953
0.65 0.6420 2.2 0.9981
0.70 0.6778 2.4 0.9993
0.75 0.7112 2.6 0.9998
0.80 0.7421 2.8 0.9999
SOURCE: The values of erf(z) to 15 places, in increments of z of 0.0001, can be found in the Mathematical Tables Project, “Table of Probability Functions . . .”, vol. 1, Federal Works Agency, Works Projects Administration, New York, 1941. A discussion of the evaluation of erf(z), its derivatives and integrals, with a brief table is given by H. Carslaw and J. Jaeger, in Appendix 2 of “Conduction of Heat in Solids”, Oxford University Press, Fair Lawn, NJ, 1959. | textbooks/chem/Inorganic_Chemistry/Introduction_to_Solid_State_Chemistry/01%3A_Lectures/1.09%3A_Diffusion.txt |
PHASE EQUILIBRIA AND PHASE DIAGRAMS
Phase diagrams are one of the most important sources of information concerning the behavior of elements, compounds and solutions. They provide us with the knowledge of phase composition and phase stability as a function of temperature (T), pressure (P) and composition (C). Furthermore, they permit us to study and control important processes such as phase separation, solidification, sintering, purification, growth and doping of single crystals for technological and other applications. Although phase diagrams provide information about systems at equilibrium, they can also assist in predicting phase relations, compositional changes and structures in systems not at equilibrium.
GASES, LIQUIDS AND SOLIDS
Any material (elemental or compound) can exist as a gas, a liquid or a solid, depending on the relative magnitude of the attractive interatomic or intermolecular forces vs the disruptive thermal forces. It is thus clear that the stability (existence) of the different states of aggregation, which are referred to as phases, is a function of temperature and pressure (since with increased pressure the atoms, for exampled of a gas phase, are closer spaced and thus subject to increased interatomic attraction).
In general terms, a “phase” is a homogeneous, physically distinct, mechanically separable portion of a material with a given chemical composition. To illustrate this definition, let us look at a few examples of common multi-phase systems. Ice cubes in water constitute a two-phase system (ice and liquid water), unless we include the vapor above the glass in our system, which would make it a three-phase system. A mixture of oil and water would also be a two-phase system. Just as oil and water represent two distinct liquid phases, two regions of a solid with distinctly different composition or structure separated by boundaries represent two solid phases.
If we look at a one-component system, such as liquid water, we recognize that because of the energy distribution of the water molecules, some water molecules will always possess sufficient energy to overcome the attractive forces on the surface of $\mathrm{H}_{2} \mathrm{O}$ and enter into the gas phase. If thermal energy is continuously supplied to a liquid in an open container, the supply of high energy molecules (which leave the liquid phase) is replenished and the temperature remains constant - otherwise the loss of high energy molecules will lower the temperature of the system. The total quantity of heat necessary to completely "vaporize" one mole of a liquid at its boiling point is called its molar heat of vaporization, designated by $\Delta \mathrm{H}_{\mathrm{V}}$. Similarly, the heat required to completely melt one mole of a solid (the heat required to break the bonds established in the solid phase) is called the (latent) heat of fusion ( $\Delta \mathrm{H}_{\mathrm{V}}$ ).
Visualize a liquid in a sealed container with some space above the liquid surface. Again, some of the most energetic liquid molecules will leave the liquid phase and form a "gas phase" above the liquid. Since gas molecules will thus accumulate in the gas phase (at a constant temperature), it is inevitable that as a result of collisions in the gas phase some molecules will re-enter the liquid phase and a situation will be established whereby the rate of evaporation will equal the rate of condensation – i.e., a dynamic equilibrium between the liquid and gas phase will exist. The established pressure in the gas phase is referred to as the equilibrium vapor pressure, which is normally significantly less for solids than for liquids.
For obvious reasons it is desirable to know for any given material the conditions (P, T) under which the solid state, the liquid state and the gaseous state are stable, as well as the conditions under which the solid and liquid phases may coexist. These conditions are graphically presented in equilibrium phase diagrams, which can be experimentally determined.
ONE-COMPONENT PHASE DIAGRAM
Figure 1 illustrates the temperatures and pressures at which water can exist as a solid, liquid or vapor. The curves represent the points at which two of the phases coexist in equilibrium. At the point $T_{t}$ vapor, liquid and solid coexist in equilibrium. In the fields of the diagram (phase fields) only one phase exists. Although a diagram of this kind delineates the boundaries of the phase fields, it does not indicate the quantity of any phase present.
It is of interest to consider the slope of the liquid/solid phase line of the $\mathrm{H}_{2} \mathrm{O}$ phase diagram. It can readily be seen that if ice $-$ say at $-2^{\circ} \mathrm{C}-$ is subjected to high pressures, it will transform to liquid $\mathrm{H}_{2} \mathrm{O}$. (An ice skater will skate not on ice, but on water.) This particular pressure sensitivity (reflected in the slope of the solid/liquid phase line) is characteristic for materials which have a higher coordination number in the liquid than in the solid phase $\left(\mathrm{H}_{2} \mathrm{O}, \mathrm{Bi}, \mathrm{Si}, \mathrm{Ge}\right)$. Metals, for example have an opposite slope of the solid/liquid phase line, and the liquid phase will condense under pressure to a solid phase.
Fig. 1 Pressure/Temperature Diagram for Water. (Not drawn to scale.)
PHASE RULE AND EQUILIBRIUM
The phase rule, also known as the Gibbs phase rule, relates the number of components and the number of degrees of freedom in a system at equilibrium by the formula
$F = C - P + 2 \tag{1}$
where F equals the number of degrees of freedom or the number of independent variables, C equals the number of components in a system in equilibrium and P equals the number of phases. The digit 2 stands for the two variables, temperature and pressure.
$\mathrm{CaCO}_{3}(\mathrm{~s}) \leftrightarrows \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{~g}) \tag{2}$
There are also three different chemical constituents, but the number of components is only two because any two constituents completely define the system in equilibrium. Any third constituent may be determined if the concentration of the other two is known. Substituting into the phase rule (eq. [1]) we can see that the system is univariant, since $F=C-P+2=2-3+2=1$. Therefore only one variable, either temperature or pressure, can be changed independently. (The number of components is not always easy to determine at first glance, and it may require careful examination of the physical conditions of the system at equilibrium.)
The phase rule applies to dynamic and reversible processes where a system is heterogeneous and in equilibrium and where the only external variables are temperature, pressure and concentration. For one-component systems the maximum number of variables to be considered is two - pressure and temperature. Such systems can easily be represented graphically by ordinary rectangular coordinates. For two-component (or binary) systems the maximum number of variables is three pressure, temperature and concentration. Only one concentration is required to define the composition since the second component is found by subtracting from unity. A graphical representation of such a system requires a three-dimensional diagram. This, however, is not well suited to illustration and consequently separate two-coordinate diagrams, such as pressure vs temperature, pressure vs composition and temperature vs composition, are mostly used. Solid/liquid systems are usually investigated at constant pressure, and thus only two variables need to be considered – the vapor pressure for such systems can be neglected. This is called a condensed system and finds considerable application in studying phase equilibria in various engineering materials. A condensed system will be represented by the following modified phase rule equation:
$F = C - P + 1 \tag{3}$
where all symbols are the same as before, but (because of a constant pressure) the digit 2 is replaced by the digit 1, which stands for temperature as variable. The graphical representation of a solid/liquid binary system can be simplified by representing it on ordinary rectangular coordinates: temperature vs concentration or composition.
PHASE DIAGRAM
With the aid of a suitable calorimeter and energy reservoir, it is possible to measure the heat required to melt and evaporate a pure substance like ice. The experimental data obtainable for a mole of ice is shown schematically in fig. 2. As heat is added to the solid, the temperature rises along line "a" until the temperature of fusion $\left(T_{f}\right)$ is reached. The amount of heat absorbed per mole during melting is represented by the length of line " $\mathrm{b}$ ", or $\Delta \mathrm{H}_{\mathrm{F}}$. The amount of heat absorbed per mole during evaporation at the boiling point is represented by line " $d$ ". The reciprocal of the slope of line "a", $(d H / d T)$, is the heat required to change the temperature of one mole of substance (at constant pressure) by $1^{\circ} \mathrm{CF} .(\mathrm{dH} / \mathrm{dT})$ is the molar heat capacity of a material, referred to as " $\mathrm{C}_{\mathrm{p}}$ ". As the reciprocal of line "a" is $C_{p}$ (solid), the reciprocals of lines "c" and "e" are $C_{p}$ (liquid) and $C_{p}$ (vapor) respectively.
Fig. 2 H vs T Diagram for Pure $\mathrm{H}_{2} \mathrm{O}$. (Not to scale.)
From a thermodynamic standpoint, it is important to realize that fig. 2 illustrates the energy changes that occur in the system during heating. Actual quantitative measurements show that 5.98 kJ of heat are absorbed at the melting point (latent heat of fusion) and 40.5 kJ per mole (latent heat of evaporation) at the boiling point. The latent heats of fusion and evaporation are unique characteristics of all pure substances. Substances like Fe, Co, Ti and others, which are allotropic (exhibit different structures at different temperatures), also exhibit latent heats of transformation as they change from one solid state crystal modification to another.
ENERGY CHANGES
When heat is added from the surroundings to a material system (as described above), the energy of the system changes. Likewise, if work is done on the surroundings by the material system, its energy changes. The difference in energy $(\Delta \mathrm{E})$ that the system experiences must be the difference between the heat absorbed (Q) by the system and the work (W) done on the surroundings. The energy change may therefore be written as:
$\Delta \mathrm{E}=\mathrm{Q}-\mathrm{W} \tag{4}$
If heat is liberated by the system, the sign of $Q$ is negative and work done is positive. $Q$ and $\mathrm{W}$ depend on the direction of change, but $\Delta \mathrm{E}$ does not. The above relation is one way of representing the First Law of Thermodynamics which states that the energy of a system and its surroundings is always conserved while a change in energy of the system takes place. The energy change, $\Delta \mathrm{E}$, for a process is independent of the path taken in going from the initial to the final state.
In the laboratory most reactions and phase changes are studied at constant pressure. The work is then done solely by the pressure $(P)$, acting through the volume change, $\Delta \mathrm{V} .$
$\mathrm{W}=\mathrm{P} \Delta \mathrm{V} \quad and \quad \Delta \mathrm{P}=0 \tag{5}$
Hence:
$\mathrm{Q}=\Delta \mathrm{E}+\mathrm{P} \Delta \mathrm{V} \tag{6}$
Since the heat content of a system, or the enthalpy $\mathrm{H}$, is defined by.
$\mathrm{H}=\mathrm{E}+\mathrm{PV} \tag{7}$
$\Delta \mathrm{H}=\Delta \mathrm{E}+\mathrm{P} \Delta \mathrm{V} \tag{8}$
so that:
$\Delta \mathrm{H}=\mathrm{Q}-\mathrm{W}+\mathrm{P} \Delta \mathrm{V} \tag{9}$
or
$\Delta \mathrm{H}=\mathrm{Q} \tag{10}$
Reactions in which $\Delta \mathrm{H}$ is negative are called exothermic since they liberate heat, whereas endothermic reactions absorb heat. Fusion is an endothermic process, but the reverse reaction, crystallization, is an exothermic one.
ENTROPY AND FREE ENERGY
When a gas condenses to form a liquid and a liquid freezes to form a crystalline solid, the degree of internal order increases. Likewise, atomic vibrations decrease to zero when a perfect crystal is cooled to $0^{\circ} \mathrm{K}$. Since the term entropy, designated by $\mathrm{S}$, is considered a measure of the degree of disorder of a system, a perfect crystal at $0^{\circ} \mathrm{K}$ has zero entropy.
The product of the absolute temperature, $\mathrm{T}$, and the change in entropy, $\Delta \mathrm{S}$, is called the entropy factor, T $\Delta \mathrm{S}$. This product has the same units (Joules/mole) as the change in enthalpy, $\Delta \mathrm{H}$, of a system. At constant pressure, $\mathrm{P}$, the two energy changes are related to one another by the Gibbs free energy relation:
$\Delta \mathrm{F}=\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S} \tag{11}$
where
$\mathrm{F}=\mathrm{H}-\mathrm{TS} \tag{12}$
The natural tendency exhibited by all materials systems is to change from one of higher to one of lower free energy. Materials systems also tend to assume a state of greater disorder whereby the entropy factor $\mathrm{T} \Delta \mathrm{S}$ is increased. The free energy change, $\Delta \mathrm{F}$, expresses the balance between the two opposing tendencies, the change in heat content $(\Delta \mathrm{H})$ and the change in the entropy factor $(T \Delta S)$.
If a system at constant pressure is in an equilibrium state, such as ice and water at $0^{\circ} \mathrm{C}$, for example, at atmospheric pressure it cannot reach a lower energy state. At equilibrium in the ice-water system, the opposing tendencies, $\Delta \mathrm{H}$ and $\mathrm{T} \Delta \mathrm{S}$, equal one another so that $\Delta \mathrm{F}=0$. At the fusion temperature, $\mathrm{T}_{\mathrm{F}}$:
$\Delta \mathrm{S}_{\mathrm{F}}=\dfrac{\Delta \mathrm{H}_{\mathrm{F}}}{\mathrm{T}_{\mathrm{F}}} \tag{13}$
Similarly, at the boiling point:
$\Delta S_v=\frac{\Delta H_v}{T_v} \tag{14}$
Thus melting or evaporation only proceed if energy is supplied to the system from the surroundings.
The entropy of a pure substance at constant pressure increases with temperature according to the expression:
$\Delta S=\dfrac{C_p \Delta T}{T} \quad \text { (since }: \Delta H=C_p \Delta T \text { ) } \tag{15}$
where $\mathrm{C}_{\mathrm{p}}$ is the heat capacity at constant pressure, $\Delta \mathrm{C}_{\mathrm{p}}, \Delta \mathrm{H}, \mathrm{T}$ and $\Delta \mathrm{T}$ are all measurable quantities from which $\Delta \mathrm{S}$ and $\Delta \mathrm{F}$ can be calculated.
$\mathbf{F} \quad \mathbf{v s} \quad \mathbf{T}$
Any system can change spontaneously if the accompanying free energy change is negative. This may be shown graphically by making use of $F$ vs $T$ curves such as those shown in fig. 3.
Fig. 3 Free energy is a function of temperature for ice and water.
The general decrease in free energy of all the phases with increasing temperature is the result of the increasing dominance of the temperature-entropy term. The increasingly negative slope for phases which are stable at increasingly higher temperatures is the result of the greater entropy of these phases.
PHASE DIAGRAMS (TWO-COMPONENT SYSTEMS)
SOLID SolutionS
A solution can be defined as a homogeneous mixture in which the atoms or molecules of one substance are dispersed at random into another substance. If this definition is applied to solids, we have a solid solution. The term “solid solution” is used just as “liquid solution” is used because the solute and solvent atoms (applying the term solvent to the element in excess) are arranged at random. The properties and composition of a solid solution are, however, uniform as long as it is not examined at the atomic or molecular level.
Solid solutions in alloy systems may be of two kinds: substitutional and interstitial. A substitutional solid solution results when the solute atoms take up the positions of the solvent metal in the crystal lattice. Solid solubility is governed by the comparative size of the atoms of the two elements, their structure and the difference in electronegativity. If the atomic radii of a solvent and solute differ by more than 15% of the radius of the solvent, the range of solubility is very small. When the atomic radii of two elements are equal or differ by less than 15% in size and when they have the same number of valency electrons, substitution of one kind of atom for another may occur with no distortion or negligible distortion of the crystal lattice, resulting in a series of homogeneous solid solutions. For an unlimited solubility in the solid state, the radii of the two elements must not differ by more than 8% and both the solute and the solvent elements must have the same crystal structure.
In addition to the atomic size factor, the solid solution is also greatly affected by the electronegativity of elements and by the relative valency factor. The greater the difference between electronegativities, the greater is the tendency to form compounds and the smaller is the solid solubility. Regarding valency effect, a metal of lower valency is more likely to dissolve a metal of higher valency. Solubility usually increases with increasing temperature and decreases with decreasing temperature. This causes precipitation within a homogeneous solid solution phase, resulting in hardening effect of an alloy. When ionic solids are considered, the valency of ions is a very important factor.
CONSTRUCTION OF EQUILIBRIUM PHASE DIAGRAMS OF TWO-COMPONENT SYSTEMS
To construct an equilibrium phase diagram of a binary system, it is a necessary and sufficient condition that the boundaries of one-phase regions be known. In other words, the equilibrium diagram is a plot of solubility relations between components of the system. It shows the number and composition of phases present in any system under equilibrium conditions at any given temperature. Construction of the diagram is often based on solubility limits determined by thermal analysis – i.e., using cooling curves. Changes in volume, electrical conductivity, crystal structure and dimensions can also be used in constructing phase diagrams.
The solubility of two-component (or binary) systems can range from essential insolubility to complete solubility in both liquid and solid states, as mentioned above. Water and oil, for example, are substantially insoluble in each other while water and alcohol are completely intersoluble. Let us visualize an experiment on the water-ether system in which a series of mixtures of water and ether in various proportions is placed in test tubes. After shaking the test tubes vigorously and allowing the mixtures to settle, we find present in them only one phase of a few percent of ether in water or water in ether, whereas for fairly large percentages of either one in the other there are two phases. These two phases separate into layers, the upper layer being ether saturated with water and the lower layers being water saturated with ether. After sufficiently increasing the temperature, we find, regardless of the proportions of ether and water, that the two phases become one. If we plot solubility limit with temperature as ordinate and composition as abscissa, we have an isobaric [constant pressure (atmospheric in this case)] phase diagram, as shown in fig. 4. This system exhibits a solubility gap.
Fig. 4 Schematic representation of the solubilities of ether and water in each other.
COOLING CURVES
Fig. 5 Cooling curves: (a) pure compound; (b) binary solid solution; (c) binary eutectic system.
SOLID Solution EQUILIBRIUM DIAGRAMS
Read in Jastrzebski: First two paragraphs and Figure 3-4 in Chapter 3-8, "Solid Solutions Equilibrium Diagrams," pp. 91-92.
Read in Smith, C. O. The Science of Engineering Materials. 3rd ed. Englewood Cliffs, NJ: Prentice-Hall, 1986. ISBN: 9780137948840. Last two paragraphs and Figure 7-8 in Chapter 7-3-1, "Construction of a Simple Equilibrium Diagram," pp. 247-248.
Fig. 6 Plotting equilibrium diagrams from cooling curves for Cu-Ni solid solution alloys. (a) Cooling curves; (b) equilibrium diagram.
Fig. 7
INTERPRETATION OF PHASE DIAGRAMS
From the above discussion we can draw two useful conclusions which are the only rules necessary for interpreting equilibrium diagrams of binary systems.
Rule 1 - Phase composition: To determine the composition of phases which are stable at a given temperature we draw a horizontal line at the given temperature. The projections (upon the abscissa) of the intersections of the isothermal line with the liquidus and the solidus give the compositions of the liquid and solid, respectively, which coexist in equilibrium at that temperature. For example, draw a horizontal temperature line through temperature $T_e$ in fig. 7. The $T_e$ line intersects the solidus at $f$ and the liquidus at $g$, indicating solid composition of $f \%$ of $B$ and $(100-f) \%$ of $A$. The liquid composition at this temperature is $\mathrm{g} \%$ of $\mathrm{B}$ and $(100-\mathrm{g}) \%$ of $\mathrm{A}$. This line in a two-phase region is known as a tie line because it connects or "ties" together lines of one-fold saturation - i.e., the solid is saturated with respect to $B$ and the liquid is saturated with respect to $A$.
Rule 2 - The Lever Rule: To determine the relative amounts of the two phases, erect an ordinate at a point on the composition scale which gives the total or overall composition of the alloy. The intersection of this composition vertical and a given isothermal line is the fulcrum of a simple lever system. The relative lengths of the lever arms multiplied by the amounts of the phase present must balance. As an illustration, consider alloy I in fig. 7. The composition vertical is erected at alloy I with a composition of $e \%$ of $B$ and $(100-e)\%$ of $A$. This composition vertical intersects the temperature horizontal $\left(T_e\right)$ at point $e$. The length of the line " $f-e-g$ " indicates the total amount of the two phases present. The length of line "$e-g$" indicates the amount of solid. In other words:
\begin{aligned}
&\dfrac{\mathrm{eg}}{\mathrm{fg}} \times 100=\% \text { of solid present } \
&\dfrac{\mathrm{fg}}{\mathrm{fg}} \times 100=\% \text { of liquid present }
\end{aligned}
These two rules give both the composition and the relative quantity of each phase present in a two-phase region in any binary system in equilibrium regardless of physical form of the two phases. The two rules apply only to two-phase regions.
ISOMORPHOUS SYSTEMS
An isomorphous system is one in which there is complete intersolubility between the two components in the vapor, liquid and solid phases, as shown in fig. 9. The $\mathrm{Cu}-\mathrm{Ni}$ system is both a classical and a practical example since the monels, which enjoy extensive commercial use, are $\mathrm{Cu}-\mathrm{Ni}$ alloys. Many practical materials systems are isomorphous.
Fig. 9 Schematic phase diagram for a binary system, A-B, showing complete intersolubility (isomorphism) in all phases.
INCOMPLETE SOLUBILITY
Read in Smith, C. O. The Science of Engineering Materials. 3rd ed. Englewood
Cliffs, NJ: Prentice-Hall, 1986. ISBN: 9780137948840.
Chapter 7-3-5, "Incomplete Solubility," pp. 252-253.
EUTECTIC SYSTEMS
Read in Smith: Chapter 7-3-6, "Eutectic," pp. 253-256.
EQUILIBRIUM DIAGRAMS WITH INTERMEDIATE COMPOUNDS
Read in Jastrzebski, Z. D. The Nature and Properties of Engineering Materials. 2nd ed.
New York, NY: John Wiley & Sons, 1976. ISBN: 9780471440895.
Chapter 3-11,"Equilibrium Diagrams with Intermediate Compounds," pp. 102-103.
Fig. 15 Binary system showing an intermediate compound. C is the melting point (maximum) of the compound AB having the composition C’. E is the eutectic of solid A and solid AB. E’ is the eutectic of solid AB and solid B.
TWO-COMPONENT SYSTEMS
A pressure-temperature diagram is extremely useful when one wishes to discuss a one-component system. A component in this case is an element or compound (for example, Fe, MgO, or $\mathrm{H}_{2} \mathrm{O}$). As long as there is only one component present, the system may be adequately described by the two variables, pressure and temperature. In a two-component system, a third variable is needed; this is composition. The two-component system may be described by a three dimensional pressure - temperature composition diagram. It is however in the framework of this course not appropriate to deal with the complexities of the 3D phase diagram at this point and, since most materials are used under isobaric (constant pressure) conditions, we will first discuss an isobaric two-dimensional diagram of temperature versus composition. At a later point we will explain how the one component pressure-temperature and the isobaric two-component temperature-composition diagrams may be combined to construct the three-dimensional pressure-temperature-composition diagram of a two-component system.
Temperature-composition diagrams
A two-component temperature-composition diagram at constant pressure is called a binary phase diagram or equilibrium diagram. Temperature is plotted as the ordinate and composition as the abscissa. In the system $A B$, the composition is usually expressed in terms of the mole fraction of $B, x_B$, or the weight percent of $B$, w/o $B$. (The mole fraction of $B$ is equal to the number of moles of $B$ divided by the sum of the number of moles of $A$ plus the number of moles of $B$. The weight percent of $B$ is equal to the product of 100 and the weight of $B$ divided by the sum of the weights of $A$ and $B$.) The composition need not be specified in terms of $A$ since the sum of the mole fraction of $B$ and the mole fraction of $A$ is equal to one, and the sum of the weight percent of $A$ and the weight percent of $B$ is equal to 100. For pure $A$, both $x_B$ and w/o $B$ are zero; and for pure $B$ they are one and 100, respectively.
A typical binary phase diagram (Figure 1) indicates those phases present in equilibrium at any particular temperature and composition at the constant pressure for which the phase diagram was determined. At low temperatures, the only phase present is the solid designated by $S$ in the phase diagram. Pure $A$ melts at the temperature $T_{f A}$ and pure $B$ melts at the temperature $T_{fB}$ Alloys of composition between pure $A\left(x_B=0\right)$ and pure $B\left(x_B=1\right)$ exist only in the solid state of aggregation until the temperature of the solidus line is reached. The solidus is represented in the phase diagram by the lower curve extending from $\mathrm{T}_{\mathrm{fA}}$ to $\mathrm{T}_{\mathrm{fB}}$. Below the solidus temperature only solid exists; but above the solidus, there is a two-phase region $(L+S)$ in which both liquid and solid phases are in equilibrium. The liquid-plus-solid region extends from the solidus temperature over a finite temperature interval for all alloys of the system $A B$. The extent of the temperature interval reduces to zero for the pure components $A$ and $B$. since according to the one-component pressure-temperature diagram, at any particular pressure there is only one temperature at which liquid and solid are in equilibrium.
The upper boundary of the liquid-plus-solid region is the liquidus temperature. Above this temperature, any alloy of the system exists as a liquid phase until the temperature increases to that level at which vaporization begins. The pure compositions A and B vaporize completely at $T_{v A}$ and $T_{v B}$, because the liquid and vapor phases for a pure component at constant pressure can be in equilibrium at only one temperature. Alloys of intermediate composition do not completely vaporize at any one temperature but over a finite temperature interval.This is shown in the phase diagram of Figure 1 by the liquid-plus-vapor $(L+V)$ region. At temperatures above this region, only the vapor phase exists.
Metallurgists or materials scientists are more often concerned with the liquid and solid phases and less often with the vapor phase. For this reason, and the fact that it is difficult to determine vapor phase diagrams, most binary diagrams extend no higher in temperature than the all-liquid region.
Cooling through a two-phase field
Solidification or melting of an intermediate alloy may be illustrated by the enlarged view of a two-phase field shown in Figure 2.
If an alloy of composition $\mathrm{x}_{\mathrm{o}}$, originally liquid at temperature $\mathrm{T}_1$, is cooled, it remains liquid until the temperature reaches the liquidus temperature at point 2. At this temperature $T_2$, the first particle of solid appears. This solid does not have the same composition as the parent liquid. The composition of the first solid formed at $\mathrm{T}_2$ is found by constructing an isothermal line from 2 to $2^{\prime}$. This isotherm is called a tie-line. The composition of the solid and liquid phases in equilibrium at a particular temperature is given by the intersection of the tie-line at that temperature with the solidus and liquidus curves, respectively. Thus at temperature $T_2$, the solid has the composition $\times 2^{\prime}$ and the liquid the composition $x_2$ (where $x_2=x_0$ ).
Upon further cooling to $T_3$, the composition of the liquid has shifted to the left along the liquidus to $\times 3$, and the composition of the solid has shifted to the left along the solidus to $x 3^{\prime}$ because at $T_3$, the only liquid and solid compositions that can be in equilibrium with one another are $x 3$ and $x 3^{\prime}$ respectively.
Despite the fact that both liquid and solid have compositions different from the alloy composition, the overall alloy (liquid and solid together) retains its original composition $x_0$. As the temperature approaches the solidus temperature $\mathrm{T}_5$, the solid composition approaches $x^{\prime}$, (where $x_5{ }^{\prime}=x_0$ ) and the last quantity of liquid of composition $x_5$ freezes. At temperatures below the solidus, the solid composition remains unchanged, for example, at $\mathrm{T}_6$ the solid composition is $\mathrm{x}_0$.
The above analysis applies only to solidification at such a slow rate of cooling that equilibrium is achieved at every temperature. This is not always the case, however.
lever law
To determine exactly how much liquid and solid are present at any given temperature, let us derive a relation known as the "lever law." If one mole of alloy of composition $x_0$ is chosen as a basis, then, at any temperature $T$. the fraction of liquid $f_L$ is found as follows: The sum of the fractions of liquid and solid must equal one, that is,
$f_S+f_L=1$
The number of moles of B present in the alloy must be the same as the number of moles of B present in the solid phase plus the number of moles of B present in the liquid phase, that is,
$x_O=x_S f_S=x L f_L$
Since
$f_S=1-f_L$
we may substitute and obtain
$x_O=x_S-x_S f L+x_L f_L$
On rearranging and solving for $\mathrm{f}_\mathrm{L}$ we find that
$f_L=\left(x_S-x_0\right) /\left(x_S-x L\right) .$
This relation is called the "lever law" because the fraction of one of the two phases present is equal to the opposite side of the tie-line $\left(\mathrm{x}_{\mathrm{S}}-\mathrm{x}_{\mathrm{O}}\right)$ divided by the entire tie-line $\left(x_S-x_L\right)$ In this way the tie-line acts as though it were a lever whose fulcrum is at the point $x_0$. In a similar way it is found that
$f_s=\left(x_O-x_L\right) /\left(x_S-x L\right) .$ | textbooks/chem/Inorganic_Chemistry/Introduction_to_Solid_State_Chemistry/01%3A_Lectures/1.10%3A_Phase_Equilibria_and_Phase_Diagrams.txt |
Problem #1
(a) Cerium has many isotopes (8 to be exact), but only ${}^{140}\mathrm{Ce}$ and ${}^{142}\mathrm{Ce}$ are present in substantial amounts. Which isotope of cerium is the most abundant?
Answer
from the periodic table, you see that the atomic mass of $\mathrm{Ce}$ is 140.115, which must be the weighted sum of the isotope masses here we assume that it is necessary to consider only ${ }^{140} \mathrm{Ce}$ and ${}^{142} \mathrm{Ce}$ so, $(x$ mass of ${ }^{140} \mathrm{Ce}$ $)+(1-x)$ mass of ${}^{142} \mathrm{Ce}$ $=140.115$ for the purposes of this decision, we can approximate the atomic masses of the isotopes as $\sim 140$ for ${ }^{140}$ and $\sim 142$ for ${}^{142} \mathrm{Ce}$ and solve for $\mathrm{x}$ $\mathrm{x}=0.94$ which means that ${ }^{140} \mathrm{Ce}$ is the most abundant isotope of cerium
Production of chromium in an electric arc furnace would involve the reaction of carbon with chromium sesquioxide according to the following reaction:
$\mathrm{Cr}_2 \mathrm{O}_3+a \mathrm{C}=b \mathrm{CO}+c \mathrm{Cr}$
(i) Balance the equation, i.e., specify the values of $a, b$, and $c$. Insert the correct values below.
Answer
$\mathrm{Cr}_2 \mathrm{O}_3+3 \mathrm{C}=3 \mathrm{CO}+2 \mathrm{Cr}$
(ii) Calculate the minimum amount of chromium (in $\mathrm{kg}$ ) produced if the reaction consumed $333 \mathrm{~kg} \mathrm{C}$ and produced the stoichiometric amount of Cr. Assume $100 \%$ efficiency.
Answer
$333 \mathrm{~kg} \mathrm{C}=333000 / 12.011=27725$ moles $\mathrm{C}$ the stoichiometric amount of $\mathrm{Cr}$ is $2 / 3$
amount of carbon on a molar basis. Therefore, amount of $\mathrm{Cr}=27725 \times 2 / 3$ moles of $\mathrm{Cr} =(27725 \times 2 / 3) \times 51.996=961 \mathrm{~kg} \mathrm{Cr}$
Problem #2
(a) Antimony has two isotopes, ${ }^{121} \mathrm{Sb}$ and ${ }^{123} \mathrm{Sb}$. Which isotope has the higher natural abundance?
Answer
from the periodic table, you see that the atomic mass of $\mathrm{Sb}$ is $121.757$, which must be the weighted sum of the isotope masses
so, $\left(x\right.$ mass of $\left.{ }^{121} \mathrm{Sb}\right)+(1-\mathrm{x})$ mass of ${ }^{123} \mathrm{Sb}=121.757$
for the purposes of this decision, we can approximate the atomic masses of the isotopes as $\sim 121$ for ${ }^{121} \mathrm{Sb}$ and $\sim 123$ for ${ }^{123} \mathrm{Sb}$ and solve for $\mathrm{x}$
$\mathrm{x}=0.62$ which means that ${ }^{121} \mathrm{Sb}$ is the more abundant isotope of antimony
(b) Production of hafnium by the Kroll Process would involve the reaction of magnesium with hafnium tetrachloride according to the following reaction:
$\mathrm{HfCl}_4+a \mathrm{Mg}=b \mathrm{MgCl}_2+c \mathrm{Hf}$
(i) Balance the equation, i.e., specify the values of $a, b$, and $c$. Insert the correct values below.
Answer
$\mathrm{HfCl}_4+2 \mathrm{Mg}^2 \quad 2 \mathrm{MgCl}_2+\mathrm{Hf}$
(ii) Calculate the minimum amount of magnesium (in $\mathrm{kg}$ ) needed to convert $111 \mathrm{~kg} \mathrm{HfCl}_4$ into elemental hafnium.
Answer
$111 \mathrm{~kg} \mathrm{HfCl}_4 = 111000/[178.49 + (4 \times 35.45)] = 347 \text{ moles } \mathrm{HfCl}_4$
the stoichiometric amount of $\mathrm{Mg}$ is twice the amount of $\mathrm{HfCl}_4$ on a molar basis
$\therefore$ amount of $\mathrm{Mg}=347 \times 2$ moles of $\mathrm{Mg}=(347 \times 2) \times 24.305=16.9 \mathrm{~kg} \mathrm{Mg}$
Problem #3
(a) Show by means of a calculation that blue light of wavelength, $\lambda=444 \mathrm{~nm}$, is not capable of exciting electrons in $\mathrm{Li}^{2+}(\mathrm{g})$ from the state $n=2$ to $n=4$.
Answer
let's equate the energy required to excite electrons in $\mathrm{Li}^{2+}(\mathrm{g})$ from the state $n=2$ to $n=4$ with the minimum energy needed from an incident photon to cause the excitation
\begin{aligned}
\frac{h c}{\lambda} &=K Z^2\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right) \
\therefore \lambda &=\frac{h c}{K Z^2\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)}=\frac{6.6 \times 10^{-34} \times 3.00 \times 10^8}{2.18 \times 10^{-18} \times 3^2\left(\frac{1}{2^2}-\frac{1}{4^2}\right)} \
&=5.38 \times 10^{-8} \mathrm{~m}=53.8 \mathrm{~nm}<444 \mathrm{~nm}
\end{aligned}
$\therefore$ since $\mathrm{E}$ scales with $1 / \lambda$, blue light of wavelength $\lambda=444 \mathrm{~nm}$ does not have enough energy per photon to cause the excitation
(b) Is the value of the energy of transition from the state $n=2$ to $n=4$ in $\mathrm{Li}^{2+}, \Delta E_{2 \rightarrow 4}$, greater than or less than the value of the energy of transition from the state $n=1$ to $n=2$ in $\mathrm{Li}^{2+}$, $\Delta E_{1 \rightarrow 2}$? Explain with the use of an energy level diagram. There is no need to calculate the values of the two quantities.
Answer
Problem #4
(a) In a gas discharge tube what is the minimum frequency $(v)$ of a photon capable of ionizing ground-state electrons in $\mathrm{Li}^{2+}$?
Answer
here is the central concept: the energy of the incident photon must be at least as great as the ionization energy (I.E.)
$\mathrm{Li}^{2+}$ is a one-electron atom, so we can calculate the I.E. using the Bohr Model
$I . E .=E_{\infty}-E_1=0-\left(-\dfrac{K Z^2}{n^2}\right)$, where $Z=3$ and $n=1$ for the ground-state of $\mathrm{Li}^{2+}$ and $\mathrm{K}$ is the ground-state energy of atomic hydrogen
energy of incident photon is given by $\mathrm{E}=\mathrm{h} v$
$v=\dfrac{K Z^2}{h} \dfrac{\left(2.18 \times 10^{-18} J\right)(3)^2}{6.6 \times 10^{-34}}=2.97 \times 10^{16} \mathrm{~Hz}$
(b) Explain with reference to the relevant physical forces why the value of the $1^{\text {st }}$ ionization energy of $\mathrm{Li}$ is less than the $3^{\text {rd }}$ ionization energy of $\mathrm{Li}$.
Answer
the $1^{\text {st }}$ ionization represents the removal of one of the 3 electrons from neutral $\mathrm{Li}$
the $3^{\text {rd }}$ ionization represents the removal of the single electron from the $\mathrm{Li}^{2+}$ ion
in the second case, the single electron alone feels the pull of the positive charge of the nucleus
in the first case the same positive charge is felt by three electrons; hence, each electron feels a weaker pull than is the case with a lone electron under the influence of the same positive charge | textbooks/chem/Inorganic_Chemistry/Introduction_to_Solid_State_Chemistry/02%3A_Assessments/2.01%3A_Self-Assessment-_Structure_of_the_Atom__Answer.txt |
Problem #1
(a) Draw the energy level diagram that shows that the linear combination of atomic orbitals from two atoms of oxygen $(\mathrm{O})$ results in the formation of the stable molecule, $\mathrm{O}_2{ }^{2-}$. The molecular orbitals in $\mathrm{O}_2{ }^{2-}$ increase in energy according to the sequence $\sigma_{2 s}, \sigma_{2 s} *, \sigma_{2 p_z}, \pi_{2 p_{x, y}}, \pi_{2 p_{x, y}}^*, \sigma_{2 p_z} *$.
Answer
(b) Indium phosphide $(\operatorname{InP})$ is a semiconductor with a band gap, $E_{\mathrm{g}}$, of $1.27 \mathrm{eV}$. Calculate the value of the absorption edge of this material. Express your answer in meters.
Answer
for absorption of incoming radiation, the following must be true:
$E_{\text {radiation }}=E_g \text {. }$
using the Planck relationship gives the wavelength of the absorption edge
\begin{aligned}
&E_{\text {radiation }}=\frac{h c}{\lambda} \
&\therefore \lambda=\frac{h c}{E_g}=\frac{6.6 \times 10^{-34} \times 3 \times 10^8}{1.27 \times 1.6 \times 10^{-19}}=9.74 \times 10^{-7} \mathrm{~m} .
\end{aligned}
Problem #2
Chemical analysis of a silicon (Si) crystal reveals boron (B) at a level of 0.0003 atomic percent.
(a) Assuming that the concentration of thermally excited charge carriers from the $\mathrm{Si}$ matrix is negligible, calculate the density of free charge carriers (carriers $/ \mathrm{cm}^3$ ) in this $\mathrm{Si}$ crystal.
Answer
each B atom will attract an electron and thus create a "mobile hole"; we only have to determine the number of $\mathrm{B}$ atoms $/ \mathrm{cm}^3$ of $\mathrm{Si}$. The atomic volume of the host crystal ($\mathrm{Si}$) is given on your $\mathrm{PT}$ as $12.05 \mathrm{~cm}^3 / \mathrm{mole}$.
$\# \mathrm{Si}$ atoms $/ \mathrm{cm}^3=\frac{6.02 \times 10^{23} \text { atoms }}{1 \text { mole }} \times \frac{1 \text { mole }}{12.05 \mathrm{~cm}^3}=5.00 \times 10^{22}$ atoms $/ \mathrm{cm}^3$
$\therefore \# \mathrm{~B} \text{ atoms} / \mathrm{cm}^3=5.00 \times 10^{22} \times 0.0003 \times 10^{-2}=1.50 \times 10^{17} \mathrm{~B} / \mathrm{cm}^3$
thus, the number of free charge carriers ("holes") is $1.50 \times 10^{17} / \mathrm{cm}^3$; they are created through the acquisition of one electron by each $\mathrm{B}$ atom from the valence band of the host $\mathrm{Si}$ crystal.
(b) Draw a schematic energy band diagram for this material and label the valence band, conduction band, band gap, and the energy level associated with the B impurity.
Answer
2.03: Self-Assessment- Electronic Materials Answer
Problem 1
Indium phosphide (InP) is a semiconductor with a band gap, $\mathrm{E}_g$, of $1.27 \mathrm{eV}$. Calculate the value of the absorption edge of this material. Express your answer in meters.
Answer
For absorption of incoming radiation, the following must be true:
$\mathrm{E}_{\text {radiation }}=\mathrm{E}_{\mathrm{g}} \nonumber$
Using the Planck relationship gives the wavelength of the absorption edge:
\begin{aligned}
&\mathrm{E}_{\text {radiation }}=\frac{\mathrm{hc}}{\lambda} \
&\lambda=\frac{\mathrm{hc}}{\mathrm{E}_{\mathrm{g}}}=\frac{6.6 \times 10^{-34} \times 3 \times 10^8}{1.27 \times 1.6 \times 10^{-19}}=9.74 \times 10^{-7} \mathrm{~m}
\end{aligned}
Problem 2
Chemical analysis of a germanium ($\mathrm{Ge}$) crystal reveals antimony ( $\mathrm{Sb}$ ) at a level of $0.0002$ atomic percent.
a. Assuming that the concentration of thermally excited charge carriers from the Ge matrix is negligible, calculate the density of free charge carriers (carriers $/ \mathrm{cm}^3$ ) in this $\mathrm{Ge}$ crystal.
Answer
Each $\mathrm{Sb}$ atom will donate an electron to the conduction band; we have only to determine the number of $\mathrm{Sb}$ atoms $/ \mathrm{cm}^3$ of $\mathrm{Ge}$. The atomic volume of the host crystal ($\mathrm{Ge}$) is given on your PT as $13.57 \mathrm{~cm}^3 / \mathrm{mole}$.
$\# \mathrm{Ge}$ atoms $/ \mathrm{cm}^3=\frac{6.02 \times 10^{23} \text { atoms }}{1 \text { mole }} \times \frac{1 \text { mole }}{13.57 \mathrm{~cm}^3}=4.44 \times 10^{22}$ atoms $/ \mathrm{cm}^3$
$\# \mathrm{Sb}$ atoms $/ \mathrm{cm}^3=4.44 \times 10^{22} \times 0.0002 \times 10^{-2}=8.87 \times 10^{16} \mathrm{Sb} / \mathrm{cm}^3$
Thus, the number of free charge carriers is $8.87 \times 10^{16} / \mathrm{cm}^3$; they are created by the donation of one electron by each $\mathrm{Sb}$ atom to the conduction band of the host $\mathrm{Ge}$ crystal.
b. Draw a schematic energy band diagram for this material and label the valence band, conduction band, band gap, and the energy level associated with the $\mathrm{Sb}$ impurity.
Answer
2.04: Self-Assessment- Crystalline Materials Answer
Problem 1
Calculate the acceleration potential that will result in electron diffraction from the (311) plane of platinum ($\mathrm{Pt}$) at an angle, $\theta$, of $33.3^{\circ}$. The lattice constant of platinum, $a$, has a value of $3.92 \AA$.
Answer
\begin{aligned}
\lambda_e &=2 d_{(311)} \sin \theta=2 \dfrac{a}{\left(h^2+k^2+l^2\right)^{1 / 2}} \sin \theta \
&=2 \times \dfrac{3.92}{\sqrt{11}} \times \sin 33.3^{\circ}=1.30 \AA \
q \bar{V} &=\dfrac{1}{2} m v^2 \therefore v=\sqrt{2 e\bar{v} / m} \
\lambda_e &=\dfrac{h}{m v}=\dfrac{h}{(2 m e \bar{v})^{-1 / 2}} \
\therefore V &=\dfrac{h^2}{2 \lambda^2 m e} \
&=\dfrac{\left(6.6 \times 10^{-34}\right)^2}{2\left(130 \times 10^{-10}\right)^2 \times 9.11 \times 10^{-31} \times 1.6 \times 10^{-19}} \
&=88.4 \bar{v}
\end{aligned} | textbooks/chem/Inorganic_Chemistry/Introduction_to_Solid_State_Chemistry/02%3A_Assessments/2.02%3A_Self-Assessment-_Bonding_and_Molecules__Answer.txt |
(a) A melt of magnesium borate glass with the composition $20 \% \mathrm{MgO}-80 \% \mathrm{~B}_2 \mathrm{O}_3$ is cooled at a rate of $r_1$. The glass transition temperature, $T_g$, is measured to be $T_1$. In order to raise the value of $T_g$ to $T_2>T_1$ while keeping the cooling rate equal to $r_1$, how should the $\mathrm{MgO}$ content of the glass be changed? Explain with reference to atomic structure.
Answer
Decrease the $\mathrm{MgO}$ concentration to less than $20 \%$. The addition of $\mathrm{MgO}$ to $\mathrm{B}_2 \mathrm{O}_3$ modifies the melt by promoting chain scission of the borate network. The resulting decrease in viscosity of the melt confers greater mobility than would be the case at comparable undercooling at a lower $\mathrm{MgO}$ content and therefore reduces the excess volume of the solidified glass. We know from the relationship between volume and temperature that the value of $T_g$ is given by the break in the $V$ vs. $T$ trace, and furthermore that $T_g$ tracks with $V^{\mathrm{xs}}$. To raise $T_g$ is tantamount to raising $\mathrm{V}^{\mathrm{xs}}$ which then argues for changing the composition of the melt in a way that raises its viscosity, i.e., decreasing the amount of modifier.
(b) On a plot of molar volume, $\mathrm{V}_{\mathrm{m}}$, versus temperature, $\mathrm{T}$, sketch cooling curves for a borate melt that solidifies to form (1) crystalline $\mathrm{B}_2 \mathrm{O}_3$; and (2) amorphous $\mathrm{B}_2 \mathrm{O}_3$. Indicate which material was cooled more quickly. No calculation necessary. Label the melting point of (1) and the glass transition temperature of (2). Indicate the excess molar volume, $\mathrm{V}^{\mathrm{xs}}$, and describe why it is a measure of atomic disorder.
Answer
The glassy solid was cooled more quickly.
For a given compound, the crystal represents the tightest packing of atoms in order to establish the highest number of bonds and thereby achieve maximum decrease in energy. In the liquid state, the interatomic spacing is large in comparison to that of the solid state. Glass formation is the result of the inability of the atoms to reach the proper positions they would occupy in the crystal lattice owing to inadequate time to do assume said positions during cooling. In effect, the system is quenched en route from the loose packing of the melt to the tight packing of the crystal. The molar volume compares, on the basis of a constant mass of material (1 mole!) the degree to which the atomic packing of the solid reaches that of the crystal. The greater the deviation from this tight packing, the greater the deviation from crystallinity, and hence the greater the degree of disorder.
2.06: Self-Assessment- Aqueous Answer
(a) The value of $K_{\mathrm{a}}$ for perchloric acid, $\mathrm{HClO}_4(a q)$, is $1 \times 10^8$. Calculate the $p \mathrm{H}$ and the $p \mathrm{OH}$ of $1.11 \mathrm{~M} \mathrm{~HClO}_4(a q)$ in water.
Answer
with a value of $K_a=10^8, \mathrm{HClO}_4$ is a strong acid $\Rightarrow$ complete dissociation
\begin{aligned}
&\therefore 1.11 \mathrm{M} \mathrm{HI}(\mathrm{aq}) \Rightarrow 1.11 \mathrm{M}=\left[\mathrm{H}^{+}\right]=\left[\mathrm{ClO}_4^{-}\right] \
&\therefore \mathrm{pH}=-\log _{10}\left[\mathrm{H}^{+}\right]=-\log _{10} 1.11=-0.0453 \
&\because \mathrm{pOH}+\mathrm{pH}=14 \Rightarrow \mathrm{pOH}=14.0453
\end{aligned}
(b) The compound, yttrium iodate, $\mathrm{Y}\left(\mathrm{IO}_3\right)_3$, upon dissolution in water dissociates into $\mathrm{Y}^{3+}$ and solubility product, $K_{\mathrm{sp}}$, of $\mathrm{Y}\left(\mathrm{IO}_3\right)_3$.
Answer
$\mathrm{Y}\left(\mathrm{IO}_3\right)_3=\mathrm{Y}^{3+}+3 \mathrm{IO}_3^{-}$, from which we get $K_{\mathrm{sp}}=\left[\mathrm{Y}^{3+}\right]\left[\mathrm{IO}_3^{-}\right]^3$
$\mathrm{c}_{\mathrm{s}}=2.22 \times 10^{-3}=\left[\mathrm{Y}^{3+}\right]=1 / 3\left[\mathrm{IO}_3^{-}\right]=\left[\mathrm{IO}_3^{-}\right]=3\left[\mathrm{Y}^{3+}\right]$
$\therefore K_{\mathrm{sp}}=\mathrm{c}_{\mathrm{s}}\left(3 \mathrm{c}_{\mathrm{s}}\right)^3=27 \mathrm{c}_{\mathrm{s}}{ }^4=27\left(2.22 \times 10^{-3}\right)^4=6.56 \times 10^{-10}$
2.07: Self-Assessment- Solid Solutions
(a) Construct the phase diagram $(\mathrm{T}, \mathrm{c})$ for $\mathrm{Ag}-\mathrm{Cu}$ given the following data. (Assume all phase lines to be straight.)
$\begin{array}{ll}\mathrm{T}_{\mathrm{M}} \mathrm{Ag}: & 960^{\circ} \mathrm{C} \ \mathrm{T}_{\mathrm{M}} \mathrm{Cu}: & 1080^{\circ} \mathrm{C} \ \mathrm{T}_{\mathrm{E}} \text { (Eutectic) } & 780^{\circ} \mathrm{C}: \alpha[9 \mathrm{wt} . \% \mathrm{Cu}] ; \beta[92 \mathrm{wt} . \% \mathrm{Cu}] ; \text { Eutectic comp. } 28 \mathrm{wt} . \% \mathrm{Cu} \ & 400^{\circ} \mathrm{C}: \alpha[1 \mathrm{wt} . \% \mathrm{Cu}] ; \beta[100 \mathrm{wt} . \% \mathrm{Cu}]\end{array}$
Answer
(b) Determine the liquidus temperature for a $60 \mathrm{wt} . \% \mathrm{Ag}-40 \mathrm{wt} . \%$ Cu alloy.
Answer
From the phase diagram in (a) liquidus $\mathrm{T}$ for $40 \mathrm{wt} . \% \mathrm{Cu}$ alloy is $\approx 840^{\circ} \mathrm{C}$
(c) Determine which other Ag-Cu alloy composition has the same liquidus temperature as the one determined in (b).
Answer
From phase diagram in Prob. 3, other composition with same liquidus $\mathrm{T}$ is $\approx 20 \text{wt}.\% \mathrm{Cu}$
(d) $26 \mathrm{~g}$ of sterling silver ($92.5 \text{wt}.\% \mathrm{Ag}-7.5 \mathrm{wt} . \% \mathrm{Cu})$ are melted together with $376 \mathrm{~g}$ of pure copper $(\mathrm{Cu})$. Given the phase diagram for $\mathrm{Ag}-\mathrm{Cu}$, determine:
(i) the liquidus temperature for the alloy formed;
(ii) the solidus temperature for this alloy;
(iii) the composition of the alloy formed.
Answer
(i) From phase diagram in part (a), liquidus $\mathrm{T} \approx 1060^{\circ} \mathrm{C}$
(ii) From phase diagram in part (a), solidus $\mathrm{T} \approx 870^{\circ} \mathrm{C}$
(iii) $26 \mathrm{~g}$ of Sterling Silver has $(26)(0.925)=24.05 \mathrm{~g} \mathrm{Ag}$ and $(26)(0.075)= 1.95 \mathrm{~g} \mathrm{Cu}$. Total $\mathrm{Cu}=1.95+376=378 \mathrm{~g}$.
$\text { wt. } \% \mathrm{Cu}=\frac{378 \mathrm{~g} \mathrm{Cu}}{24 \mathrm{~g} \mathrm{Ag}+378 \mathrm{~g} \mathrm{Cu}}=94 \nonumber$
2.08: Self-Assessment- Reactions and Kinetics Answer
1. Urbium (Ur) is an upscale element found in big cities. Its oxide $\left(\mathrm{UrO}_2\right)$ is not very stable and decomposes readily at temperatures exceeding $666^{\circ} \mathrm{C}$. The figure below shows how the rate of reaction varies with the concentration of $\mathrm{UrO}_2$ at $777^{\circ} \mathrm{C}$. The rate, $r$, is in units of $\mathrm{M} / \mathrm{s}$ and the concentration of $\mathrm{UrO}_2, c$, is in units of $\mathrm{M}(\mathrm{mole} / \mathrm{L})$. The slope has a value of $1.77$ and the intercept has a value of $1.46$.
(a) What is the order of reaction?
Answer
The order is the slope: 1.77
(b) Calculate the value of the rate constant. Pay strict attention to the units.
Answer
$r=k c^n \rightarrow \log r=\log k+n \log c ; \text{when } c=1, r=k=10^{1.46}=28.8$
$\text{Units of } k=r / c^{\mathrm{n}}=(\mathrm{M} / \mathrm{s}) /\left(\mathrm{M}^{1.77}\right)=\mathrm{M}^{-0.77} / \mathrm{s} \rightarrow k=28.8 \mathrm{M}^{-0.77} / \mathrm{s}$
(c) On the graph above, draw the line showing how the rate of reaction varies with the concentration of $\mathrm{UrO}_2$ at $888^{\circ} \mathrm{C}$. No calculation necessary. Pay attention to relative values and slopes.
Answer
The upper line on the graph represents the isotherm at $888^{\circ} \mathrm{C}$. Note same slope as $777^{\circ} \mathrm{C}$ but greater value of $r$-intercept.
2. Show by a calculation that the diffusion length of boron (B) in germanium (Ge) is less than $1.0 \mu \mathrm{m}$ at a temperature of $1200 \mathrm{~K}$ for a diffusion time of 30 minutes. The diffusion coefficient of B in Ge at $1200 \mathrm{~K}, D_{\mathrm{B}}$, has the value of $2.0 \times 10^{-17} \mathrm{~m}^2 / \mathrm{s}$.
Answer
The diffusion length is approximated by the relationship $x=\sqrt{D t}$ or $x=2 \sqrt{D t}$
$\therefore \sqrt{D t}=\sqrt{2.0 \times 10^{-17} \frac{\mathrm{m}^2}{\mathrm{~s}} \cdot 30 \mathrm{~min} \cdot 60 \frac{\mathrm{s}}{\mathrm{min}}}=1.90 \times 10^{-7} \mathrm{~m}<1.0 \mu \mathrm{m}$ | textbooks/chem/Inorganic_Chemistry/Introduction_to_Solid_State_Chemistry/02%3A_Assessments/2.05%3A_Self-Assessment-_Amorphous_Materials__Answer.txt |
1) Consider the combustion of a candle $\left(\mathrm{C}_{25} \mathrm{H}_{52}\right)$ with oxygen $\left(\mathrm{O}_2\right)$ to form carbon dioxide $\left(\mathrm{CO}_2\right)$ and water $\left(\mathrm{H}_2 \mathrm{O}\right)$. A typical candle has $100 \mathrm{~g}$ of $\mathrm{C}_{25} \mathrm{H}_{52}$. Please answer the following questions:
a) (2 pts) Write the balanced reaction for the combustion of a candle (2 points).
b) (2 pts) You are in a closed room with a mole of $\mathrm{O}_2$ molecules. If you light 5 candles, what will be the limiting reagent and how much excess (in grams) of either the $\mathrm{O}_2$ or $\mathrm{C}_{25} \mathrm{H}_{52}$ will remain?
2) Your student ID is made of a plastic called polyvinyl chloride. The molecular unit in this material is $\mathrm{C}_2 \mathrm{H}_3 \mathrm{Cl}$, and for this question you can assume the card is made of only this molecule, with a density of $1.4 \mathrm{grams} / \mathrm{cm}^3$. Please answer the following questions:
a) (1 pt) Use your $3.091$ ruler to determine the mass of your student ID card in grams (assume a thickness of $2 \mathrm{~mm}$ and that the card is perfectly rectangular).
b) (2 pts) How many moles of $\mathrm{C}_2 \mathrm{H}_3 \mathrm{Cl}$ are in the ID card?
c) $\left(2\right.$ pts) There are only 2 stable isotopes of chlorine, ${ }^{35} \mathrm{Cl}$ and ${ }^{37} \mathrm{Cl}$. What is their relative abundance?
d) (1 pt) You take all the chlorine out of your card and us it to disinfect a 100,000-gallon swimming pool which requires $1 \mathrm{~kg}$ of $\mathrm{Cl}$. How many ID cards do you need to disinfect the pool?
3.02: Quiz 2
1) Below is the energy level diagram showing the transitions made by an electron in a hydrogen atom according to the Bohr model.
a) Identify the highest energy absorption indicated among the four electron transitions indicated above and calculate the energy of the photon required to cause the transition in eV. (2 pts)
b) Will a transition starting from n=3 to n=2 emit a photon in the visible light range (390nm-700nm)? Can this photon ionize an electron in the n=3 level? ( 2pts)
c) Use your spectroscope to look at the lights above you. What is the minimum number of energy levels in a Bohr atom needed to produce these lines? (1 pt)
2) When humans are exposed to sunlight, Ultraviolet-B (UVB) light of wavelength $\sim \mathbf{2 9 5} \mathbf{n m}$ reacts in our skin to make vitamin $D\left(C_{28} \mathrm{H}_{44} \mathrm{O}\right)$. The recommended daily dose of vitamin $D$ for adults is $\mathbf{0.1 \mathrm{mg}}$ which amounts to $1.52 * 10^{17}$ molecules.
a) It takes one photon to create one vitamin D molecule. How many Joules of UVB energy does our body absorb per day to create the necessary amount of vitamin $D$? ( 3 pts)
b) About $2.5 \mathrm{~mW}\left(2.5 \times 10^{-3} \mathrm{~J} / \mathrm{s}\right)$ of vitamin $D$ - producing UVB light strike our exposed skin when we're outside (with a t-shirt and shorts on).
If $5 \%$ of incident UVB light is utilized by our skin, how long should people spend outside to ensure they get the required dose of vitamin $D$? (2 pts)
3.03: Quiz 3
1) A scientist acquired the following photoelectron spectrum from a pure elemental sample.
a. What is the element? (1 pts)
b. Write the electronic configuration of the element in noble gas notation. (1 pts)
c. Write the electronic configuration of this element in box notation (you only need to show the electrons for the shells marked with stars). (2 pts)
d. This element is ionized to its +1 ion. What are the principal ($n$) and angular momentum ($l$) quantum numbers of the electron that’s lost? (2 pts)
2) The solution provided contains potassium chloride ($\mathrm{KCl}$) dissolved in water: use it to answer the following questions.
a. What is the conductivity of the solution? Assume the baseline conductivity of the water is negligible. (1pt)
b. Is the solid ionic or covalent? Why? (1pt)
c. Does magnesium oxide ($\mathrm{MgO}$) have a higher or lower lattice energy than potassium chloride? Justify your answer. (2 pts)
d. One mole of potassium chloride ($\mathrm{KCl}$) and one mole of nickel(II) chloride ($\mathrm{NiCl}_2$) are dissolved in equal volumes of water. Which would have a higher conductivity? (1 pt)
3.04: Quiz 4
1) Carbonic acid, $\mathrm{H}_2 \mathrm{CO}_3$, is a product of $\mathrm{CO}_2$ and $\mathrm{H}_2 \mathrm{O}$ and has contributed to ocean acidification. Please answer the following questions.
a) Draw the Lewis dot structure, with correct molecular geometry, that carbonic acid forms. (1 pt)
b) What geometry does the carbon in carbonic acid adopt according to VSEPR? (1 pt)
c) Will all the lone pairs lie in the same plane when the molecule is in the lowest energy state? (1 pt)
d) Where is the largest angle in the molecular structure? Please give your answer as $X-Y-Z$, where $Y$ is the atom the angle is formed (i.e., the vertex). For example, $\mathrm{H}-\mathrm{O}-\mathrm{H}$ or $\mathrm{O}-\mathrm{Fe}-\mathrm{O}$. (1 pt)
e) Look at the 3D molecular model you built from last Wednesday's Goodie Bag #4: VSEPR for the molecule $\mathrm{H}_3\mathrm{Si} - \mathrm{SF}_2 - \mathrm{CN}$. What is the smallest angle in your structure? (1 pt)
2)
a) Draw two possible resonance structures for $\mathrm{CH}_3 \mathrm{NCO}$. (4 pts)
b) Which of your two resonance structure contributes more to the resonance hybrid? Explain in one sentence. (1 pt)
VSEPR Geometries by Boundless Chemistry. License: CC BY-SA. This content is excluded from our Creative Commons license. For information, see https://ocw.mit.edu/fairuse. | textbooks/chem/Inorganic_Chemistry/Introduction_to_Solid_State_Chemistry/03%3A_Quizzes/3.01%3A_Quiz_1.txt |
1. Consider the molecule $\mathrm{OBr}^{-}$.
a. Fill in the MO diagram for the molecule using arrow notation. Label each atomic orbital side with the correct atom. (2 pts)
b. What is the bond order? (1 pt)
2. Allene is a compound with formula $\mathrm{C}_3 \mathrm{H}_4$ with the following Lewis dot structure:
a. Give the hybridization for all the carbons in the allene. (1 pt)
b. Label the individual bonds as sigma or pi bonds. (1 pt)
c. List the intermolecular interactions that could happen between only an allene and an allyl alcohol molecule. Allyl has the following structure: (1 pt)
d. Rank the following in order of ascending boiling point: pure allene, a mixture of allene and allyl alcohol, and pure allyl alcohol. (1 pt)
3. Given the visible spectrum below, answer the following questions.
a. If you shine a violet light onto a red small LED, how much heat energy will a promoted electron in the red small LED dissipate? (2 pts)
b. Rank the amount of heat energy dissipated in the mystery LED, a red LED, and a blue LED if you were to shine a violet light onto them. (Hint: use your GB to help LEaD you to the answer ) (1 pt)
3.06: Quiz 6
1. For p-type and n-type $\mathrm{Si}$... (2 points)
a. What is a possible dopant atom?
p-type: _______________ n-type: _______________
b. What are the charge carriers?
p-type: _______________ n-type: _______________
c. How many bonds does the dopant atom form once it has donated its charge carriers?
p-type: _______________ n-type: _______________
d. Label the structures below as p-type and n-type $\mathrm{Si}$. Label conduction and valence bands, donor level, and acceptor level. (2 points)
e. For $100 \mathrm{~g}$ of $\mathrm{Si}$, calculate the mass of As needed in order to have $3.091 * 10^{17}$ carriers $/ \mathrm{cm}^3$. (2 points)
2. Lattice structures (You may want to use your pre-built FCC structure :) )
a. The radius of a nickel atom is $r=1.97 \AA$ What is the volume packing fraction of the FCC unit cell? (1 point)
b. What is the direction of closest packing? (1 point)
c. Consider one face of your FCC lattice. How many nearest neighbors does the central atom have in the same plane? (1 point)
d. Under sufficient pressure, some elemental metals transitions from BCC to another cubic structure. What is the cubic structure that it transitions to? (1 point)
3.07: Quiz 7
1. You shoot energetic beams of electrons at molybdenum and copper samples and receive the emitted x-ray spectra shown below.
a. The samples were irradiated with beams with different amounts of energy. Draw an arrow to the curve that was irradiated with higher energy. (1 point)
b. Unfortunately, you forgot to label which spectrum came from which metal. Please fill in the boxes above with the correct metal. (1 point)
c. On the x-ray spectra above, draw in the characteristic peak corresponding to the copper L$\alpha$ energy transition. (1 point)
2.
a. What planes would produce the first three peaks you would see for a BCC crystal structure? What about for FCC? Give the planes in the form of (hkl). (2 points)
b. Express $\frac{\sin ^2 \theta_n}{\sin ^2 \theta_1}$ in terms of Miller indices $h_1, k_1, l_1$ and $h_n, k_n$, and $I_n$ where $\theta_n$ is the angle of the $n$th peak; $h_1, k_1$, and $I_1$ are the Miller plane indices of the first peak; and $h_n, k_n$, and $I_n$ are the miller plane indices of the nth peak. Hint: take the ratio of two Bragg's law equations. (2 points)
You are given an XRD plot. Let’s try to find the lattice parameter of the mystery pure metal. (We do not have polonium in the lab.)
c. Fill out the table below in which the peaks are at $2 \theta=20^{\circ}, 28.4^{\circ}$, and $35.04^{\circ}$. (1 points)
Peak 2$\theta$ $\dfrac{\sin ^2 \theta_n}{\sin ^2 \theta_1}$
1 $20^{\circ}$
2 $28.4^{\circ}$
3 $35.04^{\circ}$
d. Using the relationship derived in part b, what crystal structure is your mystery pure metal? (1 points)
e. Calculate the lattice parameter if the $x$-ray source has a wavelength of $1.7 \AA$. $\left(10^{10} \AA=1 \mathrm{~m}\right)$. (1 points) | textbooks/chem/Inorganic_Chemistry/Introduction_to_Solid_State_Chemistry/03%3A_Quizzes/3.05%3A_Quiz_5.txt |
1) A pteropod shell (made of $\mathrm{CaCO}_3$ ) weighs $0.01 \mathrm{mg}$. By the year 2100 the ocean will be $126 \%$ more acidic than pre-industrial levels if we continue on our current path. This will lead to the dissolution of the shell: $\mathrm{CaCO}_3(\mathrm{~s}) \rightarrow \mathrm{Ca}^{2+}(\mathrm{aq})+\mathrm{CO}_3{ }^{2-}$ (aq)
You measure this reaction by weighing the shell vs. time every 1 hour and you record your results:
Experiment $[\mathrm{CaCo}_3](\mathrm{~mg})$
1 ($t=0$ hr) 0.01
2 ($t=1$ hr) 0.008
3 ($t=2$ hrs) 0.006
4 ($t=3$ hrs) 0.004
a) If the shell fully dissolves in 0.1 liters of water, what is the molar concentration of the resulting $\mathrm{CO}_3{ }^{2-}$? (1.5 points)
b) What is the order of the reaction? (1.5 points)
c) What is the rate of the reaction (expressed in M/s)? (1.5 points)
d) What is the rate constant for the reaction and what units does it have? (1.5 points)
2) You find a sheet of glass in the Glass Lab and want to increase its resistance to fracture, so you can throw baseballs at it. This glass is soda glass, meaning $\mathrm{Na}_2 \mathrm{O}$ was added in the melt.
a) A schematic of a $2 \mathrm{D}$ cross-section of amorphous $\mathrm{SiO}_2$ is shown below in one box. Draw how adding $\mathrm{Na}_2 \mathrm{O}$ changes this structure.
b) After adding $\mathrm{Na}_2 \mathrm{O}$, does the glass transition temperature increase, decrease, or stay the same?
c) In order to strengthen the glass, you decide to use an ion exchange method. What is one ion that you could exchange in, and which ion leaves the glass during this process?
d) Describe how ion exchange increases the toughness of the glass, with reference to molar volume and the stress field.
3.09: Quiz 9
1a) Circle the monomer (the smallest repeat unit) in the following two polymers (2 points):
Polyethylene glycol
Polystyrene
1b) What is the primary intermolecular force in polystyrene (1 point)?
1c) Which polymer would you expect to contain more crystalline regions? Explain in one sentence (1 point).
1d) Which polymer would you expect to be soluble in water? Explain in one sentence (1 point).
1e) One of these polymers is made via a radical process and the other is made via a condensation process. Which is made through a radical process (1 point)?
2. In your goodie bag, you mixed Elmer’s glue and Borax. The relevant chemical components are shown below:
a) Draw the Borax component on the following strands of Elmer’s glue (2 point).
b) The below graph contains the elastic regions of two stress-strain curves, corresponding to regular Elmer’s glue and Elmer’s glue treated with Borax. Clearly label the two graphs as treated or untreated (2 points).
3.10: Quiz 10
You want to know how quickly your MIT acceptance balloon will run out of helium gas ($\mathrm{He}$) inside of it. Since you tied the balloon so tightly, you neglect the pinhole at the neck and model the balloon as a closed sphere of latex.
Note: Assume that helium concentration inside the balloon remains constant.
1. The balloon is $0.1 \mathrm{~mm}$ thick. The concentration of He gas inside the balloon is $0.04462 \mathrm{M}$ and outside the balloon is $2.32 \times 10^{-7} \mathrm{M}$. If the diffusivity constant of helium through latex is $7 \times 10^{-9} \mathrm{~m}^2 / \mathrm{s}$ at room temperature, what is the flux of helium gas through the latex balloon (Hint: $1 L=0.001 \mathrm{~m}^3$ )?
2. You’re shocked to discover that the helium gas is diffusing through the balloon so quickly! In order to preserve the cherished memories of your MIT undergrad, you decide to slow the diffusion of helium through the balloon.
a. What concentration of helium gas would you have to have outside the balloon to halve the flux?
b. At what temperature would you have to store the balloon to halve the flux, compared to question $1\left(E_a=0.09 \mathrm{eV}=1.44 \cdot 10^{-20} \mathrm{~J}\right)$?
c. Your beloved balloon is exposed to the Boston winter! It lowers the temperature of the balloon below the glass transition temperature of the latex. How does this affect the diffusion of helium through the balloon?
d. Your friend cross-linked the latex in her balloon before filling it with helium. How would this impact the diffusion of helium through the balloon?
e. Name another way that you could decrease the flux of helium through the balloon. | textbooks/chem/Inorganic_Chemistry/Introduction_to_Solid_State_Chemistry/03%3A_Quizzes/3.08%3A_Quiz_8.5.txt |
Problem 1 (17 POINTS)
a) A Lithium ion $\left(\mathrm{Li}^{2+}\right)$ laser is emitting light of wavelength $450 \mathrm{~nm}$ due to electrons relaxing to energy level $n=4$ from some unknown energy level. What is the unknown energy level? [4 points]
b) Against all lab safety protocol (do. not. do), you grab a different laser and observe the light through your spectrometer. You observe 10 spectral lines. This means that there are at least how many possible energy levels? [2 points]
c) You purchase two new lasers that each emit one wavelength only: one emits $200 \mathrm{~nm}$ light, and the other emits $900 \mathrm{~nm}$ light. However, you mix up the lasers and don't know which is which. You shine each of these lasers onto a slab of metal. One laser ionizes electrons from the metal but the other doesn't. Which laser doesn't ionize the metal? Give one sentence explaining why. [2 points]
d) You begin to increase the frequency of the laser you named in part c). If you continue doing this, will you eventually observe electrons being emitted? Give one sentence explaining why or why not. [2 points]
e) You begin to increase the intensity of the laser you named in part c). If you continue doing this, will you eventually observe electrons being emitted? Give one sentence explaining why or why not. [2 points]
f) You find an article that says that electrons in your metal slab will start to be ionized at $500 \mathrm{~nm}$. If you switch to the laser from part c) that DOES ionize the metal (i.e. the one you didn’t give in your answer to c)) and shine it on said metal, what will be the velocity of the emitted electrons? [5 points]
Problem 2 (24 POINTS)
NOTE: For each of the following questions, assume the atom or ion is in the ground state.
a) Write the ground state electron configuration for the following atoms and ions. [6 points]
• $\mathrm{O}^{2-}$
• $\mathrm{Si}$
b) For the following atoms and ions, draw a box-and-arrow diagram to represent the last completely or partially filled subshell, following Aufbau’s rules for filling. You may use noble gas abbreviations for filled inner shells. [6 points]
• $\mathrm{O}^{2-}$
• $\mathrm{Si}$
c) Paramagnetic materials are attracted by a magnetic field, but do not retain a magnetization when the external field is removed. Paramagnetism is caused by unpaired electrons in a material's atoms. Based on this information, which atom and/or ion in part $\mathrm{A}$ are paramagnetic, if any? [2 points]
d) Write the complete set of quantum numbers (the values for $n, l, m_l, m_s$) for each of the valence electrons in $\mathrm{Si}$. [8 points]
e) How many of the electrons in $\mathrm{Si}$ have the same set of quantum numbers ( $n, 1$, $\left.\mathrm{m}_{\mathrm{l}}, \mathrm{m}_{\mathrm{s}}\right)$? [2 points]
Problem 3 (19 POINTS)
Below is the PES spectrum for the outer electrons of element $\mathrm{X}$.
a) Label the axes and the direction of increasing ionization energy on this graph. The graph is plotted so that the highest peak corresponds to the electrons that are ionized first during the experiment. [3 points].
b) Element X has the same inner electron configuration as neon. Identify element $\mathrm{X}$ and label the peaks by subshell, including the number of electrons in each peak.
Note: only the peaks corresponding to valence electrons are shown in the graph. [6 points].
c) Consider an ion with the same electronic configuration as $\mathrm{X}$ but one more proton. On the graph above, draw the new PES spectrum for this ion. [4 points].
d) You were taking PES spectrum data for carbon in a vacuum. However, the seal on the vacuum machine broke and let in an unknown gas. The spectrum for carbon is still present, but you need to identify the unknown peaks. You observe the following on the contaminated spectrum:
i. There are 6 peaks total (including peaks from the carbon) on the spectrum
ii. Five peaks are of the same height
iii. One peak is three times higher than the others
What is the contaminating element? Explain your reasoning. [6 pts]
Problem 4 (17 POINTS)
a) You are trying to synthesize chocolate in the lab. The first step is to produce theobromine $\left(\mathrm{C}_7 \mathrm{H}_8 \mathrm{~N}_4 \mathrm{O}_2\right)$, a component of cacao. The reaction in the cacao plant creates theobromine and $\mathrm{O}_2$ as byproducts. You have $\mathrm{N}_2, \mathrm{SO}_2, \mathrm{H}_2 \mathrm{O}, \mathrm{CO}_2$ available in the lab as potential reactants. Select your reactants and write the balanced reaction. [4 points]
b) Now you need sugar to make chocolate. Your reaction produced 2 moles of theobromine, and you have 5 moles of glucose $\left(\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6\right)$ available in your lab. The recipe for chocolate calls for a $3: 1$ ratio of theobromine to glucose by volume. Is theobromine or glucose limiting, and how many $\mathrm{cm}^3$ of the chocolate mixture can you produce? [10 points]
Density of theobromine: $1.52 \mathrm{~g} / \mathrm{cm}^3$ at room temperature
Density of glucose: $1.54 \mathrm{~g} / \mathrm{cm}^3$ at room temperature
c) Chocolate by itself it fine, but you conclude that the ultimate form of chocolate is chocolate-covered strawberries. It takes $0.5 \mathrm{~cm}^3$ of melted chocolate to cover an average strawberry. Now using the chocolate you made in part b, how many moles of strawberries could you cover in chocolate? Assume that the density change with temperature change is negligible. [3 points]
Problem 5 (23 POINTS)
a) Two structures with the formula $\mathrm{CH}_2 \mathrm{~N}_2$ are drawn below. Draw the missing lone pairs (where applicable) and write the formal charge on each element. [8 points]
b) Cyanamide is made industrially from calcium carbide $\left(\mathrm{CaC}_2\right)$, which in turn is made from calcium oxide $(\mathrm{CaO})$. Write the charges on each element for calcium chloride $\left(\mathrm{CaCl}_2\right)$, sodium chloride $(\mathrm{NaCl})$, calcium carbide $(\mathrm{CaO})$, and magnesium carbide $(\mathrm{MgO})$. [6 points]
$\begin{array}{lll}\text { NaCl } & \text { CaO } & \text { MgO } \ \text { Charge on Na: } & \text { Charge on Ca: } & \text { Charge on Mg: } \ \text { Charge on Cl: } & \text { Charge on O: } & \text { Charge on O: }\end{array}$
c) Rank the compounds in terms of decreasing lattice energy. Explain in one sentence. [5 points]
d) Write the Lewis structure for dichloromethane ($\mathrm{CH}_2 \mathrm{Cl}_2$). [4 pts] | textbooks/chem/Inorganic_Chemistry/Introduction_to_Solid_State_Chemistry/04%3A_Exams/4.01%3A_Exam_1.txt |
Problem 1 (17 POINTS)
a. Use the fact that $\mathrm{SeOF}_2$ is trigonal pyramidal to write the Lewis Dot Structure, including lone pairs, for $\mathrm{SeOF}_2$. (3 points)
b. Below is a drawing of two different amino acids, Methionine and Lysine, without any lone pairs drawn.
Name the VSEPR geometry around the following central atoms:
i.The sulfur in methionine (2 points)
ii.The circled carbon in lysine. (2 points)
iii.Which of these angles is smaller? Explain why or why not in ONE sentence. (2 points)
$\mathrm{CH}_2-\mathrm{S}-\mathrm{CH}_2 \quad \mathrm{OR} \quad \mathrm{CH}_2-\mathrm{CH}_2-\mathrm{CH}_2 \nonumber$
c. Below is one structure for the molecule $\left[\mathrm{CH}_6 \mathrm{~N}_3\right]^{+}$
i. There are three other resonant structures for this molecule; what are they? (6 points)
ii. The structure that we gave you for $\left[\mathrm{CH}_6 \mathrm{~N}_3\right]^{+}$ contributes the least to the overall bonding of the molecule (it is the least stable). Why? (2 points)
Problem 2 (23 POINTS)
$\mathrm{NaCl}$ is a compound that forms a rock salt structure (unit cell below). The radius of the $\mathrm{Na}^{+}$ atoms is $1.16 \AA$; the radius of the $\mathrm{Cl}^{-}$ atoms is $1.67 \AA\left(1 \AA=10^{-10} \mathrm{~m}\right)$.
a. Name the Bravais lattice (lattice type) of $\mathrm{NaCl}$. What is the basis? (2 points)
b. How many $\mathrm{Na}^{+}$ ions are in the unit cell? How many $\mathrm{Cl}^{-}$ ions? (4 points)
c. Determine the close-packed direction of this structure and use it to calculate the lattice parameter, a. ( 3 points)
d. Draw (211), [0$\bar{2}$1], [301], and (1$\bar{1}$0) in the boxes below. Make sure to label which is which. (8 points)
e. Which of the planes in part d is closer in distance to its equivalent plane in a neighboring cell? Show how you arrived at this answer. (4 points)
f. Using atomic packing factor, explain why calcium is denser than potassium. (2 points)
Problem 3 (18 POINTS)
a. Rank the following intermolecular forces in terms of their average relative strength: London dispersion, hydrogen bonding, dipole-dipole, and dipole-induced-dipole. (4 points)
1. ___________________ (strongest)
2. ___________________
3. ___________________
4. ___________________ (weakest)
You have the four following molecules in liquid form in your lab cabinet.
b. Which has the highest boiling point? (2 points)
c. You make a mixture of two of the liquids.
i. If you wanted your mixture to have the LOWEST possible boiling point, which two of the four molecules would you choose? (4 points)
ii. If you want your mixture to have the HIGHEST possible boiling point, which two of the four molecules would you choose? (4 points)
d. You have two samples of polyethylene, a polymer made of chains of carbon and hydrogen, as shown below. For one sample, each molecule chain is 50 units long and in the next sample, each chain is 500 units long. Which sample of plastic will be stronger? Why? (4 points)
Problem 4 (21 POINTS)
a. The molecular orbitals of a second-row diatomic molecule are shown in this figure. Label them each as $\sigma, \sigma^*, \pi$, or $\pi^*$ by filling in the boxes provided. (The $2 \mathrm{p}$ and $2 \mathrm{~s}$ atomic orbitals are labelled for reference). (6 points)
b. Circle which of the following this skeleton MO diagram could correspond to. (2 points)
$\mathrm{C} - \mathrm{C}$ $\mathrm{N} - \mathrm{O}$ $\mathrm{H} - \mathrm{F}$
c. Rank the following molecules in terms of stability: $\mathrm{Li}_2, \mathrm{Be}_2 \mathrm{~B}_2, \mathrm{C}_2$ Be sure to show your work (no need to show MO diagrams). (8 points)
d. Acrylonitrile is a feedstock chemical used for the manufacture of “nitrile” polymers. Pictured below is its chemical structure. What is the hybridization of the circled atoms? (3 points)
d. Give the total number of $\sigma$ bonds and total number of $\pi$ bonds in acrylonitrile. ( 2 points )
Problem 5 (21 POINTS)
a. UV light corresponds to wavelengths from $10-400 \mathrm{~nm}$. What is the range of band gaps corresponding to UV light? (3 points)
b. You only have two materials in the lab to build an LED: GaAs (band gap $1.42 \mathrm{eV}$ at $300 \mathrm{~K}$) and $\mathrm{GaN}$ (band gap $3.2 \mathrm{eV}$ at $300 \mathrm{~K}$ ). Which would you choose as a UV emitter? (3 points)
c. UV light can damage eyes, so you decide that for safety, your LED should emit in the visible range too. You recall that you can tune the band gap by alloying your material. Determine the fraction $\mathrm{x}$ of GaAs in your alloy such that $\mathrm{GaA} \mathrm{s}_{\mathrm{x}} \mathrm{N}_{1-\mathrm{x}}$ will emit red light (wavelength $700 \mathrm{~nm}$ ). Remember that the band gap of an alloy is the weighted average of the band gaps of its components. (3 points)
d. If you place the $\mathrm{GaAs}_{\mathrm{x}} \mathrm{N}_{1-\mathrm{x}}$ adjacent to your answer from (B) your LED will emit both UV light and red light. However, you observe that it is really dim: you decide to dope it with $\mathrm{Mg}$.
i. Is $\mathrm{Mg}$ an $\mathrm{n}$ - or $\mathrm{p}$-type dopant in $\mathrm{GaAs}_{\mathrm{x}} \mathrm{N}_{1-\mathrm{x}}$? (2 points)
ii. If you dope $1 \mathrm{~g} \mathrm{GaN}$ with $\mathrm{Mg}$ such that $1 \mathrm{Mg}$ atom replaces 1 in every $10^6 \mathrm{Ga}$ atoms, how many extra carriers are produced? What kind of carriers are these? (4 points)
e. In order to get enough current to $\mathrm{GaAs}_{\mathrm{x}} \mathrm{N}_{1-\mathrm{x}}$ and your answer from (B) processing requires that the two are stacked. Which should be on top to ensure no light is blocked? (3 points)
f. Do you expect a larger temperature-dependence of conductivity in the doped or the undoped $\mathrm{GaAsN}$, at moderate temperatures? (3 points) | textbooks/chem/Inorganic_Chemistry/Introduction_to_Solid_State_Chemistry/04%3A_Exams/4.02%3A_Exam_2.txt |
Problem 1 (23 POINTS)
a) In the 2 crystalline solids shown below, there are circled SIX different point defects TOTAL. Label each defect with its name. (6 points)
You decide to buy a ring from a jewelry store down on Mass Ave. Drawing on your deep knowledge of materials science from 3.091, you decide that a platinum ring is definitely the way to go. Unfortunately, the jeweler graduated from That School Down the Street; you are upset by their claim that the ring is “defect-free crystalline platinum.” Let’s prove them wrong!
b) First, you weigh the ring: it’s exactly 2 grams. How many platinum atoms are in the ring? (2 points)
c) How many atomic sites are in the ring, $\mathrm{N}_{\mathrm{Pt}}$? (2 points)
d) It's a little chilly in the shop: $11^{\circ} \mathrm{C}(284 \mathrm{~K})$. Assuming the ring is in thermal equilibrium, calculate the total number of vacancies, $\mathrm{N}_{\mathrm{v}}$, that you would expect at the current temperature. The constant, $A$, has a value of 1.11. Assume the vacancy formation energy for $\mathrm{Pt}$ is $0.96 \mathrm{eV}$. (4 points)
e) The jeweler is stressed out by your frantic calculations and eager to prove they’re not cheating you; they hastily point out that the atoms that came from those vacancies are probably sitting in interstitial sites in the lattice. You quickly pull up a paper that has measured interstitial concentration in $\mathrm{Pt}$.
Note: the units of interstitial defect concentration are mol/L.
i) Estimate the formation energy (in J) of an interstitial defect from the plot. How does this compare to the vacancy formation energy in $\mathrm{Pt}$, which was given above as $0.96 \mathrm{eV}$? Hint: assume that interstitial defect concentration follows an Arrhenius relationship. (5 points)
ii) Could the jeweler be right, based on the activation energies for interstitial formation and vacancy formation? In other words, could the atoms from the vacant sites have become interstitial defects? Explain your reasoning in one sentence. (4 points)
Problem 2 (20 POINTS)
a) Sketch the following on the axes provided:
i) A cooling curve (molar volume vs. temperature) for amorphous $\mathrm{SiO}_2$ (3 points)
ii) A cooling curve (molar volume vs. temperature) for crystalline $\mathrm{SiO}_2$ (3 points)
Please label curves $\mathrm{i}$ and $\mathrm{ii}$.
b) On your plot above, label:
i) the melting point(s) (2 points)
ii) the glass transition temperature(s) (2 points)
c) List TWO ways you could change the glass transition temperature. (4 points)
d) Tempered glass exhibits increased strength compared to regular glass. Consider a slice taken across a sheet of tempered glass:
With reference to your plot in a), explain how the molar volume changes across the slice. (4 points)
e) Which of the following stress-strain curves could correspond to the tempered glass above? Circle the letter corresponding to your chosen curve. (2 points)
Problem 3 (20 POINTS)
In GB8, we explored the impact of ocean acidification on pteropods by considering the dissolution of $\mathrm{CaCO}_3$ in a citric acid solution. Now, we'll consider a similar process, but using hydrofluoric acid instead. Hydrofluoric acid, $\mathrm{HF}$, has a higher acid dissociation constant than citric acid.
a) Write out the acid dissociation reaction for hydrofluoric acid. Label the conjugate acid/base pairs. (5 points)
b) What is the concentration (M) of a solution containing $0.05 \mathrm{~g}$ of $\mathrm{HF}$ in $0.1 \mathrm{~L} \mathrm{H}_2 \mathrm{O}$? (3 points)
c) If the acid dissociation constant for hydrofluoric acid is $\mathrm{K}_{\mathrm{a}}=7.2 \times 10^{-4}$, what is the $\mathrm{pH}$ of the solution from part b)? (4 points)
e) If we had given you hydrofluoric acid $\left(\mathrm{K}_{\mathrm{a}}=7.2 \times 10^{-4}\right)$ rather than citric acid $\left(\mathrm{K}_{\mathrm{a}}=1.8 \times 10^{-5}\right)$ in the goodie bag, would you expect the shells to have dissolved faster or slower? Briefly explain why. (4 points)
f) Explain what would happen to the $\mathrm{pH}$ of the hydrofluoric acid solution if it were to sit in a $\mathrm{CO}_2$ environment for a few hours, and why. (4 points)
Problem 4 (21 POINTS)
To compensate for some of the oxygen consumed by burning the candle on one of your many Friday night dates, you decide to run a reaction that produces $\mathrm{O}_2$ gas as fast as possible. You recall two options:
$(p) 2 \mathrm{NO}_2 \rightarrow 2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_2(\mathrm{~g}) \quad \mathrm{OR} \quad (q) \mathrm{O}_3+\mathrm{Cl} \rightarrow \mathrm{ClO}(\mathrm{g})+\mathrm{O}_2(\mathrm{~g})$
You are able to find some information about the reaction $(p)$ : its rate law is
$\text { rate }=\mathrm{k}\left[\mathrm{NO}_2\right]$
(Note: reactions $p$ and $q$ have the same numerical value for their reaction constants)
Trial #
[$\mathrm{O}_3$]
$\mathrm{M}$
[$\mathrm{Cl}$]
$\mathrm{M}$
rate
$\mathrm{M}/\mathrm{s}$
1 0.07 0.25 10
2 0.14 0.25 20
3 0.07 0.5 20
a) Use the experimental data in the table above about reaction ($q$) to determine its:
i. Rate law (5 points)
ii. Overall order (2 points)
iii. Value of $\mathrm{k}$, the reaction constant (5 points)
b) Plot the linear version of the integrated rate law that describes $2 \mathrm{NO}_2 \rightarrow 2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_2(\mathrm{~g})$. Label the axes, $y$-intercept, and slope. (5 points)
c) If you have 2 mols of $\mathrm{NO}_2, 1 \mathrm{~mol} \mathrm{of} \mathrm{Cl}$, and $1 \mathrm{~mol}$ of $\mathrm{O}_3$ available (but you can only run one reaction), which reaction would you choose to produce oxygen the fastest? Justify your answer with ONE sentence. (4 points)
Problem 5 (16 POINTS)
Your research supervisor wants you to characterize two pure elemental powders she found in the back of the lab. Unfortunately, time on the XRD machine is very limited; to save time, you mix BOTH POWDERS together and perform XRD on the mixture.
You get the following plot of intensity (counts) vs. $2 \theta$ (degrees):
From the plot, you have the following data to help identify the samples. You may find it helpful to add columns to assist your thinking (ungraded).
$2 \theta$ (degrees) $(\sin \theta)^2$
25.6 0.049
29.6 0.065
30.7 0.070
35.6 0.093
42.4 0.131
51.3 0.187
a) Use selection rules to determine the crystal structures of the samples (8 points) Note: Assume sample 1 corresponds to the element with the lowest 2-theta peak.
Sample 1: ___________________________ Sample 2: ___________________________
b) What are the lattice constants of the samples? Copper k-alpha radiation (wavelength=1.54 Angstroms) was used. Show your thinking! (8 points)
Sample 1: ___________________________ Sample 2: ___________________________ | textbooks/chem/Inorganic_Chemistry/Introduction_to_Solid_State_Chemistry/04%3A_Exams/4.03%3A_Exam_3.txt |
Problem 1 (21 POINTS)
a) i. Fill in atomic and Molecular orbitals for the $\mathrm{MO}$ diagram of Carbon Monoxide ($\mathrm{CO}$) below (3 pts):
ii. Label the molecular orbitals as sigma or pi (2 pts)
iii. Label the molecular orbitals as bonding or antibonding (2 pts)
iv. What is the bond order of $\mathrm{CO}$? (2 pts)
b) Methylamine $\left(\mathrm{H}_3 \mathrm{C}-\mathrm{NH}_2\right)$ is an organic compound that's used to synthesize lots of chemicals, ranging from photographic developers to cold medicine.
i. Draw a methylamine molecule, showing its 3D VSEPR geometry. Include any lone pairs. ( 3 pts)
ii. What is the shape predicted by VSEPR of the atoms around (2 pts) ...
$\mathrm{C}$:_____________________ $\mathrm{N}$:____________________
iii. Consider the two bond angles in methylamine below. Which is larger? (1 pt)
1) $\mathrm{H}-\mathrm{C}-\mathrm{N}$
2) $\mathrm{C}-\mathrm{N}-\mathrm{H}$
$\overline{\text { Smaller }} \quad \quad \quad \overline{\text { Larger }}$
c) If we add an $\mathrm{H}^{+}$ ion to methylamine, we get methylammonium $\left(\mathrm{CH}_3 \mathrm{NH}_3^{+}\right)$, an organic ion often used in perovskite solar cells.
i. Draw a methylammonium ion using its VSEPR geometry. Include any lone pairs. (2 pt)
ii. Label the formal charges on each atom on your drawing above (2 pts)
iii. Does methylammonium have any stable resonance structures? If so, draw one. If not, why not? (2 pts)
Problem 2 (17 POINTS)
You are a new scientist at a fancy spacesuit company. To test your polymer knowledge, your boss has asked you to design a material for their newest spacesuit. Unfortunately, in spite of the critical nature of the spacesuit, the chemistry lab seems to be horribly underfunded, and you have only two polymers to choose from.
a) Polymer 1 is made from the monomer tetrafluoroethylene $\left(\mathrm{C}_2 \mathrm{~F}_4\right)$. Draw the Lewis Structure for a single tetrafluoroethylene molecule. What is the hybridization of one of the carbon atoms? (4 pts)
b) Draw the repeat unit for a polymer made from polymerization of this monomer. What polymerization technique should you use? (3 pts)
c) Polymer 2 is a polymer made from the following monomers. Draw the corresponding polymer. What polymerization technique should you use? (5 pts)
d) The space suits must deflect water, i.e. the polymer shouldn't interact with $\mathrm{H}_2 \mathrm{O}$. Which polymer should you choose? (Polymer 1 or 2 ) Give a one-sentence explanation making sure to include intermolecular forces. ( $3 \mathrm{pts})$
e) Your boss is worried that your space suit will tear too easily. How could you modify your polymer to improve its durability? (2 pts)
Problem 3 (21 POINTS)
3. Iron ($\mathrm{Fe}$), which is an important component of many parts on the Apollo 3.091, forms a BCC structure at room temperature.
a) Draw the (110) plane of $\mathrm{Fe}$ on the graph provided below. (3 pts)
Modified image of astronaut working. Original image is in the public domain (source: NASA).
b) What is the close-packed direction of $\mathrm{Fe}$? Indicate where this is in relation to the (110) plane drawn above by DRAWING it on the graph provided in part (a) and LABEL the direction with the correct crystallographic nomenclature, $<h k l>$. (5 pts)
c) It is difficult to get 100% pure iron because there are often some impurity elements in the material. One common impurity element is hydrogen, which is present as an interstitial element in iron.
CIRCLE the image below that accurately shows an interstitial hydrogen atom in the (110) plane of iron. (3 pts)
d) Iron has a radius of $1.24 \AA$. Calculate the density of $\mathrm{Fe}$ in $\mathrm{g} / \mathrm{cm}^3$? ( 1 atomic mass unit $= 1.66 \times 10^{-24}$ grams, $\left.1 \AA=10^{-8} \mathrm{~cm}\right)(6 \mathrm{pts})$
You decide to run a tensile test on an $\mathrm{Fe}$ wire. After the experiment, you look at the part of the sample that has deformed and see the following:
e) In ONE sentence, explain what is special about the shaded planes in the diagram above. (4 pts)
Problem 4 (20 POINTS)
We’ve talked a lot about semiconductors, but we haven’t mentioned one interesting class of semiconductors - organic semiconductors. Professor Grossman will be using organic LEDs for horticultural lighting on the spaceship, so it’s important to understand them!
LEDs can be used to grow plants in space!
a) One example of an organic semiconductor is polyethyne ($\mathrm{PE}$). A portion of its structure is shown below. Circle the repeating unit. (2 pts)
b) Polyethyne is a semiconductor because its resonance structures allow electrons to move. One is structure shown above; draw one repeating unit of another polyethyne resonance structure. (3pts)
c) To enhance conductivity to useful levels, polyethyne needs to be doped — but the dopants behave differently from inorganic semiconductors. When doping with bromine, two electrons are removed from the polymer per bromine molecule:
$\mathrm{PE}+\mathrm{Br}_2 \rightarrow(\mathrm{PE})^{2+}+2 \mathrm{Br}^{-} \nonumber$
In this case, is bromine acting as a $\mathrm{p}$-type dopant or $\mathrm{n}$-type? Briefly explain your answer. (Conduction occurs through the polymer chains; the bromine ions do not contribute significantly, and for this question you can assume that they are present only to keep total charge neutral) $(3 \mathrm{pts})$
d) The band diagram for neutral polyethyne is shown below. Sketch the band diagram for $\mathrm{Br}_2-$ doped polyethyne on the same axes. (4 pts)
e) Certain organic semiconductors follow the same trend of temperature vs. conductivity as inorganic semiconductors. A schematic is shown below.
i. What causes the near-zero slope of temperature dependence in the intermediate temperature regime? (4 pts)
ii. Theoretical studies have shown that polyethyne could exhibit near-zero temperature dependence from 100 $\mathrm{K}$ up to about 450 $\mathrm{K}$; however, the glass transition temperature of your PE is near 400 $\mathrm{K}$. What is one thing you could do to increase the glass transition temperature of your PE semiconductor so that it can be used over a wider temperature range? (4 pts)
Problem 5 (21 POINTS)
An alien sword of a PURE METAL has fallen from space onto MIT’s campus! As an expert materials scientist, you’re called in to investigate its properties.
An alien sword! Who knows what it’s made of… oh wait, we can find out using x-rays. :)
a) First thing’s first - let’s figure out what crystal structure the material in this sword has! You stick the sword into your x-ray diffraction (XRD) machine and wait for some results.
(Note: It’s a little different in real life.)
Unfortunately, the alien sword releases a blast of ~alien energy~ during the experiment! Not only does it damage your XRD machine, but your results get all ripped up. You manage to salvage some PIECES of the XRD graph, shown below:
Based on the salvaged XRD graph, what are ALL of the possible cubic crystal structures the sword could have? (4 pts)
b) You still don’t know what the sword is made of! Luckily, the x-ray generator part of your million-dollar XRD machine hasn’t been damaged. Undeterred, you decide to make the sword the source of x-rays by accelerating electrons at it to generate an x-ray spectrum.
Draw an x-ray spectrum. Label the horizontal and vertical axes, the Brehmsstrahlung radiation, $\mathrm{K}_\alpha, \mathrm{K}_\beta, \mathrm{L}_\alpha$, and $\mathrm{L}_\beta$. (6 pts)
c) You can figure out what material your sword is by analyzing its x-ray spectrum!
In the resultant $\mathrm{x}$-ray emission spectrum, you observe a $\mathrm{L}_\alpha$ peak at a wavelength $\lambda=2.383 \times 10^{-9}$ meters. What is the metal the sword is made of? Assume the screening factor, sigma $(\sigma)$, for $\mathrm{L} \alpha$ radiation is $7.4$.
Use the equation $E=13.6 *(Z-\sigma)^2\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)$, where $\sigma$ is the screening factor. (3 pts)
d) Unfortunately, you got so absorbed in your work that you don’t notice shooting accelerated electrons at the sword has heated it up! The sword has been at 600$\mathrm{K}$ for the 3 hours you spent analyzing your data. In other words, the sword has been ANNEALED.
In ONE sentence, explain how this might change at least two mechanical properties of the sword. (4 pts)
e) The graph below has a stress-strain curve for the sword BEFORE your experiments.
i. Draw the stress-strain curve for the sword AFTER your experiments, i.e. after it was ANNEALED and has been returned to room temperature. (4 pts)
Problem 6 (19 POINTS)
You're working in Professor Grossman's lab and have been given the following solutions:
Solution 1: $0.0334$ grams of $\mathrm{CaF}_2$ in $2 \mathrm{~L}$ of water
Solution 2: $0.0334$ grams of $\mathrm{CaF}_2$ in $3 \mathrm{~L}$ of water
Solution 3: Solution $2+1 \mathrm{~L}$ of $5^* 10^{-5} \mathrm{M} \mathrm{PbF}_2$
a) Assuming the ionic solid completely dissolves to create a saturated aqueous solution, what is the $\mathrm{K}_{\mathrm{sp}}$ of $\mathrm{CaF}_2$ in Solution 1 ? (3 pts)
b) The concentration of Solution 2 is $1.43^* 10^{-4} \mathrm{M}$, and the $\mathrm{K}_{\mathrm{sp}}$ of $\mathrm{PbF}_2$ is $3.7^* 10^{-8}$.
i. What are the concentrations of each ion in Solution 3? (4 pts)
ii. Write out the $\mathrm{K}_{\mathrm{sp}}$ expressions for both $\mathrm{CaF}_2$ and $\mathrm{PbF}_2$. (2 pts)
iii. Use the expressions from b)ii. to show why there wouldn't be any precipitates. (4 pts)
c) Now, consider Solution 2 alone. You want to precipitate the $\mathrm{CaF}_2$ out of solution using one of the following chemicals in your lab: $\mathrm{PbI}_2, \mathrm{CaS}$, and $\mathrm{MgF}_2$.
i. Name one chemical you could use to precipitate $\mathrm{CaF}_2$. (there may be multiple correct answers) ( 2 pts)
ii. How many moles of the chemical that you chose for part c) i. are necessary to precipitate the original $4.28 * 10^{-4}$ moles of $\mathrm{CaF}_2$ out of the solution? (4 pts)
PROBLEM 7 (25 POINTS)
Above are sketches (from three different textbooks) of three samples of silicon dioxide. One of the samples contains $\mathrm{Na}_2 \mathrm{O}$ as a network modifier.
a) Which sample contains $\mathrm{Na}_2 \mathrm{O}$? (2 pts)
b) Which sample is crystalline quartz? (2 pts)
c) Plot the molar volume vs temperature curves of the three materials on the same plot. Be sure to label your axes. (6 pts)
d) $\mathrm{Na}_2 \mathrm{O}$ and $\mathrm{CaO}$ are both network modifiers. Network modifiers are added when the glass is in a liquid state. You know that the rate law for the dissolution of $\mathrm{CaO}$ in liquid silica is rate $_{\mathrm{CaO}}=4[\mathrm{CaO}]$.
Write a similar rate law for the dissolution of $\mathrm{Na}_2 \mathrm{O}$ in liquid silica and use it to fill in the following table. The $\mathrm{Na}_2 \mathrm{O}$ reaction is second order in terms of $\mathrm{Na}_2 \mathrm{O}$ and independent of the silica concentration. Please determine the value of the rate constant, and write the units for $\left[\mathrm{Na}_2 \mathrm{O}\right]$ and rate.
Fill in the units row and find the rate of trial 2 $(8 \mathrm{pts})$:
$\left[\mathrm{Na}_2 \mathrm{O}\right]$ rate
Unit
Trial 1 $2 \times 10^{-5}$ $1.2 \times 10^{-9}$
Trial 2 $4 \times 10^{-5}$
$\text{Rate law:}$ $\text{rate}_{\mathrm{Na}2\mathrm{O}}$ = ________________________
e) What are the units of the rate constant in:
i. the $\mathrm{CaO}$ rate law ( $\left.2 \mathrm{pts}\right)$
ii. the $\mathrm{Na}_2 \mathrm{O}$ rate law ( $\left.2 \mathrm{pts}\right)$
f) Thinking about the size of $\mathrm{Na}$ vs. $\mathrm{Ca}$ ions, explain in ONE sentence why it makes sense that the sodium oxide reaction occurs at a faster rate than the calcium oxide reaction. (3 pts)
Problem 8 (24 POINTS)
You’re interested in understanding the diffusion of hydrogen gas through a palladium ($\mathrm{Pd}$) filter to make sure the concentration of $\mathrm{H}$ doesn’t get too high in the living quarters of the rocket.
Modified astronaut sleeping. Original image is in the public domain (source: NASA).
a) Calculate the original number of $\mathrm{Pd}$ vacancies in a $1 \mathrm{~mol}$ sample of $\mathrm{Pd}$, assuming a vacancy activation energy $=0.7 \mathrm{eV}, \mathrm{T}=300 \mathrm{~K}$, and the entropic constant $\mathrm{A}=1(5 \mathrm{pts})$
b) Does hydrogen diffuse through the palladium lattice through interstitial sites or through substitutional sites? Use Hume-Rothery rules to justify your answer in ONE sentence. ( 3 pts)
c) The diffusion coefficient of hydrogen through palladium at $300 \mathrm{~K}$ is $3.2 \times 10^{-10} \mathrm{~m}^2 / \mathrm{s}$. You expose one side of the wafer to a hydrogen atmosphere with concentration $0.0446 \mathrm{M}$ and the other to air with a hydrogen concentration of $2.23 \times 10^{-8} \mathrm{M}$. What is the flux of hydrogen atoms through this Pd crystal at $300 \mathrm{~K}$? The Pd wafer is $2.7 \mathrm{~mm}$ thick. Hint: 1 liter is $0.001 \mathrm{~m}^3$. (5 pts)
d) Draw the concentration profile of hydrogen atoms across this crystal at 300$\mathrm{K}$. (3 pts)
e) At $400 \mathrm{~K}$, the diffusion coefficient is $\mathrm{D}(400 \mathrm{~K})=3.51 \times 10^{-5} \mathrm{~m}^2 / \mathrm{s}$.
i. Assume that diffusion follows the Arrhenius law: $\mathrm{D}=\mathrm{D}_0 * \exp \left(-\mathrm{E}_{\mathrm{a}} /\left(\mathrm{k}_{\mathrm{B}} \mathrm{T}\right)\right)$. What is the activation energy $\mathrm{E}_{\mathrm{a}}$ of the diffusion of $\mathrm{H}$ through $\mathrm{Pd}$? (4 pts)
ii. At what temperature should we hold the Pd such that the flux of hydrogen atoms through the Pd filter is $1.38 \times 10^{-4} \mathrm{~mol} / \mathrm{m}^2 \mathrm{~s}$? Hint: use the information from part c. or part e. to determine $\mathrm{D}_0$.(4 pts)
Problem 9 (19 POINTS)
The following diagram shows the pH of some common substances:
a) From the scale above, name:
i. The strongest acid (1 pt)
ii. The weakest base (1 pt)
iii. A neutral substance (1 pt)
The acidic ingredient in vinegar is acetic acid $\left(\mathrm{CH}_3 \mathrm{COOH}\right)$. You bought a bottle of vinegar and measured the $\mathrm{pH}$ of your solution is around $2.4$, and the molar concentration of acetic acid in vinegar is around $0.85 \mathrm{M}$.
b) Write the acid reaction of acetic acid with water. Label the conjugate acid and base pairs. (4 pts)
c) What is the hydronium ion $\left(\mathrm{H}_3 \mathrm{O}^{+}\right)$ concentration of our $0.85 \mathrm{M}$ solution of acetic acid? (3 pts)
d) Based on the information given and your answer to parts b) and c), determine the value of the acid ionization constant, $\mathrm{K}_{\mathrm{a}}$, for acetic acid. (3 pts)
e) What is the hydronium ion $\left(\mathrm{H}_3 \mathrm{O}^{+}\right)$ concentration of a $0.010 \mathrm{M}$ solution of acetic acid? (3 pts)
f) What is the $\mathrm{pH}$ of a $0.010 \mathrm{M}$ solution of acetic acid? (3 pts)
Problem 10 (13 POINTS)
As the universe cooled after the big bang, ionized hydrogen $(\mathrm{H}+)$ combined with free electrons ($\mathrm{e}-$) to form hydrogen gas.
a) Write a balanced equation for this reaction. (2 pts)
b) When $\mathrm{H}+$ and $\mathrm{e}-$ combine, a photon is emitted as the electron relaxes to the ground state, essentially reversing the process of ionization from the ground state.
i. What is the energy of this electronic transition? (2 pts)
ii. What is the wavelength of the photon that is emitted? (3 pts)
c) We can learn about the elemental composition of stars by observing which wavelengths are absorbed by the star.
i. Explain in ONE sentence why absorption spectra have dark lines (as shown in the figure above). [Note – to save money at CopyTech we printed this in greyscale, but it was originally a spectrum going from blue to red.] (3pts)
ii. Explain in ONE sentence why the shorter-wavelength dark lines are closer together than the longer-wavelength dark lines (3pts)
Modified image of astronaut waving © original source unknown. All rights reserved. This content is excluded from our Creative Commons license. For information, see https://ocw.mit.edu/fairuse. | textbooks/chem/Inorganic_Chemistry/Introduction_to_Solid_State_Chemistry/04%3A_Exams/4.04%3A_Final_Exam.txt |
What is This?
The purpose of this document is to serve as a guide and resource that gives you a quick overview of each lecture. For each lecture, there is a summary of the main topics covered, the Why This Matters moment, and the new Why This Employs section, plus a few example problems. These handouts will be given by “Unit,” which is the content that goes into each exam. There’s also a Concept Map at the beginning of each Unit, to show how the various materials are connected together.
So why did we make this? We hope it’s useful to get a good snapshot of any given lecture. Whether you couldn’t make it to a lecture or you couldn’t stop thinking about a lecture, this is a way to quickly get a sense of the content. It also gives me a chance to provide additional details that I may not have time for in the Why This Matters example, and also it lets me try out the Why This Employs section, which I certainly will not have time to discuss much in the lecture. Hopefully you find it useful, and as with anything else in this class, any and all feedback is most welcome!
One point about these lecture summaries. Please note that the lecture summaries are not meant to be a substitute for lecture notes. If you were to only read these summaries and not go to lecture, yes you’d get a good sense of the lecture from a very high level view, but no, you wouldn’t get enough out of it for it to be your only resource to learn the material!
How This Connects: Unit 1, Lectures 1-9
In this class we have 36 lectures, 20 recitations, 9 goodie bags, and 12 problem sets, all tightly integrated and each one designed to contribute to your overall 3.091 learning experience. These different resources come together in the Celebrations of Learnings throughout the semester, including 9 quizzes, 3 midterms, and a final exam. For each unit, defined as the content that goes into a midterm, we have created a Concept Map to elucidate connections between the different class components so that you can see clearly the context of each.
Below is an image of the Exam 1 Concept Map. This demonstrates how each of the aspects of the course fit together: you have lots of resources! The Practice Problems, Recitations, Goodie Bags, and Lectures are ungraded resources to help you prepare for the quizzes and exams. All of the material listed on this concept map is fair game for Exam 1.
Lecture 1: Introduction and the chemistry of the periodic table
Summary
“Never trust an atom, they make everything up,” is a classic chemistry joke that also happens to be so true, but if we want to know which elements comprise our materials of interest, then we need to know how these elements interact with one another, how they are structured, and how the material was processed to achieve this structure. To truly answer these questions, investigators in the last few centuries have made use of the scientific method. This involves asking questions, gathering and examining evidence, identifying explanations, and re-testing.
One of the most exciting questions in the history of chemistry was that of the atom—Democritus and Leucippus hypothesized indivisible building blocks of the universe (though now we know that atoms can be broken down into even smaller “subatomic particles”). Robert Boyle, in his quest to identify different atoms, studied metal ores (naturally occurring rocks containing metal impurities) and succeeded in separating a large number of elements, including many metals. As the list of discovered atoms grew, scientists also began studying their categorization. For example, Antoine Lavoisier named four property-based categories for the 33 elements he identified: Gases, Non-Metals, Metals, and Earths. John Dalton identified 36 different elements, and designed a graphical symbol to represent each of them.
In this lecture we introduced the concept of chemical reaction balancing, which involves adding coefficients to the reactants and/or products in the reaction to ensure that there are equal numbers of each type of atom on the left- and right-hand sides of the arrow. This is necessary because of the law of conservation of mass, which states that nothing is created or destroyed at the atomic level, so the mass of the reactants equals the mass of the products. Rather, atoms can be rearranged to form different molecules. The first chemical reaction we balanced was a combustion reaction, which is a reaction in which a carbonaceous material burns in the presence of oxygen. We found the limiting reagent for this reaction, which is the reacting species (atom or molecule) that is used up first. The yield of the reaction is the amount of product formed.
Dalton stated that atoms of a given element have a particular weight and other distinguishing characteristics. He proposed the law of multiple proportions, which says that when two elements react to form a series of compounds, the ratios of masses of the 2nd element per gram to the 1st can be expressed as ratios of integers.
Why this matters
In this first lecture, we started talking about the discovery of the elements. Well, I don’t mean to give anything away, but there aren’t going to be all that many of them in the end. In fact, to date we’ve got 118 on the official roster, but if we consider which ones we can use to make stuff for humans that number is lower, more like 80 or 90. How many elements do you think there are in your cell phone? It depends on the phone, but in mine there are 64 different elements!
That’s more than two thirds of all of the ones we could even have used and it’s incredible to think about. Because of this massive design potential, the world we live in has completely changed.
As a Materials Scientist, I love that we have this tradition (shout out to Danish scholar Christian J. Thomsen) of naming the age we live in by the material that mattered most during that age. But as a Materials Scientist, I also love that we’ll never be able to do that again. The reason is that we truly live in a new era, one where we can realize the dream of Richard Feynman, who in his famous speech “There’s Plenty of Room at the Bottom” put forth the vision for atomicscale design and nanotechnology 30 years ahead of its time. Today, we routinely control both the choice of elements as well as their structure as they come together to make materials do things we never thought possible even just 10 years ago! In other words, we control their chemistry at the atomic scale.
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This gets to why this is so cool (and, you guessed it, why it matters). You see, so many of our current global challenges – whether in energy, health, or the environment – have a bottleneck that occurs because of the material that makes up a central part of the technology (in some devices it’s called the “active layer”). That means that by “simply” choosing or making a different material, one has the potential to completely change the properties, cost, and manufacturability of the technology. This is unlike a previous revolution around silicon where we went from a cost of $1 per transistor in 1965 to$0.00000001 per transistor in 2015. Talk about a learning curve! That was primarily due to advances in processing of the same element, silicon, to get to smaller and smaller components. More than a billion transistors now fit nicely on a single chip. Unlike this dramatic age of silicon, many of the next technology revolutions will rely on changing the material itself: in some cases, completely. We now live in the Materials Design Age, which has come just in time since it converges with very pressing challenges on a global scale that will rely on new materials to be solved. And this is why the discovery of these elements matters so much. They are our playground and the basis from which we will make the next set of technologies to change the world for the better.
Why this employs
In this section, we’ll discuss how the material in each lecture connects to real jobs. I’m trying it out this year for the first time, and since I likely won’t have time to spend on it in lecture, this document is the only place you’ll find it. In Why This Matters, I passionately believe that each lecture can and should be connected to some larger-scale challenge, or innovation, or inspiration. In Why This Employs, my goal is to connect each lecture with a possible field of employment. Similar to Why This Matters, this is not meant to be exhaustive but rather illustrative, to give you another way in which the material you’re learning has value beyond the classroom and connects to something, in this case a job.
We spent half of this first lecture talking about the class structure, not just administratively (what’s graded, what’s not, etc) but also pedagogically (the different components, how they come together to provide a learning experience, etc.). Thinking about how to teach better, and how students learn, and how new technologies should or should not be a part of this, is a wonderfully rich area of employment. If you’re interested in learning more, check out MIT’s Teaching and Learning Lab or some of the cool initiatives in education like EdX, NEET, or the “superUROP” programs.
Another employment direction related to this lecture is the discovery of new elements. Here, we examined how the earliest chemists attempted to identify elements. This came down to smashing, burning, boiling, reacting, etc. It turns out that’s still a thing, only in modern times all this reacting occurs with much bigger pieces of equipment and much larger teams of people. The next element yet to be discovered (at the time of writing, that would mean the 119th known element) will likely not have a very long lifetime, and will be very difficult to make in large quantities. But pushing the boundaries between what we have found naturally and what we can make synthetically as it relates to the fundamental building blocks of the universe. . . well, that’s a pretty exciting thing to do. And given the prediction by some that an “island of stability” exists for heavier elements that have yet to be made, it’s possible that an entirely new era of chemistry awaits such discoveries.
The last example I’ll give for employment relates to the idea of the metal ores that I mentioned earlier in this lecture. Remember that this is where Dalton and many others were discovering those new elements. Basically, you take a rock and carefully break it up and viola, you get Fe out of it (ok, it’s a bit more complicated than that, but that’s the idea). It turns out that quite often the metal atom is locked up pretty tightly in the form of an “oxide” (yes, that means bonded to oxygen, but we’ll get to that in a few lectures). The point is that extracting metals like aluminum from rocks takes an enormous amount of energy, which means it requires a lot of burning of fossil fuels. What if you could come up with a new, much “greener” way to get metals or minerals in their pure form out of the rocks from which they originate? This would be a very big deal. Mining metals and minerals involves massive teams of engineers who cover a wide array of jobs from figuring out where to mine, to characterization of the starting materials, to optimizing extraction chemistry, to working on new ways to do that more efficiently.
Example Problems
1. In this lecture, you witnessed my passion for pyrotechnics.
a) Write and balance the chemical reaction for a similar process, the combustion of propane ($\mathrm{C}_3 \mathrm{H}_8$).
Answer
$\mathrm{C}_3 \mathrm{H}_8+5 \mathrm{O}_2 \rightarrow 3 \mathrm{CO}_2+4 \mathrm{H}_2 \mathrm{O}$
b) If you have $500 \mathrm{g}$ of propane, how many moles do you have? How many molecules?
Answer
First, find the molar mass of propane.
$3 * 12 \mathrm{~g} / \mathrm{mol}+8 * 1 \mathrm{~g} / \mathrm{mol}=44 \mathrm{~g} / \mathrm{mol} \nonumber$
Use the grams given and the molar mass to convert to moles:
$500 / 44=11.36 \mathrm{~mol} \nonumber$
Lecture 2: Counting Atoms and Organizing the Elements
Summary
This chapter, we introduced the concept of the mole. This is a constant number, also referred to as Avogadro's number, equal to $6.0221409 \times 10^{23}$. The atomic mass units (AMUs) assigned to an element are equal to the mass in grams of a mole of atoms of that element. Just as Avogadro's number is just a number, the $\mathrm{AMU}$ is just a mass, and $1 \mathrm{AMU}$ is equal to $1.67377 \times 10^{-24}$ grams. Just like the mole, an AMU is another special unit that chemists use to connect the very small world of atoms with the comparatively large world of things we work with. This link between the macro and atomic worlds means that by simply measuring the mass of a substance, we can know how many atoms of that material we have.
In class we used the example of the combustion of a candle, otherwise known as the nicely balanced reaction $\mathrm{C}_{25} \mathrm{H}_{52}+38 \mathrm{O}_2 \rightarrow 25 \mathrm{CO}_2+26 \mathrm{H}_2 \mathrm{O}$. We used this example to emphasize how the concept of the mole gives us a link between the atomic and macroscopic worlds, between grams and atoms. We also used this example to re-emphasize the concepts of limiting reagent and yield.
Next, we focused on organizing the periodic table. Unlike in the last chapter, here organization means more than coming up with a naming scheme or a simple categorization of the elements, as Dalton, Lavoisier, and many others had already done. In this lecture what we mean by organization is really pattern recognition. Scientists were starting to see patterns emerging among the elements, and the more elements were discovered, the more effort was put to understanding these patterns.
Dmitri Mendeleev developed the precursor to today's periodic table. His big breakthrough was that he arranged the then-known (63) elements not just by their mass, but also by the repeating (periodic!) patterns he observed in their properties. This led him to leave gaps in the periodic table that would later be filled with newly discovered elements. Because of the periodicity in properties, he could even predict the properties of many as-of-yet undiscovered elements. The rows of your periodic table are called periods, and the columns are called groups. The Main Group Elements are those in periods two and three. There is a general trend from metallic to non-metallic elements along the periods. The Transition Elements are in groups 3-12 and periods 4 and 5 .
Why this matters
The population of humans on this planet has seen a dramatic, exponential growth. Let’s zoom in on where the growth really started to kick into high gear. See that uptick, around the early 1900’s? Notice the massive change in slope. That’s when we figured out how to make nitrogen in a way that plants can use.
People vs. time plot © Population Reference Bureau. All rights reserved. This content is excluded from out Creative Commons license. For information, see https://ocw.mit.edu/fairuse.
In other words, with the discovery that is now known as the Haber-Bosch process (named after the German chemists Fritz Haber and Carl Bosch), the world was able to make fertilizer in the abundance needed to feed billions. The key comes down to something called "fixing" nitrogen. Plants need nitrogen to grow, but most plants cannot use any of that plentiful $78 \% \mathrm{~N}_2$ that's in the air. Instead, they need it in a form that serves up single $\mathrm{N}$ atoms, like ammonia $\left(\mathrm{NH}_3\right)$. SO how can we go from $\mathrm{N}_2$ to $\mathrm{NH}_3$? Easily, with the following chemical reaction:
$\mathrm{N}_2+\mathrm{H}_2 \rightarrow \mathrm{NH}_3$
Ah, sorry, let’s balance that:
$\mathrm{N}_2+3 \mathrm{H}_2 \rightarrow 2 \mathrm{NH}_3$
That’s better. What Haber and Bosch actually did was develop a catalyst and process that allowed this reaction to be carried out much more efficiently than previously. But catalysis is getting a little ahead of ourselves. For now, let’s focus on the balanced reaction and count some atoms.
In 2019, 500 million tons of nitrogen is produced for fertilizer each year using the Haber-Bosch process. Half of the protein in all human beings on this planet comes from nitrogen that was fixed by this process! So here's my question: how long can we continue fixing nitrogen this way? And since the reaction tells us we need both $\mathrm{N}_2$ and $\mathrm{H}_2$, which one would be the limiting reagent?
From all of our discussion so far in this lecture we know that we can use the periodic table to determine that 1 mole of $\mathrm{NH}_3$ is 17 grams. Just for simplicity, suppose in the future we continue needing to make the same amount we're making today, namely 500 million tons of $\mathrm{NH}_3$ per year. There are roughly a million grams per ton, so 500 million tons per year $500 \times 106 \times 106$ grams per year. Dividing by 17 grams per mole, this means we need to make roughly $30 \times 10^{12}$ moles of $\mathrm{NH}_3$ /year. Given the balanced reaction above, this means we'll need as input to the reaction $15 \times 10^{12}$ moles of $N_2$ per year. Now, the mass of the atmosphere is about $5.15 * 10^{21}$ grams, of which $78 \%$ is $\mathrm{N}_2$, so that makes about $4 \times 10^{21}$ grams of $\mathrm{N}_2$ in the air. And since the atomic mass for a nitrogen atom is 7 , then it's 14 for $\mathrm{N}_2$ which means there are 14 grams per mole of $\mathrm{N}_2$. So the number of moles available in the atmosphere is $4 \times 10^{21}$ grams of $\mathrm{N}_2 / 14$ grams $/$ mole $=3.5 \times 10^{20}$ moles of $\mathrm{N}_2$ in the atmosphere. Well that's good news! Taking $3.5 \times 10^{20}$ moles of $\mathrm{N}_2$ total and dividing that by $15 \times 10^{12}$ moles of $\mathrm{N}_2$ needed per year, we find that we could keep at it with the Haber-Bosch process of fixing nitrogen from the atmosphere for more than 20 million years or so before using it all up.
That’s just cool. Think about this: we just went from atom to macroscopic world and it’s all because we have this concept of the mole. But we never really answered the question, namely which one is the limiting reagent in the reaction, the $\mathrm{N}_2$ or the $\mathrm{H}_2$? We now know how long $\mathrm{N}_2$ will last us if we were to use it all up in this reaction, so what’s left is to do the same thing for $\mathrm{H}_2$. Speaking of which, where will we even get $\mathrm{H}_2$? Certainly not from the air. I’ll let you all work through this but here’s a hint: the oceans contain a whole lot of water, like more than $10^{18}$ tons. Maybe you don’t even need to do any calculations to figure this out!
Why this employs
How does counting or organizing atoms lead to a job? Well, for one thing we’re talking about order, precision, and measurement. After all, Mendeleev’s success relied on combining careful measurement of properties with precise ordering. This skill is needed for many different jobs. One place with measurement as its mission is the National Institute of Standards and Technology (NIST). Check out the tag line in their Materials Measurement Laboratory: “MML serves as the national reference laboratory for measurements in the chemical, biological and material sciences. Activities range from fundamental and applied research to the development and dissemination of certified reference materials and data to assure the quality of measurement results.” And that’s just one of their many divisions.
Another job that comes to mind is one related to the fertilizer discussed in our Why This Matters. The three fundamental constituents of commercial fertilizers used throughout the world are nitrogen, phosphorus, and potassium. As we know, nitrogen requires fixing. Potassium must be obtained in other ways. For example, it comes nicely with chlorine as potassium chloride ($\mathrm{KCl}$), but we’d prefer not to have chlorine in the soil since it can lead to toxicity in the crops and high acidity and salinity in the soil. However, getting potassium in other forms is expensive and requires, as you might guess, chemistry. To make matters more complicated, both the natural sources for potassium and the specific soil needs are highly dependent on geography. For farmers in tropical regions in Brazil and some countries in Africa, the soil and rock compositions require fertilizers that better match the chemical needs of plants than are currently available. As a solution, more fertilizer is added to achieve as much crop yield as counterparts in the north, but many of the basic ingredients, especially the potassium, are produced in more northern regions and therefore are much more expensive in the Southern Hemisphere. The need for local, sustainable fertilizing solutions to increase yields is urgent! But this is starting to sound more like a second edition of Why This Matters and less like a Why This Employs. My point is that the development of fertilizers is a problem of chemistry, and the world is in desperate need of new solutions. In terms of jobs, this could mean becoming something called an “Analytical Chemist” which means you apply the principles of chemistry to figure out what stuff is made of, what the precise compositions are, how to monitor it, modify it, etc. This could also mean working at a company or research lab that develops cutting edge ideas (like recent work on using electrochemistry to make potassium).
Extra practice
1. Solar cells sometimes contain a titanium ($\mathrm{Ti}$) coating that stops the cell from short-circuiting. The Kroll Process for making titanium metal out of titanium oxide is:
$\mathrm{TiCl}_4+\mathrm{Mg} \rightarrow \mathrm{MgCl}_2+\mathrm{Ti}$
a) Balance the reaction.
Answer
$\mathrm{TiCl} _4+2 \mathrm{Mg}->2 \mathrm{MgCl}_2+\mathrm{Ti} \nonumber$
b) If you have $50 \mathrm{~g}$ of both $\mathrm{TiCl} 4$ and $\mathrm{Mg}$, what is the limiting reagent in the reaction?
Answer
Molar mass of Mg: $24.3 \mathrm{~g} / \mathrm{mol}$
Molar mass of $\mathrm{TiCl}_4: 189.9 \mathrm{~g} / \mathrm{mol}$
Find the moles of each reactant by dividing $50 \mathrm{~g}$ by the individual molar masses:
$50 \mathrm{~g} / 24.3=2 \mathrm{~mol} \mathrm{Mg} \quad \quad 50 \mathrm{~g} / 189.9=0.3 \mathrm{~mol} \mathrm{TiCl}_4 \nonumber$
Use the mole ratio from the balanced reaction $\left(1 \mathrm{~mol} \mathrm{TiCl}_4 / 2 \mathrm{~mol} \mathrm{Mg}\right)$ to see how much $\mathrm{TiCl}_4$ could be used up with the amount of $\mathrm{Mg}$ you have available:
$(2 \mathrm{~mol} \mathrm{Mg} \text { available }) /\left(2 \mathrm{~mol} \mathrm{Mg} / 1 \mathrm{~mol} \mathrm{TiCl}_4\right)=1 \mathrm{~mol} \mathrm{TiCl}_4 \nonumber$
$1 \mathrm{~mol} \mathrm{TiCl}_4$ would be used up if we had an endless supply of it with the amount of Mg we have. This amount of $\mathrm{TiCl}_4$ is more than the amount we actually have available, meaning that the $\mathrm{TiCl}_4$ will run out before the $\mathrm{Mg} . \mathrm{TiCl}_4$ is therefore the limiting reagent.
c) What is the yield of the reaction in grams?
Answer
To calculate the $\mathrm{Ti}$ yield of the reaction, we use the ratio of $\mathrm{Ti}$ to $\mathrm{TiCl}_4$ (because these two have a 1:1 molar ratio) and then convert from moles to grams of $\mathrm{Ti}$:
$0.3$ moles of $\mathrm{TiCl}_4$ are being used (it's the limiting reagent, so it will be used in its entirety) and because of the 1: 1 molar ratio we know that $0.3$ moles of $\mathrm{Ti}$ will be produced. Multiply this number by the molar mass of $\mathrm{Ti}$:
$0.3 * 47.9 \mathrm{~g} / \mathrm{mol}=14.37 \mathrm{g ~Ti} \nonumber$
d) Plot the yield of $\mathrm{Ti}$ (g) as a function of increasing $\mathrm{Mg}$ available for reaction.
Answer
The plot of $\mathrm{Ti}$ grams (on the $\mathrm{y}$ axis) vs. $\mathrm{Mg}$ grams (on the $\mathrm{x}$ axis) starts at 0 and increases linearly until the $\mathrm{Ti}$ grams reach $14.37 \mathrm{~g}$, at which point we run out of $\mathrm{TiCl}_4$ and the reaction will stop.
Lecture 3: The Discovery of the Electron and the Structure of the Atom
Summary
This lecture began with discussion of ground-breaking experiments that brought the scientific community closer to understanding the structure of the atom. These included J.J. Thomson’s experiments with the cathode ray tube (CRT), which allowed him to find the charge-to-mass ratio of electrons, which he realized were negatively charged particles found in all atoms. Robert Millikan was able to ascertain the mass and the charge of the electron using his oil drop experiment. The atom was not the most fundamental particle after all, rather it was made of smaller particles like electrons, and as we learn next those smaller things that make up atoms have a structure to them. The electronic structure.
Building on the radiation studies of Marie and Pierre Curie, Ernest Rutherford was able to identify three types of particles found in radiation: alpha particles, which were large and positively charged, beta particles, which were small and negatively charged, and gamma rays, which didn’t deflect or interact with the photographic plate in Rutherford’s experiment at all. Rutherford then conducted the gold foil experiment, which showed that atoms have positive charge that is localized in the core of the atom rather than dispersed among the negative charge, as Thomson had proposed. This positively charged center of the atom is called the nucleus. The positively charged particles in the nucleus are called protons. For any neutral atom, the number of protons present must equal the number of electrons present. This number is called the element’s atomic number, which is denoted by the letter $\mathrm{Z}$. Except for the case of a neutral hydrogen atom composed of only one proton in its nucleus and one electron, all atoms also contain neutrons in their nuclei, which are, as the name might suggest, neutrally charged particles. To obtain the mass number of a given atom, the number of protons and neutrons in that specific nucleus are added together. Atoms of the same element can have different numbers of neutrons. This makes them different isotopes of the same element. To obtain the mass given for each element in your periodic table, a weighted average is taken of the mass numbers of that element’s naturally occurring stable isotopes. The weight is assigned according to the isotopes’ percent abundance in nature.
After Rutherford’s experiments, there remained the question of how an atom structured in this way could be stable. Niels Bohr was the one who postulated that the angular momentum of the electrons in atoms must be quantized, which means that it takes on discrete values. This meant that the electron’s energy and distance from the nucleus were quantized as well—the electron can only occupy certain orbits.
Why this matters
The discovery of the electron led to a profound new understanding of the atom, which in turn led to an entirely new theory (quantum mechanics), which in turn led to orbitals, bonding, and the building blocks for all of modern chemistry, as we’ll see soon. But the discovery of the electron also led to a new capability that revolutionized the world: the ability to paint with electricity. The phosphor screen that was on the end of Thomson’s cathode ray tube was the precursor to television, and the beginning of a technological revolution that changed almost every aspect of our lives. TV wasn’t invented by Thomson himself, but Thomson’s discovery of this new fundamental paint brush – the electron itself – was crucial for the concept to take off. Look at what the man on the left in this image is holding: it’s an early version of a TV, which you can now recognize as a giant cathode ray tube! John Logie Baird was one of the first to commercialize television sets, selling his first one in 1925. You can see the inside of an early TV in the middle picture: note that it was comprised of a CRT plus a coil to induce a magnetic field – essentially the same setup that Thomson used in his groundbreaking experiments.
Man with CRT & open TV images © sources unknown. This content is excluded from our Creative Commons license. For information, see https://ocw.mit.edu/fairuse.
If electrons were scanned across the phosphor screen fast enough, an image could be painted so many times per second that it looked like motion to the human eye. This was easy to do, because changing a magnetic field quickly and precisely (to position electrons on the screen) can be achieved with simple electronic circuitry. Over the next generation, electrons became a dominant way to exchange visual information. The chemistry of the screen itself is also essential. A ‘screen’ is actually a phosphor coating that was often put on a piece of glass and was the target of the electron beam. Phosphor is a broad term for materials that emit light in response to a stimulus. This act of emitting light is called luminescence. It can get confusing since the term “phosphorescent” means that the material slowly emits light over time after being stimulated. A good example of a phosphorescent material would be glow-in-the-dark toys, where the stimulus is light, and the phosphor can keep glowing for many hours. On the other hand, a “fluorescent” material also emits light in response to a stimulus, but as a quick flash lasting nanoseconds. The term phosphor can refer to either of these, but the coating on the CRT screen and in those early televisions is the fluorescent kind. You can imagine why: if it didn’t emit light quickly,, the image would have afterglows that lasted for hours, which wouldn’t exactly make for a nice movie-watching experience.
To make matters even more confusing, the term phosphor (which as we just established can refer to materials that are phosphorescent or materials that are fluorescent), is not to be confused with the element phosphorus. Phosphorous also emits light, but of a different nature entirely, since its glow comes from a chemical reaction (a process called “chemiluminescence”). So, to sum up: the screen we’re talking about uses a phosphor which is fluorescent, not phosphorescent, and doesn’t have any phosphorus in it. Great.
Chemically speaking, what are these phosphors? There are so many different materials that are phosphors that it would take an entirely separate book to go through them all. They can be small molecules, complex solids, or liquids. They can emit light in response to many different stimuli, from light to heat to electric fields to the electrons we’re discussing here in our CRTs. The first television screens were black and white, so the goal was simply to use phosphors that glowed white when they were struck by electrons. Powders made from zinc, cadmium, and sulfur and a dab of silver were early ingredients. In order to achieve color TV, blue, green, and red phosphors were needed. The electron beam could be pointed at whichever ones were needed to make a given color, again using the magnetic field to guide the electrons with fantastic precision. It turned out the green and blue phosphors were pretty straightforward to make, zinc-sulfide with a touch of copper for green, and silver for blue. But red was a different story: for red phosphors it took additional decades’ worth of research to find anything that worked. This is the reason why color television didn’t come on the scene until the mid-1960’s, a full 40 years after black white television was introduced! The red phosphor challenge was finally solved by using a complex mixture of yttrium, oxygen, sulfur, and a bit of europium for good measure.
Since we’ve listed yttrium, I have to mention that it is one of four – yes four! – elements that were discovered in the same cave in the same town, Ytterby, Sweden. Yttrium, terbium, erbium, and ytterbium are all named after this town. We’re still waiting for cambrium or bostonium.
Now, it’s only your parents or grandparents who will remember what watching TV on a CRT was like. That’s because TVs nowadays work on a different principle, namely light emitting diodes (LEDs). But in the end, it’s still painting with electrons! An LED emits light using the cascade of electrons from high energy to low energy in a solid material. We’ll learn all about that material (it’s called a semiconductor) and why electrons can make light this way a little later in the semester. The light that comes out of an LED in your TV (or phone or tablet) still often passes through a phosphor, since the phosphor can be used to fine-tune the color of light we see. In this case the phosphor is stimulated by light instead of electrons, but the outcome is the same: light and color are produced using electrons and chemistry.
Why this employs
How can we get a job based on the discovery of the electron? That’s kind of like asking how we can get a job related to the knowledge of atoms – in other words, electrons and atoms make up everything so anything would qualify, at least if it has any reference to matter in the employment description. But how about if we go back, specifically, to that first beam of electrons that Thompson made to discover their very existence. . . what jobs might involve beams of electrons. We already discussed painting with electrons in the topic of displays in Why This Matters, but it turns out there’s a whole lot more you can do with a beam of electrons than just make pictures. To do so, though, we’ll need more power. A lot of power.
If you crank up the power of that electron beam enough, and you keep it in a vacuum (just like Thompson did) to make sure they don’t scatter off of anything, then you can use the electron beam to create very localized heat, enough to melt any material. If it’s done to join two materials together in a precise manner we call it welding. Electrons have a lot of advantages in welding because they can be focused to very small areas using magnetic fields, and the power can be dialed in to whatever the job requires (and there’s almost no limit to how powerful you can make a beam of electrons). On the higher end, for industrial welding, electron beams can get up to much as 10,000,000 Watts per $\mathrm{mm}^3$ which can heat up a metal at a rate of 1,000,000,000 Kelvin per second!
Electron beam welding is a big industry, and there are many jobs associated with it. You could weld metal parts together at GE for its aviation business, and I mention that company specifically because it was James Russel who invented electron beam welding while he was at GE, back in the 1950’s (as a fun side note, he also invented the first CD). If you do a search for electron beam welding you’ll find huge numbers of listing for companies that provide it as a service to other companies. Bodycote, for example, states that they are, “the world’s leading provider of heat treatments and specialist thermal processing services,” and with 5,700 employees in 26 countries they have a lot of jobs. But there are many smaller scale operations that hire people who like to shoot massively high energy electrons at metal parts, like the family-owned Roark in Brownsburg, IN, which boasts one of the largest electron beam welding chambers in the U.S., or Precision Technologies, Inc., which has a cool-sounding 5-axis electron beam welder that goes up to 150 kV, “capable of welding parts as small as a few grams up to ones weighing several tons,” or how about Fraunhofer FEP which develops customized electron beam technologies and states, “Using electrons we improve your materials and products!” Cool.
Extra practice
1. Your friend Wonder Woman tells you that her bracelets are made of a metal, Feminum (this is legit, it was in the '$70 \mathrm{~s}$ TV show), only found on her home island. Feminum (symbol Fm) has atomic number 120 and atomic mass $285.47$. The element has 2 isotopes, ${ }^{285} \mathrm{Fm}$ and ${ }^X \mathrm{Fm}$. The natural abundance of ${ }^{285} \mathrm{Fm}$ on the island is $77 \%$.
a. What is the other isotope, and what is its natural abundance?
b. How many protons and neutrons does each isotope have?
Answer
$77 \%$ abundance of one isotope implies 23 percent abundance of the other, since we know there are only two isotopes. The atomic mass is the weighted avg. of the two isotopes. Solve the weighted average equation to obtain the missing mass number:
$(0.77)(285)+(0.23)(x)=285.47 \quad \quad x=287.04, \text { rounds to } 287 \nonumber$
2. The number above the $\mathrm{Li}$ symbol on the periodic table (on the right) is the number of protons in a lithium nucleus. What is the number below the symbol? How do you calculate it?
Answer
The number below the atomic symbol is the atomic mass. It's not specific to the atom-it's a weighted average of the element's stable (non-radioactive) isotopes, with weights given according to the isotopes' abundance in nature.
For example, lithium's atomic mass is $6.94$, and its two stable isotopes are $7 (92.4\%$ abundance) and $6(7.6 \%$ abundance )
3. Assuming all the protons (red) and neutrons (blue) are visible, use proper notation to write out the atom that corresponds to each of the following nuclei:
Image courtesy of CK-12 Foundation, License: CC BY-NC.
Answer
a) mass : 13 , atomic : 6 , atomic symbol: $\mathrm{C}$
b) mass : 7 , atomic : 3 , atomic symbol: $\mathrm{Li}$
Lecture 4: The Bohr Model and Electronic Transitions
Summary
The Bohr model is a framework used to describe the quantized nature of atoms. By assuming that electrons orbited around atomic nuclei-like planets around the sun-Bohr showed that the electrons bound to atomic nuclei can only exist in discrete energy levels. By equating the attractive force of the Coulomb interaction between the negatively charged electron and the positive nucleus, $F_{C o u l o m b}=-Z e^2 / r^2$, and the repulsive force due to rotation, $F=m v^2 / r$, and substituting the quantized angular momentum, $L=m v r=n h / 2$, we find formulas for the quantized radius and the quantized energy. The energy of an electron in the nth energy level of an atom which has $\mathrm{Z}$ protons is
$E=-13.6 Z^2 / n^2 \quad[e V] \nonumber$
The integer $\mathrm{n}$, which can have any positive value $\geq 1$, is called a quantum number because it makes it so that the energy can only take on certain, discrete values as opposed to a continuum of possibilities. The state $n=1$ is the ground state, and higher energy levels are called excited states. When an electron goes from a state with a lower quantum number to one with a higher quantum number, the energy of the electron becomes less negative: it gains energy. We know that energy must be conserved, so the energy gained in such a transition must come from somewhere. In fact, a photon with exactly the $\Delta \mathrm{E}$ between two energy states in the atom can be absorbed, exciting the electron. Similarly, if the transition happens in the opposite direction, the excess $\Delta \mathrm{E}$ can be emitted in the form of a photon.
The kinds of photons capable of being absorbed or emitted by a Bohr model atom or molecule can be equivalently characterized by their energy, frequency, or wavelength. Einstein taught us that light can also be quantized in his work on the photoelectric effect: the energy of the photon can take discrete values:
$E=h f=h c / \lambda \nonumber$
Where $\mathrm{f}$ is frequency, $\lambda$ is wavelength, $\mathrm{h}$ is Planck's constant, $\mathrm{c}$ is the speed of light, and the second equality is true because $\mathrm{c}=\mathrm{f} \lambda$. The wavelength of light that can interact with a particular energy transition can be found by equating the two expressions above. With this in mind, we see that light can be used to probe what electronic transitions are happening in an atom or molecule, and therefore what energy levels are present in a material. The Bohr model fit the spectral lines of stars observed previously by astronomers; in particular, the discrete wavelengths emitted by hydrogen gas that had been observed by spectrometer finally had an explanation.
There are a few limitations to the Bohr model. We now know that electrons are not really orbiting the nucleus (although their angular momentum is quantized as predicted). Further, the Bohr model can only be used to describe electronic transitions with a single electron, such as $\mathrm{H}_{\mathrm{He}} \mathrm{He}^{+}$, or $\mathrm{Li}^{2+}$. However, the key takeaways hold: if the energy states in an atom or molecule are known, the wavelengths of light that can be emitted or absorbed by the material are also determined. Likewise, if the discrete wavelengths emitted or absorbed by an atom or molecule are observed, the energy levels available to an electron are also known.
Why this matters
How does absorption of photons by electrons that are transitioning from one energy level to another matter? One application this principle is connected to is the refrigerator(!) That might not seem obvious at first, but let’s see why. Take a look at the average energy used by a fridge in the U.S. over a 55-year period. It’s interesting to note that the average size of a refrigerator grew steadily for 35 years and then plateaued in the 1980’s.
Average energy use vs. time © source unknown.
This, it turns out, was not because people didn’t want more fridge, but rather because they couldn’t fit anything larger through the kitchen door. More to the point, note that the energy use per unit increased along with the size for 30 years, until suddenly it started decreasing rapidly in the 1970’s, even though the size continued to increase. This is because during that time there was also an oil shortage and people were starting to get concerned about energy use. Some of these people had important titles like President of the United States (Nixon, Ford, and Carter), and a whole lot of legislation aimed at energy conservation happened because of them. For example, President Ford signed the first law ever on this topic in 1975, the Energy Policy and Conservation Act, which led to the establishment of fuel economy standards for cars and efficiency standards for appliances. After that, more laws were signed that created incentives to lower energy consumption (towards a goal of making the U.S. an energy independent nation by 1985), with Carter presenting a plan to congress stating that, “conservation is the quickest, cheapest, most practical source of energy,” requiring federal agencies to develop energy conservation plans, and creating the Department of Energy (DOE).
In this Why This Matters, I want to emphasize the importance of government policy. It was the government stepping in that led to the technological innovation that led to greater efficiency and lower costs. In 1992, the Energy Star program was introduced which led to even greater reduction in the energy consumption of appliances and critically increased consumer awareness. Not only did refrigerators themselves become cheaper because of this innovation, but the energy they used continued to decrease. In 2019 in the U.S., we save about $20 billion annually on energy consumption alone for our refrigerators compared to the 1970’s. Sadly, since 2017, the Trump administration has been trying to eliminate the Energy Star program, which many consider to be one of the most successful voluntary energy efficiency programs in the world. It has saved Americans$430 billion in energy costs across hundreds of appliances, not to mention the lower CO2 emissions that have resulted.
Now we still haven’t connected back to our chemistry lesson, and while I could do that for the energy savings alone, there’s another example involving a fridge-related policy decision that very clearly brings Bohr back to the table. That is the Montreal Protocol, which was signed in 1987 and was the first treaty to be ratified by all countries in the world. It represented universal agreement to protect the stratospheric ozone layer by phasing out both production and consumption of ozone-depleting chemicals, like chlorofluorocarbons (CFCs). Taking this action led to the recovery of the earth’s ozone layer that protects life from harmful UV radiation. Millions of lives have been saved as a result. But what is it about CFC’s that makes them deplete ozone? And why is ozone so important for blocking UV radiation? The answer, of course, comes from chemistry.
Ozone is a molecule consisting of three oxygen atoms, $O_3$, and a common CFC molecule consists of a carbon atom, three chlorine atoms, and a fluorine atom, $\mathrm{CCl}_3 \mathrm{~F}$. Occasionally, the CFC molecule breaks down and gives up one of its $\mathrm{Cl}$ atoms, like this:
$\mathrm{CCl}_3 \mathrm{~F}(\mathrm{~g}) \rightarrow \mathrm{CCl}_2 \mathrm{~F}(\mathrm{~g})+\mathrm{Cl}(\mathrm{g}) \nonumber$
Note the "(g)" subscripts are included to show that all this is happening in the gas phase. Once a chlorine atom is freed up, it can then react with ozone, like this:
$\mathrm{O}_3(g)+\mathrm{Cl}(g) \rightarrow \mathrm{ClO}(g)+\mathrm{O}_2(g) \nonumber$
And so each CFC molecule can deplete one ozone molecule. But it doesn't stop there. A third reaction takes place, since there are also oxygen atoms in the stratosphere. When an oxygen atom interacts with the $\mathrm{ClO}$ molecule, it reacts with it to make $O_2$, like this:
$\mathrm{ClO}(g)+\mathrm{O}(g) \rightarrow \mathrm{Cl}(g)+\mathrm{O}_2(g) \nonumber$
And therein lies the real problem: the Cl atom is now free again and ready to attack another ozone molecule. Because of this catalytic cycle, one single CFC molecule can lead to the destruction of 100,000 ozone molecules!
Solar radiation spectrum by Robert A. Rohde as part of the Global Warming Art project. License: CC BY-SA. Source: Wikimedia Commons.
That brings us to the second question from above: why does losing ozone lead to more UV radiation? This is where electron transitions come in. Check out the plot in this figure of the solar spectrum. The x-axis is the wavelength of light coming from the sun, covering only the UV, visible and infrared (so it is just a sliver of the full EM spectrum we showed a few pages back). The y-axis is the intensity of light at a given wavelength, in Watts per area. A Watt is a measure of power, which is energy per time, or Joules/s. In plain words, this is a plot showing the power of sunlight as a function of its color.
The first thing to point out on the plot is that solid black curve, which corresponds to a theoretical prediction of the black body radiation emitted by an object at the temperature of the sun. This temperature varies a lot depending on where in the sun it's measured, but near the surface, $\mathrm{T}=5250^{\circ} \mathrm{C}$ is a decent approximation. Note that it's a nice continuous spectrum, and it has a shape typical of black body radiation (remember, that's the kind of radiation we mentioned earlier - the kind that Planck introduced the concept of quantization to explain).
Next, there's the yellow shaded parts that correspond to the actual measured power from the sun before it interacts with our atmosphere. It follows the black body radiation quite well, which validates the prediction. And then, there's the red shaded region. This is the power from the sun as measured on earth, so the only difference between the red and the yellow is that the red is a measurement of photons from the sun that have passed through the atmosphere. It follows the yellow pretty well in general, except that now there are gaps in the spectrum: discrete wavelengths where there is no light observed on earth at all! The plot shows the reason why: next to some of the larger gaps, there is a molecule. $\mathrm{H}_2 \mathrm{O}$ is listed in several spots. One gap is attributed to $\mathrm{O}_2$ and another to $\mathrm{CO}_2$, and one of those gaps is due to our nice, friendly, no-longer-getting-depleted ozone molecule, $\mathrm{O}_3$. It's the gap all the way over on the left side of the plot, and it's right where the UV light would have been shining down on us. We can see that it does indeed shine down on us outside of the atmosphere (the yellow), but not at the earth's surface (the red).
This tiny sliver of absence of UV light due to the $O_3$ molecule is absolutely essential for life on earth as we know it, and it's all thanks to electrons being pumped up in energy from one state to another. In this case it happens in a molecule, not an atom, but the same principle we learned in this lecture for electron transitions according to the Bohr model applies. There are discrete energy states where electrons are allowed to be in the $\mathrm{O}_3$ molecule, and transitions between these energy levels can happen when an electron either absorbs or emits a photon. Just like the Bohr model for the hydrogen atom, the discrete nature of electronic transitions in molecules means that absorption of light occurs only at distinct points in the spectrum. For ozone, there just happens to be an energy difference between two electron states that has the energy of a UV photon, which is why this particular molecule absorbs well in the UV.
Why this employs
I was thinking about mentioning jobs in spectral astronomy, since the Bohr model explained the observation of spectral lines from hydrogen and other light elements out in space. The same type of analysis, using basically just a (much) fancier version of the spectroscope you got in your Goodie Bag, is used today to understand the stuff that makes up distant stars, nebulae, galaxies, quasars, all sorts of other intergalactic matter, including whether exoplanets have earth-link atmospheric compositions.
But instead of going down that outer space route, I’d like to pick up here on the point I made in Why This Matters, regarding how solving the ozone depletion problem is a beautiful example of how important science policy can be in making the world a better place. What employment opportunities are there along such lines? “Science policy” is a broad term that can apply to a wide array of jobs and activities, from deciding how science is funded to how it can (or cannot) be translated into commercial products, to how science impacts human health to protecting the health of the environment. The jobs that result from this direction often involve a connection to politics and Washington (if in the U.S.) but they don’t necessarily have to. One way to get involved is through fellowships, where students can engage with policy-makers in DC for a summer or a semester.
There are a number of options along these lines, like the AAAS Diplomacy fellow (check out https://www.aaas.org/programs/scienc...cy-fellowships), or fellowships at the Science and Technology Policy Institute, or the White House Office of Science and Technology Policy (OSTP). And then beyond such general ones there are many specialized fellowships in science policy that go by the sub-field, like the American Geophysical Union Congressional Science Fellowship, or the various NOAA Sea Grants for ocean-related policy, or the John Bahcall Public Policy Fellowship for astronomy. And many more.
Beyond fellowships there are tons of jobs out there, like at science policy centers, think tanks or other organizations, from the Union of Concerned Scientists, to the RAND Corporation, to UNESCO, to the Federation of American Scientists, to the Center for Science in the Public Interest, to University-based centers like Arizona State’s Consortium for science policy and outcomes. These are only a very few examples – there are so many more. You could get involved directly with the science funding agencies like the NSF, DOE, or NIH (if you become a program manager at any of those please give me a call). Science policy is an incredibly important part of what science is, what it can become, and how it can be used most effectively, wisely, and for the good of the world.
Extra practice
1. What is the energy associated with the ground state in a lithium ion?
Answer
$E_n=-R_y \dfrac{Z^2}{n^2}=-13.6 \dfrac{3^2}{1^2}=-122.4 \mathrm{eV} \nonumber$
2. If a light wave has the same magnitude of energy as the lithium ground state, what are the frequency and wavelength of this light?
Answer
\begin{gathered}
E=h \nu=\dfrac{h c}{\lambda} \
122.4[\mathrm{eV}]=\dfrac{\left(4.14 x 10^{-15}[\mathrm{eVs}]\right)\left(3 \times 10^8[\mathrm{~m} / \mathrm{s}]\right)}{(\lambda[\mathrm{m}])} \
\lambda=9.9 \times 10^{-7}[\mathrm{~m}]
\end{gathered}
Lecture 5: Wave-Particle Duality and Quantum Mechanics
Summary
This chapter, we continued our discussion of ionization. We restricted our examples to ionization of hydrogen atoms in order to use Bohr's model of the atom. An electron in the ground state, or lowest energy state, of a hydrogen atom has an energy of $-13.6 \mathrm{eV}$. Therefore, this electron's ionization energy, or the energy required to free it from the nuclear pull and effectively transition it to energy level $\mathrm{n}=\infty$, is $13.6 \mathrm{eV}$. The ionization energy of an electron in an atom with atomic number $\mathrm{Z}$ can be found with the following formula:
$\left.\Delta E_{\text {ionization }}=-13.6 Z^2 \quad \dfrac{1}{n_f^2}-\dfrac{1}{n_i^2}\right)(e V] \nonumber$
Ionization energy can be thought of in terms of Einstein's photoelectric effect. In his groundbreaking experiment, Einstein shone light (streams of photons) of different energies onto slabs of metal. The photons had to reach a minimum threshold energy in order to ionize the metal atoms. It makes sense that metals with higher first ionization energies will be better able to hold on to their electrons. In other words, light with higher frequency will be necessary to ionize metals with higher first ionization energies. Photo-Electron Spectroscopy (PES) is a characterization technique that involves ionizing materials with photons and measuring the removed electrons' kinetic energies. The energy of the bound state of each removed electron can be calculated by finding the difference between the incoming photon's energy and the resulting kinetic energy of the freed electron.
The Bohr model couldn't explain quantization of electron levels; they were imposed in the model, but not understood. Through a series of experiments in the early 20th century, wave-particle duality was demonstrated. This meant that light waves were shown to behave as particles, and particles were shown to behave as waves. The most famous of these experiments was the doubleslit experiment, where electrons were shown to act like both waves and particles. Louis de Broglie showed that all matter acts like a wave and has a corresponding wavelength $(\lambda)$. This wavelength can be found using the de Broglie relation, where $\mathrm{p}$ is the object's momentum, which is equal to its mass times its velocity, and $\mathrm{h}$ is Planck's constant:
$\lambda=\dfrac{h}{p} \nonumber$
Erwin Schrödinger, with The Schrödinger Equation, described the relation between a wavelike particle's spatial distribution and its allowed energies. In the Schrödinger equation, quantization is a natural consequence of solving for a standing wave. This wavefunction $\psi$ does not have direct physical meaning, at least that we know of, but its square $|\psi|^2$ corresponds to the probability of finding the particle at any given location. Solving the Schrödinger equation for an electron in an atom gives a set of wavefunctions that correspond to the electron's energy levels and spatial probability distributions. Note the dramatic difference from the Bohr picture: now the electron can be anywhere there is a nonzero $|\psi|^2$ which is almost everywhere. The concept of exact position does not apply in the quantum world, since particles cannot have their position and momentum specified exactly, in accordance with the Heisenberg Uncertainty Principle.
Why this matters
Let's apply the de Broglie wavelength equation to an electron. We know the mass of an electron is $9 \times 10^{-31} \mathrm{~kg}$. Now put the electron in an electric field caused by a voltage of $100 \mathrm{~V}$. We know this will cause the electron to accelerate and gain some momentum. Specifically, the energy of the electron is: charge* voltage $=\mathrm{e}^* \mathrm{~V}=100$ volts $* 1.6 \times 10^{-19}$ Coulombs $=1.6 \times 10^{-17}$ Joules. This is because the unit volts is equal to units of $\mathrm{J} / \mathrm{C}$. So now we have an energy of the electron accelerated across $100 \mathrm{~V}$, and from that we can get its velocity, $v 6 \times 10^6 \mathrm{~m} / \mathrm{s}$. Next we can get its wavelength: $\lambda=h / m v=6.6 \times 10^{-34} J^* s /\left(9 \times 10^{-31} \mathrm{~kg} * 6 \times 10^6 \mathrm{~m} / \mathrm{s}\right)$, which when we realize the units of $J=[\mathrm{kg}]\left[\mathrm{m}^2\right] /\left[\mathrm{s}^2\right]$ we see gives us $\lambda=1.2 \times 10^{-10} \mathrm{~m}=0.12 \mathrm{~nm}$ for this electron.
This figure is in the public domain (created by NASA).
Remember the electromagnetic spectrum from last chapter? That’s light covering orders of magnitude of wavelength, only a sliver of which lies in the visible range (400-750 nm). But in order to use this light to see things, we must use a wavelength of light that is either equal or smaller in size to what we’re trying to see. Otherwise the light simply won’t pick it up (you can check out the physics of diffraction for more info on this, but for now, you can take my word on it). The correlation between the limit of what we can see and the wavelength of light we use to see it is why I like this other electromagnetic spectrum picture. On it is superimposed examples of objects the size of the given wavelength. If we want to use a type of light to see, say, a human, then we’d need to stay at or below wavelengths of 1 m since humans are about 1 m in size. That doesn’t mean we can see in the microwave region, but that’s just a limitation of our eyes not of what is possible with a type of light. With better eyes than ours, we’d be able to use microwaves to see people, but we wouldn’t, for example, be able to use radio waves since they have a longer wavelength than the thing we’re trying to see.
Now let’s go to the atomic scale: at the x-ray part of the spectrum, we’d need to use x-rays or gamma rays to see objects at that scale. But it turns out that not only can our eyes not see in that part of the spectrum, but even the devices and detectors we’ve built cannot see all that well. We can take x-rays and do so all the time, but only to see images on the mm resolution as opposed to nm. For the latter, we’d need to see an x-ray reflecting off of a nm-sized object and be able to measure it with precision, something we’re not able to do. That said, we do use x-rays to get atomic-scale information, but that’s a different thing all together and it has to do with the spacing between atoms in solids. We’ll learn all about that later in the semester. But here we’re still trying to actually “see” the object and the object is an atom. We can’t use x-rays—even though they’re at the right wavelength—and forget about gamma rays, because they’re even harder to detect. But what else could we use?
That’s why we did the electron wavelength calculation to start this Why This Matters section – using the fact that an electron is a wave, and by giving it the right velocity, we can make an electron have exactly the wavelength needed to see at the atomic scale. Electrons can be our “flashlight”! It also turns out they’re a lot easier to capture and use to construct images than x-rays. This discovery jump-started the whole field of electron microscopy and changed the way we see matter. It was Don Eigler and Erhard Schweizer who showed in 1989 that by seeing individual $\mathrm{Xe}$ atoms, they could arrange 35 of them to spell their company’s logo. This is definitely an expensive way to write a logo, but the point is that the accomplishment realized Richard Feynman’s dream of being able to put an atom wherever you want.
Speaking of Feynman, he was an incredible scientist and a masterful teacher. If you want to see an amazing lecture entirely devoted to the double-slit experiment, find and watch Feynman’s 1959 speech, “There’s Plenty of Room at the Bottom,” and you won’t regret it. This is where he laid out the dream of putting atoms anywhere you want, and that was truly the first vision of the field we now call nanotechnology. The reason it took 30 years to realize Feynman’s dream is that we were working in the dark, literally. If you can’t see atoms and molecules, it’s hard to control them or at least know that something you’ve done led to an atomic-scale control that you wanted. This type of control led to the nanotechnology revolution and it was entirely enabled by the ability of scientists to see materials at the nm scale, which as you now know was made possible by the wave nature of the electron.
Why this employs
It is true that we put humans on the human without resorting to quantum mechanics, evidence to how powerful and useful the classical view of the world is. But in order to truly understand the world, from the atom to the oceans and mountains to the universe to life itself, the quantum nature of it all must be understood and in some cases quantum effects completely dominate. Electrons and their interaction with light fall into this category: without a quantum description there’s no hope in explaining the interactions that take place. So let’s talk about jobs related to lasers.
Stimulated emission (first discussed in a paper by Einstein in 1917) involves the interaction between a photon and an atom that has electrons in an excited state. If the photon has just the right wavelength when it hits the excited atom, then the atom emits a second photon identical to the first. The LASER is “Light Amplification by Stimulated Emission of Radiation,” and it is entirely and only described by quantum mechanics since it relies on knowing the discrete energies levels of electrons in the atom. And by the way, any time you’re online, you’re using a laser since that’s how information travels through fiber optic cables.
There are so many job postings for “laser engineer” that it wouldn’t be possible to even begin to list (one site I just checked has 7000+ listing!). The reason is that lasers are used today in every walk of life. There are jobs with lasers in medicine, like in eye surgery, to industry where lasers cut and weld, to tracking systems for all sorts of applications, to research in new types of spectroscopy, to so many many more. So if you want a job that uses wave-particle duality to make coherent light, then that’s an awesome first step, but the next step would be to get more specific.
Take LIDAR as an example: this stands for Light Detection and Ranging, which is a remote sensing method that uses laser pulses to measure distances. LIDAR has completely revolutionized the ability of robots, like self-driving cars for example, to see. It's inside those bulky boxes on top of a self-driving car that spins continuously, giving 360 degrees of visibility as well as extremely accurate depth information (to $+/-2 \mathrm{~cm}$ ). It's still expensive ( $\ 75 \mathrm{~K}$ a pop for the car-top systems!), but because of how enabling LIDAR is, and how much the demand for this technology will grow, there are many jobs opportunities. These range from jobs at big player companies like Velodyne that's already selling systems to many markets, to smaller companies trying to innovate the technology like Luminar that promises a $\ 500$ LIDAR system in the size of a soda can. There's a whole range in between including self-driving car companies themselves like Waymo that have decided to build their own LIDAR systems. And this was just one tiny example of the vast number of LASER-based job opportunities.
Example Problems
1. You observe hydrogen electrons transitioning from $\mathrm{n}=3$ to $\mathrm{n}=1$.
a) What is the energy change in this transition? Is it absorption or emission?
Answer
The electron is falling to a lower energy state from a higher energy state, so it must be emission.
$\Delta E=-13.6 Z^2\left(\frac{1}{3^2}-\frac{1}{1^2}\right)=-12.09 \mathrm{eV} \nonumber$
We can double check our intuition: since the energy value we found is negative, the transition must correspond to emission.
b) What color light do you observe?
Answer
$\lambda=\frac{h c}{10.2[e V]}=1.22 \times 10^{-7} \mathrm{~m}=102.6 \mathrm{~nm} \nonumber$
This transition is in the UV.
c) How many possible transitions could an electron falling from $=3$ to $n=1$ complete? Which transition is responsible for the orange color?
Answer
There are two possible paths: straight from $n=3$ to $n=1$ or $n=3$ to $n=2+n=2$ to $n=1$.
Next, calculate the wavelength of the three possible photon emissions. We already know that from $\mathrm{n}=3$ to $\mathrm{n}=1$, the wavelength of the emitted photon was $102.6 \mathrm{~nm}$.
From $\mathrm{n}=2$ to $\mathrm{n}=1$:
\begin{aligned}
&\Delta E=-13.6 Z^2\left(\frac{1}{2^2}-\frac{1}{1^2}\right)=-10.2 \mathrm{eV} \
&\lambda=\frac{h c}{10.2 \mathrm{eV}}=1.22 \times 10^{-7} \mathrm{~m}=122 \mathrm{~nm}
\end{aligned}
This transition is also in the UV.
From $\mathrm{n}=3$ to $\mathrm{n}=2$:
\begin{aligned}
&\Delta E=-13.6 Z^2\left(\frac{1}{3^2}-\frac{1}{2^2}\right)=-1.89 \mathrm{eV} \
&\lambda=\frac{h c}{1.89 \mathrm{eV}}=6.57 \times 10^{-7} \mathrm{~m}=657 \mathrm{~nm}
\end{aligned}
This transition corresponds to orange light.
2. A red laser, a green laser, and a blue laser shine on different pieces of the same kind of metal, and detectors are set up to measure the presence of emitted electrons. For the red laser, no electrons are detected. For the green laser and the blue laser, electrons are observed flying off of the metal. Which of the following statements must be true?
i) The number of electrons emitted from the metal using the blue laser is greatest
ii) The kinetic energy of electrons emitted from the metal using the blue laser is highest
iii) The red laser would lead to electron emission if the laser intensity was increased
Answer
ii) is true, because the blue light has higher energy than green and red light, and the energy of electrons emitted is proportional to the energy of the photons hitting the metal surface
i) and iii) are incorrect because the number of electrons is related to the number of photons = light intensity, and light intensity does not affect energy of electrons emitted.
Lecture 6: The Atomic Orbital and Quantum Numbers
Summary
One of the main limitations of the Bohr model is that it can only describe an atom with a single electron. To find the properties of multiple electrons in an atom, it is necessary to solve the Schrodinger equation, $(K+V) \psi=E \psi$, where $\mathrm{K}$ is kinetic energy, $\mathrm{V}$ is potential energy $(-1 / r$ for an electron in an atom), $\mathrm{E}$ is the total magnitude of energy, and $\psi$ is the wavefunction, or orbital of the electron. The solution to the Schrodinger equation for an electron in a hydrogen atom has three separate components: $\psi=R(r) P(\theta) F(\psi)$. The full form of the solution form has three big implications:
1. Electrons don't really orbit around atoms, at least in the sense that planets orbit around the sun. The solution to the Schrodinger equation yields a wavefunction, which when squared, gives an expression $|\psi|^2=1$ for the probability distribution of the electron's location in relation to the nucleus. Further, though each electron has some radius it is most likely to be found, there is also a chance that it is much closer to or much further from the nucleus. For example, while Bohr model electrons with $n=1$ live at exactly $0.529 \AA$ from the nucleus, the wavefunction for a real electron in the hydrogen atom has occupational probability peaked at $0.529 \AA$ and smeared out on either side of that value. As the principle quantum number increases, the number of nodes - forbidden regions - increases, giving distinct bands around the atom where the electron is likely to be.
2. Four quantum numbers are necessary to fully describe a particular solution. The principle quantum number, $n$ is primarily responsible for the energy level of the electron. Just like in the Bohr model, $n$ can take any integer value from 1 to infinity. As $n$ gets larger, the total energy gets less negative (bigger). An electron is equivalently referred to as being in the $n^{t h}$ energy level and the $n^{t h}$ shell; the remaining quantum numbers describe subshells. The angular momentum quantum number, $l$, is primarily responsible for the shape of the orbital. The values that $l$ can take on depend on which shell the electron is in: the range goes from 0 to $(n-1)$. The magnetic quantum number, $m_l$, is responsible for the orientation of the orbital in space. Its values depend on both $n$ and $l$, and the range of $m_l$ goes from $(-l)$ to $(l)$. Finally, the spin quantum number, $m_s$ can have one of two values: $+1 / 2$ or $-1 / 2$, which is equivalently referred to as up or down.
3. Electrons that solve the Schrodinger equation are distinct from Bohr electrons. While there is only one electron per shell in the Bohr model, the Schrodinger equation gives degenerate solutions: as $\mathrm{n}$ increases, there are more and more orbital levels. Further, the Bohr model gives a fixed allowed radius for the electron to live, while the Schrodinger equation gives a distribution of probability that an electron is at a given distance from the center (and some forbidden zones!). Finally, though the Bohr model could only describe the behavior of one-electron atoms, the Schrodinger equation is much more broad: it can describe the electronic structure of an atom with any number of electrons.
We also discussed how the values for the $l$ quantum number get the letter names $\mathrm{s}, \mathrm{p}, \mathrm{d}$, and $\mathrm{f}$ corresponding to $l=0,1,2,3$, respectively. If we say an electron is in the $2 s$ orbital, then we know it corresponds to the quantum numbers $n=2, l=0, m_l=0$, and spin could be up or down. Other examples of sets of quantum numbers and the corresponding orbitals were covered and these hydrogen orbitals were shown as the building blocks of chemistry. That's because to describe an atom and its properties, we fill these orbitals from lowest energy up, with the number of electrons $\mathrm{Z}$ in the atom. This occupation follows a general trend that is the topic of the next lecture, but in this one we covered a very important rule: the Pauli Exclusion Principle. This rule states that no two electrons can have the same set of quantum numbers, meaning that two electrons and only two electrons can occupy any given orbital.
Why this matters
Quantum weirdness is one of the most important reasons why nanotechnology has gotten so much attention recently. Now, there has been quite a bit of press about nanotechnology over the last two decades, but one of my all-time favorite articles appeared long ago in the U.K. Telegraph: “How Super-Cows and Nanotechnology will Make Ice Cream Healthy.” I always remember that piece because of the way they define nanotechnology: they describe a company as “experimenting with nanotechnology, or the science of invisibly tiny things.” I suppose I like this quote because of how ridiculous it is, since first of all there are many different realms of science and engineering associated with tiny things we cannot see (nanotechnology is one of them, but how about nuclear physics with its quarks and gluons, or solar wind with its hot plasma particles that reach our earth only to be deflected by our planet’s magnetic field, or dark matter?). Second of all, I like this quote because it shines a spotlight – albeit incorrectly – on one hold-up in nanotechnology development, which was not being able to observe what was happening. This changed once we could use electrons as a way to illuminate matter, as discussed in last chapter’s Why This Matters.
The main point is that nanotechnology holds great promise. Not so much for the reasons that gave it fame in the early 2000’s (healthy ice cream being the least of it, we were supposed to have a space elevator by now!), but rather because of the chemistry that’s “under the hood” of nanotechnology. What does nanotech even mean, and why do we have a whole name for it? Let’s answer that with an example directly related to today’s lecture.
Images of bulk semiconductors, ice cream scoop, quantum dot © sources unknown.
We’ve been talking about orbitals of electrons in atoms and how the orbital is the probability cloud for the electron. Well what if we take that cloud, and we squeeze it? Or in other words, confine it so that it can’t take up the same volume as it typically would. There’s a name for that: quantum confinement. And when we confine quantum things, like in this case an electron in an atom, new properties emerge. Take color as an example. For a piece of bulk material, where “bulk” here just means it’s not nanosized, I might have something that’s kind of optically boring. It’s just a piece of gray material. But then I take out my nano-ice-cream-scooper and I make a little tiny chunk of the material. Suddenly, without changing anything except the size, I can tune the color to pretty much anything I want!
The fact that changing the size of this material allows us to tune a given property is quite incredible. Imagine if I were to take one of those strips of metal from your first Goodie Bag and break it into two pieces: you wouldn’t expect the color of the metal to change. But if you could break just a tiny piece of it off, like one with just 10-20 atoms in it, then that piece would change color. How weird is that? It’s because of this electron cloud running out of real estate when the amount of matter is small enough. When it’s nanoscale, the confined probability cloud changes the energy of the electron, which in turn changes the color of the material. We talked about electrons changing their energy levels in atoms as a way to either absorb or emit a photon corresponding to that energy shift. Well here we’ve got a new way to shift the starting and terminating levels themselves, simply by changing the size of the material. That’s an example of why nano is such a big deal.
The ability to control the properties of a material is one of the foundations of our modern era. As we’ve already learned, understanding the periodic table of the elements and all of the differences between the behavior of these elements gave scientists and engineers knowledge of the basic ingredients that are used to build our world. Going nano is kind of like adding a whole other dimension to the periodic table: it’s like each element can now do new things, take on new properties, and be utilized in new applications.
Why this employs
Quantum computing has gotten a lot of attention recently, and for good reason. Over the past decade, we have reached the point where quantum bits, or qubits, can be made and controlled experimentally. The computers we use today manipulate information in the form of individual bits, which are the famous 1’s and 0’s of the last 50+ years. But in a quantum computer, the quantum mechanical phenomena that we’ve been learning about in this chapter are used to manipulate information.
It’s qubits that hold the key since those are the keepers of quantum states, which have quantum numbers just like the electron states in an atom. When two qubits interact with one another we get much more than just 1’s and 0’s, instead we get all possible superpositions. It’s like being able to harness the probability cloud as a computing cloud where all possible answers are computed at once.
Quantum computing is getting a lot bigger and a lot more exciting, with many companies predicting commercial quantum computers by 2025. The Why This Employs idea for this lecture is to take those quantum states we’ve just discussed and run with them. . . all the way to Google’s Artificial Intelligence Lab, or to the team at IBM Q, or to D-Wave, or to the folks at Intel working on their Qubit Chips, or to any of the 63 companies currently listed at Wikipedia under “Companies worldwide engaged in the development of quantum computing.” A list that is growing rapidly. And that doesn’t even include doing research on this topic in a lab here at MIT :)
Example problems
1. Write the quantum numbers for each electron in a neutral carbon atom.
Answer
Note: the spin signs are selected arbitrarily, so it would be just as correct to flip them all. For
$\begin{array}{c|c|c|c} \mathrm{n} & \mathrm{l} & \mathrm{m} & \mathrm{m}_s \ \hline \hline 1 & 0 & 0 & 1 / 2 \ 1 & 0 & 0 & -1 / 2 \ 2 & 0 & 0 & 1 / 2 \ 2 & 0 & 0 & -1 / 2 \ 2 & 1 & -1 & 1 / 2 \ 2 & 1 & 0 & 1 / 2 \end{array}$
the $2 \mathrm{p}$ electrons, they must have the same spin due to Hund’s rule.
Lecture 7: Filling the Periodic Table and Our First Bond (it’s Ionic)
Summary
The Aufbau principle is a tool used to remember the (typical) order in which subshells are filled in a multielectron atom. Electron interactions complicate which subshells are filled first: shielding and orbital penetration affect which combination of electron energy and shape (quantum numbers $n$ and $l$, respectively), is the next-most-stable state. The Aufbau principle is often represented visually. Though the Aufbau principle generally gives the order that electrons fill in the subshells, it doesn't tell the full story. Previously, we learned that each orbital can hold two electrons: one spin up and one spin down. We also learned that each value of $l$, written here as $\mathrm{s}, \mathrm{p}, \mathrm{d}$, and $\mathrm{f}$, can have a different number of subshells based on the allowed values of $m_l$. The electronic configuration of any atom can be written using these guidelines: for example, potassium is $1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 4 s^1$. Often, the electronic configuration is shortened to include only the nearest fully occupied noble gas atom and the valence electrons: in this case, potassium is [Ar] $n= 4 s^1$.
When an atom has only a partially filled shell, Hund's rule comes in to play: all of the orbitals in the partially-filled subshell must be singly occupied before any are doubly occupied. For $\mathrm{C}$, which has 6 electrons and electronic configuration $1 s^2 2 s^2 2 p^2$, Hund's rule tells us that the electrons are distributed as shown in the diagram below.
If the elements are arranged by increasing number of electrons (and protons), periodic trends start to emerge, and electron filling explains the organization of the Periodic Table with blocks that correspond to the subshell being filled. The $\mathrm{s}$- and $\mathrm{p}$-blocks are “main group” elements while those in the $\mathrm{d}$-block are “transition elements” and “f-block” are “inner transition elements.” A larger principal quantum number $n$ corresponds to bigger orbitals and therefore bigger atoms: from this, we can predict that atomic radius should increase down a period in the periodic table. On the other hand, having more protons pulls electrons in towards the atom, so atomic radius gets smaller going from left to right across a row. If an atom loses an electron, its radius decreases: this kind of atom is called a “cation.” Similarly, if an atom gains an electron, the “anion” is bigger.
Finally, we learned that ionic bonds are the electrostatic attraction between two oppositely charged ions: they're formed by the transfer of one or more electrons between the ions. The strength of an ionic bond can be approximated by a Coulomb interaction: $E=k Q Q / r$. If the distance between two ionically bonded atoms is greater, the bond will generally be weaker! Just by looking at the size of the ions that make up the bond, we can infer a lot of information about the corresponding ionic solid.
Why this matters
We just made our first bond in $3.091$, and it's a strong one! Take a look at this table of lattice energies (in $\mathrm{kJ} / \mathrm{mol}$ ) for a bunch of ionic solids. The range of ionic bond strength spans an order of magnitude in this chart, but even the weakest solid listed (potassium iodide, $\mathrm{KI}$, at $649 \mathrm{~kJ} / \mathrm{mol}$ ) is still really strong with a melting temperature of $681^{\circ} \mathrm{C}$.
The strongest ionic solid listed in this table is aluminum oxide, formed by combining $2 \mathrm{Al}$ atoms for every 3 $\mathrm{O}$ atoms (this balances charge with $\mathrm{Al}$ being $3^+$ and $\mathrm{O}$ being $2^-$). Aluminum oxide melts at a whopping $2,072{}^{\circ} \mathrm{C}$, and its widely used for its hardness and strength. For example, $\mathrm{Al}_2\mathrm{O}_3$ is used as an abrasive, including as a much less expensive substitute for diamond, and in many types of sandpaper or cutting tools. Because of its strength, it's used as an additive in many applications, including toothpaste! For this Why This Matters moment, I'd like to focus on an unfilled need rather than an established solution. This is a problem where a harder material with the right properties and in the right structure could change millions of lives.
Hemodialysis is a process to purify the blood of a person who has kidneys that don't work correctly. A typical patient will have to go in 3 times per week, sit in a chair that looks something like this for about 4 hours per session, totaling 600 hours of their time and a cost of $\ 80,000$ undergo hemodialysis, and this number has been growing steadily over the past 30 years. It is expected that, as the incidence of cardiovascular diseases such as diabetes and hypertension increase, the population of individuals needing dialysis will grow as well. That machine on the left of the chair in the picture is a hemodialysis machine. During the process blood is removed from the body through an access point in a patient's vein, and passed through the machine which removes toxins and excess fluids, after which the filtered blood is returned to the body through another access point back into circulation. Here's Why This Matters.
If that machine could be shrunk down to a much smaller size and made portable, then patients would be able to do this at home or while traveling or wherever they are, which would seriously improve their quality of life. But current filters don’t work well enough to make the device small and portable, so people who need hemodialysis treatment are stuck. That is, unless a new filter could be invented, one that satisfies three key criteria: 1) the material must be biocompatible, 2) it must allow for selective filtration so it can remove some things but not others, and 3) it must provide efficient and fast filtration. Such a filter does not currently exist. Current filters are cellulose-based porous membranes with highly variable pore size, meaning that molecular selectivity is difficult to control. The pores in current filters also overlap a lot, making tortuous paths that slow the fluid motion, increasing the amount of time it takes for the blood to flow through. Current filters are easily clogged, they’re delicate and tough to clean, so at best they’re single-use only and at worst they get clogged so much so quickly that even achieving a single use is difficult.
Image of alumina filter by Assaud, L. et al. Beilstein Journal of Nanotechnology 5 (2014): 162-72. License: CC BY.
Here we can consider the strong ionic bonds we learned about in this lecture. Here's a picture of a new filter made out of $\mathrm{Al}_2 \mathrm{O}_3$. Note from the scale bar that the pores are seriously small, only 10 s of nanometers in diameter. They're also aligned in the same direction (so none of that tortuous flow slowing things down), which means the flow through them will be fast and efficient. And note that the pores are all roughly the same size, which gives the filter the ability to sharply reject the targeted toxins. And of course, since it's a strong ionic solid, this filter can be cleaned very aggressively with pretty much whatever cleaning chemicals or thermal treatment you care to throw at it. There's still much work to be done, like reducing the cost of making such filters, and they're still a bit too brittle, but in general ionic solids hold tremendous potential as next-generation filters!
Why this employs
That example of a new type of ultra-resilient filter is only one of many where new high strength materials are needed. And it’s not just the material itself but also the way it’s processed. And because of how badly resilient new materials are needed, many of the larger chemical companies like Dow, Saint-Gobain, BASF, Sinopec, LG Chem, Arkema, or Mitsubishi Chemical, to name just a few of the giants (all of these have annual sales in the $\ 10$’s of billions.
And all of these companies have large teams working on making high strength materials. Sometimes it’s high strength steel, sometimes it’s plastic, and of course other times, ionic materials. All of these companies have job postings, but an excellent way to explore career interest is to do internships, and all of these companies also have internships for students. You could go directly to the company web site, or you could also check out the resources we have here at MIT. There are a lot of them.
For example, take a look at the web page capd.mit.edu, where if you click on “Jobs and Internships” you’ll get all sorts of support and resources. Definitely worth checking out!
Example Problems
1. Rank the following atoms in terms of size, ionization energy, and electronegativity:
$\mathrm{Ge}$ $\mathrm{Rb}$ $\mathrm{Sn}$ $\mathrm{I}$
Answer
As the period increases down the periodic table, electrons are more and more shielded from the nucleus, so atomic size increases. Additionally, as the group number increases across the periodic table, there are more and more protons and therefore increased attraction from the nucleus, so atomic size decreases.
(smallest) $\mathrm{Ge}$ $\mathrm{I}$ $\mathrm{Sn}$ $\mathrm{Rb}$ (largest)
2. Below is a PES spectrum for the outer electrons of a mystery element. Label the axes and the direction of increasing ionization energy.
a) If the configuration of the inner electrons of the mystery element are the same as neon, identify the element.
Answer
Phosphorous: its electronic configuration is $[\mathrm{Ne}] 2 \mathrm{~s}^2 3 \mathrm{p}^3$
b) Separately, while you were taking PES spectra for carbon in a vacuum, the seal on the vacuum machine broke and let in an unknown gas. You observe the following on the contaminated spectrum:
i. There are 6 peaks total (including the carbon peaks)
ii. Five peaks are the same height
iii. One peak is three times higher than the others
What is the contaminating element? Explain your reasoning.
Answer
The electronic configuration of carbon is $1 \mathrm{~s}^2 2 \mathrm{~s}^2 2 \mathrm{p}^2$. Since there are 2 electrons in each shell, these three peaks must be the same height. The additional peaks must therefore belong to $1 \mathrm{~s}^2 2 \mathrm{~s}^2$ $2 \mathrm{p}^6$, since one of the peaks is three times higher than the rest. The contaminating gas is $\mathrm{Ne}$.
Lecture 8: Ions, Ionization, and Valence
Summary
This lecture we continued our discussion of ionization energy. We saw that, similar to $\mathrm{Li}$ and $\mathrm{Be}$ in the last lecture, the ionization energy for the elements along the periods (from left to right on the periodic table) in the $\mathrm{p}$ and $\mathrm{d}$ block increases due to a higher number of protons leading to an increased nuclear pull. It makes sense that more energy is necessary to free them. On the other hand, going down the groups (from the top to bottom on the periodic table) the ionization energy decreases because the increased shielding and distance from the nucleus is more significant than the increased nuclear pull due to added protons.
The reason for increasing ionization energies for electrons closer to the nucleus is that these electrons are more tightly bound. For example, Li’s first ionization energy is 5.4 $\mathrm{eV}$, while its second ionization energy is 122 $\mathrm{eV}$. This second electron experiences less shielding than the first. This ionization energy information can be viewed on a Photo-electron Spectroscopy (PES) plot. This is a plot showing ionization energy versus relative intensity for a given element. PES involves shining enough energy on an atom to ionize all of its electrons, measuring their kinetic energy, with the difference being equal to how much energy the electron was bound by. PES is a simple, powerful technique to directly measure electron occupation.
Another periodic trend discussed was electron affinity, or the change in energy that accompanies an additional electron. The halogens (group 17 on the periodic table) have the highest electron affinity, because an additional electron would result in a full octet and complete filling of the subshell. Electron affinity increases along the periods of the periodic table up until the halogens.
This lecture also laid out the idea of valence electrons. These are the electrons that occupy orbitals beyond the highest filled noble gas core. For example, carbon and silicon both have 4 valence electrons. Carbon's highest filled noble gas core is [$\mathrm{He}$], while silicon’s is [$\mathrm{Ne}$]. Valence electrons are what drive chemical bonds. Valence electrons on atoms are depicted using Lewis Dot Diagrams, which are the chemical symbols of the elements with dots surrounding them that represent the valence electrons. Lewis Dot Structures are helpful in visualizing the octet rule, which states that atoms (ignoring the transition metals for now) are at their lowest and most stable energy state with 8 electrons in their valence.
In this lecture we began applying Lewis Dot Structures to ionic bonds. Cations are depicted with plus signs on their top right corner, and anions with negative signs, as shown below.
Why this matters
In this lecture, we’ve been focusing on atoms that lose or gain electrons, also called ions. Now we’ll discuss how this topic relates to renewable energy solutions.
Take a look at this plot of the demand of electricity (in blue) vs. supply of electricity from a wind farm (in red), recorded over a two and a half month period in 2013 in Denmark. Notice that on the demand side, there’s some variability, but there’s also a lot of regularity. The 5-day work week draws more energy than the weekends, and the amount of energy needed doesn’t really vary much from week to week. Now contrast that with the supply of electricity from a wind turbine farm. Even if you put the turbines in a known windy area, there’s still a huge amount of variability. Wind energy is not that expensive to produce, but the biggest challenge isn’t cost, it’s steadiness of supply. If you think about how to build enough of this type of unreliable supply to meet the consistent demand, you’d need to overbuild capacity by a factor of at least 10 (and even that wouldn’t be enough on some days)! That of course isn’t going to be economically feasible, so the real solution, if we want to expand how much energy we get from intermittent renewables like wind and solar, is to be able to store the energy for when it’s needed. If we have reliable storage, then the supply can be quite unreliable and we’d still be able to meet a steady demand, as long as on average the supply can meet it.
One of the easiest ways to store energy is to push something up a hill and then let it roll back down. Push a ball up a hill and then hold it there, and when you want the stored potential energy back as kinetic energy, simply let it go. We pump water up hills, called “pumped hydro,” which is currently one of the only viable technologies that can scale up to the size (meaning, power) of actual electricity grids. A massive amount of water is pumped up a hill when there’s excess energy supply, and then rolled back down to turn a turbine to generate electricity when there’s excess demand. Although this is a relatively cheap storage solution, there are all sorts of problems with scaling up any further than it is today, primarily because of the low energy we get from gravitational energy and therefore how much land is required. There’s a great post that explains why it will be hard to increase pumped hydro as an energy storage technology at dothemath, which is a cool site in general if you haven’t seen it: https://dothemath.ucsd.edu/2011/11/p...p-the-storage/.
So what else can we pump up a hill to store energy? This is where our ions come into the picture! Ions are the key ingredient in batteries. Think of a battery as an electronic hill, and when the battery is charged the ions are getting pumped up. Cations in particular, like $\mathrm{Li}+$, get pushed up the energy hill, and the lower energy state (down at the bottom of the hill) would be one in which it can get that electron back and become neutral again. When the circuit is closed, the cation falls down the energy hill to meet up with the electron, which goes around the circuit doing work along the way. In this case instead of mechanical work like the water, the electron is doing electrical work directly. This is one of the great advantages of electrochemical storage (a nice way of saying a battery), namely that we’re storing directly the kind of energy we mostly need, electrical energy. Stored mechanical or thermal energy later often needs to be converted into electrical at great efficiency loss.
But how good are batteries today, really? Could they ever be used to store the massive amounts of energy needed at "grid scales"? For reference, the U.S. uses 3 TW on average, or $3 \times 10^{12}$ Watts $= 3 \times 10^{12}$ Joules/second of energy, so you get a sense of what it means to be at the scale of the grid. Let's make sure we're straight on energy vs. power: a Watt is a Joule/second which is energy per time, which is power. That's an important metric since we need to be able to access energy at a given pace. The total amount of energy is power multiplied by time, so for the U.S. if our average power draw as a nation is $3 \mathrm{TW}$, then our daily energy use is $72 \mathrm{TW}$-hours (TWh) and our yearly energy use is $26,280 \mathrm{TWh}$. Just to give you a sense of how big this number is, a typical AAA battery stores about a single Wh of energy. That means if we wanted to store the U.S. energy needs all we would need to do is build a storage facility containing $26,280,000,000,000,000$ AAA batteries! In case you're wondering, that's about a million times the number of AAA batteries sold in the U.S. each year.
And therein lies the problem with batteries, namely their low energy density. But as I hope you’re seeing in this class, such problems are often opportunities for chemistry to come in and make a game-changing difference.
This figure shows the energy per weight vs. energy per volume for a few different storage materials. Note that batteries are in the lower left, which isn’t so good. You can see why gasoline is so unique: it’s got a very high energy density both by weight and by volume, and on top of that it’s safe and easy to transport and extremely cheap. The competition is tough.
One liter of gasoline stores 33 megajoules ($\mathrm{MJ}$) of energy, or about 9 $\mathrm{kWh}$ if you prefer those units. That liter costs about a dollar. Now take one MIT Professor, who can output about 60 $\mathrm{W}$ or 60 $\mathrm{J/s}$ on a good day. This means that the professor can generate the same amount of energy as in a liter of gasoline, $33 \mathrm{MJ}$, in roughly 153 hours. Given the average pay of an MIT professor of $\ 10 / \mathrm{hr}$, it would cost $\ 1530$ to get the energy from the professor. Compared to about $\ 1$ to get the same amount of stored energy from the gasoline. Batteries need to start looking a lot more like gasoline and a lot less like an MIT Professor if they're going to be competitive grid-scale energy storage technologies.
The good news is that there has already been tremendous progress in just the last 10-20 years. Check out this plot of energy density vs. year for batteries. Notice the very slow progress for about 150 years, with a doubling of energy density occurring roughly every 60 years. Then the slope of progress changes, and the reason is simple: chemistry. Specifically, the rapid improvements are happening because we’ve reached an era wherein we can make energy hills for ions with many more types of materials than ever before. I’ve added to the plot the number of materials cathodes are made of: historically only about 10 different cathode materials were used during those slow-paced first 150 years, but now we can and do make cathodes out of over 80 different materials, and the number is growing by the week. This explosion of materials possibility is due to a deeper understanding of ions and new ability to design materials with more efficient energy landscapes for them to traverse. So yes, there’s a lot of work to do to make batteries competitive in grid-scale energy storage, but the time is absolutely ripe to get it done.
Why this employs
Knowledge of ions can get you all sorts of jobs. But since I focused on the battery field in Why This Matters, let’s focus on that same topic here. The global battery market is already enormous and its projected growth is stunning. Here’s a chart from a report from Deutsche Bank from 2016 showing the projected near-term growth in battery production by some of the top companies that make batteries. All of these companies are hiring thousands and thousands of scientists and engineers, techno-economists (predicting tech trends is not easy!), packaging and mechanical design experts, and much more, all to work on various aspects of one thing: the ion. And there are many more jobs than that. Big battery companies like those listed in the chart are investing heavily in hiring people to work on batteries, but so are little companies. And so are many research teams at universities, who are spinning out companies based on new concepts.
Not only are there many different battery chemistries that have been around for a while and are being greatly improved, from lithium-ion to nickel-metal-hydride to lead-acid, but there are also so many exciting completely new directions for batteries. All of these future battery technologies involve jobs, and while it’s hard to list them all because of how much the field has grown, here are a few cool directions companies are pursuing:
• Silicon-based anodes (replacing the currently used graphite) which would increase energy density by a factor of 3
• Using metal nanowires in the electrolyte to prevent breakdown, making the battery last 10 or 100$\mathrm{x}$ longer
• Making the electrolyte a solid instead of the current liquid, which would make the battery more stable and allow much faster charge times •
• Throwing graphene into the battery since, anyway, what problem can graphene not solve? (ok, but in all seriousness, graphene has been shown to allow for much faster charge/discharge)
• Making batteries out of foam metals
• Making batteries foldable (paper-thin but that won’t break) and also stretchable
• Aluminum air batteries that have demonstrated a 1,100 mile range in an electric car
• Powering the charging of batteries with various means, from wi-fi signals (not inductive wireless, which is something else), to ultrasound, to amino acids, to urine. Batteries made from sodium, zinc, sulfur, magnesium, and so many other materials!
• Making batteries that can never catch fire
Example problems
1. Choose the larger atom from the following pairs:
a) $\mathrm{Rb}, \quad \mathrm{Rb}^+$
Answer
$\mathrm{Rb}$ is larger, because it has an additional shell $\left(\mathrm{Rb}^{+}\right.$ loses its $5 \mathrm{~s}^1$ electron)
b) $\mathrm{Rb}^+, \quad \mathrm{Kr}$
Answer
$\mathrm{Kr}$ is larger, because it has one less proton than $\mathrm{Rb}^{+}$, so the nucleus pulls less on the electrons
2) Draw the Lewis dot structures of $\mathrm{FeCl}_3$ and $\mathrm{MgF}_2$.
Answer
Lecture 9: Lewis Structures, Covalent Bonds, and Resonance
Summary
Covalent bonds are created via the sharing of electrons rather than through electrostatic attraction between cations and anions, as an ionic bonds. The shared electrons in covalent bonds achieve a lower and more stable energy state by interacting with two nuclei instead of one. Lewis Dot Structures represent the two electrons in a covalent bond with lines, as shown for $\mathrm{H}_2$ to the right.
1. Connect atoms, central often less electronegative
2. Determine total number of valence electrons
3. Place bonding pair of electrons between adjacent atoms
4. Starting with terminal atoms, add electrons to each one to form octet (2 for $\mathrm{H}$)
5. If electrons are left over, place on the central atom
6. If central atom hasn’t reached octet, use lone pairs from terminal atoms to form multiple bonds to the central atom to achieve octet
It is possible to have more than one Lewis structure that doesn’t valuate the octet rule. For these instances, we calculate the formal charge. The sum of the formal charge must equal the overall charge on the molecule or ion. To calculate formal charge, follow these steps:
1. Nonbonding electrons are assigned to the atom on which they are located.
2. Bonding electrons are divided equally between the bonded atoms.
3. Then, the formal charge can be computed for each atom as follows:
$\text { formal charge }=\text { valence } e^{-}-\left(\text {nonbonding } e^{-\prime} s+\frac{\text { bonding } e^{-\prime} s}{2}\right) \nonumber$
In addition to stability, formal charge tells us about the structure’s electronegativity—atoms with negative formal charge should be more electronegative, and those with positive formal charge less electronegative.
It’s also possible to have two different Lewis structures with the same formal charge. This indicates that the actual molecule is a combination of its most stable resonance structures, as with the ozone example in class, where we have on average 1.5 bonds on either side of the center oxygen.
We also discussed the concept of polarity - a measure of how unequally electrons are shared between two atoms in a bond. In nonpolar covalent bonds, the electrons are shared equally, as in $\mathrm{H}_2$.
In polar covalent bonds, the sharing is unequal. Electronegativity $(\chi)$, the tendency of an atom to attract a shared pair of electrons toward itself, can be used as a measure of polarity in a bond. In a completely nonpolar bond, the two atoms will have identical electronegativities, meaning $\Delta \chi=0$. In a strong ionic bond, there is a large difference in electronegativities. In $\mathrm{CsF}$, using Pauling's scale, $\Delta \chi=3.98-.79=3.19$.
Finally, we named some ways that the octet rule can be broken. Some elements, like boron, can have fewer than 8 electrons. Others, like sulfur, can have an expanded octet.
Why this matters
Speaking of the $\mathrm{CO}_2$ molecule and its formal charges and stable structure, that's the focus of this Why This Matters. Let's get this done up front: after a century and a half of experiments and data collection, the scientific community has established with enormous consensus that carbon dioxide emissions are the primary cause of climate change. Other theories such as solar cycles or volcanic activity do not correlate with observed temperature changes, and there is now only one consistent explanation for rising global temperatures and our changing climate: anthropogenic greenhouse gases, led by carbon dioxide.
By the way, if anyone (say, I don't know, a politician?) were to tell you something like, "There is tremendous disagreement as to whether the earth is round. The debate over the earth's shape should be encouraged - in classrooms, public forums, and the halls of Congress," you would call such a remark either obliviousness or a lie. This is the same kind of lie told by many people about the facts of climate change. A great deal of media coverage focuses on how to define such "alternative facts," but categorizing the remark is beside the point. Regardless of what we call it, this type of statement is a threat to the very foundations of science, to reason-based argument, and therefore to society. Politicians may live in a "post-fact" world, but scientists do not, and cannot. There is no "post-fact science."
Ok, back to our molecule. Why is this particular molecule so important for climate change and global warming? $\mathrm{CO}_2$ is called a "greenhouse gas" because of its ability to absorb infrared radiation, which in turn warms the planet. Take a look back at the plot of intensity of solar radiation hitting the planet as a function of frequency, from Why This Matters in lecture 4. Yellow is what strikes the upper atmosphere and red is what hits earth's surface, so the difference between the yellow and red curves is where some of that solar energy is being absorbed. Small molecules in the atmosphere, like ozone, water, and carbon dioxide, are what absorb that incoming solar energy. You can see where $\mathrm{CO}_2$ is absorbing, out towards the right in the infrared part of the spectrum, while ozone is over on the left in the UV, and water is everywhere. It's the strength of IR absorption that makes $\mathrm{CO}_2$ such an effective greenhouse gas.
About $25 \%$ of the solar radiation that hits the earth (so $25 \%$ of the integrated red in the plot) gets reflected back out to the universe. But what is reflected is not what is absorbed: instead, most of the energy that is reflected is in the longer wavelength, infrared part of the spectrum. Since $\mathrm{CO}_2$ absorbs efficiently in that part, it acts like a blanket, keeping some of that reflected energy from escaping and redirecting it back to the atmosphere and surface. This blanket keeps us at a cozy $15^{\circ} \mathrm{C}$ instead of $-18^{\circ} \mathrm{C}$, which is what the average temperature on earth would be if all of that reflected solar energy were able to escape.
As an aside, the term “greenhouse gas” is a bit of a confusing name that has unfortunately stuck. The effect of absorbing long wavelength radiation on the planet’s temperature has nothing to do with the effects of warming in a greenhouse used to grow plants, the latter relying mostly on preventing air convection from carrying heat away.
Here's the connection back to our Lewis structures and formal charge. That efficient absorption of $\mathrm{CO}_2$ in the infrared I mentioned, that is directly related to how the atoms in the molecule vibrate. The photon is absorbed because it gets transferred into vibrational energy, so the specific way atoms jiggle around is extremely important to what photons it can absorb and how efficiently it does so. If the $\mathrm{CO}_2$ molecule were to have a triple bond and a single bond, which is a valid Lewis structure but with higher formal charge, then its patterns of vibrations would change dramatically compared to the more stable double-bond case (with formal charge $=0$), which in turn would make it much less efficient at absorbing all the infrared energy!
Why this employs
Today’s lecture has been about the covalent bond, a convenient way to represent it with Lewis structures, and what we can learn about molecules from these structures. It’s hard to come across a job in any engineering or science field that doesn’t require at least some understanding of the covalent bond since it’s so ubiquitous. This also makes it hard to zero in on a specific job market for Why This Employs. That’s why I’m going to focus way down to a single material – carbon – and the jobs you could find related to its covalent bonds.
In a few lectures we’ll be learning about molecular orbitals and as an example you’ll see just how flexible carbon can be in its bonding environment. It’s one of the most flexible elements in the whole PT in that it likes single, double, and triple bonds as well as anything in between (read: resonance structures and delocalization). Let’s pick one of the many forms carbon takes to focus on here: the aromatic ring. The term aromatic does have its origins in the fact that aromatic molecules can have a sweet fragrance to them, but in fact benzene—the classic aromatic molecule— has no such smell. Instead, the way the term is used in chemistry is that the electrons in a ring of carbon have lowered their energy by delocalizing (yes, resonance structures!). Most often this occurs for a ring that has alternating single and double bonds, which then form an average over its resonance structures. Benzene is the classic example of this, as it is the simplest aromatic ring one can make.
And you’ve probably heard of benzene before, but what you may not know is just how many ways benzene can be modified to make or do something useful. Check out this chart of only a very small fraction of benzene derivatives, where the ring has some addition(s) to give it a specific property or functionality. This massive variation of the chemistry of a single molecule is possible largely because of that covalent bond that carbon makes.
Many materials we encounter on a daily basis are manufactured from derivatives of benzene. Polystyrene, for example, is a polymer formed by connecting styrene molecules together, and billions of kilograms of polystyrene are produced each year. So, certainly jobs in companies that involve manufacturing, recycling, or utilizing plastics like polystyrene are related (I’m thinking a senior chemist position at BASF, a chemical manufacturing company, for example). But the uses of benzene derivatives go far beyond plastics. For example, industries use benzene to make resins, nylon and many different synthetic fibers, as well as lubricants, rubbers, dyes, detergents, drugs, and pesticides. Take medicine as another example: here’s the acetaminophen (Tylenol) molecule – it’s a benzene derivative!
So the pharmaceuticals industry is also a great place to take your knowledge of benzene (I’m thinking Johnson and Johnson, Pfizer, Novartis, and so on). These companies invest billions and billions of dollars every year to fund research on the development and testing of new medicines, many of which have at their core a resonant Lewis structure.
Example Problems
1. Draw the resonance for $\mathrm{CHO}_2^{~-}$ and assign formal charges to each atom.
Answer
2. True or false: The following picture shows two resonance structures of a molecule
Answer
False. The two structures aren’t resonant structures because the atoms move around.
Further Reading
Lecture 1: Introduction and the chemistry of the periodic table
• History of atoms (not tested):
https://courses.lumenlearning.com/trident-boundless-chemistry/chapter/history-of-atomic-structure/
• Balancing reactions and limiting reagent:
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_(Inorganic_ Chemistry)/Chemical_Reactions/Limiting_Reagents
Lecture 2: Counting Atoms and Organizing the Elements
• The mole and molecular weight:
https://opentextbc.ca/chemistry/chapter/3-1-formula-mass-and-the-mole-concept/
• More info on careers in food science and agricultural chemistry:
https://www.acs.org/content/acs/en/careers/college-to-career/chemistry-careers/agricultural-and-foodhtml
Lecture 4: The Bohr Model and Electronic Transitions
• More on Black Body Radiation, quantization, and a cool video on the photoelectric effect:
https://chem.libretexts.org/Courses/University_of_Arkansas_Little_Rock/Chem_1402%3A_ General_Chemistry_1_(Belford)/Text/6%3A_The_Structure_of_Atoms/6.2%3A_Quantization% 3A_Planck%2C_Einstein%2C_Energy%2C_and_Photons*
• Summary of emission and absorption spectra:
http://physicsnet.co.uk/a-level-physics-as-a2/electromagnetic-radiation-quantum-phenomena/ line-spectra/
• Science in context—Einstein and other prominent scientists fled Nazi Germany (not tested):
https://physicstoday.scitation.org/do/10.1063/PT.6.4.20180926a/full/
Lecture 6: The Atomic Orbital and Quantum Numbers
• Quantum numbers:
https://courses.lumenlearning.com/cheminter/chapter/quantum-numbers/
• The Schrodinger Equation and Particle in a Box (not tested):
http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/schr.html
• Quantum mechanics crash course:
https://www.youtube.com/watch?v=qO_W70VegbQ&list=PL8dPuuaLjXtN0ge7yDk_UA0ldZJdhwkoV& index=45
• Introduction to Quantum Mechanics by David J. Griffiths, available here or at your local MIT library:
https://www.cambridge.org/core/books/introduction-to-quantum-mechanics/990799CA07A83%20/%20FC5312402AF6860311E
Lecture 9: Lewis Structures
• Drawing Lewis Dot Diagrams: Extra practice:
http://www.chem.ucalgary.ca/courses/351/Carey5th/Ch01/ch1-3depth.html
• Video tutorial:
https://www.youtube.com/watch?v=ulyopnxjAZ8 | textbooks/chem/Inorganic_Chemistry/Introduction_to_Solid_State_Chemistry/05%3A_CHEM_ATLASes/5.01%3A_CHEM_ATLAS_1.txt |
Lecture 10: Shapes of Molecules
Summary
This lecture focused on the Valence Shell Electron Pair Repulsion (VSEPR) model. This model allows us to predict the shapes of molecules—a precursor to more fully understanding their properties. VSEPR is based around electron repulsion, and gives the most stable structures as those that minimize these repulsions. To find the VSEPR representation of a molecule, follow these steps:
1. Write Lewis structure
2. Classify each electron pair as bonding or nonbonding
3. Maximize separation between domains
4. Give more space to non-bonding domains and to bonding domains with higher bond order
A bonding pair (BP) of electrons is any two electrons that take part in a bond. A double bond is made up of 2 BPs, and a triple bond is made up of 3 BPs. A lone pair (LP) of electrons is any two electrons that are not part of a bond. The strength of repulsion between electron pairs, in ascending order, goes as follows: BP-BP, BP-LP, LP-LP. Also, the repulsion of a single BP is less than that of 2 BPs or 3 BPs.
VSEPR Geometries by Boundless Chemistry. License: CC BY-SA. This content is excluded from our Creative Commons license. For information, see https://ocw.mit.edu/fairuse.
Why this matters
The average human can discriminate between 4,000 and 10,000 different odors. (Before we get excited about how awesome that is, consider that a dog can smell between 10 and 100 thousand times better than a human!). But what is smell, from a chemistry perspective? Taste and smell are related, and thinking about what’s behind them goes back to the ancient Greeks. None other than our friend Democritus (atomism!) speculated that the taste of a substance was due to the shape of its component particles. He thought that acidic particles would be sharp, since they felt like they attacked your mouth, while sweet stuff was made of nice cuddly soft shapes. While his reasoning was quite simple, the idea that taste and smell were governed by shape was amazingly prescient.
Fast forward 2500 years and we now know that the ability to taste and smell works through “receptor sites” in the tongue and nose. Here’s the tongue broken down, showing the receptor site on the right, and below that is a larger view of receptor site with various parts labeled. Note that the receptor site is also what we call a taste bud. Signals that get sent from this receptor site to the brain through the nerve fibers determine what you taste, and the signal is deeply dependent not just on the composition of the molecule itself, but also on its shape.
Take a look at the two molecules below, glucose which tastes sweet, and quinine which tastes bitter. Their chemistries are different and the way the chemistry of the molecule bonds to the sensory cell is crucial, but the way the shape fits into the pore itself and impacts the orientation of the molecule on the sensory cell can be equally important. If we didn’t know the shapes of these molecules, we’d write them out as simple 2D Lewis structures, but it’s those 3D shapes that you see in the figure that distinguish their tastes!
In the case of the carvone molecule, or $\mathrm{C}_{10}\mathrm{H}_{14}\mathrm{O}$, we have an even more striking example of the role of shape on smell. This molecule forms two mirror images, denoted $\mathrm{R}$ and $\mathrm{S}$ in the figure below. The $\mathrm{R}$ form smells like spearmint while the $\mathrm{S}$ form smells like caraway seeds. Same exact chemistry, different shape, different smell. Many molecules can take on two forms like this with mirror symmetry but that aren’t the same, and such pairs are called enantiomers. Actually, beyond molecules, enantiomers can be anything. Your hands, for example, are enantiomers. If you hold them up facing one another you’ll see they have mirror symmetry, and if you try to rotate one around the other you’ll see you can’t superimpose them. For this reason, the property of having mirror symmetry but not being superimposable is called handedness, or chirality. The fact that the two enantiomers are perceived as smelling different shows that those receptor cells must contain chiral groups, allowing them to respond more strongly to one enantiomer than to the other. Thus, both for the molecule being smelled and for the molecules used to do the smelling, molecular shape holds the key!
Why this employs
It’s about time we talk about enzymes. These are molecules, often proteins, folded into a specific (and complicated!) shape, that speed up chemical reactions in your body. Enzymes are absolutely essential for so many crucial functions of our body, including respiration, digestion, muscle and nerve function, and many more. In digestion, the role of an enzyme molecule can speed up a needed reaction by a factor of a million times! That lets you digest your dinner in hours rather than, well, a thousand years.
Enzymes work by binding to molecules in a specific way, and you may have already guessed that shape is crucial. In fact, way back in 1894 it was the Nobel Laureate Emil Fischer who came up with the ”lock and key” model to explain how enzymes work. The idea in this model is pretty much how it sounds: an enzyme’s active site is a specific shape, and only the substrate will fit into it, like a lock and key. Here’s a cartoon to illustrate how the enzyme perfectly fits, because of its shape, into a substrate, giving the substrate the power to make reactions happen faster (also called catalysis). The model has been updated over the past 100 years, for example to include the fact that the substrate and enzyme itself are dynamic and can change shape when they interact, or that the effects of the surrounding solvent are important, but the key feature that shape is crucial remains the foundational principle upon which enzymes operate.
As you may have noticed by just going to the grocery store, there’s a big market for new foods with new enzymes. There are so many enzymes involved in digestion, like lipases that help digest fats in the gut. amylase helps change starches into sugars. Maltase breaks the sugar maltose into glucose (it’s in potatoes, pasta, and beer, for example). Trypsin breaks proteins down into amino acids. Lactase breaks down lactose, the sugar in milk, into glucose and galactose, and on and on. Which brings me to the job market: here, specifically the food industry. The synthesis and use of new enzymes in the preparation of food have seen tremendous growth not only in digestion but also in the taste and texture of the food, as well as possible economic benefits. There are many jobs related to food science (check out this article in the NatureJobs pages, which is a cool site in case you haven’t seen it: https : //www.nature.com/naturejobs/science/articles/10.1038/nj7422 - 149a?WT.ecid = NATUREjobs - 20121106). But jobs related to enzymes having to do with food also relate to the future of humanity itself. How will we feed our population in the future? Will all protein need to be plant-based? Or will we eat insect-meat? Or meat grown in a lab? Will food be 3D-printed, and will it be served by robots? All of these topics have recently received vast amounts of attention, research funding, led to start-ups, and a lot of interest from larger companies, and all of that spells jobs. Specifically, jobs involving knowledge about enzymes, which in the end only function because of their shape (VSEPR!).
Example Problems
1. Draw the Lewis structure of $\mathrm{H}_2 \mathrm{~N}-\mathrm{SH}$ and determine the VSEPR geometry around (a) the nitrogen atom, and (b) the sulfur atom.
Answer
a) Geometry around the nitrogen atom: trigonal pyramidal
b) Geometry around the sulfur atom: bent
2. Determine the VSPER geometry for each of the following and predict whether each will be polar or nonpolar
a) $B F_3$
Answer
$B F_3$: Difference in N-F electronegativity: polar bonds. Trigonal planar: dipoles cancel! Nonpolar
b) $F O F$
Answer
$FOF$: Difference in $\mathrm{O}-\mathrm{F}$ electronegativity: dipole moment: polar
c) $\mathrm{CCl}_4$
Answer
$\mathrm{CCl}_4$: Difference in $\mathrm{C}-\mathrm{Cl}$ electronegativity: dipole moments; tetrahedral structure: cancel out, so nonpolar!
Lecture 11: Molecular Orbitals
Summary
Molecular orbital theory is a tool used to predict the shape and behavior of electrons that are shared between atoms. Two or more atomic orbitals are added together to make a linear combination of atomic orbitals—the LCAO method—allowing for quick characterization of the kinds of bonds formed between two atoms.
When atomic orbitals are added, they can either constructively interfere, forming a bond, or they can destructively interfere, forming an antibonding state. In a molecule, the bonding state is always lower in energy than the corresponding atomic orbital, while the antibonding state is always higher. Molecular orbital diagrams, or $\mathrm{MO}$ diagrams, are a convenient visualization tool to see how electrons are distributed between two atoms: hydrogen is shown to the right as an example.
The two hydrogen $1 \mathrm{~s}$ orbitals combine to form a $\sigma_{1 s}$ bond, which is lower in energy and therefore more stable. If we had started with helium instead, there would have been four electrons to distribute: two would form a $\sigma_{1 s}$ bond, but the other two would go in the $\sigma_{1 s} *$ antibonding state, effectively cancelling out the bond.
After placing electrons in the $\mathrm{MO}$ diagram, bond order can be calculated:
$B O=\dfrac{1}{2} \text { ( of } \mathrm{e}^{\prime} \text { in bonding orbitals }-\mathrm{e}^{\prime} \text { in anti-bonding orbitals) }$
Stronger bonds have higher bond orders, and the bond order must be $>0$ for a bond to exist at all. For our example above, a hydrogen dimer has $\mathrm{BO}=1$, but a helium dimer has $\mathrm{BO}=0$. This explains why hydrogen gas exists as $\mathrm{H}_2$, but helium gas consists of individual atoms!
If an atom has p-orbitals as well as s-orbitals, additional bonds called bonds can form. The three p-orbital domains yield three kinds of bonds, one $\sigma$ and two : $\sigma_{n p z}, \pi_{n p x}$, and $\pi_{n p y}$, where $\mathrm{n}$ refers to the particular energy level being considered. Note that the $\sigma$ bond forms for both $\mathrm{s}$ orbital overlap as well as one of the $\mathrm{p}$ orbital overlaps: the term $\sigma$ bond means that the bond has cylindrical symmetry around the bond axis, something that is not the case for the bonds. Anti-bonding $\sigma^*$ and $\pi^*$ states also form from $\mathrm{p}$-orbital $\mathrm{MO}$s. A generic $\mathrm{MO}$ diagram for $\mathrm{p}$-orbitals is shown to the left.
$\mathrm{MO}$s can also be formed from heterogenous dimers: the same rules apply. Since each of the atoms has distinct energy levels associated with its $\mathrm{s}$- and $\mathrm{p}$-orbitals, the $\mathrm{MO}$ diagram formed using two different atoms is usually skewed, and the bonds that result are polar covalent bonds. The atom that is more electronegative is lower in energy. It can be helpful to count the number of electrons in the initial atomic orbitals and make sure that they are all used when filling up the $\mathrm{MO}$s. Remember that if a dimer is positively charged, it has lost an electron, and if it is negatively charged, it has gained an electron: these have to be accounted for as well! If one of the atoms has more electrons than the other, like a bond between $\mathrm{H}$ and $\mathrm{Cl}$, the excess electrons form nonbonding pairs - just like the lone pairs in the Lewis diagrams we drew before. If all of the electrons in the $\mathrm{MO}$ diagram are paired, then the dimer is diamagnetic, but if there are any electrons left unpaired, it is paramagnetic. For low molecular weight dimers, the $2 \mathrm{~s}$ and $2 \mathrm{p}$ atomic orbitals are very close in energy, and they can interact with each other. For dimers with lower $\mathrm{MW}$ than $\mathrm{O}_2$, the $\pi_{2 p x, y}$ and $\sigma_{2 p z}$ orbitals switch, and the $\pi_{2 p x, y}$ states are filled up first.
Why this matters
There are two ways to separate pasta from the boiling water once it’s cooked. One way is to pour the mixture through a colander, or a filter, to retain pasta on one side while letting the boiling water pass through. Another is to leave it on the stove and let all of the water boil off. Each of these separation techniques leads to the same outcome, but as you can imagine they are very different. I’m not talking about how the pasta would taste (let’s ignore that part), but rather how much energy it takes to carry out this separation task. Pouring through a filter is easy, fast, and efficient, while leaving the pot on the stove would take longer and require a lot more energy.
Now shrink down from pasta to nano-pasta. In other words, the size scale of molecules. We separate molecules from one another all the time for a wide variety of industrial processes. In fact, if we look at the U.S. energy consumption, about 1/3 of it goes into what is vaguely labeled “Industry.” But did you know that 40 percent of this energy for industry serves just one process, separating molecules? The reason that number is so large is that we only use the slow and inefficient thermal approach, not the much more efficient filter approach, to perform the separations. That’s a whopping 12 percent of all energy used in the U.S. that goes into boiling one chemical off of another.
So why is it that we don’t just pour these chemicals through a nano-filter, just like we do with pasta and colanders? If we did switch from thermal separation to a filter, we could save up to 90 percent of that energy! The reason we don’t, is the filter itself. Current filters aren’t yet up to standard. On the one hand, we’ve got filters made from polymers (materials we’ll learn about later in the semester) that can separate the tiniest of molecules very well, but they’re so delicate that they can’t be used in the harsh chemical and thermal environments of most industrial processes. On the other hand, we’ve got filters made from ceramics (materials with strong ionic and/or covalent bonds) that are super resilient and can handle the conditions, but they can’t get down to the sizes of the small molecules that need to be separated.
So it is that we come to our molecular orbitals, orbital filling, and the specific example of $\mathrm{N}_2$ vs. $\mathrm{O}_2$ covered in class. Because one of the big separations that we need to do on a massive scale involves exactly these two molecules. $\mathrm{O}_2$ is plentiful in the air (unless you go into a closed room, light a candle, and wait 12 hours as we learned in lecture 2!), but for many applications we need $\mathrm{O}_2$ in much higher concentrations than its naturally occurring 21 percent in air. Take combustion as an example: the 78 percent $\mathrm{N}_2$ in the air has a negative impact on combustion processes since the nitrogen molecules get heated up during the reaction to very high temperatures, which is not only inefficient, but breaks down into toxic nitrogen oxide ($\mathrm{NOx}$) gasses. Increasing the amount of oxygen and decreasing the amount of nitrogen leads to much higher combustion efficiencies, lower harmful emissions, and higher processing temperatures. Now, you may be thinking that I'm only talking about fossil fuel processing, and yes that is certainly a prime example of our use of combustion as a society, but getting purified $\mathrm{O} 2$ molecules reaches far beyond, to applications in the medical industry to sewage treatment to metal manufacturing, to name only a few.
The separation of $\mathrm{O}_2$ from $\mathrm{N}_2$ is in so much demand globally that it is performed at a quantity of 31 billion kilograms per year. Because the separation is done inefficiently using thermal processes, in this case going cryogenic, which is cooling instead of boiling but for the same purpose, 47 teraBtu (British thermal units, or the amount of thermal energy needed to raise one pound of water by one degree Fahrenheit) of energy are used per year just to do the separation! Just to be clear, that's $47,000,000,000,000$ Btu of energy, or if you prefer $13,774,340,298,094$ Watt-hours. A typical household in the U.S. uses about 900,000 Watt-hours of energy per moth, just for reference.
How do we save up to 90 percent of the massive energy used to separate $\mathrm{O}_2$ from $\mathrm{N}_2$ by switching from thermal-based separation to a filter-based approach? The answer lies in those molecular orbitals! They tell us about the bonding of each molecule and the interactions of the molecules with yet-to-be-invented filter materials that could combine the best of both worlds from polymer to ceramic materials. The $\mathrm{MO}$s of $\mathrm{N}_2$ and $\mathrm{O}_2$ also tell us that these molecules respond differently to externally applied magnetic fields, which may in turn be useful in boosting the separation efficiency. There is a tremendous opportunity for new filters that take these two most abundant molecules in the air and put them into separate compartments, but it all has to start with knowing how the electrons in the molecules behave. And that, of course, we get from molecular orbital theory.
Why this employs
The problem with fusion energy is that it’s uncontrolled. Way back in the 1950’s, Disney was making feel-good movies (grab some popcorn and check out, “Our Friend the Atom”), about how energy was very soon going to be, "too cheap to meter." 70 years later, why is this not the case? Fusion is so attractive on so many levels: unlike fission, which is the stuff of current nuclear reactors, fusion has no radioactive waste or byproducts, with the only outcome of a fusion reaction being ridiculous amounts of energy and helium. Fusion is the engine of the stars, so you know this energy is serious. Let's break it down: remember the battery vs. gasoline comparison I made in Lecture 8? Roughly, the best Li-ion batteries today can store about $1 \mathrm{MJ}$ of energy per $\mathrm{kg}$. Gasoline, on the other hand, can store $45 \mathrm{MJ}$ per kg. But let's keep going. The explosive TNT stores around $4160 \mathrm{MJ} / \mathrm{kg}$. Uranium when used in nuclear fission stores a whopping $3,456,000 \mathrm{MJ}$ $/ \mathrm{kg}$ ! This incredible energy density is a strong argument for nuclear energy generation and is often invoked. But when we move to fusion, when we move to the stuff of the stars, all of these numbers feel tiny. The energy density of the fuel for fusion, which is a combination of tritium and deuterium, is an incredible $576,000,000$ $\mathrm{MJ} / \mathrm{kg}$. And that fuel is highly abundant and cheap. That's why the dream of fusion is still alive even after 70 years of trying, and in fact today there is a huge resurgence in fusion energy. For more information you don't need to go very far, check out the PSFC (Plasma Science and Fusion Center) right here at MIT. Or the new MIT spin-out called Commonwealth Fusion Systems.
If nuclear fusion is to become a reality (in just 10 years according to some, but we've also been hearing that since the Disney movie so we need to approach with careful optimism), one of the single most important ingredients to making it work will be magnets. Lots and lots of magnets and very powerful ones. That's because one of the most likely ways to get fusion to work is to contain the massive energy released (which, by the way, gets to temperatures up to 100 million degrees), is to confine the energy using magnetic fields. Now, in fusion reactor designs the magnetic fields are often generated with superconducting coils, so it's different from our unpaired electrons in the $\mathrm{MO}$ diagram of $\mathrm{O}_2$. But the general idea that a material can be responsive to an external magnetic field comes from its electron filling, and the more electrons that are unpaired like in the $\mathrm{O}_2$ molecule, the more responsive it can be.
So what are the jobs related to developing new magnets? You could go to work at a nuclear fusion start-up like Commonwealth Fusion, or to the large government-led fusion operations like the ITER program in France. But so many other industries need stronger, cheaper, lighter magnets that the job market extends far beyond just fusion. You could look for jobs in companies that manufacture magnets (there are many), or companies pursuing new ideas for recycling magnets (like Urban Mining Company), or as a scientist at a U.S. lab pushing the frontiers of magnets (like Florida's National High Magnetic Field Laboratory, or the one at Los Alamos), or at companies trying to make magnets that don't rely on rare-earth elements (like Toyota for example, among many).
Extra practice
1. Consider $O_1 512$ and $O_2^{+}$
a) Draw the $\mathrm{MO}$ diagram of each molecule.
b) Find the bond order of each.
c) Label each one as paramagnetic or diamagnetic.
Answer
$\mathrm{O}_2$ is paramagnetic because it has unpaired electrons.
$\mathrm{O}_2^{+}$ is also paramagnetic, because it has an unpaired electron. Its bond order is $2.5$
2. Draw the $\mathrm{MO}$ diagram of $\mathrm{HCl}$.
Answer
Lecture 12: Hybridization in Molecular Orbitals
Summary
If multiple atomic orbitals within an atom have similar energy levels, they can hybridize, combining to form equal orbitals that have a lower average energy. Consider methane, $\mathrm{CH} 4$, as an example: the carbon atom has two $2\mathrm{s}$ electrons and two $2\mathrm{p}$ electrons. The $2\mathrm{s}$ and $2\mathrm{p}$ states hybridize, yielding four equal-energy, unpaired electrons that are ready to bond with hydrogen atoms. As hybridization occurs, carbon's four electrons are redistributed so as to be maximally spaced apart, lowering the energy of the system and yielding the most stable state: a tetrahedral methane molecule. This kind of hybridization is called $\mathrm{sp}^3$, because one $\mathrm{s}$-orbital energy level combined with three $\mathrm{p}$-orbital energy levels.
For the case of ethane, $\mathrm{C}_2\mathrm{H}_4$, the hybridization process is slightly different, as shown here (this time as an energy level diagram). Each carbon forms only three bonds: two with hydrogen atoms, and one with the other carbon. The $\mathrm{s}$-orbital energy level combines with two $\mathrm{p}$-orbital energy levels to form three equal $mathrm{sp}^2$ bonds; the remaining $2\mathrm{p}$ electrons form a higher-energy $\pi$ bond between the two carbon atoms and yield a double bond between the carbon atoms.
The next logical molecule logical molecule to consider is acetylene, $\mathrm{C}_2\mathrm{H}_2$. In this case, each carbon is only bonded to one hydrogen, so only one $2 \mathrm{p}$ energy level hybridizes with the $2\mathrm{s}$ energy level to form a $\mathrm{sp}$ hybridized bond. The remaining $2\mathrm{p}$ orbitals form two bonds, yielding a triple bond between the carbon atoms.
In summary, hybridization occurs to lower the overall energy of the system: atomic orbitals combine with each other to form mixed states with a lower average energy. Knowing the hybridization of the molecule is equivalent to knowing the molecular shape: VSEPR gives the geometric name corresponding to the specific combination of bond angles that minimize the overall energy of the system.
Why this matters
More than 3 billion people on this planet live in water-stressed regions. 1.8 billion people drink fecally contaminated water. 600 million people boil their water to clean it. In places where water scarcity is a serious issue, by some estimates 70% of all disease and 30% of all death can be attributed to the lack of water or water quality. Fresh water makes up just 2.5% of all water on earth, but more than 2/3 of that is tied up in glaciers. This means that only 1% of all water on the planet is drinkable, and the balance of this precious resource is way offscale.
Given the level of global crisis that access to freshwater has become, it makes a lot of sense to turn our attention to the other 97% of water on the planet, saltwater in the oceans. The problem, of course, being that it’s not drinkable (or useful in most agriculture), unless the salt is removed. The good news is that desalination is growing in terms of use and installed capacity, but the bad news is that it still costs far too much to become a ubiquitous substitute for groundwater.
How can we work on lowering the cost and increasing efficiency in desalination? First of all, we need to know what the current cost breakdown is, and as you can see in the pie chart in this image, the major cost of desalination is in the energy it takes to pump water through the system. By “system” I mean a pretreatment facility where seawater is run through sand to filter out large impurities (shells, rocks, seaweed, etc.), followed by the actual “desal” part of desalination, where the water is run through a set of membranes (40,000 of them in the plant shown in the picture) that remove the salt and allow freshwater to pass. Pumping the saltwater through these membranes requires by far the biggest part of the plant’s energy consumption, so improving the membranes and making them more energy efficient is crucial. In fact, it’s not just that the membranes used today are energy inefficient, but it’s also that they get fouled up (bacteria and other organic matter grow on their pores) and are extremely delicate so can’t be cleaned very well. This means for most of the time the plant is paying higher energy costs than it needs to (by a factor of 2 or 3 sometimes!) because it has to pump water through membranes that are filthy and blocked and can’t be cleaned. Now that’s what I call a materials design opportunity! Make a better membrane for the salt removal step of desalination, and make the process cheaper. That’s how we come to our Why This Matters and connection to today’s lecture.
We learned about the sigma and pi bonds that can form when the $\mathrm{AO}$s of carbon atoms hybridize.
In the examples in class we had carbon bonded to hydrogen and sometimes also to itself. If we only have $\mathrm{sp2}$ hybridization and we only have $\mathrm{C}$ atoms and no $\mathrm{H}$ atoms, then we arrive at a very, very cool material: graphene. It's bonded together in a honeycomb lattice sheet (a 2D sheet) with $\mathrm{sp}^2$ bonds, and those extra $\mathrm{p}$ electrons form pi bonds across the plane to give it a huge stability boost. It's a very cool material and its isolation from graphite (which is just stacks of graphene) won the Nobel Prize in 2010. (The scientists who discovered graphene were able to separate it from a chunk of graphite using only run-of-the-mill tape! So remember, sometimes all you need for a Nobel Prize is pencil lead, tape, and determination). I can't possibly go into all the details for why graphene is cool, but if you search online you'll see right away. Another consequence of graphene is that it launched a whole field of " $2 \mathrm{D}$ materials" where researchers have realized that so many other materials can be made into these sheets that are only one or a few atoms thick. It's now even possible to make completely new stacks of 3D materials by mixing and matching $2 \mathrm{D}$ sheets (check out for example the paper by Geim and Grigorieva, "Van der Waals heterostructures," Nature volume 499, pages 419-425 (25 July 2013). By the way, Geim was one of the two scientists who won the Nobel prize for graphene's discovery. The other is Novoselov, but Geim is a bit special as he is the only person to ever have won both the Nobel and the IgNobel prizes, the latter for levitating frogs. But I digress.
The point is that graphene may just be the ultimate membrane. It’s only 1 atom thick, so in terms of viscous loss it’s hard to beat. Plus, it’s much more resilient than today’s polymer membranes so it can be cleaned easily. It’s no wonder graphene’s been considered as a potential for desalination since 2012, even by people you might already know, for example, “Water Desalination across Nanoporous Graphene,” by Cohen-Tanugi and Grossman, Nanoletters volume 12, pages 3602- 3608 (2012). Because of its massive potential in water desalination and also water treatement and purification in general, there are many research groups and even already a number of companies working towards commercializing graphene-based membranes (Via Separations, for example). This is all extremely exciting, but it’s also only possible because of the hybridization that occurs in carbon atoms, which combined with those pi bonds allows them to take the form of graphene.
Why this employs
This is a tough one since hybridization in chemistry is what enables so many molecules to exist at all, and this impacts nearly all job sectors, not to mention life itself. But since we covered graphene in Why This Matters let’s stick to graphene here in the Employment category too. There are companies directly manufacturing graphene and investing a lot of into making it cheap, at large scales, and at high quality (meaning very few defects if possible), or with tailored functional chemistries. There are also different versions of graphene, from pure graphene to graphene-oxide to reduced graphene oxide and so on. Graphene Supermarket, ACS Material, or Graphenea are all examples of companies making graphene products and all of them have job openings for students who know about hybridization.
Why this employs
Let’s keep going with the carbon fibers concept from the Why This Matter section above. As I mentioned, carbon fibers could completely revolutionize structural materials and make so many aspects of our lives lighter, stronger, and more efficient. But a lot of work is yet to be done on the fibers themselves to lower their cost, as well as to integrate them into other products, from fabrics to plastics to concrete to metal. A lot of companies are doing extremely interesting work related to these problems. And yes, it all starts from a solid understanding of those Miller planes and crystallographic directions.
Hexcel has annual revenue over $2B and their flagship product is carbon fiber and composites. Note that both their name as well as logo are all about those hexagons in graphite! Hexcel is also cool because on their website they directly list student internship opportunities. Another company in this space is Solvay which makes thousands of products, many of which involve crystal planes of graphite. There’s Toray, with its origins in race cars, which lists carbon fibers as one of their 5 core businesses, and there’s Nippon Steel which has a whole subsidiary devoted to Granoc, a light-weight fabric or yarn made from carbon fibers. Its proprietary carbon fiber chemistry, they promise, is, “spinning the future.” Example Problems 1. Give the Miller indices of the shaded plane: Answer 2. Draw the following in a cubic unit cell: a) ($\bar{2}10$) Answer b)[$1\bar{2}1$] Answer Lecture 20: X-ray Generation Summary Atoms are too small to see with visible light, but x-rays have just the right range of wavelengths to image at the atomic scale. Wilhelm Rontgen first discovered x-rays by observing electrons striking a metal target in a cathode ray tube. Electrons were emitted from a heated filament and directed by an electric field inside the tube to very high velocities. When the high-speed electrons strike metal atoms, two kinds of xrays are generated. Continuous x-ray radiation is caused by the deflection of an incoming electron as it interacts with a metal target atom: as the name suggests, the allowed wavelengths for the Bremsstrahlung (which translates to ”braking radiation”) are a continuous spectrum above a lower bound. The lower bound on wavelength (upper bound on energy) is set by the energy of the incoming electron, because the energy of the emitted x-ray cannot be greater. The second type of x-ray that can be generated is a product of the electronic structure of the metal target atoms. When a high-energy electron is incident on a metal atom, it can knock out an electron in the inner core of the target atom, allowing a higher energy core electron to cascade down. If the electron cascades down from the $n=2$ to the $n=1$ energy level, the photon that’s emitted is called an alpha photon. These x-rays are discrete: they occur at a single wavelength that corresponds to the difference in energy levels. Crystallographers use a slightly different notation for the energy levels within an atom: the ground state ($n=1$) is called the $\mathrm{K}$ shell, $n=2$ is the $\mathrm{L}$ shell, and $n=3$ is the $\mathrm{M}$ shell. Using this notation, a photon that is emitted via the process of an electron falling from $n=2$ to $n=1$ is called a $\mathrm{K}$-alpha photon. The energy difference between inner-shell electrons is huge- that’s why the photons that are emitted are x-rays. Heavier elements have bigger differences between the energy levels, so the $\mathrm{K}$-alpha wavelength is smaller. It is therefore possible to determine which kind of atom the x-ray was emitted from, just by knowing the wavelength: these x-rays are called characteristic xrays. In this plot, the Brehmsstrahlung and the characteristic x-rays are shown together, as they’d be measured. The first two peaks are the $\mathrm{K}$-beta (M to K) and $\mathrm{K}$-alpha (L to K) lines, since those correspond to the biggest energy transitions within the atom. Then come the $\mathrm{L}$-beta (N to L) and $\mathrm{L}$-alpha (M to L) lines, which are much smaller energy gaps. The intensity can be changed by changing the number of electrons that are incident on the target. The lower-bound Brehmsstrahlung wavelength can be changed by adjusting the kinetic energy of the incident electrons, though this wouldn’t change the location of the characteristic x-ray peaks at all. The only way to shift the characteristic x-rays is to change the target metal used to generate the x-rays. If the energy of the incident x-rays is too low, though, no characteristic x-rays would appear: the incident electrons must have enough energy to knock out a core electron to kick off the cascade. Why this matters The discovery of x-rays, and the subsequent ability to fully control what wavelength of x-ray is generated, has had so much impact on modern life that it’s really tough to choose only one example for Why This Matters. But I guess since I have to pick, I may as well go back to the roots of Rontgen himself and one of the very first ways he used his new rays. That would be to take a picture of his wife! More precisely, her hand - it’s the first x-ray picture ever taken and it started a revolution (The black circle on the second finger from the left is her ring, by the way). The key thing about x-rays, which Rontgen noticed right away, is that they are able to pass through soft body tissue like skin and flesh, but they get absorbed by the denser material like bone. This means that if you shine x-rays on a body part, the bones inside cast shadows, which can be captured with a photographic plate. This was an incredible advance for medicine because for the first time a doctor could see inside a living patient without having to cut the patient open. Some of the earliest users of x-rays in medicine were military doctors, who could use them to optimize removal of bullets from a wounded soldier. As doctors saw how useful they could be, x-rays became central to diagnosis and by the 1930s they had led to the specialized field of radiology. Not only are x-rays currently utilized routinely in many aspects of medicine, but the technology of x-ray imaging continues to improve dramatically. Beyond improved contrast in the images, one very recent breakthrough applies techniques originally developed for the Large Hadron Collider in Geneva, which is still the world’s largest particle accelerator at 27 km long. With a combination of advanced shutters, cameras, and software, a completely new type of x-ray imaging is possible, with 3D and color images. Take a look at this very first x-ray picture of teeth, taken in 1896 with a 9-minute exposure time (!). Now look at a modern-day x-ray picture of teeth next to it. That’s an incredible improvement, and even more so when you realize that the picture on the right is fully 3D and a dentist can spin it around to see all angles. Why this employs There’s a whole lot of X-ray generation that goes on in the world, and the medical imaging example I gave above in Why This Matters is only one part of it. For example, X-rays are also used as scanners at airports, or as ways to do quality control for many industrial parts, or to analyze the degradation of valuable paintings, or in the field now called “X-ray astronomy,” to name only a few examples. X-rays are also used not only to see an illness like cancer, but also to treat it (“radiation therapy”). Just the medical market for X-rays alone is set to surpass$16.5B globally by 2025. And all these uses for X-rays spells jobs.
There are many companies out there that make X-ray equipment. Siemens for example sells 6 different types of “digital X-ray equipment,” but that’s just one category – they’ve also got a robotic X-ray machine, as well as two types of CT scanners (yes those also use X-rays), and a fluoroscopy machine which uses X-rays to examine movement. And that’s just one company! Siemens is a very large company, with the subsidiary “Siemens Healthineers” making this type of equipment. 41% of their US workforce of 13K people is in RD, engineering, IT, or tech support. And there are many others: Philips, GE Healthcare, Hitachi Healthcare America, Medtronic, Samsung Medical Devices, Eizo, Xoran Technologies, and United Imaging, Canon Medical Systems, and I’m still just getting started. All of these companies are selling wide ranges of X-ray equipment and they’re all competing with one another for customers, which means they’re constantly trying to improve their technology and differentiate what they’re selling from their competitors. And beyond these large companies, there are many small-scale companies looking to innovate the field. For example the company MARS Bioimaging, based in New Zealand, is commercializing the 3D color X-ray technology that came out of the Hadron Collider.
And that’s just the equipment side. There are also all the companies working on new software to analyze the images produced by all this X-ray equipment. For example, there are many start-ups applying AI to analyze the X-ray image data and in some cases AI has been shown to perform better than human doctors!
Example problems
1. Look at the x-ray spectrum below. You can see that $\lambda_{\kappa \beta}$ is lower than $\lambda_{\kappa \alpha}$. What about $\lambda_{\Lambda \alpha}$? Draw this peak on the spectrum.
Answer
\begin{aligned}
E_{\kappa \alpha}&=13.3(6-1)^2\left(\dfrac{1}{1^2}-\dfrac{1}{2^2}\right)=255 \mathrm{eV} \
&L_\alpha: n_i=3 ; n_f=3+1=4 \
&L_\alpha: n_i=3 ; n_f=3+1=4 \
E_{L \alpha}&=13.6(6-7.4)^2\left(\dfrac{1}{2^2}-\dfrac{1}{3^2}\right)=3.7 \mathrm{eV} \
&\quad \quad \quad E_{\kappa \alpha}>E_{L \alpha} \
&\quad \quad \quad \lambda_{\kappa \alpha}<\lambda_{L \alpha}
\end{aligned}
2. What is the wavelength of $\kappa_\alpha \mathrm{x}$-rays produced by a copper source?
Answer
\begin{gathered}
E=\dfrac{h c}{\lambda}=-13.6(Z-1)^2\left(\dfrac{1}{n_f^2}-\dfrac{1}{n_i^2}\right) \
Z=29, n_f=1, n_i=2 \lambda=1.54 A
\end{gathered}
Further Reading
Lecture 11: Molecular Orbitals
• Video breaking down Molecular Orbital Theory:
https://www.youtube.com/watch?v=P21OjJ9lDcs
• Cool tool for visualizing molecular orbitals of water:
http://www.bcbp.gu.se/~orjan/qc/h2o/ index.html
Lecture 12: Hybridization in Molecular Orbitals
• More detailed reading on hybridization:
https://opentextbc.ca/chemistry/chapter/8-2-hybrid-atomic-orbital
Lecture 13: Intermolecular interactions
• Reviewing intra- vs. inter-molecular interactions:
https://www.khanacademy.org/test-prep/ mcat/chemical-processes/covalent-bonds/a/intramolecular-and-intermolecular-forces
• Hydrogen bonds–essential for life:
https://www.cbsnews.com/news/nature-up-close-water-and-life-as-we-know
• How Geckos Stick to Der Waals:
https://www.sciencemag.org/news/2002/08/how-geckos-stick-der-waals
Lecture 14: Phase diagrams
• Tracing a phase diagram:
http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch14/ phase.php
• More on phase changes, including interactive boiling point example:
https://courses.lumenlearning. com/boundless-chemistry/chapter/phase-changes/
• On the thermodynamics of tempering chocolate:
https://acselementsofchocolate.typepad. com/elements_of_chocolate/TEMPERINGCHOCOLATE.html
Lecture 15: Electronic bands in solids
• More on band theory:
http://hyperphysics.phy-astr.gsu.edu/hbase/Solids/band.html
• More on semiconductor materials:
https://www.pveducation.org/pvcdrom/pn-junctions/semiconductor-materia
• Britney Spears’ Guide to Semiconductor Physics (one of the all-time best websites on the Internet):
http://britneyspears.ac/lasers.htm
Lecture 18: The perfect solid: crystals
• Review of SC, BCC, and FCC crystals:
https://www.youtube.com/watch?v=_h-Xv9nsJLc
• More on unit cells and packing lattices:
https://chem.libretexts.org/Bookshelves/General_ Chemistry/Map%3A_A_Molecular_Approach_(Tro)/12%3A_Solids_and_Modern_Materials/12.3% 3A_Unit_Cells_and_Basic_Structures
Lecture 19: Slicing a crystal: the Miller Planes
• Step-by-step Miller:
https://www.youtube.com/watch?v=9-us_oENGoM
• Beyond 3.091–crystallography and reciprocal space:
https://www.youtube.com/watch?v=DFFU39A3fPY
Lecture 20: X-ray generation
• More details on x-ray generation:
http://xrayweb.chem.ou.edu/notes/xray.html
• X-rays in medical imaging:
https://www.hopkinsmedicine.org/health/treatment-tests-and-therapies/ xrays | textbooks/chem/Inorganic_Chemistry/Introduction_to_Solid_State_Chemistry/05%3A_CHEM_ATLASes/5.02%3A_CHEM_ATLAS_2.txt |
How This Connects: Unit 3, Lectures 21-30
The purpose of this document is to serve as a guide and resource that gives you a quick overview of each lecture. For each lecture, there is a summary of the main topics covered, the Why This Matters moment, and the new Why This Employs section, plus a few example problems. So why did we make this? We hope it’s useful to get a good snapshot of any given lecture. Whether you couldn’t make it to a lecture or you couldn’t stop thinking about a lecture, this is a way to quickly get a sense of the content. It also gives me a chance to provide additional details that I may not have time for in the Why This Matters example, and also it lets me try out the Why This Employs section, which I certainly will not have time to discuss much in the lecture. Hopefully you find it useful!
One point about these lecture summaries. Please note that the lecture summaries are not meant to be a substitute for lecture notes. If you were to only read these summaries and not go to lecture, yes you’d get a good sense of the lecture from a very high level view, but no, you wouldn’t get enough out of it for it to be your only resource to learn the material!
Below is an image of the Exam 3 Concept Map. This demonstrates how each of the aspects of the course fit together: you have lots of resources! The Practice Problems, Recitations, Goodie Bags, and Lectures are ungraded resources to help you prepare for the quizzes and exams. All of the material listed on this concept map is fair game for Exam 3.
Lecture 21: Bragg’s Law and x-ray Diffraction
Summary
X-ray diffraction (XRD) is a method used for characterizing solids. It relies on the diffraction of x-rays upon striking crystal planes (the Miller planes we’ve learned about!) By assuming that each plane of atoms is continuous, and that they reflect the incoming x-rays such that the incident angle and the reflected angle are equal, the Braggs derived the equation that bears their name and relates the distance between repeating planes ($\mathrm{d}$) and the x-ray angle of incidence ($\theta$) to the x-ray wavelength:
Two x-rays striking equivalent Miller planes with the same angle of incidence will constructively interfere if the additional distance that one of the travels is equal to the wavelength of the x-ray. Quantitatively, if $\lambda=2 d \sin \theta$, the intensity of the outgoing x-rays with wavelength $\lambda$ are enhanced. the constructive interference will occur whenever the path length distance is an integer multiple of the wavelength: $2 d \sin \theta=n \lambda$ for integer $\mathrm{n}$. For $3.091$, we'll assume $n=1$. When constructive interference occurs, a signal will reach the detector in the XRD machine and a peak will be observed in a plot of the x-ray intensity. For destructive interference, no peak will be observed. Knowing the angle that gives rise to a peak as well as the wavelength of the incident x-rays allows us to obtain the distance between the planes that produced the reflection. This is known as the Bragg condition:
$2 d_{h k l} \sin \theta_{h k l}=\lambda$
For a given Miller plane, denoted by $(h k l)$, the Bragg condition is satisfied by a pair $(d, \theta)$ of inter-planar spacing and incident angle.
For each of the crystal structures (SC, BCC, or FCC), there are reflections that even when the Bragg Condition is met lead to destructive interference, due to crystal symmetry. The pattern of peak absences was used to derive a set of rules called selection rules, which allow us to know, or at least narrow down the possibilities of, the crystal structure of a material based on its XRD peaks. For the case of SC, there are no rules and any plane is fine. There are no forbidden reflections.
For the case of $\mathrm{BCC}$, allowed reflections are those where $\mathrm{h}+\mathrm{k}+\mathrm{l}$ is an even number. Forbidden reflection are those for which $\mathrm{h}+\mathrm{k}+\mathrm{l}$ is an odd number. For the case of $\mathrm{FCC}$, allowed reflections are those where $\mathrm{h}, \mathrm{k}, \mathrm{l}$ are all odd or $\mathrm{h}, \mathrm{k}, \mathrm{l}$ are all even. Forbidden reflections of FCC are those where h,k,l is mixed odd/even.
Why this matters
For solids, structure can be as important as the chemistry itself, and they are deeply connected. When I look up the crystal structure of a element in the Periodic Table I see what it is for the ground or lowest energy state of that element. This is the overall “happy place,” energetically speaking, of the material. But materials can take on other, metastable structures and be very happy there, too. And Why This Matters is because the properties can be completely different depending on which crystal structure the material takes, and XRD is the single most important characterization method we have to determine crystal structure. We’ve already seen the difference between graphite and diamond, which contains the same exact element (carbon) but just arranged in a different structure. The same is true for, well, pretty much everything. Take another element, iron, as an example.
Here’s a phase diagram for iron. As you may remember from Lecture 14, a phase diagram is a plot of the different phases of a material as a function of some variables, in this case pressure and temperature. Notice that at normal, or “ambient,” temperature ($\approx 300\mathrm{K}$) and pressure ($\approx 1 \mathrm{bar}$) conditions, iron is a $\mathrm{BCC}$ crystal. This is also what you’ll find if you look up its crystal structure in the PT. But notice from reading the phase diagram that if we raise the temperature it becomes $\mathrm{FCC}$, and if we raise the pressure it goes into the HCP (it’s not cubic so we haven’t covered it) phase. Fun fact: if you keep raising the temperature eventually it will go back to being $\mathrm{BCC}$.
The reason all of this matters is that the structure changes the properties. In the case of iron, the element’s magnetic properties are affected. If it’s FCC then it will be ’antiferromagnetic’ as opposed to the BCC ’ferromagnetic’ behavior. I recommend taking a magnetics course in the future to learn more about these terms! This has huge implications in magnetic technologies. And I know you’re probably thinking: sure, but we don’t build too many iron-based technologies that operate at 1000$\mathrm{K}$! You’re right, but the trick is that often times we can coax these materials to get stuck in one of those metastable phases and then use it for technologies while it’s in that phase. Again, diamond is a great example: it’s not the ground state of carbon so it is metastable as diamond instead of graphite, but we know that it stays stuck there for a long time so for most technologies that use diamond (say, a piece of jewelry), we don’t worry about it changing out of its metastable phase.
Coming back to the topic of the lecture, when we make something, whether that something is as old as elemental iron or as new as a nanostructured perovskite, the simplest and most common way we have to tell its crystal structure is by $\mathrm{XRD}$. In some cases, the use of $\mathrm{XRD}$ can unravel the structural mystery of a material, as in the case of the double-helix for DNA or the many proteins since. In other cases, it’s used to not unlock the secret to a completely new structure, but rather to classify a material into one or the other well-known structures. Sometimes the reason is to understand a material, sometimes it’s to engineer the material properties, and often times it’s both. But whatever the motivation, this incredibly powerful characterization tool has revolutionized what we know about solids.
Why this employs
We’ve been referencing these crystallographers (who are very picky about notation!) for several lectures now. But who are these people? And more to the point: who hires them? A whole lot of X-ray crystallographer jobs are out there in the biotech industry, in companies of all sizes. Blueprint Medicines, which looks like a Harvard spin-out and is just down the street, has an opening now with the title, “Senior Scientist/Principal Scientist, X-Ray Crystallography,” with the first job function description being, “Provide x-ray crystallographic and protein structure-function support, including structure-based drug design, to on-going drug discovery projects and new target discovery initiatives.” And the larger pharmaceutical and biotech companies have even more jobs. Take Novartis, which also has a big headquarters near MIT, just down Mass Ave. They’re hiring people to perform ”crystallography experiments including crystallization screening using automated liquid handling. GlaxoSmithKline has an opening for someone to “enable higher throughput x-ray crystallography.” Johnson and Johnson is hiring people with X-ray crystallography expertise to do, “automated chemistry.” And I have to mention one last example because of the title of their current opening, Bristol-Myers-Squibb is looking for a, “Research Investigator, Solid-State Chemistry.” Gotta love it! They want, “an entry level scientist with background in X-ray crystallography, X-ray diffraction, and solid-state characterization.” That’s now you!
And it’s not all about pharma. Hospitals are hiring X-ray crystallographers too (these are not the same position as a radiologist), to work on research projects for example with openings at Mass General and Dana Farber. And many research positions in X-ray diffraction are out there too, from positions at the Howard Hughs Medical Institute to university labs and centers all across the country.
Example Problems
1. Determine the structure (simple cubic, body centered cubic, or face centered cubic) to which this $\mathrm{XRD}$ pattern most likely corresponds (copper $\kappa_{\alpha}$ x-rays were used).
Answer
Given that the indices of each plane are either all odd or all even, using the selection rules we are able to determine that this structure is $\mathrm{FCC}$.
Lecture 22: From x-ray Diffraction to Crystal Structure
Summary
This lecture we finished analyzing the $\mathrm{XRD}$ spectrum of an $\mathrm{Al}$ sample, shown below.
The plot was obtained by shining $\mathrm{K}$-alpha x-rays from a $\mathrm{Cu}$ target onto our $\mathrm{Al}$ sample. What we want to do is figure out the crystal structure and the lattice constant of $\mathrm{Al}$. To answer these questions, we need our handy Miller plane separating distance equation (where ' $\mathrm{d}$ ' is the distance between two repeating Miller planes with indices $\mathrm{hkl}$ in a cubic system, and ' $a$ ' is the lattice constant):
$d_{h k l}=\dfrac{a}{\sqrt{h^2+k^2+{ }^2}}$
and the Bragg condition:
$2 d_{h k l} \sin \theta_{h k l}=\lambda$
Notice that both of these equations include $d_{h k l}$. We can use this to our advantage and substitute one equation for $d_{h k l}$ into another to obtain the following:
$\left(\dfrac{\lambda}{2 a}\right)^2=\dfrac{\left(\sin \theta_{h k l}\right)^2}{h^2+k^2+l^2}$
We know the value of the wavelength, because it is fixed by the $K_a$ x-rays from the copper source. These x-rays have a wavelength of $1.54 \AA$. So the expression for our example becomes:
$\left(\dfrac{1.54 \AA}{2 a}\right)^2=\dfrac{\left(\sin \theta_{h k l}\right)^2}{h^2+k^2+l^2}$
Now we have constants on both sides of the equal sign, because the lattice parameter does not change. We can make an educated guess of the $\mathrm{hkl}$ value by following the procedures outlined in the chart on the following page.
From the selection rules we know that $\mathrm{Al}$ is an $\mathrm{FCC}$ metal, since the ($\mathrm{hkl}$) combinations are always either all even or all odd. We can also take a value of $\theta$ and a value of h, k, and l and plug these into the equation above to find the lattice parameter, given in the rightmost column of the chart.
Why this matters
Photo of Henry Moseley is in the public domain.
This is Henry Moseley (image, Royal Society of Chemistry). He’s pictured there in his lab, holding in his hands, of course, a modified cathode ray tube. He was experimenting with X-rays. But Moseley’s interests were less about the crystal structures and Bragg conditions and more about the X-ray lines themselves and what they meant. He carried out a systematic study of the metals used to generate the Xrays, comparing the X-ray emission from 38 different chemical elements.
Some work had already been done that led to our understanding of the characteristic and continuous parts of the X-ray generation spectrum, as we discussed a few lectures ago. But a full systematic study had not been carried out until Moseley’s work. Take a look at the difference between the two X-ray spectra generated with two different targets: $\mathrm{Mo}$ and $\mathrm{Cu}$. Note the $\mathrm{K}_{\alpha}$ and $\mathrm{K}_{\beta}$ lines for each one, and that they’re shifted to lower wavelength for $\mathrm{Mo}$ compared to $\mathrm{Cu}$. As we know, this is because of difference in energy between the shells of $n=1$ and $n=2$ (for $\mathrm{K}_{\alpha}$) or $n=1$ and $n=3$ (for $\mathrm{K}_{\beta}$), and it makes sense this energy difference is greater (corresponding to lower wavelength) for $\mathrm{Mo}$ since it’s heavier than $\mathrm{Cu}$. But what exactly is the dependence, and why is it present? Let’s look at the data for a sequence of targets, directly from a subset of Moseley’s data.
These are characteristic X-ray lines for $\mathrm{Ca}$ up through $\mathrm{Zn}$, so going across the $\mathrm{d}$-block elements of the fourth row in the PT. Note that it’s not actually $\mathrm{Zn}$ but rather brass, which we’ve already learned is a mixture of $\mathrm{Zn}$ with $\mathrm{Cu}$ – that’s because $\mathrm{Zn}$ would melt under the high energy electron bombardment before it could give off any characteristic X-rays, so Moseley gave it the extra strength it needed by making brass, and then subtracted out the emission from $\mathrm{Cu}$. Nice trick!
What Moseley found was that if the characteristic emission lines were plotted as the square of their energy vs. atomic number, that you'd get a straight line. He fit the data by considering the lines to come from a core excitation, so a difference in energy levels from Bohr's model:
$E_{x-r a y}=13.6[e V](Z-1)^2\left(\dfrac{1}{1^2}-\dfrac{1}{2^2}\right)=\dfrac{3}{4}(13.6[\mathrm{eV}])(Z-1)^2$
This is now called Moseley's law. Can you see why it's $\mathrm{Z}-1$ instead of $\mathrm{Z}$ as in the Bohr model? It's because the electron cascading down to generate the X-ray is "seeing" a 1-electron screening of the nucleus. That's because one of those core 1s electrons was knocked out, but there's one left there that screens out a positive charge, hence the $\mathrm{Z}-1$. So what did this trend in the data mean? Here's what Moseley said in his 1913 paper, "We have here a proof that there is in the atom a fundamental quantity, which increases by regular steps as one passes from one element to the next. This quantity can only be the charge on the central positive nucleus, of the existence of which we already have definite proof."
The reason this is such a big deal, and why I'm making it the Why This Matters for this chapter, is that even through the Mendeleev periodic table had been around and more elements were being discovered and added, there was a major flaw in the periodic table: the position predicted by an element's atomic weight did not always match the position predicted by its chemical properties. Remember that the positioning by Mendeleev was based on weight and properties and when the periodicity called for it, he chose to order the elements based on their properties, rather than their atomic weight. But was there something more fundamental than atomic weight?
Moseley's data only made sense if the positive charge in nucleus increased by exactly one unit as you go from one element to the next in the PT. In other words, he discovered that an element's atomic number is identical to how many protons it has! I know this seems kind of obvious to us now, but back then "atomic number" was simply a number with no meaning, other than the element's place in the periodic table. The atomic number was not thought to be associated with any measurable physical quantity. For Mendeleev, periodicity was by atomic mass and chemical properties; for Moseley, it was by atomic number. This led to a much deeper understanding of the periodic table and his insights immediately helped to understand some key mysteries, for example where to place the lanthanides in the PT $(\mathrm{La}=\# 57, \mathrm{Lu}=\# 71)$, or why $\mathrm{Co}$ comes before $\mathrm{Ni}$. And the gaps that Mendeleev brilliantly left open in his PT to create periodicity now made sense by missing atomic numbers in a sequence, for example elements $43,61,72$, and $75$ were now understood to contain that many protons (they were discovered later by other scientists: technetium, promethium, hafnium and rhenium).
Moseley died tragically in 1915 at age 27 in a battle in WW1. In 1916 no Nobel Prizes were awarded in physics or chemistry, which is thought to have been done to honor Moseley, who surely deserved one.
Why this employs
We've covered X-ray generation and machine manufacturers two chapters ago, and in the last chapter we looked at jobs related to X-ray crystallography. For this last Why This Employs related to Xrays, it's time to go big or go home. And when I say big, I mean really big. When electrons or for that matter any charged particles are accelerated to near light speeds, then the acceleration they experience simply to stay in a loop produces massively energetic radiation. The wavelength can vary dramatically, but very often these enormous accelerators are used to make super-high-energy X-rays. The intensity of these rays is dazzlingly bright, millions of times brighter than sunlight and thousands of times more intense than X-rays produced in other ways. This level of brightness makes them useful for pretty much any and all areas of research and fields of science. Some types of measurements are only possible when synchrotron light is used, and for other types one can get better quality information in less time than with traditional light sources. They've been shown to be useful in so many areas it's impossible to list them all, but certainly in biology, chemistry, physics, materials, medicine, drug discovery, and geology, to name only a few fields, synchrotrons have made a dramatic impact. They're in such demand that often one needs to book time on them many months in advance. A typical synchrotron can have as many as 50 "beam lines" that grab the high energy X-rays out of the loop and focus them in a beam where experiments are done. These lines are usually booked and put to use 24 hours per day, 7 days per week, all year round. Here’s a photo of one of them, the Advanced Photon Source at Argonne National Laboratory in Illinois. Their tagline overview statement reads, “The Advanced Photon Source (APS) at the U.S. Department of Energy’s Argonne National Laboratory provides ultra-bright, high-energy storage ring-generated x-ray beams for research in almost all scientific disciplines.”
The Employment part of this is pretty cool. These facilities, which are called “synchrotrons,” are all over the world, and they require thousands of people to build and then run. The international nature of them is astounding: just do a image search for synchrotron and you’ll find pictures of them all over the planet. And that means jobs in many different locations. Some are old, like the one at Berkeley National Lab (but it’s still kicking!), some are medium-sized like that APS pictured above, and some are huge like the Hadron Collider I referenced in the last lecture as a place where 3D X-ray imaging was invented. Colliders are also synchrotron light sources, since they’re built to accelerate particles at very high speeds. Even though colliders may be used to smash particles together at these speeds, they’re also often used simply as a way to generate high intensity light.
I don’t have a specific job title in mind, but if you look at a list like this one: https://en.wikipedia.org/wiki/List_of_synchrotron_radiation_facilities you’ll see where these synchrotrons are, and for each one of them there’s an “employment” link you can click on to explore possible jobs.
Extra practice
1. Determine the element that made up the sample from Lecture 21 Extra Practice Problem 1. The XRD pattern is reproduced below (copper $k_a$ x-rays were used).
Answer
To know which element was used as the sample, find the lattice parameter (a) using the equation for interplanar spacing $\left(d_{h k l}\right)$, Bragg's law, and Moseley's law. For $\mathrm{h,k,l}$ and $\theta_{h k l}$, pick a plane and a corresponding angle from the chart we developed last chapter, shown below:
\begin{gathered}
d_{h k l}=\dfrac{a}{\sqrt{h^2+k^2+l^2}} \
\lambda=2 d_{h k l} \sin \left(\theta_{h k l}\right)
\end{gathered}
The energy corresponds to $\mathrm{Cu}(\mathrm{Z}=29) k_\alpha$ radiation:
\begin{gathered}
E=\dfrac{h c}{\lambda}=13.6[e V](Z-1)^2\left(\dfrac{1}{n_f^2}-\dfrac{1}{n_i^2}\right) \
\lambda=\dfrac{h c}{-13.6(Z-1)^2\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)} \
d_{h k l}=\dfrac{h c}{-13.6(Z-1)^2\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right) 2 \sin \theta_{h k l}} \
a=\dfrac{h c \sqrt{h^2+k^2+l^2}}{-13.6(Z-1)^2\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right) 2 \sin \theta_{h k l}}=3.53 A
\end{gathered}
This lattice parameter corresponds to $\mathrm{Ni}$.
2. You would like to perform an XRD experiment, but you don't know what target is used in the diffractometer in your lab. You put in a calibration sample of iron, which is BCC and has a lattice parameter of $2.856$ angstroms. If you observe the following XRD pattern, what material is the target? You are pretty sure that there is a filter that prevents anything with lower energy than $k_a$ radiation from hitting your sample.
The peaks observed are as follows:
counts 10 1000 20 2200 8 5 1200 2500
$2\theta$ 17.38 20.87 24.67 29.62 30.30 35.15 36.45 42.37
a) What kind(s) of x-rays are hitting the sample?
Answer
$k_\alpha \operatorname{AND} k_\beta$
b) How many planes are represented by the data? Which planes are they?
Answer
4 planes are represented: $(110),(200),(211)$, and $(220)$
c) What are the interplanar spacings associated with these planes?
Answer
$\dfrac{2.856}{\sqrt{h^2+k^2+l^2}} \nonumber$
Plugging in each ($\mathrm{hkl}$), the spacings are 2.02, 1.43, 1.17, and 1.01 A
d) Which element was used as the target?
Answer
\begin{gathered}
\lambda=2 d_{h k l} \sin \left(\theta_{h k l}\right) \
\lambda_{k_\alpha}=0.73 A \
\lambda_{k_\beta}=0.61 A
\end{gathered}
Answer
$E=\dfrac{h c}{\lambda}=13.6(Z-1)^2\left(\dfrac{1}{n_f^2}-\dfrac{1}{n_i^2}\right) \nonumber$
For $k_\alpha, n_i=2$ and $n_f=1$. For $k_\beta, n_i=3$ and $n_f=1$.
\begin{aligned}
&h=4.135 \times 10^{-15} \mathrm{eV} . \mathrm{s} \
&c=3 x 10^8 \mathrm{~m} / \mathrm{s} \
&\mathrm{Z}=42
\end{aligned}
Lecture 23: Point Defects
Summary
A point defect is a localized disruption in the regularity of the crystal lattice. There are four types of point defects: vacancies, interstitial impurities, self-interstitials, and substitutional impurities.
Arrhenius determined a law for the temperature dependence of the rate at which processes occur:
$k=A e^{-E_a / R T}=A e^{-E_a / k_B T}$
where $\mathrm{R}$ is the gas constant (or $k_B$ is the Boltzmann constant) and $E_a$ is the activation energy. The term in the exponent should be unitless: therefore, if the activation energy is given in $\mathrm{J} / \mathrm{mol}$, use the version with $R$, but if the activation energy is given in $J$, use the $K_B$ version. Recall that the gas constant is just $R=K_B * N_A$. The units are determined by the prefactor $\mathrm{A}$, which can be thought of as an average kinetic energy of the system.
Vacancies are always present in every solid because they're a result of thermally-activated processes. We can consider the rate of formation of a vacancy and the rate of removal of that same vacancy as two thermally-activated processes, each with their own rate. At any given temperature, when the rate of forming the vacancy is the same as the rate of "de-forming" the vacancy then the vacancy concentration in the crystal will be in equilibrium. Since each rate is thermally activated we can use an Arrhenius equation to describe both the forward and back process, and setting them equal for equilibrium one arrives at a formula describing how the vacancy concentration depends on temperature:
$N_v=N e^{-E_a / k_B T}$
where $N_v / N$ is the fractional concentration of vacancies and $E_a$ is the activation energy $[\mathrm{J}]$ required to remove one atom. If the vacancy occurs in an ionic solid, charge neutrality must be maintained. Therefore, the defect either forms as a Schottky defect, where a pair of charges (one cation and one anion) is removed, or as a Frenkel defect, where the vacant atom sits elsewhere in the lattice on an interstitial site. For a Schottky defect in an ionic solid like $\mathrm{CaCl}_2$, two anions $\left(\mathrm{Cl}^{-}\right)$ must be removed for each cation $\left(\mathrm{Ca}_2^{+}\right)$ vacancy to maintain charge neutrality. Frenkel and a smaller cation, like $\mathrm{AgCl}, \mathrm{AgBr}$, and $\mathrm{AgI}$, for example.
Interstitial defects can occur in covalent solids as well: in this case, an extra atom occupies a site that is not part of the lattice, but the charge neutrality requirement doesn't necessitate the creation of a vacancy provided the interstitial atoms have the same charge as the lattice atoms. For example, a $\mathrm{C}$ interstitial in $\mathrm{Fe}$ is charge neutral. If the interstitial atom is the same type of atom as the lattice, like a $\mathrm{Si}$ atom in a $\mathrm{Si}$ lattice but not on a lattice site, the defect is called a self-interstitial. The energy required to form a self-interstitial $(2-5 \mathrm{eV})$ is much higher than for a vacancy $(0.5-1 \mathrm{eV})$, so these defects are much less common: this can be rationalized by thinking about how hard it would be to squeeze an atom between similarly-sized atoms arranged in a closely-packed lattice.
Atoms which take the place of another atom in a lattice are called substitutional defects. Generally, the Hume-Rothery rules provide guidelines to which atoms can be a substitutional defect: the atomic size must be within $+/- 15\%$, the crystal structure must be the same, the electronegativity must be similar, and the valence must be the same or higher.
Why this matters
Let’s pick up on the interstitial defect of carbon in iron, otherwise known as steel. This particular defect is one that has positive benefits if it’s controlled carefully and the right amount of carbon (not much, it turns out) is placed in the right positions within the iron lattice (the tetrahedral holes, for example, in bcc $\mathrm{Fe}$). In fact, the change in iron’s properties are absolutely tremendous and represent a spectacular example of how defects can be used beneficially. If you Production vs. time figure removed due to take a piece of pure iron and ap- copyright restrictions. ply sideways strain on it, then its resolved shear stress is quite low, around 10 MPa. That means that if you push sideways on a piece of pure iron it will deform under 10 MPa of pressure. But with just 1% $\mathrm{C}$ on interstial sites, the $\mathrm{C}$-doped iron can have a resolved shear stress as high as 2000 MPa, 200 times larger than the undoped case!
Now, this phenomenon has been known and practiced for over 2500 years, when people first observed the mechanical strength imparted on iron when it was heated by a charcoal fire (the charcoal was the carbon source). But that’s just it: 2500 years ago, or 1000 years ago, or even just 100 years ago, not a whole lot of steel was being made each year. This is now changing, and it’s changing dramatically, and it’s Why This Matters.
Industrial Carbon Emission Chart
Take a look at the chart above (from US Geological Survey, UN, FAO, World Aluminium Association) of the production amount for some of the materials humans make on a scale massive enough to require enormous chunks of the world’s energy consumption. Cement and steel are the top two, and estimates put them at 10-15% combined of annual global $\mathrm{CO}_2$ emissions. If we look at just $\mathrm{CO}_2$ emissions from industrial processes, steel has the biggest share at 25% of the total. But even more important are those slopes in the production trends: note that the use of these products is growing and will continue to grow dramatically into the future. Back 2500 years ago it didn’t matter how steel was made. They also had no idea why the charcoal gave iron those properties. Today, not only do we need to find new ways to make steel more efficiently, but we also know what’s happening in the material at the atomic and bonding scale. In other words, we understand its solid state chemistry.
How can we make steel in a more energy efficient manner? Answering that question relies on knowledge of the point defects in the material, and specifically on the energy it takes to get carbon into the interstitial lattice. And it’s not always obvious. For example, if we compare the atomic size of $\mathrm{C}$ with the sizes of the available interstitial volumes in $\mathrm{Fe}$, it’s clear that it doesn’t quite fit, and some type of lattice distortion will have to take place in order to accommodate the interstitial defect, even as small as a $\mathrm{C}$ atom. But that means it’s not as simple as occupying the defect site with the most room, since we need to know how atoms get strained in response to the defect being there. Take the example of -iron: in that phase you would think a $\mathrm{C}$ atom occupies the larger tetrahedral hole, but in fact it prefers to go to the octahedral interstitial site. The reason for this preference is that when the $\mathrm{C}$-atom goes into the interstitial, strain gets relieved for the octahedral site by two nearest neighbor iron atoms moving a little bit, while for tetrahedral site, four iron atoms are nearest-neighbor and the displacement of all of these requires more strain energy. This is just one phase of iron and two different sites. There may be ways to move $\mathrm{C}$ atoms into other phases more easily, that would take less energy, and give the same strength. Or perhaps there are other ways beyond fire (which is why steel-making takes so much energy) to get the defect chemistry just right. This is a hard problem, but it’s a critical one: take a look at this chart from a recent paper published in Science that breaks down which sectors will be hardest to make “green.” Note the prominence of steel and cement! To solve such hard problems, we will need advances in defect chemistry.
Why this employs
This is an easy one: there’s actually a job position called “Defect Engineer”! At Global Foundries, they care about defects for the basic manufacture of semiconductor materials while at Intel, there are openings for Defect Engineers to work on 3D XPoint which is a new non-volatile memory technology. Intel also has an opening for a “Defect Reduction Engineer.” These and so many more similar openings are for industries where devices are made in clean rooms, often very clean cleanrooms. A “Class 1” cleanroom, for example, means that if you take a meter cubed of air at any given place in the room, you would count less than 10 particles of size 100 nanometers, less than 2 particles size less than 200 nanometers, and zero particles larger than that. The reason fabrication facilities need such a high level of air purity is that they’re making features on the order of 10’s of nanometers and so any small defect can be a big problem.
Why this employs
In the last lecture for this section I listed glass manufacturing and companies working on innovating in glass chemistry. For this chapter on engineered glass, let’s talk about smarts. In particular, “smart glass.” For now, that label means one specific type of silicate-based glass: switchable glass. It has been around for a long time, as even in the 1980’s you may have noticed (ok, your parents may have noticed) people wearing the sunglasses that automatically tinted and de-tinted in response to the sun (that used what are called “thermochromic” materials embedded in the glass, which change color based on temperature. They never worked all that well, staying a little too shaded inside and a little too unshaded outside, but the idea was there. But now we’ve gone from thermo- to electro-chromic glass, and the possibilities are seriously exciting. With a tiny applied voltage, glass can be engineered to go back and forth between near full transparency and near full-opacity. Apart from being extremely cool, this type of technology can have a lot of positive sustainability-related benefits, since the glass can be programmed to automatically dim and brighten in response to outdoor light conditions — that can in turn dramatically reduce a building’s energy needs.
This type of smart glass is still on the early side although a number of companies are taking off, and that means jobs. These will be jobs at either mid or early-range start-ups, but in some cases they’ve closed mega (>\$100M) fund-raising rounds so definitely growth is strong. Some companies in this space include Kinestral, Smartglass, View, Suntuitive, Gentex, Intelligent Glass, or Glass Apps. A lot of the investment in these companies is coming from the bigger ones like Asahi Glass or Corning, which have of course also started their own smart glass programs. Taken together all of this spells jobs in the future of glass. And its future looks very bright, far beyond switching the color or transparency, as the thermal, electronic and optical properties of the material continue to be engineered. We may or may not be living in the “Age of Glass,” as Corning likes to say, but we sure are living in an exciting time for this material.
Example Problems
1. The 2-D structure of soda-lime glass (used in windows) is shown below.
a) What compounds were used to make this glass? Do these compounds serve as network formers or network modifiers?
Answer
$\mathrm{SiO}_2$ : network former
$\mathrm{CaO}$ : network modifier
$\mathrm{Na}_2 \mathrm{O}$ : network modifier
b) How do each of the added compounds impact the bond structure in the glass?
Answer
$C a_O$ : breaks one bond/creates two network modifiers (coordinated with $1 \mathrm{Ca}^{2+}$ ion)
$\mathrm{Na}_2 \mathrm{O}$ : breaks one bond/ creates two network modifiers (coordinated with $2 \mathrm{Na}^{+}$ ions)
3. If they are cooled at the same rate, would you expect silica glass with 14% $\mathrm{Na}2\mathrm{O}$ or 25% $\mathrm{Na}2\mathrm{O}$ to have a:
Higher molar volume?
Higher glass transition temperature?
Higher viscosity?
Answer
$14 \%$ would have the higher molar volume
$14 \%$ would have the higher glass transition temperature
$14 \%$ would have the higher viscosity
4. If a silica glass is doped with $\mathrm{MgO}$, and then ion exchange is performed such that $\mathrm{Ca}$ ions replace the $\mathrm{Mg}$ ions, how would the mechanical properties of the glass change?
Answer
$\mathrm{Ca}$ ions take up more space than the $\mathrm{Mg}$ ions, so the glass will be under internal compression (like the Prince Rupert's drop)
Lecture 27: Reaction Rates
Summary
Chemical kinetics means the study of reaction rates, which correspond to changes in concentrations of reactants and products with time. Some terms to know: concentration $=$ moles / liter $=$ molarity $=[]$, rate $=\mathrm{d}[] / \mathrm{dt}$, a rate law is some equation that relates the rate to [], an integrated rate law relates the [ ] to $\mathrm{t}$ (ime), and the Arrhenius equation gives us the rate vs. $\mathrm{T}$(emperature)
Take a simple reaction where $a A \rightarrow b B:$ since mass is conserved, A disappears no faster than $\mathrm{B}$ appears, so the actual reaction rate is $=1 / \mathrm{b} \mathrm{d}[\mathrm{B}] / \mathrm{dt}=-1 / \mathrm{a} \mathrm{d}[\mathrm{A}] / \mathrm{dt}$. In other words, the change in the concentration of $\mathrm{B}$ must equal the opposite of the change in concentration of $\mathrm{A}$ weighted by one over the molar coefficient a or b. We can have more than one reactant product and the same idea holds. For example, suppose we have 2 of each: $a \mathrm{~A}+\mathrm{bB} \rightarrow \mathrm{cC}+\mathrm{dD}$. In this case the reaction rate would be:
$\text { rate }=\dfrac{-1}{a} \dfrac{d[A]}{d t}=\dfrac{-1}{b} \dfrac{d[B]}{d t}=\dfrac{1}{c} \dfrac{d[C]}{d t}=\dfrac{1}{d} \dfrac{d[D]}{d t}$
The general way to write an equation for the rate for the equation above is: rate $=\mathrm{k}[\mathrm{A}]^m[\mathrm{~B}]^n$, where $\mathrm{k}=$ rate constant and is dependent on conditions ($\mathrm{T}, \mathrm{P}$, solvent), m and n are exponents determined experimentally, $m+n$ is called the reaction order. Note that the rate units must always be M/s by definition, so this means that units of $\mathrm{k}$ depend on $\mathrm{n}$ and $\mathrm{m}$. For this class we’ll cover three different orders of reactions: $0^{th}, 1^{st},$ and $2^{nd}$.
To know the order of a reaction based on data tables like the one below, take any two rows of data: say the $\mathrm{t}=26 \mathrm{~min}$ and $\mathrm{t}=70 \mathrm{~min}$ rows. The concentration ratio between these two times is $0.0020 / 0.0034=0.5882$. The rate ratio is $1.8 / 5.0=0.36$
First of all the rate is changing so it can’t be 0th order. Second of all, at two different times the ratio of concentrations is not equal to the ratio of rates, so it can’t be 1st order. But if we square the ratio of concentrations, $(0.0020/0.0034)^2 = 0.35$ which is very close to $0.36$, so now we have our answer: from the data we can say the reaction is 2nd order!
To know the role of temperature in determining reaction rates, we must first learn about collision theory. Collision theory frames the reaction between molecules, say A and B, as follows: 1) a reaction can only occur when A and B collide, 2) not all collisions result in the formation of product, 3) there are two factors that matter most: the energy of the collision, and the orientation of molecule A with respect to B at the time of collision.
We can think of the energy required for $\mathrm{A}$ to react with $\mathrm{B}$ to be a kind of “activation energy” or $\mathrm{E}_a$. As we learned in chapter 14 (phases), molecules at a given temperature have a distribution of kinetic energies with that temperature being the average. That means some molecules have much more energy than the average, while other have less. Reactions are similar in that it’s the part of the distribution higher than the activation energy that matters. This plot shows how this works: the distribution of energies for a given molecule at two different temperatures shows that for higher temperature more molecules will have energies above the activation energy than for the lower temperature.
The Arrhenius equation gives us an expression that summarizes the collision model of chemical kinetics. It goes as follows: rate = (collision frequency)*(a steric factor)*(the fraction of collisions with $\mathrm{E} > \mathrm{E}_a$). In math terms, that’s shown here for the equation for rate ($\mathrm{k}$).
$\mathrm{A}=$ the frequency factor, and its units depend on the reaction order. For example if the reaction is first order then the frequency factor must have units of $s^{-1}$ The activation energy, $\mathrm{E}_a$, we've already discussed. If it's given in units per mole, like $\mathrm{J} / \mathrm{mol}$, we use $\mathrm{R}$ as it's written, where $\mathrm{R}$ is the ideal gas constant $\mathrm{R}=8.314 \mathrm{~J} / \mathrm{K}^*$ mol. If the activation energy is given in units of $\mathrm{eV}$, then the constant used would be the Boltzmann constant in units of $\mathrm{eV}\left(8.61733 \times 10^{-5} \mathrm{eV} / \mathrm{K}\right)$.
This relationship means that if we plot the natural $\log$ of the rate vs. $1 / \mathrm{T}$ then it should be a straight line with slope $=-\mathrm{Ea} / \mathrm{R}$ and intercept $=\ln (\mathrm{A})$, as shown in the plot above.
So we’ve covered concentration, and now temperature. The last example of a way to change the rate of a reaction that we’ll mention (and unlike those other two, we’ll really just mention it and not go into detail) is the catalyst. A catalyst is a way to increase the rate of a reaction without having anything consumed as part of it. It’s a material that, in the language of our discussion on Arrhenius above, lowers the activation energy for the reaction.
Why this matters
Let's keep going with the catalyst theme for this section. It is estimated that $\approx 90 \%$ of all commercially produced chemical products involve catalysts at some stage in the process of their manufacture! Some of these processes I've already highlighted in other Why This Matters moments, like the Haber-Bosch process for fixing $\mathrm{N}_2$, or the depletion of $\mathrm{O}_3$ by CFCs. At the time, we hadn't learned about reaction rates or catalysts, so I didn't go into it. But in both cases the role of the catalyst is absolutely essential (in fact, the big innovation of Haber-Bosch was not to discover the reaction (which had been known) but rather to discover a catalyst that lowers the temperature needed to make the reaction happen economically and at large scales.
Let's discuss another world-changing catalytically enhanced reaction: namely, the removal of most toxic emissions from cars and trucks. I know, you may be thinking that the tailpipe of a car smells pretty toxic. And that's because it is, but it's a whole lot better than it used to be, and the reason is the catalysts that are now part of every tailpipe in the form of what is called the catalytic converter. Why did we need these in the first place? It all goes back to the very first reaction we wrote on the first day of lecture: combustion. One example I gave was the combustion of methane:
$\mathrm{CH}_4+2 \mathrm{O}_2 \rightarrow 2 \mathrm{H}_2 \mathrm{O}+\mathrm{CO}_2$
It's true that $\mathrm{CO}_2$ is harmful to the environment for reasons of climate change, but there's nothing toxic in those products... so what's the problem?
Ah, if only cars burned pure methane! But gasoline is far, far away from a pure fuel source. And furthermore, even modern car engines are far, far away from being able to burn the fuel perfectly without side-reactions. Gasoline is a mixture of about 150 different chemicals, and these include not just those hydrocarbons that combust, but also a host of additives that range in purpose from corrosion inhibitors to lubricants to oxygen boosters. Since this complex chemical soup doesn’t burn cleanly, we get both direct products and by-products that go far beyond the pure case of $\mathrm{H}_2\mathrm{O}$ and $\mathrm{CO}_2$. Many of these products are pollutants and some are really bad ones. An incomplete list would include: carbon monoxide ($\mathrm{CO}$) which is poisonous, nitrogen oxides (like $\mathrm{NO}$ and $\mathrm{NO}_2$, or “$\mathrm{NOX}$” as they’re called) which cause smog and have many adverse health effects, sulfur oxides (yes, you guessed it, “$\mathrm{SOX}$”) which cause acid rain, and unburned hydrocarbons or volatile organic compounds (VOCs) which cause cancer.
Similar to the removal of CFC's from refrigerants, cleaning up the tailpipe represents a fantastic example for how policy and regulation can make the world better. It was the Clean Air Act that Congress passed in 1970 that gave the newly-formed EPA the legal authority to regulate this toxic mess that came out of cars. As a result, today's cars are $\approx 99 \%$ fewer emissions compared to the 1960s! The fuels are also cleaner (lead was removed and sulfur levels lowered), and taken together cities have much healthier air. Take a look at this picture of New York city from 1973 (left) compared to 2013 (right). In the 1970's, smog from car exhaust overwhelmed most major U.S. cities.
The key technology that enabled this dramatic clean-up is the catalytic converter. Inside a catalytic converter there are actually multiple catalysts, each enhancing different reactions. Most cars today use what is called a "three-way" catalyst, which just means that it tackles all three of the biggest pollutants: NOX, hydrocarbons, and $\mathrm{CO}$. The key materials used as the catalysts are a combination of platinum, palladium, and rhodium. Inside a catalytic convertor one typically has a honeycomb mesh just to get a large surface area. Since the temperature gets quite high (and in fact needs to be high for the catalysts to operate, which is why cold engines pollute more than hot ones), the honeycomb mesh is made out of a ceramic material like alumina so it can handle $\mathrm{T}=500^{\circ} \mathrm{C}$ without cracking or degradation. The $\mathrm{Pt}$, $\mathrm{Pd}$, and $\mathrm{Rh}$ metals coat the $\mathrm{Al}_2 \mathrm{O}_3$ mesh, and the exhaust flows through.
Take a look at this catalytic converter schematic. You can see two different chambers, one where the metal acts as a "reduction catalyst" for the NOX removal and the other where a different metal (or combination of metals) acts as an "oxidation catalyst" to treat $\mathrm{CO}$ and unburned hydrocarbons. In the reduction catalyst chamber, the reaction we're trying to accelerate is $2 \mathrm{NO} \rightarrow \mathrm{N}_2+\mathrm{O}_2$. In order to do so, the catalyst binds the $\mathrm{NO}$ molecule to it, and is then able to pull off the nitrogen atom from $\mathrm{NO}$ and hold it in place. Then another $\mathrm{N}$ atom that also got pulled off of a different $\mathrm{NO}$ molecule gets stuck to the catalyst somewhere nearby, and those Likewise, the oxygen can form $\mathrm{O})2$. The whole point is that the catalyst finds a different way to carry out the same reaction with much lower barriers. The oxidation catalyst burns $\mathrm{CO}$ and hydrocarbons using remaining $\mathrm{O} 2$ gas, for example to get this reaction to go: $2 \mathrm{CO}+\mathrm{O}_2 \rightarrow 2 \mathrm{CO}_2$. Again, that reaction would not occur at a high rate normally, but the catalyst breaks it down into steps (like splitting $\mathrm{CO}$) that occur much more easily.
Why this employs
The rate of a reaction is of course of central importance to anything one does that involves a reaction. And how many different types of jobs do anything involving a reaction? A ton! We could be talking about any sort of chemical synthesis job, or in drug design where reaction rate is crucial to the drug’s effectiveness and safety, or how about food making (even beyond beer) where rates determine anything from how long to let dough rise to how fast that dough browns in an oven vs. the apples inside turning mushy (I just made an apple pie in case you couldn’t tell), to how quickly to food goes bad. The point is that making stuff is inherently about rates.
And so instead of focusing on any one of these things that we make, I’d like to use this Why This Employs section to mention the fact that in nearly everything we make today, far too much natural capital is expended, and it doesn’t have to be that way. A single computer chip takes more than 600 times as much mass to make it and tremendous amounts of both water and energy (including annealing multiple times at temperatures over 1000°C. Cement is a wonderful material (“liquid stone”) but it takes baking calcium silicate at very high temperatures to make the tiny particles called “clinkers” that give it the ability to take on any shape in an instant and then dry and set so quickly after. Why is it that, time and time again, we fail to do what nature does so well? Animals can build incredibly strong and complex structures without going a single degree over room temperature. Spiders spin silk stronger than steel without lighting a fire. How can we do better? Much of it comes down to reaction rates. We use these massive amounts of energy and high temperatures and harsh chemicals because we want to make stuff quickly and because we need to make stuff in almost unimaginable quantities. China alone produced 2.5 trillion kilograms of cement last year!
There are more sustainable ways to make stuff, especially if we can get a handle on the reaction rates. Ten years ago I remember reading articles about the new revolution of “green chemistry” (check out for example this 2010 article in Scientific American, https://www.scientificamerican. com/article/green-chemistry-benign-by-design/. I remember thinking then that this field was ready to take off. Well, it didn’t really happen with a bang, but slow and steady this idea of benign design is taking hold. If you search for green chemistry jobs, you’ll find many companies now investing in this area substantially and that spells new jobs. The idea of green chemistry on the job market ranges quite broadly, from finding ways to synthesize materials without toxic chemicals, to making the drug discovery process biodegradable, to lowering the temperatures needed in a manufacturing step. Whatever it is, it’s all about the rates.
Example problems
1. Acetic acid is made from carbon monoxide and methanol according to the following equation:
$CO(g)+MeOH(g) \rightarrow A c O H(g)$
Your company wants to know how to improve this reaction: they present you, a chemical consultant, with the following data. Propose a rate law for the reaction.
$\mathrm{CO}$ pressure $\mathrm{MeOH}$ pressure Acetic acid formed
21.5 atm 15.0 atm 980 mol/hour
11.1 atm 15.0 atm 499 mol/hour
11.1 atm 10.0 atm 502 mol/hour
Answer
The rate is fairly insensitive to changing $\mathrm{MeOH}$, but the rate drops by about half when the pressure of $\mathrm{CO}$ is halved. Therefore, the rate is $0^{th}$ order in methanol, but $1^{st}$ order in $\mathrm{CO}$:
$rate = k[CO]$
Lecture 28: Equilibium and Solubility
Summary
Many of you have probably heard the expression, “like dissolves like”. . . but let’s add a little bit more chemistry to that phrase. What is really meant by the word “like” is “similar bonds,” so the expression could also be said, “similar bonds dissolve similar bonds.” Solubility is a metric that tells us how much the solute (the solid being dissolved) can break apart and dissolve into the solvent (the liquid in which the solid is dissolving). When the solubility is miscible that means that it can be full mixed together at any concentration.
Intermolecular forces play a huge role in solubility. Water is called a polar solvent (also called hydrophilic) because it’s a solvent with large dipole moments, while hydrocarbons are examples of nonpolar solvents (also called hydrophobic) since the dominant IMF is London. Polar solvents are in general immiscible with nonpolar solvents.
What’s really going on at the scale of the molecule and its various IMFs, is that if a molecule can lower or keep similar its energy by welcoming a different molecule into its bonding environment, then it will do so. If on the other hand the other molecule cannot bond in the same way (as in the example of the longer hydrocarbon chaining having much more London forces than H-bond ability) then it would raise the energy of the system for the solvent to bond to it over itself.
Particles in a solution that re-join the solid are said to be precipitating. both dissolution and precipitation occur simultaneously and at the point of saturation, or maximum solubility, they happen at exactly the same rate. This is called dynamic equilibrium.
We know from our previous lectures on ionic solids that these are cations and anions in the solid, which is the basis of the ionic bond. When dissolved in water, they also remain ions, just now in solution, where they interact strongly with water. We can understand why, since we know from our IMF lecture that the ions will interact with the dipole of the water molecule. The reason there is a trend in the solubility of these salts is related to the strength of the bond between ions in the solid relative to those the liquid. As the ionic bonding becomes stronger (from weakest at $\mathrm{NaI}$ to strongest at $\mathrm{NaF}$), less of it dissolves.
Suppose the general reaction has already reached its dynamic equilibrium:
$a A+b B \leftrightarrow c C+d D$
This means that the four different concentrations, $[\mathrm{A}],[\mathrm{B}],[\mathrm{C}]$, and $[\mathrm{D}]$ are not changing even though the reaction is still going in both directions. Below is a nice picture of what might happen to these concentrations as a function of time. At first during the reaction the concentrations change, and then they all flatten out. At that point, when they're all flat, the system has reached dynamic equilibrium. We define the reaction quotient, $\mathrm{Q}$, as the ratio of
$\dfrac{[C]^c[D]^d}{[A]^a[B]^b}$
This expression for $\mathrm{Q}$ can be evaluated at any point during the reaction, whether the system has reached equilibrium or not. $\mathrm{BUT}$, we have a special way to call $\mathrm{Q}$ once the system is in dynamic equilibrium, and that's with the letter $\mathrm{K}$, or the equilibrium constant for the reaction. Often, we put a subscript "eq" on the $K$ just to make sure we know what it refers to, so $\mathbf{K}_{e q}=\mathbf{Q}$ (for the reaction quotient in equilibrium). The Keq for a solid dissolving in water is the solubility product $\left(\mathbf{K}_{s p}\right)$
By comparing the value of $\mathrm{Q}$ to the equilibrium constant, $\mathrm{Keq}$, for the reaction, we can determine whether the forward reaction or reverse reaction will be favored.
If $\mathrm{Q}=\mathrm{K}_{e q}$, the reaction is at equilibrium. If $\mathrm{Q}<\mathrm{K}_{e q}$, like at point $\mathrm{D}$, then the reaction will move to the right (in the forward direction) in order to reach equilibrium. If $\mathrm{Q}>\mathrm{K}_{\text {eq }}$, like in point $\mathrm{E}$, then the reaction will move to the left (in the reverse direction) in order to reach equilibrium. Le Chatelier coming at us in graphical form!
Why this matters
From examining the tiny bubbles trapped in ice cores, we know that the level of $\mathrm{CO}_2$ currently in the atmosphere is more than it has been in over a million years, and the amount is growing exponentially. But while this incredibly important greenhouse gas has received tremendous attention for its role in climate change, there is a much less frequently discussed impact that $\mathrm{CO}_2$ is having on our planet. Maybe that’s because it’s happening in the oceans, which we don’t live in so it’s harder to make as big of a deal about the changes occurring. But we will feel these changes soon enough. The ocean has absorbed more than 500 billion tons of $\mathrm{CO}_2$ from the atmosphere, as it quietly captures roughly a quarter of what’s being emitted day in and day out (today we humans crank out 1,200,000 grams of $\mathrm{CO}_2$ per second). So why does this matter?
The problem with all that extra $\mathrm{CO}_2$ in the oceans is that it makes the water more acidic. Over the past 200 years alone, ocean water has become 30 percent more acidic. This change in acidity is faster than any known change in ocean chemistry over the last 50 million years! In geological terms, this is an extremely abrupt perturbation, and without dramatic changes in human behavior we are only seeing the beginning.
Unfortunately, one of the most important defenses for ocean creatures - the shell - cannot survive increased acidity. Here's a picture of what happens to the shell of a pteropod, a type of mollusk only a few $\mathrm{cm}$ long, after 45 days in a solution of $\mathrm{pH}=7.8$. Now, the oceans aren't quite that acidic yet (current ocean $\mathrm{pH}$ is 8.1), but they're well on their way. Pteropods are one of only a few basic types of sea creature that sits at the very bottom of the ocean food chain, right above plankton and seaweed, which means that if their numbers are reduced, everything higher on the food chain is impacted, from krill to salmon, to herring and birds, to seals and polar bears, and of course, to us.
In the goodie bag that goes with these lectures, you’ll be able to see this dissolution in action. We make it go faster than 45 days since I don’t want you to wait that long, but the chemistry at play is the same. I wanted you to touch and feel this precious calcium carbonate loss, so that you can get a glimpse into what is beginning to happen to 2/3 of our planet.
Why this employs
In order for drugs to have a physiological effect on the human body, they must be in solution. In particular, they must be dissolved in solution if they start out in solid form, like tablets do. Now, the rate of the dissolution is extremely important since determines how fast the drug is absorbed in the body. Rates were discussed in the last chapter. But solubility itself (i.e. regardless of how fast it dissolves, what is its ability to fully dissolve at all?) is also of crucial importance in drug design. If the drug is not very soluble, formulations can lead to difficulties in the use of the drug which may otherwise have very beneficial effects. If a drug has low solubility that means it will likely have a low level of what is called “bioavailability,” which basically just means that it won’t get enough exposure in the body.
In pharmaceutical companies that do most of the new drug development, it is estimated that a whopping 40% of all new chemicals (possible new beneficial drugs) are practically insoluble in water, making them effectively useless. And what is it that these companies do to try to increase solubility, especially for otherwise promising candidates? Chemical modification, of course! And so it is that all of the large pharmaceutical companies are looking for people who know about solubility! In terms of where to look for jobs, the thing about drug design companies is that you don’t have to look very far. Within just 1 mile of campus, there are dozens of both large and small companies specializing in drug design. More broadly, Kendall Square now has the largest biotechnology industry per square foot, with over 120 companies within a mile.
In terms of which ones of these companies have jobs related to solubility, well because of how important solubility is, it’s pretty much all of them. Check out for example, the job postings at Takeda, Novartis, Alnylam, Sanofi, Pfizer, and Genentech just to get started, and then walk down the street to go learn more about the companies, what they do, and meet with them in person!
Example Problems
1. Calcium fluoride has a $\mathrm{K}_{sp}$ of $5.3×10^{-9}$. How much calcium fluoride can dissolve in 1$\mathrm{L}$ of water?
Answer
$k_{s p}=[C a][F]^2 \nonumber$
For every mole of $\mathrm{Ca}$ that dissolves, twice as many moles of $\mathrm{F}$ must dissolve since the chemical formula is $\mathrm{CaF}_2$. Assuming $\mathrm{x}$ moles of $\mathrm{Ca}$ dissolve:
\begin{gathered}
K_{s p}=[x][2 x]^2 \
5.3 \times 1-^{-9}=4 x^3 \
x=0.00109 M \
78 \dfrac{g}{\mathrm{~mol}} \dfrac{0.00109 \mathrm{~mol}}{L}=85.9 \mathrm{mg} / 1 \mathrm{~L}
\end{gathered}
Lecture 29: Common Ions and Acids/Bases
Summary
Le Chatelier’s principle states that the position of equilibrium moves to account for changes to the system, such as the introduction of a new ion to a solution. But what happens if we add a new compound that has one ion in common with an existing solution? The equilibrium curve for the $\mathrm{AgCl}$ dissociation is shown to the right. Suppose we add $\mathrm{NaCl}$: it will dissociate completely into ions $\mathrm{Na}^{+}(\mathrm{aq})$ and $\mathrm{Cl}^{-}(\mathrm{aq})$. We can use an ICE table to determine the shift in equilibrium: if we start at point $\mathrm{B}$ and add $0.1 \mathrm{M} \mathrm{NaCl}$, we can use the equilibrium constant for $\mathrm{AgCl}, \mathrm{K}_{s p}=1.7 \times 10^{-10}$. As $\left[\mathrm{Cl}^{-}\right]$ increases, $\left[\mathrm{Ag}^{+}\right]$ must decrease to keep $\mathrm{K}_{s p}$ constant.
[$AgCl$] [$Ag^+$] [$Cl^-$]
I (all solid) $1.3 \times 10^{-5}$ $1.3 \times 1^{-5}$
C + x precipitates $-x$ $+0.1 - x$
E More solid $1.3 \times 10^{-5} - x$ $+0.1 - x$
\begin{gathered}
K_{s p}=\left[\mathrm{Ag}^{+}\right]\left[\mathrm{Cl}^{-}\right]=(0.1-x)\left(1.3 \times 10^{-5}-x\right) \
K_{s p}=0.1\left(1.3 \times 10^{-5}-x\right)=1.7 \times 10^{-10} \quad x=1.7 \times 10^{-9}
\end{gathered}
This is called the common ion effect: the solubility of one constituent is repressed by the addition of a second solute.
For a generic case of any acid, call it "A" we write its dissolution reaction as follows: $\mathrm{HA}+\mathrm{H}_2 \mathrm{O} \rightarrow$ $\mathrm{H}^{+}+\mathrm{A}^{-}+\mathrm{H}_2 \mathrm{O}$. Although we write $\mathrm{H}^{+}$ to make the proton dissociation explicit, we never actually mean that if it's in water. That's because the $\mathrm{H}^{+}$ion is not stable in $\mathrm{H}_2 \mathrm{O}$, instead it always becomes $\mathrm{H}_3 \mathrm{O}^{+}$. The first person to define acids and bases in any sort of more concrete chemical way was good old Svante Arrhenius. According to him, an acid is a substance that dissolves in water to produce $\mathrm{H}+$ ions, and a base is a substance that dissolves in water to produce hydroxide $\left(\mathrm{OH}^{-}\right)$ions.
Arrhenius defined acids and bases in terms of the presence of $\mathrm{H}^{+}$ or $\mathrm{OH}^{-}$ions in solution: for example, $\mathrm{HCl}(a q) \rightarrow \mathrm{H}^{+}(a q)+\mathrm{Cl}^{-}(a q)$ is an "Arrhenius acid" while $\mathrm{NaOH}(s) \rightarrow \mathrm{Na}^{+}(a q)+ O H^{-}(a q)$ an "Arrhenius base."
Sorenson came up with the "power of hydrogen" scale, or "pH" scale for short. $\mathrm{pH}=-\log \left[\mathrm{H}^{+}\right]$. The $\mathrm{pH}$ scale runs from 0 to 14 , with 0 being extremely acidic and 14 being extremely basic. We can do the same thing for the concentration of $\left[\mathrm{OH}^{-}\right]$: the power of $\mathrm{OH}^{-}$ in solution, or the "pOH" $=-\log \left[\mathrm{OH}^{-}\right]$, with the opposite definition.
A strong acid like $\mathrm{HCl}$ fully dissociates: $\mathrm{HCl}+\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{H}^{+}+\mathrm{Cl}^{-}+\mathrm{H}_2 \mathrm{O}$, so $0.1 \mathrm{M} \mathrm{HCl}$ has $\left[\mathrm{H}^{+}\right]=0.1$, so $0.1 \mathrm{M} \mathrm{HCl}$ solution has $\mathrm{pH}=-\log [0.1]=1.0$.
Some materials can be both acids and bases. None other than water itself is one such material! When a molecule can either give $[\mathrm{H}+]$ ions or $[\mathrm{OH}-]$ ions into solution, it's called amphoteric: $\mathrm{H}_2 \mathrm{O}(l)+\mathrm{H}_2 \mathrm{O}(l) \leftrightarrow \mathrm{H}_3 \mathrm{O}^{+}(a q)+\mathrm{OH}^{-}(a q) . \mathrm{K}_{s p}$ for water is called $\mathrm{K}_w: K_w=\left[\mathrm{H}_3 \mathrm{O}^{+}\right]\left[\mathrm{OH}^{-}\right]= 1.0 \times 10^{-14}$. If an acid is added to pure water the hydrogen ion concentration increases (and the $\mathrm{OH}$ ion concentration decreases) A base goes the other way, adding hydroxide ions into the pure water solution. All of the tricks we learned for $\mathrm{K}_{s p}$ apply to $\mathrm{K}_w$, which is especially useful in the context of acids and bases.
Arrhenius' definition of acids and bases has two key limitations: first, because acids and bases were defined in terms of ions obtained from water, the Arrhenius definition applied only to molecules in aqueous solution. Second, and more important, the Arrhenius definition predicted that only materials that dissolve in water to give $\mathrm{H}^{+}$ and $\mathrm{OH}^{-}$ions can have the properties of acids and bases. But there are many examples where this is not true! We need to go beyond Arrhenius to understand some acids and bases.
Why this matters
Now that we're armed with the concepts of the Common Ion Effect and Le Chatelier's principle, we can go into more detail on the chemistry that's causing the possible catastrophe discussed in the previous Why This Matters: ocean acidification and its impact on the fate of calcium carbonate. It's all about equilibrium and how $\mathrm{CO}_2$ is shifting it in the oceans. The first reaction we consider is what happens when the oceans encounter more $\mathrm{CO}_2$ from the atmosphere: namely, the $\mathrm{CO}_2$ dissolves in the $\mathrm{H}_2 \mathrm{O}$ to form carbonic acid:
$\mathrm{CO}_2+\mathrm{H}_2 \mathrm{O} \leftrightarrow \mathrm{H}_2 \mathrm{CO}_3$
So more $\mathrm{CO}_2$ getting dissolved in the ocean means there's more carbonic acid in the ocean, which then produces dissociated ions as follows:
$\mathrm{H}_2 \mathrm{CO}_3 \leftrightarrow \mathrm{HCO}_3^{-}+\mathrm{H}^{+}$
But on the other hand, we've got solid $\mathrm{CaCO}_3$ which is the calcium carbonate, that is in a nice equilibrium with the following dissociation reaction:
$\mathrm{CaCO}_3 \leftrightarrow \mathrm{Ca}^{2+}+\mathrm{CO}_3^{2-}$
The equilibrium constant for the dissolution of $\mathrm{CaCO}_3$ is $\mathrm{k}_{s p}=5 \times 10^{-9}$. That means not a whole lot of it will dissociate in "normal" (i.e. last $>50$ million years) ocean conditions. The reason this equilibrium is getting thrown off is that there are now extra $\mathrm{H}^{+}$ions that come from the $\mathrm{CO}_2$ reaction above. These $\mathrm{H}^{+}$ions consume (react with) $\mathrm{CO}_3^{2-}$ to form $\mathrm{HCO}^{3-}$ which lowers the concentration of $\mathrm{CO}_3^{2-}$. Ah, but we just learned about that with the Common Ion Effect: if we add or take away some molecule that's part of a balanced equilibrium, then we will drive that equilibrium to counter whatever we've done. In this case , $\mathrm{CO}_3^{2-}$ is getting consumed more than before, which in turn drives more dissolution of the $\mathrm{CaCO}_3$. That's why the shells are dissolving, and it's why I want you to experience this chemistry directly in your goodie bag. What you're seeing is those experiments is an accelerated version of what is happening in the oceans.
Acidity is measured by " $\mathrm{pH}$," which is a logarithmic scale. Since the Industrial Revolution, the $\mathrm{pH}$ of the oceans has decreased by $\approx 0.1$ to $8.07$, which is equivalent to a $\approx 30 \%$ increase in the oceans' acidity. I showed that trend in last lecture's Why This Matters. Estimates are that at our current rate of $\mathrm{CO}_2$ emissions, the acidity of the oceans will reach a whoppingly more acidic state of $\mathrm{pH}=7.7$ by 2100 . That would represent a $>5$-fold increase in acidity compared to pre-industrial levels, and that would also pretty much do it in terms of wiping out most ocean life.
The ocean has absorbed roughly 525 billion tons of $\mathrm{CO}_2$ from the atmosphere over the past 200 years, at a pace of $\approx 22$ million tons per day. The change that this induces is the fastest known chemical change the ocean has experienced over the past 50 million years and probably much longer. Around 250 million years ago, at the end of what is known as the "Permian era," records show that there was a sudden change in the oceans, and geologists refer to what happened as, "The Great Dying." This is because more than $90 \%$ of all marine life disappeared from the fossil record. Most scientists who study this kind of thing relate the mass extinction to a huge spike in volcanic activity, which put vast clouds of acidic dust into the atmosphere, which then fell into the oceans and raised their acidity.
The reason the dissolution of calcium carbonate is so important is that it is essential not just to the pteropods I showed before, but to so many other “base” elements of the ocean’s food chain, like phytoplankton, shell fish (mussels, oysters, lobsters, crayfish, etc), and of course coral, to name only a few examples. The calcium carbonate structures in these sea creatures are highly sensitive to slight changes in acidity, and without them the health of the oceans will fail. I’ll end this Why This Matters with a thought-provoking quote from Ken Caldeira who is a climate scientist at Stanford. In trying to help people get perspective on what we’re doing, he said, “Well, if the Romans had industrialized, it now would be two thousand years later. The seas would still be rising and still be completely acidified, and, yes, maybe they would have got a century or so of higher G.D.P. than they would have otherwise.” The point being: the deleterious changes we’re on track to make in the oceans won’t just impact our kids, and our kids’ kids; it will impact tens and more likely hundreds of generations to come.
Why this employs
I talked about ocean acidification in this lecture since we’re introducing acids and bases, and what’s going on in the oceans is a good example of the effects of chemistry happening on a planetary scale. But what if you want to actually work on this problem, like as a job? Well, that’s not as straightforward as you might think, or for that matter may have hoped, but there are some things you can do.
For one thing, there are many, many academic or state-run programs that are studying ocean chemistry, and many of them have postings for jobs. This ranges from a random ocean acidification project at the University Di’ Bologna, Italy, to the National Oceanic and Atmospheric Administration Cooperative Science Center for Earth System Sciences and Remote Sensing Technologies (yes, thankfully they go by NOAA-CESSRST), to other government agencies such as the EPA or any agency working in conservation and policy, to Washington State’s Department of Ecology, to Stony Brook University who is currently seeking an, “Ocean Acidification Monitoring Associate.” There are many faculty positions in this area, and I believe there will be more in the future. So if you’re interested in both this topic as well as a career in academia, then the path of research in scientific centers is a good one. There are a number of different specializations that would lead to studying (and hopefully helping to save!) the oceans, from marine biologist to environmental engineer to a “chemical oceanographer,” who is kind of like an oceanographer but specializing in the chemical composition of the oceans rather than their ecology, biological life and geology.
Example problems
1. Hydrofluoric acid reacts with calcium carbonate according to the following equation:
$2 \mathrm{HF}(a q)+\mathrm{CaCO}_3 \rightarrow \mathrm{H}_2 \mathrm{O}(l)+\mathrm{CO}_2(g)+\mathrm{CaF}_2$
If $100 \mathrm{~mL}$ of a $10 \mathrm{mM}$ solution of $\mathrm{HF}$ is added to $100 \mathrm{~mL}$ of a $10 \mathrm{mM}$ solution of $\mathrm{CaCO}_3$, how much $\mathrm{CaF}_2$ will precipitate? Assume the reaction goes to completion.
Answer
In this problem, the first clue is that the reaction "goes to completion:" this indicates that we don't need to worry about equilibrium. When a $10 m M$ solution of $\mathrm{HF}$ is added to a $10 m M$ solution of $\mathrm{CaCO}_3$, the final solution has a different concentration.
The original solutions each contain $1 \mathrm{mmol}$ of material, since $0.1 L * 10 \mathrm{mmol} / L=1 \mathrm{mmol}$. The final solution, therefore, has $1 \mathrm{mmol} \mathrm{Hf}$ and $1 \mathrm{mmol} \mathrm{CaCO}_3$ in $0.2 \mathrm{~L}$ of water, so the concentration of the final solution is $5 \mathrm{mM}$ in both reactants.
Next, we need to think about how much $\mathrm{CaF}_2$ we can form: if we start with $1 \mathrm{mmol}$ of $\mathrm{HF}$ and 1 mmol $\mathrm{CaCO}_3$, you can check that the limiting reagent is $\mathrm{CaCO}_3$, since we need twice as much $\mathrm{HF}$ as $\mathrm{CaCO}_3$ to form $\mathrm{CaF}_2$. We'll end up with $0.5 \mathrm{mmol} \mathrm{CaF}_2$, and leave behind $0.5 \mathrm{mmol} \mathrm{CaCO}_3$.
Next, we need to determine how much will dissolve. We can use an ICE table and the common ion effect. We start with $0.5 \mathrm{mmol}$ of calcium ions in $0.2 \mathrm{~L}=0.25 \mathrm{mM}$:
$\begin{array}{c|c|c|c} & \mathrm{CaF}_2 & \mathrm{Ca}^{2+} & 2 F^{-} \ \hline \mathrm{I} & & 2.5 \mathrm{mM} & 0 \mathrm{mM} \ \hline \mathrm{C} & & +\mathrm{x} & +2 \mathrm{x} \ \hline \mathrm{E} & & 2.5+\mathrm{x} & 2 \mathrm{x} \end{array} \nonumber$
\begin{aligned}
5.3 \times 10^{-9} &=(.0025+x)(2 x)^2 \
x &=0.000649
\end{aligned}
Therefore, $0.649 \mathrm{mM}$ will be soluble, leaving $2.5 \mathrm{mM}-0.649 \mathrm{mM}=1.85 \mathrm{mM} C a F_2$ will precipitate.
$1.85 \mathrm{mmol} / L * 0.2 L=0.37 \mathrm{mmol} \nonumber$
$0.37 \mathrm{mmol} * 78.07 \mathrm{~g} / \mathrm{mol}=28.9 \mathrm{mgCaF} \mathrm{C}_2 \nonumber$
Therefore, $28.9 \mathrm{mg}$ of $\mathrm{CaF}_2$ will precipitate.
Lecture 30: Bronsted-Lowry Acids/Bases and Neutralization
Summary
In this lecture, we started by considering what happens if you combine an acid and a base directly: for example, $\mathrm{NaOH}(\mathrm{aq})+\mathrm{HCl}(\mathrm{aq}) \rightarrow \mathrm{H} 2 \mathrm{O}(\mathrm{l})+\mathrm{NaCl}(\mathrm{s})$. Each of these products, the water and the salt, find an equilibrium of their own. So the arrow goes both ways $\mathrm{H}++\mathrm{OH}-\leftrightarrow \mathrm{H}_2 \mathrm{O}(\mathrm{l})$ and water gives the usual neutral ion concentrations of $\mathrm{H}+$ and $\mathrm{OH}-$. Salt also will have its equilibrium reaction to form ions $\mathrm{Na}+$ and $\mathrm{Cl}-$ in solution. Sometime the $\mathrm{Na}+$ and $\mathrm{Cl}-$ ions called "spectators" when in solution, since they don't participate in making the solution either acidic or basic. The reaction of an acid with a base to make water and salt is called a neutralization reaction.
Bronsted and Lowry defined acids and bases more broadly: a Bronsted-Lowry acid is anything that releases $\mathrm{H}^{+}$ions. A Bronsted-Lowry base is anything that accepts $\mathrm{H}+$ ions. A "conjugate pair" is a nice way to keep track of the proton transfer that happens in acid/base chemistry. If we consider the generic reaction $\mathrm{HA}+\mathrm{H}_2 \mathrm{O}$, where we label the A for "acid" and purposefully write the $\mathrm{H}$ separately since we know as an acid it will be giving that $\mathrm{H}$ up as a $\mathrm{H}^{+}$ion, the molecule left the other conjugate pair over after giving up the $\mathrm{H}+$ ion into solution, $\mathrm{A}-$, is called its conjugate pair. Similarly, if it reacted with $\mathrm{H}_2 \mathrm{O}$, then the $\mathrm{H}_2 \mathrm{O}$ gained the $\mathrm{H}+$ to become $\mathrm{H}_3 \mathrm{O}^{+}$, and those two molecules are conjugate pairs. According to Bronsted-Lowry, an acid base reaction is essentially just a proton transfer reaction.
A general way to write an acid mixing with a base would be: $\mathrm{HA}+\mathrm{B} \rightarrow \mathrm{BH}^{+}+\mathrm{A}^{-}$ where here it is very clear that the acid transfers a proton to the base.
How do we know why and when an acid or base is "strong" vs. "weak?" This table lists common strong acids and bases. Acids other than these six are essentially all weak acids. The only common strong bases are the hydroxides of the alkali metals and the heavier alkaline earths ($\mathrm{Ca}, \mathrm{Sr}$, and $\mathrm{Ba}$; any other bases are likely to be weak. We can quantify why these molecules are considered "strong" by considering their acid dissociation constants. For $\mathrm{HCl}$, the acid dissociation constant, or in other words the equilibrium constant for the acid, $\mathrm{K}_a=\left[\mathrm{H}_3 \mathrm{O}^{+}\right]\left[\mathrm{Cl}^{-}\right] /[\mathrm{HCl}]$ $\approx 106$ - this is huge, and it means effectively full dissociation. A strong acid like $\mathrm{HCl}$ is in fact strong because it fully dissociates. And the opposite is true too: if an acid fully dissociates, then it is a strong acid. The same applies for strong bases.
For example, suppose you have a $1.0 \mathrm{M}$ solution of $\mathrm{CH}_3 \mathrm{COOH}$. That means that there's 1 mole of $\mathrm{CH}_3 \mathrm{COOH}$ in a liter of water. Since $K_a=10^{-5}$, that means that $\left[\mathrm{H}_3 \mathrm{O}^{+}\right]$ for this $1 \mathrm{M}$ solution is around $0.003$, which in turn means that the $\mathrm{pH}$ of $1 \mathrm{M} \mathrm{CH}_3 \mathrm{COOH}$ is about $2.5$. Note that if there's $0.003$ moles of $\left[\mathrm{H}_3 \mathrm{O}^{+}\right]$ that can from 1 mole of $\mathrm{CH}_3 \mathrm{COOH}$, then that means that only $0.3 \%$ of it dissolved. That's not a lot of dissolution compared to the near $100 \%$ for the strong acids!
Here's another example: will the salt formed from the following reaction have a $\mathrm{pH}$ greater than, less than, or equal to $7? \mathrm{CH}_3 \mathrm{COOH}(\mathrm{aq})+\mathrm{NaOH}(\mathrm{s}) \leftrightarrow \mathrm{Na}^{+}+\mathrm{CH}_3 \mathrm{COO}^{-}(\mathrm{aq})+\mathrm{H}_2 \mathrm{O}(\mathrm{l})$ To answer this we don't even need to do any math. That's because if a weak acid is mixed with a strong base (as is the case here), we automatically know that it will be basic.
In general, when the following are mixed, the results are:
• Weak acid mixed with weak base: $\mathrm{pH}<7$ for $K_a>K_b ; \mathrm{pH}=7$ if $K_a=K_b ; \mathrm{pH}>7$ for $K_a<K_b$
• Strong acid mixed with string base: $\mathrm{pH}=7$
• Strong acid mixed with weak base: $\mathrm{pH}<7$
• Strong base mixed with weak acid: $\mathrm{pH}>7$
Why this matters
It might feel a bit on the hot side at $470^{\circ} \mathrm{C}$, and you'd be surrounded by sulfuric acid most of the time. That's what it would be like if you lived on Venus. Here's a nice picture from a Russian space probe back in the 1970's that landed on the surface, showing what a stroll would be like over there. These temperatures and acidic atmospheres may sound crazy, but many materials require similar conditions for processing. A lot of aspects of our industrial revolution have effectively relied on reproducing the conditions of Venus! In fact, today, nearly 250 million tons of sulfuric acid is produced per year, which makes it one of the largest chemicals produced worldwide. The liberated $\mathrm{H}+$ ions that come from $\mathrm{H} 2 \mathrm{SO} 4$ are put to use in a wide range of industries, from detergents to lead-acid car batteries to dyes to metal processing, it is in the production of fertilizers where its use dominates: roughly $50 \%$ of all sulfuric acid is used to make fertilizer!
In one of our earlier Why This Matters, we discussed the Haber-Bosch process, which uses a catalyst, high temperatures (like $\approx 400^{\circ} \mathrm{C}$ ), and high pressures (like $\approx 180$ atm.) to fix $\mathrm{N}_2$ molecules and turn them into ammonia, $\mathrm{NH}_3$. But the thing is, that's only one of the three most important ingredients plants need to grow: the other two are phosphorus and potassium. Phosphorus, in particular, is typically made with vast amounts of sulfuric acid. Why is this the case? You already know the answer, it's the chemistry.
found in sedimentary deposits, meaning that they formed by deposition of phosphate-rich materials in marine environments. Here's a picture of a phosphate rock formation being mined in Utah. But if you take phosphate rock and grind it up into a powder, it still won't be very soluble in water (we now have some knowledge about what that means!). This means that if you put the mineral directly into fertilizer, it will not be useful to plants. It's just like for the case of $\mathrm{N}_2$ in the air: even though there's plenty of $\mathrm{N}_2$, plants have no way of absorbing it. Rather, they need to work with $\mathrm{NH}_3$ in order to take in nitrogen. The same thing is true for phosphorus: tricalcium phosphate, $\mathrm{Ca}_3\left(\mathrm{PO}_4\right)_2$, which is what phosphate rock mostly consists of, is not able to give phosphorus to plants because it doesn't dissolve enough in water. The $\mathrm{K}_{s p}$ for this ionic salt is a whoppingly low $2.07 \times 10^{-33}$! By the way, it's a pretty good thing that this doesn't dissolve very much since it's a key ingredient in bones and teeth.
That's where sulfuric acid comes in: it's a fairly simple way to liberate the phosphoric acid $\mathrm{H}_3 \mathrm{PO}_4$ from the phosphate rock. In particular, the $\mathrm{H}_2 \mathrm{SO}_4$ from the phosphate rock. In particular, the $\mathrm{H}_2 \mathrm{SO}_4$ gives up its hydrogen in exchange for the calcium to form calcium sulfate, as follows:
$\mathrm{Ca}_3\left(\mathrm{PO}_4\right)_2+3 \mathrm{H}_2 \mathrm{SO}_4 \rightarrow 2 \mathrm{H}_3 \mathrm{PO}_4+3 \mathrm{CaSO}_4$
Once the phosphoric acid is produced in this manner, it can then be made into different types of fertilizer that are now water-soluble, so the plants can absorb it. These are generally termed "phosphates," or salts of phosphoric acid, which in the context of fertilizers include ammonium phosphates, calcium phosphates, and sodium phosphates. Bottom line: we feed the world because of our knowledge of acid/base chemistry!
Why this employs
Acids and bases are used so much in industry and in life: it's not hard to find jobs that involve acids and bases. You could check out DuPont, for example, which sells phosphate fertilizer plants that do just what we discussed above (for example, see http://cleantechnologies.dupont.com/ industries/phosphate-fertilizer/). http://cleantechnologies.dupont.com/...atefertilizer/. BASF was one of the very first companies to make and sell sulfuric acid on large scales, and they still do and have many job openings related to acid chemistry. There's Sigma Aldrich, which is a massive company with 10,000 employees and nearly $\ 3 \mathrm{~B}$ in annual revenue: their "acids and bases" category shows hundreds of products (btw, another company, Merck, bought them recently for $\ 17 \mathrm{~B})$. Or how about Cabot Laboratories, which doesn't make acids and bases explicitly but uses them to make every one of the cool chemical products they do make, whether elastomers or aerogels or "advanced carbons."
You could take any one of many different acids, look up its name, and likely find a market for it, and then all the companies that either make it or use it, many of which will have jobs related to it. Take formic acid: it’s getting more interest from a range of different industries because it’s easier to handle than other acids (see for example this article about it in C&EN (https://cen.acs.org/ articles/93/i48/Chemical-Makers-Eyes-Formic-Acid.html), and is used heavily in the textiles industry. BASF and Eastman Chemical are big on formic acid, but a lot of other companies are getting interested. Speaking of textiles, that’s another industry that uses massive amounts of acids, including formic acid but also citric, acetic, hydrochloric, and nitric acid. Textiles also rely heavily on bases like sodium hydroxide and baking soda for dyeing, cleaning, and fire-proofing clothes to name a few uses. All just to say that acids and bases and truly a bedrock of industry, and knowing how to make them or how to use them leads to massive job opportunities.
Example Problems
1. Identify the conjugate acid/base pairs in the following reactions:
a) $\mathrm{H}_2 \mathrm{PO}_4^{-}+\mathrm{H}_2 \mathrm{O} \leftrightarrow \mathrm{HPO}_4^{2-}+\mathrm{H}_3 \mathrm{O}^{+}$
Answer
b) $\mathrm{H}_2 \mathrm{O}+\mathrm{NH}_3 \leftrightarrow \mathrm{OH}^{-}+\mathrm{NH}_4^{+}$
Answer
2. Calculate the $K_a$ of an $0.2 \mathrm{M}$ aqueous solution of propionic acid $\left(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CO}_2 \mathrm{H}\right)$ with a pH of $4.88$. The dissociation can be expressed as
$\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CO}_2 \mathrm{H}+\mathrm{H}_2 \mathrm{O} \leftrightarrow \mathrm{H}_3 \mathrm{O}^{+}+\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CO}_2^{-}$
Answer
This problem can be solved with an ICE table:
$\begin{array}{c|c|c|c} & \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CO}_2 \mathrm{H} & \mathrm{H}_3 \mathrm{O}^{+} & \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CO}_2^{-} \ \hline \mathrm{I} & 0.2 & 0 & 0 \ \hline \mathrm{C} & -\mathrm{x} & +\mathrm{x} & +\mathrm{x} \ \hline \mathrm{E} & 0.2-\mathrm{x} & \mathrm{x} & \mathrm{x} \end{array} \nonumber$
\begin{gathered}
-p H=\log \left[H_3 O^{+}\right]=-4.88 \
\left[\mathrm{H}_3 \mathrm{O}^{+}\right]=10^{-4.88}=1.32 \times 10^{-5}=x \
K_a=\frac{\left[\mathrm{H}_3 \mathrm{O}^{+}\right]\left[\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CO}_2^{-}\right]}{\left[\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CO}_2 \mathrm{H}\right]}=\frac{x^2}{0.2-x}=\frac{\left(1.32 \times 10^{-5}\right)^2}{0.2-1.32 \times 10^{-5}}=8.69 \times 10^{-10}
\end{gathered}
Further Reading
Lecture 22: From x-ray Diffraction to Crystal Structure
• Laue condition (beyond-the-scope of 3.091):
http://www.physics.udel.edu/~yji/PHYS624/ Chapter3.pdf
• Nice visualizations:
http://web.pdx.edu/~pmoeck/phy381/Topic5a-XRD.pdf
Lecture 23: Point Defects
• Schottky and Frenkel defects, plus some beyond-the-scope stuff:
http://ww2.chemistry.gatech. edu/class/6182/wilkinson/nonstoi.pdf
• How defects give different gemstones their distinctive looks:
https://www.tf.uni-kiel.de/ matwis/amat/iss/kap_6/advanced/t6_1_1.html
Lecture 26: Engineering Glass Properties
• Inside a Corning factory:
https://www.youtube.com/watch?v=gZPeyErbqz4
• 3D printing glass:
https://www.youtube.com/watch?v=IvcpbtpWpGY
Lecture 28: Equilibrium and solubility
• Solubility product practice:
https://www.chem.purdue.edu/gchelp/howtosolveit/Equilibrium/ Solubility_Products.htm
• A method for making more medicines soluble:
https://www-chemistryworld-com.stanford. idm.oclc.org/health-tech/dissolving-insoluble-drugs/3008720.article | textbooks/chem/Inorganic_Chemistry/Introduction_to_Solid_State_Chemistry/05%3A_CHEM_ATLASes/5.03%3A_CHEM_ATLAS_3.txt |
Balancing reactions
A chemical reaction involves rearranging elements in compounds to make different substances. They are usually written as a sum of reactants, which when combined yield a sum of products :
$A+B \rightarrow C+D$
Here, A, B, C, and D represent chemical compounds. The fundamental principle guiding the process of balancing a reaction is conservation of mass: a chemical reaction cannot create or destroy mass! This has several implications that can be used to determine whether a reaction is valid:
1. The mass of the reactants must equal the mass of the products
2. Every element that is in a reactant must be in a product
3. The number of each type of atom in the reactants must equal the number of each type of atom in the products
Example: Ethylene and oxygen gas are combined to make water and carbon dioxide. If you start with 4 moles of $\mathrm{O}_2$ gas, how many moles of water and carbon dioxide can you make? The unbalanced equation is given below:
$\mathrm{C}_2 \mathrm{H}_4+\mathrm{O}_2 \rightarrow \mathrm{CO}_2+\mathrm{H}_2 \mathrm{O}$
First, we must balance the reaction. One method to do this is by using a table:
$\begin{array}{c||c|c|c||c|c|c} \hline & \mathrm{C} & \mathrm{H} & \mathrm{O} & \mathrm{C} & \mathrm{H} & \mathrm{O} \ \hline \text { Initial (unbalanced): } & 2 & 4 & 2 & 1 & 2 & 3 \ \hline \begin{array}{l} \text { Need: even number of oxygens } \ \text { Try: } 2 \times \mathrm{H}_2 \mathrm{O} \text { on the right } \end{array} & 2 & 4 & 2&1 &4 &4 \ \hline \begin{array}{l} \text { Need: even number of carbons } \ \text { Try: } 2 \times \mathrm{CO}_2 \text { on the right: } \end{array} & 2 & 4 & 2 & 2 & 4 & 6 \ \hline \begin{array}{l} \text { Need: more oxygen on the left } \ \text { Try: } 3 \times \mathrm{O}_2 \text { on the left } \end{array} & 2 & 4 & 6 & 2 & 4 & 6 \end{array}$
Once there are the same number of each elements on both sides of the reaction, we’re done balancing! The final reaction is
$\mathrm{C}_2 \mathrm{H}_4+3 \mathrm{O}_2 \rightarrow 2 \mathrm{CO}_2+2 \mathrm{H}_2 \mathrm{O}$
The greatest common factor between the coefficients in front of each compound is 1, so this is the simplest form.
Yield
The yield of a reaction is the maximum amount of products that can be made with the reactants that are put in. For example, consider a s’more, which consists of a marshmallow, a piece of chocolate, and two graham crackers. If you had 5 marshmallows, 4 pieces of chocolate, and 6 graham crackers, you could make 3 full s’mores (with 2 extra marshmallows and 1 extra marshmallow). We combine chemical compounds in the same way!
Example: Ammonia $\left(\mathrm{NH}_3\right)$ is produced when nitrogen gas $\left(\mathrm{N}_2\right)$ is combined with hydrogen gas $\left(\mathrm{H}_2\right)$. Write a balanced equation for this reaction, and determine how much ammonia can be produced if you start with 5 moles of hydrogen gas.
First, let’s balance the reaction. We can try some coefficients by inspection and verify they satisfy conservation of mass:
$\mathrm{N}_2+3 \mathrm{H}_2 \rightarrow 2 \mathrm{NH}_3$
This balanced reaction tells us that for every three moles of $H_2$ gas, we can make two moles of ammonia. Therefore, if we start with 5 moles of $H_2$:
5 moles of $H_2 \times \dfrac{2 \text { moles of } \mathrm{NH}_3}{3 \text { moles of } \mathrm{H}_2}=3.33$ moles of $\mathrm{NH}_3$
Limiting reagents
If we don’t start with the right stoichiometric ratios of reagents, there might be some reactant left over after we have formed the products. If we go back to the s’mores example, we were able to make three full s’mores, with extra chocolate and marshmallow. Since all of the graham crackers were used before the other reactants, they are the limiting reagent. The limiting reagent is specific to the initial amount of reactants available.
Example: The Kroll process for making titanium metal out of titanium chloride is:
$\mathrm{TiCl}_4+\mathrm{Mg} \rightarrow \mathrm{MgCL}_2+\mathrm{Ti}$
You react $25 \mathrm{~kg}$ of $\mathrm{Mg}$ with $200 \mathrm{~kg}$ of $\mathrm{TiCl}_4$.
a) Balance the reaction, b) determine the limiting reagent, and c) determine the yield of $\mathrm{Ti}$ in this reaction.
Answer
a) To balance the reaction, we can start by looking at the $\mathrm{Cl}$ atoms: we need to double the $\mathrm{MgCl}_2$ on the right to equal the left. Then, we just need to balance the $\mathrm{Mg}$ atoms: we need double on the right to account for the extra we just created on the left. The balanced reaction is therefore
$\mathrm{TiCl}_4+2 \mathrm{Mg} \rightarrow 2 \mathrm{MgCl}_2+\mathrm{Ti} \nonumber$
b) To find the limiting reagent, we need to find the molar mass of the reactants:
\begin{gathered}
\mathrm{TiCl}_4: 47.87+4 \times 25.45=189.7 \mathrm{~g} / \mathrm{mol} \
M g: 24.3 \mathrm{~g} / \mathrm{mol}
\end{gathered}
Next, we can convert from grams to moles:
\begin{gathered}
200 \mathrm{~kg} \mathrm{TiCl}_4 \times \dfrac{1000 \mathrm{~g} \mathrm{TiCl}_4}{1 \mathrm{~kg} \mathrm{TiCl} l_4} \times \dfrac{1 \mathrm{~mol} \mathrm{TiCl}_4}{189.7 \mathrm{~g} \mathrm{TiCl} l_4}=1054 \mathrm{~mol} \mathrm{TiCl}_4 \
15 \mathrm{~kg} \mathrm{Mg} \times \dfrac{1000 \mathrm{~g} \mathrm{Mg}}{1 \mathrm{~kg} \mathrm{Mg}} \times \dfrac{1 \mathrm{~mol} \mathrm{Mg}}{24.3 \mathrm{~g} \mathrm{Mg}}=1029 \mathrm{~mol} \mathrm{Mg}
\end{gathered}
The balanced reaction tells us we need twice as many moles of magnesium as moles of titanium chloride. We don't have enough $\mathrm{Mg}$ to react with all of the $\mathrm{TiCl}_4$, so $\mathrm{Mg}$ is the limiting reagent.
c) The yield is determined by the initial amount of the limiting reagent. The balanced reaction tells us we get two moles of $M g \mathrm{Cl}_2$ and 1 mole of $\mathrm{Ti}$ per mole of $M g$ we react, so the yield is
$1029 \mathrm{~mol} \mathrm{Mg} \times \dfrac{1 \mathrm{~mol} \mathrm{Ti}}{2 \mathrm{~mol} \mathrm{Mg}} \times \dfrac{47.87 \mathrm{~g} \mathrm{Ti}}{1 \mathrm{~mol} \mathrm{Ti}}=24.7 \mathrm{~kg} \text { of } \mathrm{Ti} \nonumber$ | textbooks/chem/Inorganic_Chemistry/Introduction_to_Solid_State_Chemistry/06%3A_Recitations/6.01%3A_Balancing_Reactions_Yield_and_Limiting_Reagents.txt |
Avogadro and moles
When balancing a reaction or determining its yield, it was really important to keep track of the number of each kind of atom before and after the reaction. Atoms can be difficult to account for in real life due to their minuscule size. Instead, we usually keep track in moles : a mole consists of Avogadro's number of atoms (or molecules), or $6.022 \times 10^{23}$ atoms or molecules per mole.
Critically, we can convert from moles to a value that is easy to measure in a lab: mass. The molar mass of a a substance is defined to be the number of grams in one mole of that substance. The molar mass of a single element is also called the atomic mass; the molar mass of a compound can be obtained by summing the molar mass of the constituent elements.
Example: How many moles of nickel are in $102 \mathrm{~g}$ of nickel? How many moles of $\mathrm{H}_2 \mathrm{O}$ are in $50 \mathrm{~g}$ of water?
\begin{gathered}
102 \mathrm{~g} \mathrm{Ni} \times \dfrac{1 \mathrm{~mol} \mathrm{Ni}}{58.69 \mathrm{~g} \mathrm{Ni}}=1.74 \mathrm{~mol} \mathrm{Ni} \
50 \mathrm{~g} \mathrm{H}_2 \mathrm{O} \times \dfrac{1 \mathrm{~mol} \mathrm{H}_2 \mathrm{O}}{(15.999+2 \times 1.0107) \mathrm{g} \mathrm{H}_2 \mathrm{O}}=2.77 \mathrm{~mol} \mathrm{H}_2 \mathrm{O}
\end{gathered}
Periodic table
The periodic table is our key to solving problems! It contains a wealth of information that we can use to understand and calculate the properties of materials. The elements are organized by the number of protons: the rows are called periods, and the columns are called groups. Atoms are made up of positively charged protons, neutral neutrons, and negatively charged electrons. Protons and neutrons live at the center of the atom: together, these are the main source of mass. Electrons orbit around the protons and neutrons (more on this later). An example Bohr model of carbon is shown here:
We can read off many properties of atoms from the periodic table: An atom has the same number of protons
and electrons. The atomic mass is the sum of the number of protons and neutrons: it is the average mass of one atom of each element. The units of atomic mass are AMU (atomic mass units, $=1 / 12$ the mass of a carbon-12 isotope), or equivalently (due to a convenient convention), the number of grams in a mole of a substance.
Isotopes
You may have noticed that the atomic masses on the periodic table aren't integers, even though the number of protons $+$ neutrons must be an integer. The number represented as atomic mass is really the average atomic mass, where a weighted average is taken over the possible isotopes of an atom. An isotope is an atom that has a few extra (or missing) neutrons, and therefore a different atomic mass. Isotopes are usually written like this:
${ }_Z^A X$
Here, $\mathrm{X}$ is the element symbol, $\mathrm{Z}$ is the atomic number, and $\mathrm{A}$ is the atomic mass. The number of neutrons in the isotope is just
$\# \text { neutrons }=A-Z$
Example: Complete the following table:
Isotope Abundance Atomic mass (amu)
${ }^{28} \mathrm{Si}$ $92.18\%$ $28$
${ }^{29} \mathrm{Si}$ ? $29$
${ }^x \mathrm{Si}$ $3.12\%$ ?
From the periodic table, the average atomic mass of silicon is $28.0855$ amu. First, we can solve for the percentage of ${ }^{29} \mathrm{Si}$ by using the fact that the total abundance must sum to $100 \%$:
\begin{gathered}
100 \%=92.18 \%+3.12 \%+y \
y=4.71 \%
\end{gathered}
Then, we can use a weighted average to calculate the atomic mass and determine what the third isotope is:
\begin{gathered}
28.0855=0.9218 \times(28 a m u)+0.0471 \times(29 a m u)+0.0312 x \
x=30
\end{gathered}
Combustion
A combustion reaction occurs when a substance reacts with oxygen gas, producing light and heat.
Example: If you put $400 \mathrm{~g}$ of sugar (glucose) into a $1 \mathrm{~m}^3$ box, light it on fire, and quickly seal the box, would there be enough oxygen to completely combust the sugar? Write a balanced reaction; determine the limiting reagent and yield.
Density of air: $1.225 \mathrm{~g} / \mathrm{L}=1.225 \mathrm{~kg} / \mathrm{m}^3$
Oxygen weight percentage in air: $23.2 \%$
Chemical formula of glucose: $\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6$
We can try to match the number of moles in the glucose by adjusting the coefficients on the right side. This gives the right number of $\mathrm{C}$ and $\mathrm{H}$. Finally, we can adjust the coefficient on the oxygen to get a balanced reaction:
$\mathrm{C}_6\mathrm{H}_{12}\mathrm{O}_6 + 6\mathrm{O}_2 \rightarrow 6\mathrm{CO}_2 + 6\mathrm{H}_2\mathrm{O}$
To determine the limiting reagent, we need to figure out how many moles of each reactant we have:
400 g glucose $\times \dfrac{1 \mathrm{~mol} \text { glucose }}{(6 \times 12.0107+12 \times 1.01+6 \times 15.999) \text { g glucose }}=2.22 \mathrm{~mol}$ glucose
Next, we need to determine how much oxygen is in the box:
$1 m^3 \text { air } \times \dfrac{1.225 \mathrm{~kg} \text { air }}{1 m^3 \text { air }} \times \dfrac{1000 g}{k g} \times 0.232 \times \dfrac{1 \text { mol oxygen }}{2 \times 15.999 \text { g oxygen }}=8.88 \text { mol oxygen }$
where the multiplication by $0.232$ to account for the fact that air is $23.2 \%$ oxygen.
We need 6 times as much oxygen as glucose to fully react, and $6.66 / 2.22 \leq 6$, so oxygen is the limiting reagent. We can use the limiting reagent to calculate the yield:
$8.88 \mathrm{~g} \mathrm{O}_2 \times \dfrac{1 \mathrm{~mol} \mathrm{H}_2 \mathrm{O}}{1 \mathrm{~mol} \mathrm{O}_2} \times \dfrac{(2 \times 1.01+15.999) \mathrm{g} \mathrm{H}_2 \mathrm{O}}{1 \mathrm{~mol} \mathrm{H}_2 \mathrm{O}}=158.8 \mathrm{~g} \mathrm{H}_2 \mathrm{O}$ | textbooks/chem/Inorganic_Chemistry/Introduction_to_Solid_State_Chemistry/06%3A_Recitations/6.02%3A_Avogadro_and_Moles_Periodic_Table_Isotopes_and_Combustion.txt |
Waves and photons
Photons are quanta of light: the energy of a photon with frequency $\nu$ (or wavelength $\lambda$ ) is given by the Planck-Einstein relation.
$E=h \nu=\dfrac{h c}{\lambda}$
Here, $\mathrm{h}$ is Planck's constant, and $\mathrm{c}$ is the speed of light. The electromagnetic spectrum is shown here:
Recall that the equality above came from the fact that frequency and wavelength are related through the speed of light as $c=\lambda \nu$. It's also important to remember that the energy of a photon is independent of the amplitude of the wave:
The energy is set by the wavelength (or equivalently, frequency); the amplitude of the wave is related to its intensity: this can be thought of as the number of photons of a certain energy.
Example: A radio station broadcasts at $100.7 \mathrm{MHz}$ with a power output of $50 \mathrm{~kW}$. How many photons are emitted each second? Recall that $1 W=1 J / s$.
The energy per second that leaves the radio station is
$50 \mathrm{~kW} \times \dfrac{1000 \mathrm{~W}}{1 \mathrm{~kW}} \times \dfrac{1 \mathrm{~J} / \mathrm{s}}{1 \mathrm{~W}}=5 \times 10^4 \mathrm{~J} / \mathrm{s}$
We can calculate the energy per photon with the Planck-Einstein relation:
$E=h \nu=\left(6.626 \times 10^{-34} \mathrm{~J}^* \mathrm{~s}\right) \times\left(100.7 \times 10^6 \mathrm{~s}^{-1}\right)=6.672 \times 10^{-26} \mathrm{~J} / \text { photon }$
Finally, the number of photons per second is
$N=\left(5 \times 10^{-4} \mathrm{~J} / \mathrm{s}\right) \dfrac{1 \text { photon }}{6.672 \times 10^{-26} \mathrm{~J}}=7 \times 10^{29} \text { photons } / \mathrm{s}$
Bohr model
The photoelectric effect tells us that shining light with enough energy on an atom can cause the emission of electrons. In $3.091$, we'll use the Bohr model to model this process. In the Bohr model, an electron in a hydrogen atom can only live in discrete energy levels, which we label with $n$. The lowest energy level is the ground state, with $n=1$. The change in energy corresponding to the transition from an initial energy level, $n_i$ to a final energy level, $n_f$ is given by
$\Delta E=-13.6\left(\dfrac{1}{n_f^2}-\dfrac{1}{n_i^2}\right)[e V]$
If a photon with just the right energy impinges on the atom, the photon can be absorbed and the electron excited to a higher energy level.
Conversely, if an electron relaxes from a higher energy level to a lower one within the atom, a photon with frequency $\nu=\Delta E / h$ is released.
Example: A power source emits $8 \times 10^{18}$ photons/s at $10 \mathrm{~W}$. Determine how much energy each photon has. Then, calculate what state an electron in the ground state of a hydrogen atom would end up in if excited by a photon with this energy.
To calculate the energy of each photon, we can use dimensional analysis to get to Joules:
$E=\dfrac{10 \mathrm{~J} / \mathrm{s}}{5 \times 10^{18} \text { photons } / \mathrm{s}} \times \dfrac{1 \mathrm{eV}}{1.6 \times 10^{-19} \mathrm{~J}}=12.5 \mathrm{eV}$
If the electron starts in the ground state, we need to solve
$\Delta E=12.5 \mathrm{eV}=-13.6 \mathrm{eV}\left(\dfrac{1}{n_f^2}-\dfrac{1}{1^2}\right) \quad n_f=3.51$
Since the electron must live in an integer state, it must either go to the third or fourth level. It would take more energy than we have to get to the fourth energy level, the highest accessible energy level is $n=3$. | textbooks/chem/Inorganic_Chemistry/Introduction_to_Solid_State_Chemistry/06%3A_Recitations/6.03%3A_Waves_Photons_and_Bohr_Model.txt |
Energy, Frequency, and Wavelength
Last time, we discussed electromagnetic waves and how light is quantized as photons. We related the energy of a photon to its frequency and wavelength using the Planck-Einstein relation:
$E=h \nu=\dfrac{h c}{\lambda}$
In $3.091$, the most common units we will use for energy are joules $(\mathrm{J})$ or electron volts $(\mathrm{eV})$. A joule is equivalent to a $\operatorname{kg} \times \frac{\mathrm{m}^2}{\mathrm{~s}^2}$, which is like a force integrated over a distance. Electron volts are also a unit of energy, but they are much smaller than a joule. An $\mathrm{eV}$ is literally the charge of one electron multiplied by one volt! We can convert between joules and electron volts using
$1 \mathrm{eV}=1.6022 \times 10^{-19} \mathrm{~J}$
Depending on the problem, it may be easier to work in $\mathrm{J}$ or $\mathrm{eV}$. Either one works - double checking how your units work out is a great way to check your answer.
The units of frequency $(\nu)$ are $1 / s$, or equivalently, $\operatorname{Hertz}(\mathrm{Hz}$, and the wavelength $(\lambda)$ has units of $m$. Now, we can consider the speed of light, $c$. Last time, we said $c=\lambda \nu$ : using the units above, we see that the speed has units $m / s$ as it should!
Frequently, we will use units like nanometers $(n m)$ or microns $(\mu m)$ for small length scales: these can be converted to meters using the following conversion factors:
\begin{aligned}
&1 \mathrm{~nm}=10^{-9} \mathrm{~m} \
&1 \mu \mathrm{m}=10^{-6} \mathrm{~m}
\end{aligned}
The last piece of the Planck-Einstein relation is the $h$ : Planck's constant! Planck's constant is
$h=6.626 \times 10^{-34} \mathrm{Js}=4.136 \times 10^{-15} \mathrm{eVs}$
It is important that you use the version of $\mathrm{h}$ with the units of energy in a problem!
Example: What is the energy and wavelength of light that has a frequency of $440 \mathrm{THz}$?
\begin{gathered}
E=h \nu=\left(4.136 \times 10^{-15} \mathrm{eVs}\right)(440 \mathrm{THz})\left(\frac{10^{12} \mathrm{~Hz}}{1 \mathrm{THz}}\right)=1.82 \mathrm{eV} \
\lambda=\dfrac{c}{\nu}=\dfrac{3 \times 10^8 \mathrm{~m} / \mathrm{s}}{440 \times 10^9 1 / \mathrm{s}}=6.81 \times 10^{-7} \mathrm{~m}=681 \mathrm{~nm}
\end{gathered}
Ionization
If a photon with sufficient energy is absorbed by an atom, the atom can become ionized : it loses an electron! In fact, when we discussed Bohr's model, the factor $-13.6 \mathrm{eV}$ is actually the first ionization energy of the hydrogen atom! That means it takes $13.6 \mathrm{eV}$ of energy to ionize an electron in the ground state- that is, to excite an atom in the $\mathrm{n}=1$ state so much it leaves the atom. The final energy state of a Bohr model electron that is ionized can be thought of as the limit as $n_f \rightarrow \infty$:
$\Delta E_{\text {ionize }}=\lim _{n_f \rightarrow \infty}(-13.6[\mathrm{eV}])\left(\dfrac{1}{n_f^2}-\dfrac{1}{n_i^2}\right)=\dfrac{13.6 \mathrm{eV}}{n_i^2}$
Example: A photon with a wavelength of $4.5 \mu \mathrm{m}$ strikes a hydrogen atom with an electron in an unknown energy level. The electron is then ejected from the atom, and flies through space. Determine a) the minimum energy state the electron could have been in, and how much energy would have been leftover, b) the velocity of the electron right when it leaves. Then, c) assume a beam of $4.5 \mu \mathrm{m}$ light shines on many hydrogen atoms in the state you determined in part a). If the beam power is $25 \mathrm{~mW}$, how many photons are ejected each second?
Answer
a)
\begin{gathered}
E_{\text {photon }}=\frac{h c}{\lambda}=\frac{\left(4.14 \times 10^{-15} \mathrm{eV} \mathrm{s}\right)\left(3 \times 10^8 \mathrm{~m} / \mathrm{s}\right)}{4.5 \times 10^{-6} \mathrm{~m}}=0.2757 \mathrm{eV} \
E_{\text {photon }}=-13.6\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)=13.6 \mathrm{eV}\left(\frac{1}{n_i^2}\right)=0.2757 \mathrm{eV}
\end{gathered}
Solving this, we get $n_i=7.02 \mathrm{eV}$. The electron would have to be just above the 7 th energy level $-$ since it must be in an integer energy level, the lowest initial state is $n_i=8$. The energy to ionize from the 8th energy level is
$\Delta E_{8, \infty}=-13.6\left(-\dfrac{1}{8^2}\right)=0.2125 \mathrm{eV} \nonumber$
It takes $0.2125 \mathrm{eV}$ to ionize from $n_i=8$. The remaining energy is leftover:
$E_{\text {left }}=0.2757 \mathrm{eV}-0.2125 \mathrm{eV}=0.063 \mathrm{eV} \nonumber$
The energy leftover is not absorbed by the atom, but it must go somewhere! One place it could go is to the kinetic energy of the electron that was ionized.
b) Assuming all the leftover energy becomes kinetic ene
$E=\dfrac{1}{2} m v^2 \nonumber$
We need to convert to Joules to get a velocity in reasonable units:
\begin{aligned}
0.063 \mathrm{eV} \frac{1.602 \times 10^{-10} \mathrm{~J}}{1 \mathrm{eV}} &=1.009 \times 10^{-20} \mathrm{~J}=\frac{1}{2}\left(9.11 \times 10^{-31} \mathrm{~kg}\right) v^2 \
v &=1.5 \times 10^5 \mathrm{~m} / \mathrm{s}
\end{aligned}
c) First, let's break down the beam power:
$25 \mathrm{~mW}\left(\dfrac{1 \mathrm{~W}}{1000 \mathrm{~mW}}\right)\left(\dfrac{1 \mathrm{~J} / \mathrm{s}}{1 \mathrm{~W}}\right)=0.025 \mathrm{~J} / \mathrm{s} \nonumber$
We can then solve for the energy of each photon in joules:
\begin{aligned}
&E=\frac{h c}{\lambda}=\frac{\left(6.626 \times 10^{-34} \mathrm{Js}\right)\left(3 \times 10^8 \mathrm{~m} / \mathrm{s}\right)}{4.5 \times 10^{-6} \mathrm{~m}}=4.4 \times 10^{-20} \mathrm{~J} / \text { photon } \
&\frac{\text { photons }}{\mathrm{s}}=\frac{\frac{\mathrm{J}}{\mathrm{s}}}{\frac{\mathrm{J}}{\text { photon }}}=\frac{0.025 \mathrm{~J} / \mathrm{s}}{4.4 \times 10^{-20} \mathrm{~J} / \text { photon }}=5.68 \times 10^{17} \text { photons } / \mathrm{s}
\end{aligned} | textbooks/chem/Inorganic_Chemistry/Introduction_to_Solid_State_Chemistry/06%3A_Recitations/6.04%3A_Energy_Frequency_Wavelength_and_Ionization.txt |
Quantum numbers
Every electron localized in an atom can be described by four quantum numbers. The Pauli Exclusion Principle tells us that no two electrons can share the exact same set of quantum numbers
Principal quantum number
The principle quantum number, $n$, represents the energy level of the electron, much like the $n$ used in the Bohr model. Energy is related to the principle quantum number by
$E=-\dfrac{13.6 \mathrm{eV}}{n^2}, \quad n>0$
Orbital angular momentum quantum number
The orbital angular momentum quantum number, $l$, provides information about the shape of an orbital. Unlike the description of early models of the atom, electrons in atoms don't orbit around the nucleus like a planet around the sun. However, they do have angular momentum, and the shape of the probability cloud around the nucleus depends on the value of the angular momentum. The magnitude of the angular momentum is related to the orbital angular momentum quantum number by
$L=\hbar \sqrt{l(l+1)}, \quad 0 \leq l \leq n-1$
The orbital angular momentum quantum numbers correspond directly to letters commonly used by chemists to describe the orbital subshells: for example, $l=1$ corresponds to the $\mathrm{s}$ orbitals, which are spherical:
$l=2$ corresponds to the $\mathrm{p}$ orbitals, which are lobed: Visualizations of the higher orbitals, like the $\mathrm{d}(l-2)$
and $f (l = 3)$ can be found in Chapter 6 of the Averill textbook.
Magnetic quantum number
The magnetic quantum number, $m$, distinguishes the orbitals available within a subshell. The projection of the angular momentum onto the $\mathrm{z}$-axis is related to the magnetic quantum number by
$L_z=m \hbar, \quad-l \leq m \leq l$
Spin quantum number
The spin quantum number, $m_s$, gives the spin angular momentum of each electron. Each electron can either be spin up or spin down. The magnitude of the spin is either $+1 / 2$ (spin up) or $-1 / 2$ (spin down). The magnitude of the angular momentum associated with spin is
$|S|=\hbar \sqrt{s(s+1)}$
and the spin projection onto the $\mathrm{z}$-axis is given by
$S_z=s \hbar$
The restriction on the allowed values of spin is
$s=\pm \dfrac{1}{2}$
Example: What is the electronic configuration of carbon? Write out the quantum numbers for each electron.
Answer
Carbon has six valence electrons. We'll start filling up the quantum numbers by following the rules above. Starting with $n=1$, the only possible value of $l$ is $l=0$, since $0 \leq l \leq n-1$. When $l=0, m$ must also be 0 . There are two possible spins that could go with these quantum numbers. There aren't any more available quantum numbers with $n=1$, so we must go up to a higher shell. For $n=2$, the lowest energy state corresponds to the lowest magnitude of angular momentum, $l=0$, which again requires $m=0$.
For the last two electrons, we have to remember Hund's rule. Every orbital in a subshell must be singly occupied with electrons of one spin before adding electrons of the other. So the $l=1$ electrons must have different values of $m$, but the same value of $m_s$. The quantum numbers are summarized in the table below.
$\begin{array}{c|c|c|c} n & l & m & m_s \ \hline \hline 1 & 0 & 0 & 1 / 2 \ 1 & 0 & 0 & -1 / 2 \ 2 & 0 & 0 & 1 / 2 \ 2 & 0 & 0 & -1 / 2 \ 2 & 1 & 0 & 1 / 2 \ 2 & 1 & -1 & 1 / 2 \end{array} \nonumber$ | textbooks/chem/Inorganic_Chemistry/Introduction_to_Solid_State_Chemistry/06%3A_Recitations/6.05%3A_Quantum_Numbers.txt |
Aufbau principle
Quantum numbers are a handy way to account for the electrons that fill up atomic shells, but they can be tricky to keep track of. Instead, we often utilize shorthand to label the quantum numbers:
The Aufbau principle is a trick to keep track of the order each subshell is filled in:
The superscript after each subshell indicates the number of electrons in the subshell: each should be full except for the outermost shell, which is filled based on the number of valence electrons.
Example: Write down both the full electronic configuration and the noble gas notation of $\mathrm{Br}$.
$\mathrm{Br}$ has 35 valence electrons, so we can fill up all the way to Argon. The full configuration, which mostly follows Aufbau, is
$1 \mathrm{~s}^2 2 \mathrm{~s}^2 2 \mathrm{p}^6 3 \mathrm{~s}^2 3 \mathrm{p}^6 3 \mathrm{~d}^{10} 4 \mathrm{~s}^2 4 \mathrm{p}^5$
Note here that the $3 \mathrm{~d}^{10}$ and $4 \mathrm{~s}^2$ occur in an unexpected order: there are some exceptions to the Aufbau principle. We don't expect you to memorize these exceptions for $3.091$, but know that they are noted in the periodic table! The noble gas configuration just shows the electrons beyond the last full shell:
$[\mathrm{Ar}] 3 \mathrm{~d}^{10} 4 \mathrm{~s}^2 4 \mathrm{p}^5$
Electron filling and box notation
We can note the configuration and orientation of the electrons visually using box notation. It’s usually easiest to use the Aufbau principle (or the periodic table!) to write down the electronic configuration and then translate it to box notation. It’s important to recall Hund’s rule : every orbital in a subshell must be singly occupied before any orbital is doubly occupied.
Example: Draw the electronic configuration of $\mathrm{Br}$ and $\mathrm{C}$ in box notation.
Answer
We wrote the noble gas configuration above, so we can just give the box notation of the valence electrons here:
For carbon, the electronic configuration is $1 \mathrm{~s}^2 2 \mathrm{~s}^2 2 \mathrm{p}^2$. In box notation, we fill up the $1 \mathrm{~s}$ and $2 \mathrm{~s}$ states, and populate the $2 \mathrm{p}$ according to Hund's rule:
Photoelectron spectroscopy (PES)
Photoelectron spectroscopy (PES) is an experimental method used to determine electronic structure. First, a sample is bombarded with high energy photons to ionize the atoms. Then, the kinetic energy of the electrons that are emitted is measured. The binding energy can be determined using
$E_{b i n d i n g}=h \nu-K E_{e-}$
By plotting the relative counts of electrons emitted with various binding energies, the elemental composition of a sample can be determined using PES!
Example: Determine which element corresponds to the following PES spectrum:
Answer
On this plot, the binding energy increases to the left. The electrons closer to the nucleus should have the highest binding energy, so the left-most peak must correspond to the $1 \mathrm{~s}$ electrons. The relative height of the peak corresponds to the relative frequency at which the electrons are emitted: since there are two 1s electrons, the height of the left-most peak must correspond to two electrons. The middle peak must correspond to two $2 \mathrm{~s}$ electrons. Then, since the right-most peak is $1.5$ times taller, it must represent 3 electrons in the $2 \mathrm{p}$ subshell. The electronic configuration for the element in this PES diagram must be $1 \mathrm{~s}^2 2 \mathrm{~s}^2 2 \mathrm{p}^3$, which corresponds to nitrogen. | textbooks/chem/Inorganic_Chemistry/Introduction_to_Solid_State_Chemistry/06%3A_Recitations/6.06%3A_Aufbau_Principle_Electron_Filling_Box_Notation_and_Photoelectron_Spectroscopy.txt |
Periodic trends
Inside an atom, negatively-charged electrons are attracted to the positively-charged nucleus, but they repel each other. Electrons close to the nucleus can shield the outermost electrons from Coulomb forces between the electrons and the nucleus.
Going across a row in the periodic table, each element has one extra electron and one extra proton. The additional proton has a strong effect on the electrons, as the number of inner shielding electrons does not change. However, when going down a period, there are more and more shells of electrons with increasing energy and average radius from the nucleus. Therefore, atomic radius decreases from left to right across the rows of the periodic table, and increases from top to bottom.
When atoms become ionized, their atomic radius changes as the charge state balance changes. If an atom gains electrons to become negatively charged, it is called an anion. If an atom loses electrons to become positively charged, it is called a cation. Cations are typically smaller than the neutral version of the atom, while anions are typically larger. An ionic solid contains cations and anions in a ratio that maintains charge neutrality.
Example: Arrange $\mathrm{Al}, \mathrm{C}$, and $\mathrm{Si}$ in order of increasing atomic radius.
Answer
Carbon is in the second period of the periodic table, while $\mathrm{Al}$ and $\mathrm{Si}$ are in the third: $\mathrm{C}$ must be the smallest. $\mathrm{Al}$ is in group III, while $\mathrm{Si}$ is in group IV, so $\mathrm{Al}$ is likely larger than $\mathrm{Si}$ due to relatively strong effect of the extra proton $\mathrm{Si}$ has.
(smallest) $\quad \mathrm{C} \rightarrow \mathrm{Si} \rightarrow \mathrm{Al} \quad$ (largest)
Lewis Dot Diagrams
Lewis dot diagrams are a tool used to visualize the valence electrons in atoms. Then, by pairing up electrons, we can visualize possible ways that bonds can be formed between two atoms.
There are a couple of rules of thumb that come in handy when drawing Lewis dot diagrams. The octet rule tells us that atoms gain, lose, or share electrons in order to have a full valence shell of 8 electrons (2 for $\mathrm{H}$ and $\mathrm{He}$ ). Also, the number of valence electrons in an atom is equal to the group number. We can generally start by following these steps:
1. Determine the number of valence electrons in each atom in the compound
2. Place a bonding pair between adjacent atoms
3. Starting from terminal atoms, add electrons to form octets
Example: Draw Lewis dot diagrams for the following compounds: $\mathrm{CH}_4, \mathrm{CCl}_4, \mathrm{CO}_2$, and $\mathrm{OH}^{-}$
Answer
$\mathrm{CH}_4$: $\mathrm{C}$ has 4 valence electrons (group IV), while hydrogen just has one. Sharing four electrons between the $\mathrm{C}$ and $\mathrm{H}$'s yields an octet on the carbon atom and a complete shell of two electrons on the hydrogen. Note that it's often convenient to put the atom that has the most unpaired electrons in the middle.
$\mathrm{CCl}_4$: Again, $\mathrm{C}$ has $4$ valence electrons, and $\mathrm{Cl}$ has $7$. $\mathrm{Cl}$ can only gain one additional electron to have a full octet, so each $\mathrm{Cl}$ only makes one bond. Therefore, we can put $\mathrm{C}$ in the middle again and surround it with $\mathrm{Cl}$ atoms, and every atom has a full octet.
$\mathrm{CO}_2$: $\mathrm{C}$ has 4 valence electrons, while $\mathrm{O}$ has 6. We can start by placing the $\mathrm{C}$ in the middle and forming a bond with each $\mathrm{O}$. The structure that results is unstable, because each $\mathrm{O}$ is left with an unpaired electron. $\mathrm{C}$ also has two unpaired electrons leftover. We can use them to form double bonds between the $\mathrm{C}$ and each $\mathrm{O}$ without violating the octet rule. Note that each $\mathrm{O}$ is left with two lone pairs.
$\mathrm{OH}^{-}$: Here, the oxygen brings 6 valence electrons and the $\mathrm{H}$ brings just one, but there is an additional electron in play because of the structure's overall negative charge. After forming a single bond between the $\mathrm{O}$ and the $\mathrm{H}$, we're left with unpaired electrons, so we need to rearrange some more. We can't make a double bond to the hydrogen, as it can only support two electrons. Moving the extra electron to the oxygen to join a lone pair does the trick.
Formal charge
Above, we discussed the steps for constructing Lewis dot diagrams: by following these rules, we can determine stable electronic configurations of molecules. We can also quantify this stability by calculating the formal charge.
$\text { Formal Charge }=\# \text { valence } e^{-} \mathrm{s}-\left(N+\dfrac{b}{2}\right)$
where the number of valence electrons refers to the neutral atom in isolation, $N=$ the number of nonbonding valence electrons and $b=$ the number of valence electrons participating in bonds. Formal charge should be calculated for each atom in the molecule.
Example: Determine the formal charge of $\mathrm{CO}_2$.
Answer
Oxygen: the neutral oxygen atom in isolation has 6 valence electrons. Each oxygen in $\mathrm{CO}_2$ has 4 nonbonding electrons and 4 electrons stored in bonds.
$\text { Formal Charge (Oxygen) }=6-\left(4+\dfrac{4}{2}\right)=0 \nonumber$
Carbon: the neutral carbon atom in isolation has 4 valence electrons. Each carbon in $\mathrm{CO}_2$ has 8 electrons stored in bonds.
$\text { Formal Charge }(\text { Carbon })=4-\left(\dfrac{8}{2}\right)=0 \nonumber$ | textbooks/chem/Inorganic_Chemistry/Introduction_to_Solid_State_Chemistry/06%3A_Recitations/6.07%3A_Periodic_Trends_Lewis_Dot_Diagrams_and_Formal_Charge.txt |
Resonance
Often, there are many ways that electrons can be arranged around a molecule. The existence of multiple electronic configurations that are stable leads to resonance, which describes the delocalization of electrons in covalently bonded molecules. To be considered resonance structures, molecules must meet the following criteria:
1. Atoms must be in the same orientation relative to each other
2. All resonance structures must have the same number of valence electrons
3. The octet rule must be satisfied, with a few exceptions:
- Not followed by some elements in periods three and higher, including $\mathrm{Cl}, \mathrm{Br}, \mathrm{I}, \mathrm{P}, \mathrm{Si}$
- Sulfur can have up to $12 \mathrm{e}^{-}$
- Boron can have $6 \mathrm{e}^{-}$
4. Formal charge should be on the most electronegative atoms
Example: Draw resonance structures for ozone $\left(\mathrm{O}_3\right)$ and $\mathrm{CO}_3^{2-}$.
Answer
Ozone consists of three oxygen atoms, for a total of 18 valence electrons. These can be equivalently distributed in two ways:
$\mathrm{CO}_3^{2-}$ consists of a central carbon atom which brings 4 valence electrons, and four oxygen atoms that each bring 6 electrons, and two additional electrons that yield the additional $2^{-}$ charge. The resonance structures can be represented as follows:
These equivalent structures can be combined into one compact picture (shown above on the right) that is an average of the possible structures. All of the resonance structures are equally likely, as they equivalently satisfy the guidelines for stable Lewis structures.
Formal charge (again)
In the $\mathrm{CO}_3^{2-}$ example above, the formal charge on each atom is shown in circles. The formal charge only appears on the oxygen, as oxygen is more electronegative than carbon. In the average structure, the formal charge is averaged over all of the oxygen atoms. We can check that we have the most stable resonance structure by evaluating the formal charge on various options. The resonance structure with the lowest formal charge on each atom is the most stable and will occur with higher probability over other resonance structures. As a reminder, formal charge is calculated using
Formal Charge $=\#$ valence $e^{-} \mathrm{s}-\left(N+\dfrac{b}{2}\right)$
where the number of valence electrons refers to the neutral atom in isolation, $N=$ the number of nonbonding valence electrons and $b=$ the number of valence electrons participating in bonds. Formal charge should be calculated for each atom in the molecule.
Example: Find three resonance structures of $\mathrm{CO}_2$, and determine which is the most stable.
Answer
Last recitation, we drew one possible electronic structure of $\mathrm{CO}_2$. Two more are given here:
The formal charge on each atom in each of the three viable resonance structures is shown. Though each resonance structures satisfies the rules of valid Lewis dot diagrams and viable resonance structures, the central structure is the most stable as the formal charge is lowest.
6.09: VSEPR and Polarity
VSEPR
Lewis dot structures are a great tool to visualize how electrons can be arranged in molecules. Further, drawing resonance structures and determining the formal charge can help us determine which structures are most stable. However, neither of these tools provide much insight into the physical configuration of a molecule in 3D space. For this, we turn to Valence Shell Electron Pair Repulsion theory, or VSEPR.
Once we draw a viable Lewis structure, we can use the following chart to translate the 2D representation to a 3D geometry:
Chart courtesy of Boundless.com. License: CC BY-SA. This content is excluded from our Creative Commons license. For more information, see https://ocw.mit.edu/fairuse.
Example: Draw Lewis dot diagrams and determine the 3D VSEPR geometry of the following molecules:
$\mathrm{CH}_4, \mathrm{NH}_3, \mathrm{H}_2 \mathrm{O}, \mathrm{SO}_3, \mathrm{SO}_2, \mathrm{CO}_2$
Answer
Lewis dot diagram Electrons around central atom VESPR description Sketch of 3D model
$\mathrm{CH}_4$
- 4 groups of electrons in bonds
- 0 lone pairs
tetrahedral
$\mathrm{NH}_3$
-3 groups of electrons in bonds
-1 lone pair
trigonal pyramidal
$\mathrm{H}_2\mathrm{O}$
- 4 groups of electrons in bonds
- 0 lone pairs
bent
$\mathrm{SO}_3$
- 4 groups of electrons in bonds
- 0 lone pairs
trigonal planar
$\mathrm{SO}_2$
- 4 groups of electrons in bonds
- 0 lone pairs
bent
$\mathrm{CO}_2$
- 4 groups of electrons
- 0 lone pairs
linear
Polarity
The difference in electronegativity across a molecule can generate electric dipole moments. Dipole moments are vector quantities, and by convention point from a more positive region of charge to a more negative region. If individual dipoles within a molecule cancel, there is no net dipole.
Example: Determine whether $\mathrm{CO}_2$ and $\mathrm{H}_2\mathrm{O}$ have a net dipole moment.
Answer
In $\mathrm{CO}_2$, the two electronic dipoles are exactly opposite and cancel each other, so there isn’t a net dipole. Carbon dioxide is not a polar molecule.
In $\mathrm{H}_2\mathrm{O}$, the electronic dipoles don’t fully cancel, so there is a net dipole moment. Water is a polar molecule! | textbooks/chem/Inorganic_Chemistry/Introduction_to_Solid_State_Chemistry/06%3A_Recitations/6.08%3A_Resonance_and_Formal_Charge_%28cont.%29.txt |
Hybridization and atomic orbitals
Lewis dot diagrams and VSEPR are powerful tools to think about electronic configuration and molecular shape, but neither tool offers a way to predict bond stability. The strength of a covalent bond is proportional to the amount of overlap between electronic orbitals.
Consider $\mathrm{BeH}_2$: beryllium contributes 2 valence electrons, and each of the two hydrogen atoms contribute one valence electron.
$\mathrm{H}-\mathrm{Be}-\mathrm{H}$
As a reminder, the electronic configuration of hydrogen is $1 \mathrm{~s}^1$, while the electronic configuration of beryllium is $1 \mathrm{~s}^2 2 \mathrm{~s}^2$. The valence shell of $\mathrm{Be}\left(2 \mathrm{~s}^2\right)$ is full: it has two electrons, the maximum that can live in an $\mathrm{s}$-orbital. There aren't any singly-occupied orbitals available to form bonds with hydrogen, even though we predict from the Lewis dot diagram above that there should be single bonds between each $\mathrm{H}$ and the $\mathrm{Be}$. We can draw the electronic occupation of energy states in the $\mathrm{Be}$ atom as a function of energy, and then apply the principle of hybridization:
energy levels of Be atom $\quad \rightarrow \quad$ promotion $\quad \rightarrow \quad$ sp hybridization
To get around the issue of the full $2 \mathrm{~s}$ shell, an electron could be promoted to the $2 \mathrm{~s}$ subshell, leaving two electrons that are ready to form bonds. However, the $2 \mathrm{~s}$ and $2 \mathrm{p}$ electrons aren't equal: if this were the case, the two $\mathrm{H}-\mathrm{Be}$ bonds would be unequal, which has no physical basis. Instead, consider the third picture: by combining the $2 \mathrm{s}$ orbital with one of the $2 \mathrm{p}$ subshells, an intermediate $\mathrm{sp}$-hybridized energy level is created. This sp orbital has two equal-energy subshells that allow for two equal energy bonds to be formed with the $\mathrm{Be}$ atom. This is not unique to $\mathrm{BeH}_2$: $\mathrm{sp}$ hybridization occurs any time a $2 \mathrm{p}$ subshell combines with the $2 \mathrm{~s}$ subshell. Further, if there are more electrons available to participate in bonding, $\mathrm{sp}^2$ hybridization (3 equal bonds), $\mathrm{sp}^3$ hybridization (4 equal bonds), or higher order hybridization involving $\mathrm{d}$-orbitals can take place.
Hybridization directly correlates to molecular geometry! The following table shows the equivalence:
$\begin{array}{c|c} \text { Molecular geometry } & \text { Hybridization } \ \hline \hline \text { linear } & \mathrm{sp} \ \text { trigonal } & \mathrm{sp}^2 \ \text { tetrahedral } & \mathrm{sp}^3 \ \text { trigonal bipyramidal } & \mathrm{sp}^3 \mathrm{~d} \ \text { octahedral } & \mathrm{sp}^3 \mathrm{~d}^2 \end{array}$
Molecular orbital (MO) theory
We can use molecular orbital theory to gain a better understanding of how electrons form bonds and to predict properties such as bond stability and magnetic character. In 3.091, we’ll apply $\mathrm{MO}$ theory to dimers. Recall that all electrons are negatively charged: in free space, electrons would feel a repulsive force from other nearby electrons. However, there are additional forces within an atom that create an environment that allows for the concentration of electron density into bonds. When considering how two electrons in a solid may interact, it is useful to think about wave nature of electrons. Two electrons can constructively interfere to form a bonding $\mathrm{MO}$, or destructively interfere to form an antibonding $\mathrm{MO}$.
$\mathrm{MO}$ diagrams are a convenient tool to keep track of the bonding and antibonding orbitals, and therefore the bond strength. We can quantify the bond strength by calculating bond order:
Bond order $=\dfrac{1}{2}$ ($\#$ electrons in bonding orbitals $-\#$ electrons in antibonding orbitals)
If the bond order is greater then zero, a stable dimer forms.
If a dimer has any unpaired electrons, it is paramagnetic: these unpaired electrons align with applied magnetic fields and are weakly attracted to the applied field. If all the electrons are paired, the dimers are repelled by applied magnetic fields, and the dimers are called diamagnetic.
Example: Draw a molecular orbital diagram for the $\mathrm{F}_2$ dimer. Calculate the bond order, and determine whether it is paramagnetic or diamagnetic.
Answer
We start here by drawing the $\mathrm{AO}$ diagrams of the valence electrons of each $\mathrm{F}$ involved in the $\mathrm{F}_2$ dimer. Recall that F (Group VII) has 7 valence electrons. Then, we populate the $\sigma$ bonding and $\sigma^*$ antibonding orbitals associated with the $2 \mathrm{s}$ valence shell. Then, we draw the $2 \mathrm{p} \sigma$ bonding, $\pi$ bonding, $\pi^*$ antibonding, and $\sigma^*$ antibonding orbitals. Populating from the lowest energy to the highest energy, there are enough electrons to fill the $2 p \pi^*$ antibonding orbitals.
We can calculate the bond order using the formula from above. There are eight electrons in bonding orbitals and six electrons in antibonding orbitals:
$\text { bond order }=\dfrac{1}{2}(8-6)=1$
All of the electrons are paired, so the $\mathrm{F}_2$ dimer is diamagnetic. | textbooks/chem/Inorganic_Chemistry/Introduction_to_Solid_State_Chemistry/06%3A_Recitations/6.10%3A_Hybridization_Atomic_Orbitals_and_Molecular_Orbital_Theory.txt |
More MO Theory
So far we’ve discussed $\mathrm{MO}$ diagrams with homogeneous dimers: both atoms are the same. But there can also be heterogenous dimers involving two different atoms, and these dimers can be charged! When we construct a $\mathrm{MO}$ diagram for a heterogeneous dimer, it is the same principle as for a homogeneous dimer: draw electronic orbital diagrams for each atom, populate the electronic states, and then fil in the sigma and pi bonding and antibonding states formed. The only difference in the case of the heterogeneous dimer is that we must consider the relative energy of each set of atomic orbitals. This often yields asymmetric $\mathrm{MO}$ diagrams. Typically, the more electronegative atom is lower in energy.
Example: Draw the $\mathrm{MO}$ diagram of $\mathrm{CN}^-$. Determine the bond order.
Answer
Nitrogen is lower along the energy axis because it is more electronegative than carbon, yielding an asymmetric $\mathrm{MO}$ diagram. There are 8 electrons in bonding orbitals and 2 electrons in antibonding orbitals, so the bond order is
$\text { bond order }=\dfrac{1}{2}(8-2)=3$
A dimer with bond order 3 is triple bonded! You may recognize this molecule as cyanide.
Intermolecular Forces (IMFs)
London Dispersion Forces (LDFs) exist in all molecules. As the electrons distributed across a molecule fluctuate in time, they generate small electric fields. Then, electrons in other molecules or atoms that are in close proximity rearrange in response to the generated electric fields. Molecules with more electrons experience stronger LDFs.
Dipole-dipole interactions are IMFs characterized by electrostatic force between polar molecules. The more polar the molecule, the stronger the interaction.
Hydrogen bonds ($\mathrm{H}$-bonds) are a particular type of dipole-dipole interaction that occur in molecules with a hydrogen atom $(\mathrm{H})$ bonded directly to a fluorine, oxygen, or nitrogen atom ($\mathrm{F}$, $\mathrm{O}$, or $\mathrm{N})$.
Generally, LDFs are weaker than dipole-dipole interactions, and $\mathrm{H}$-bonds are strongest of all. However, there is no way to apply a one-size-fits-all classification of IMF strength, as they depend on specific molecular geometry, steric bulk (how big the molecule is), arrangement of atoms, polarity, and other factors. If a substance contains molecules with stronger IMFs, it requires more energy to break apart the molecules, which generally means that materials with stronger IMFs have higher melting points and higher boiling points.
Example: Will it $\mathrm{H}$-bond? Circle the molecules that form hydrogen bonds in solution.
Answer
1. This molecule has a ring of carbons, and each carbon is also bonded to one hydrogen. There are no $\mathrm{H}-\mathrm{F}$, $\mathrm{H}-\mathrm{O}$, or $\mathrm{H}-\mathrm{N}$ bonds, so this molecule does not form $\mathrm{H}$-bonds.
2. Here, the molecule is shown in a notation typical of organic chemistry: each vertex along the line connecting $\mathrm{HO}$ and $\mathrm{CH}_2$ contains a carbon atom + hydrogen atoms to complete the octet, and the lines connecting the vertices represent bonds. Written another way, this molecule is $\mathrm{HO}-\mathrm{CH}_2-\mathrm{CH}=\mathrm{CH}_2$. None of the $\mathrm{C}-\mathrm{H}$ bonds participate in hydrogen bonding, but the $\mathrm{HO}$ groups will form $\mathrm{H}$-bonds with the other $\mathrm{HO}$ groups!
3. Similar to the first one, this molecule only contains bonds between $\mathrm{C}$ and $\mathrm{H}$ atoms. It will not form $\mathrm{H}$ bonds.
4. There is a lot going on with this molecule. The rings here are the same sort of structure as the ring in 1: you can imagine filling in a $\mathrm{C}$ at each vertex and then populating with $\mathrm{H}$ to fill out the octets. This molecule contains $\mathrm{N}$, but all of the $\mathrm{N}$ atoms are bonded only to $\mathrm{C}$: these $\mathrm{N}-\mathrm{C}$ groups won't form $\mathrm{H}$ bonds. However, there are some $\mathrm{OH}$ groups: these will form $\mathrm{H}$-bonds with the $\mathrm{OH}$ groups on neighboring molecules!
5. You might recognize this molecule as ammonia. Here, the $\mathrm{N}$ atom is directly bonded to hydrogen. These $\mathrm{N}-\mathrm{H}$ end groups will form $\mathrm{H}$-bonds.
6. Finally, this hydrofluoric acid molecule is just $\mathrm{H}-\mathrm{F}$. Therefore, it will form $\mathrm{H}$-bonds. | textbooks/chem/Inorganic_Chemistry/Introduction_to_Solid_State_Chemistry/06%3A_Recitations/6.11%3A_More_Molecular_Orbital_Theory_and_Intermolecular_Forces.txt |
Band diagrams
As we saw for hydrogen in lecture, band diagrams can be thought of as the continuum limit of MO theory, allowing us to think about a long chain of bonds (or even a crystalline solid!) instead of just a dimer. To recap, let's imagine what building up a carbon solid - diamond- would be like. Start with just a few carbon atoms: with electronic configuration $1 \mathrm{~s}^2 2 \mathrm{~s}^2 2 \mathrm{p}^2$, they each bring 6 electrons, including two core electrons and four valence electrons. Remember - the core electrons live closest to the nucleus; they require the most energy to remove. The valence electrons live further away from the nucleus: these are the electrons that participate in bonding. Now, let's bring in some more atoms.
We know from previous weeks that carbon can form $\mathrm{sp}^3$ hybridized bonds - these are tetrahedral, so every carbon is coordinated to bond with four other neighboring carbon atoms. The bonding happens between (hybridized) valence electrons: if we were working at $0 \mathrm{~K}$ - absolute zero temperature - we could imagine a rigid structure of nuclei and bonds. And remember, a diamond could have moles of electrons, meaning moles of bonds! The $\mathrm{sp}^3$ hybridized valence electrons collectively form the valence band. Unlike a MO diagram, in which we drew one energy level for each type of bonding and antibonding orbital, the moles of electrons in a solid form distributions: all of the 1s electrons are distributed around a very low energy value, for example.
But what happens if we turn up the temperature? It turns out, several interesting things: for one, the thermal energy now available to the system allows for atoms to begin vibrating. Though the core electrons are too low in energy to be affected, some of the electrons in the tail of the valence band gain enough energy that they can delocalize (think back to our ionization days!), and become conducting electrons. The electrons that have been excited form the conduction band.
In a metal, it requires little thermal energy to excite a sea of electrons that readily carry charge (current). It happens at room temperature! This can occur in two ways: either a band is only partially full, so there are lots of states nearby in energy to be populated, or a full band overlaps with an empty band, so again it's easy for the electrons to hop between. In a semiconductor, the energy gap between the full valence band and the empty (at $0 \mathrm{~K}$ ) conduction band is a little bit larger: as the temperature increases, statistically, a few electrons gain enough energy to hop across the gap and conduct. Finally, in an insulator, the space between the highest full band and the lowest energy band is very large: it takes so much energy for an electron to jump to the conduction band that it doesn't happen under normal temperature and operating conditions.
Semiconductors
Semiconductors are defined by their name: they are kinda conductive. These materials have a band gap, but it’s not as big as that of an insulator. Often in the field, $3 \mathrm{~eV}$ serves as a rough cut-off: band gaps below this energy belong to semiconductors, while higher energy systems are considered insulating. Silicon is by far the most mature semiconductor technology: it is used in all sorts of applications and devices.
Semiconductors have a particularly interesting property that make them useful: they can function as transducors between electricity and light. If you shine light with sufficient energy on an semiconductor, it excites carriers inside the material: specifically, one photon excites an electron from the valance band up to the conduction band, where it is mobile. What happens to the atom that electron came from? It has essentially become electron-deficient, or ionized: it can be thought of as an extra positive charge in the valence band called a hole. Similar to electrons in the conduction band, holes in the valence band can move around: both carriers play a role in how conductive a material is! This is the operating principle of a solar cell: light shines on a material, and is converted into electricity. And what sets the relevant energy scale? The band gap, of course! Light with energy greater than the bandgap will be absorbed by the semiconductor and create carriers; light lower in energy than the bandgap will not be absorbed and will have no effect.
Similarly, semiconductors can transduce electricity to light as well: think of an LED! LEDs require a little more thought (and engineering) because while it’s easy for a passing photon to find an electron/hole pair to excite, it’s not quite as simple to bump an electron and a hole together so they recombine and emit light. The device used to engineer this process is called a PN junction.
Doping
Though we won't cover PN junction operation in $3.091$, we will talk about the ingredients to make one. A PN junction is formed by stacking a $\mathrm{p}$-type semiconductor next to an $\mathrm{n}$-type semiconductor material. A $p-type$ semiconductor contains extra holes, while an $n-type$ semiconductor contains extra electrons. You may be wondering, extra carriers compared to what? A plain semiconductor material - a single material (like $\mathrm{Si}$) or an alloy (like $\mathrm{GaAs}$ - is called intrinsic. The extra carriers are introduced via doping: adding a small fraction of a different type of atom to introduce new carriers. $\mathrm{P}$-type materials are obtained by doping with atoms with fewer electrons: think doping a group IV semiconductor with group III atoms. $\mathrm{N}$-type materials are obtained by doping with atoms with more electrons: think doping group IV with group V.
The extra carriers don't have quite the same energy structures as the lattice they are substituting into. Consider adding $\mathrm{P}$ into $\mathrm{Si}$. $\mathrm{P}$ has 5 valence electrons, compared to 4 in a $\mathrm{Si}$ atom. Remember that the electronic bands in a material are really an extension of their MO diagrams: adding just one electron to an MO diagram had a drastic effect on the bonding! In the local environment of the $\mathrm{n}$-type dopant, there is one too many electrons. This electron is much higher in energy than the other valence electrons: it forms a donor level within the band gap. The donor levels are often so close to the conduction band that thermal energy is enough to excite the extra carriers into the conduction band. Therefore, in $\mathrm{n}$-type materials, the extra carriers are electron donors.
Now, consider adding $\mathrm{Ga}$ into $\mathrm{Si}$. $\mathrm{Ga}$ is missing an electron compared to $\mathrm{Si}$, so it creates an extra hole in the conduction band. Energetically, things would be much more favorable if there were an electron populating that site: the hole creates an acceptor level close to the valence band. The holes are readily excited to the valence band, where they conduct. In both of these processes, the band gap itself is unaffected by the doping: rather, the extra states within the band gap increase the carrier population at thermal equilibrium.
Example: If you dope 1 cubic meter of $\mathrm{Ge}$ with $2.43 \mathrm{mg} \mathrm{Mg}$, how many carriers does each substitution yield and how many carriers are generated? What kind of doping is this?
Answer
We can determine how many carriers each substitution yields using the periodic table: $\mathrm{Ge}$ is group IV, while $\mathrm{Mg}$ is group II. Therefore, $\mathrm{Ge}$ has four valence electrons compared to $\mathrm{Mg}$'s two. Each substitution generates 2 extra carriers. Additionally, because we are doping with an electron-deficient material, this is $\mathrm{p}$-type doping. Finally, let's set up the unit conversion to determine how many carriers are created:
$2.43 \mathrm{mg} \mathrm{Mg} \times\left(\dfrac{10^{-3} \mathrm{~g}}{1 \mathrm{mg}}\right)\left(\dfrac{1 \mathrm{~mol} \mathrm{Mg}}{24.3 \mathrm{~g} \mathrm{Mg}}\right)=10^{-4} \mathrm{~mol} \mathrm{Mg} \nonumber$
Then, since each $\mathrm{Mg}$ provides two holes:
$\dfrac{2 \text { holes }}{\text { Mg substitution }} \times 10^{-4} \mathrm{~mol} \mathrm{Mg}=2 \times 10^{-4} \mathrm{~mol} \text { holes } \nonumber$
Finally, we can pull out Avogadro:
$\left(2 \times 10^{-4} \mathrm{~mol} \text { holes }\right)\left(6.602 \times 10^{23} \dfrac{\text { holes }}{\text { mol holes }}\right)=1.32 \times 10^{20} \text { holes } \nonumber$ | textbooks/chem/Inorganic_Chemistry/Introduction_to_Solid_State_Chemistry/06%3A_Recitations/6.12%3A_Band_Diagrams_Semiconductors_and_Doping.txt |
Bravais lattices
To describe a crystal you need two ingredients: a lattice and a basis. Put another way, the lattice describes how atoms are arranged spatially: in a crystal, it is a regular, ordered pattern that tiles 3D space. The basis is the unit that is copied over the pattern: it usually consists of one or more atoms. Crystallographers utilize a set of patters called Bravais lattices to describe the ways atoms can be arranged to form crystalline solids. In $3.091$, we will focus on the subset of the Bravais lattices that are cubic: the scale in all three dimensions is the same.
There are three cubic lattices: simple cubic (SC), body-centered cubic (BCC), and face-centered cubic (FCC).
In the figure above, the grey grid shows the outline of the unit cell, while each circle represents an atom. For the atoms along the corners of the unit cell, each atom is shared amongst eight neighboring cells, so each cell contains $1 / 8$ of an atom. Similarly, each of the face atoms in the FCC unit cell are shared between two cells, so each contains $1 / 2$ an atom. We can calculate how many atoms are in each of the cubic unit cells:
$N_{S C}=8$ corners $* \dfrac{1}{8}$ atoms per corner $=1$ atom per $\mathrm{SC}$ unit cell
$N_{B C C}=8$ corners $* \dfrac{1}{8}$ atoms per corner $+1$ central atom $=2$ atoms per BCC unit cell
$N_{F C C}=8$ corners $* \dfrac{1}{8}$ atoms per corner $+6$ face atoms $* \dfrac{1}{2}$ atoms per face $=4$ atom per $\mathrm{FCC}$ unit cell
Crystal packing
We can now define some relevant metrics to describe the atomic density in different lattices and even along different directions within the crystal. We will talk about density both in terms of area and volume. To define a few terms:
• packing density: ratio of space occupied by atoms to total space available. Units are $\left[\frac{\text { atoms }}{n m^2}\right]$ (by area) or $\left[\frac{\text { atoms }}{n m^3}\right]$ by volume
• atomic packing factor: fraction of space occupied by atoms, in 2D (area ratio) or 3D (volume ratio). It is a unitless value between 0 and 1.
We also need to remember a few things from geometry: the area of a circle is $\pi r^2$, and the volume of a sphere is $\frac{4}{3} \pi r^3$. For a cubic system with side length $a$ has face diagonal length $\sqrt{2} * a$ and body diagonal length $\sqrt{3} * a$. The trick to successfully solving packing questions is being able to relate the radius of the atoms (assuming they are hard spheres and as close packed as possible) and the side length of the unit cell. From there, it's just a bit of algebra to find the solution.
Example: Below is a section of an FCC unit cell: it is a cut-away of one face. First, find the relationship between the radius of the atoms, $r$, and the unit cell width, $a$. Then, determine how many atoms are in the $2 \mathrm{D}$ unit cell. Finally, calculate the packing density of this plane (assuming $a=3.5 \AA$ ), and the 2D atomic packing fraction.
Answer
First, we need to relate $r$ to $a$. The easiest way to do this is usually to look along the closest packed direction: here, if we go down a diagonal, the atoms are always in contact along their center.
$r+2 r+r=\sqrt{2} * a \nonumber$
$r=\dfrac{\sqrt{2}}{4} a \nonumber$
Then, we want to figure out how many atoms are enclosed in the square, $2 \mathrm{~d}$ unit cell. There is one full atom in the middle, and four corners which each contain $1 / 4$ of an atom.
4 corners $* \dfrac{1}{4}$ atoms per corner $+1$ center atom $=2$ atoms per unit cell
The packing density in two dimensions is the number of atoms per square nm. We can simply divide the number of atoms in the unit cell by the area of the unit cell:
$\text { packing density }=\dfrac{2 \text { atoms per unit cell }}{(3.5 \AA)^2 \text { per unit cell }}\left(\dfrac{100 \AA^2}{1 \mathrm{~nm}^2}\right) \approx 16.33 \text { atoms } / \mathrm{nm}^2 \nonumber$
The packing fraction in two dimensions is the fraction of space occupied by atoms:
\begin{gathered}
\text { packing fraction }=\frac{\text { area occupied by atoms }}{\text { total area }}=2[\text { atoms } / \text { unit cell }] *\left(\frac{\pi r^2[\text { area per atom] }}{a^2[\text { area per unit cell }]}\right) \
=\frac{2 \pi *\left(\frac{\sqrt{2}}{4} a\right)^2}{a^2}=\frac{\pi}{4}=0.7854
\end{gathered} | textbooks/chem/Inorganic_Chemistry/Introduction_to_Solid_State_Chemistry/06%3A_Recitations/6.13%3A_Bravais_Lattices_and_Crystal_Packing.txt |
Miller indices
In crystallography, we use Miller indices to specify locations, directions, and planes in a crystal. Standard $\mathrm{x}-\mathrm{y}-\mathrm{z}$ Cartesian coordinates use a basis consisting of three orthogonal axes in three dimensions with unit length. In Miller index notation, we will rely on the lattice vectors of the unit cell as our basis. In $3.091$, since we are working with cubic Bravais lattices, the crystallographic lattice vectors are all orthogonal and the same length (with the length specific to a particular solid). With this framework, we can specify an arbitrary point in terms of its projections onto the lattice vectors, much as you would with standard Cartesian coordinates:
$\text { location }=h \overrightarrow{a_1}+k \overrightarrow{a_2}+l \overrightarrow{a_3}$
where $\overrightarrow{a_i}$ are the unit vectors with some magnitude and direction, and the values of $h, k$, and $l$ specify how far to go along each lattice vector. We can express this in shorthand as $(h, k, l)$, just like you would write an ordered pair of Cartesian coordinates! A quick note on notation: in Miller indices, the brackets and commas matter a lot! Here's a quick list of the options and what each signifies:
$\begin{array}{c|c} \text { Notation } & \text { Representation } \ \hline \hline(h, k, l) & \text { point } \ {[h k l]} & \text { direction } \ \langle h k l\rangle & \text { family of directions } \ (h k l) & \text { plane } \ \{h k l\} & \text { family of planes } \end{array}$
Note that only points use commas! Finally, negative weights can be represented with a bar over the relevant number: travel in the negative $\overrightarrow{a_1}$ direction would be indicated by $[\bar{h} k l]$, for example.
Directions and planes can be specified algorithmically:
Directions
A crystalline direction is defined in the context of a unit cell, and the vector defining the direction always passes through the origin of the unit cell. The vector stretches from the origin to the location specified in terms of the lattice vectors
$h \overrightarrow{a_1}+k \overrightarrow{a_2}+l \overrightarrow{a_3}$
where it is our job to determine the values of $[h k l]$. To determine the crystallographic direction given a picture of a vector in a unit cell:
1. Shift the vector such that it intersects the origin of the unit cell (if it doesn't already)
2. Make a mark on the edge of the unit cell where the vector intersects the unit cell frame
3. Determine the fraction of each lattice vector traversed
4. Multiply the fractions by a common multiple to yield integer $\mathrm{hkl}$ values which define the direction
Of course you can go the other way as well, from a direction $[h k l]$ to a drawing. To represent a direction $[h k l]$ in a unit cell:
1. Divide through the vector $[h k l]$ by whichever of $h, k$, and $l$ is the largest to yield fractional values
2. Place the tip of the vector at the point on the edge of the unit cell defined by the fraction found previously
3. Connect the origin to the vector tip
Example $1$
For a) and b), determine the crystallographic direction indicated on the unit cell. Then, for c) and d), sketch the direction on the axes provided. For c), sketch [100], and for d), sketch [211].
Solution
For a), we can see that the end of the vector is located at $\frac{1}{2} \overrightarrow{a_1}+1 \overrightarrow{a_2}+\frac{1}{4} a_3$. We can convert this to a crystallographic direction by multiplying through by 4 to get rid of the fractions while preserving the ratios. The direction shown in a) is therefore [241]. For b), the end of the vector is located at $1 \overrightarrow{a_2}+1 \overrightarrow{a_3}$, which is represented as [011]. Finally, the [100] and [211] planes are shown below:
Planes
Similar to a crystalline direction a crystalline plane is usually represented in a unit cell, and determined in terms of the lattice vectors $\overrightarrow{a_1}, \overrightarrow{a_2}$, and $\overrightarrow{a_3}$. When given a picture of a plane to identify:
1. Make a mark on each edge of the unit cell where the plane intersects the unit cell frame
2. Determine the fraction of each lattice vector traversed. If the plane never intercepts a lattice vector, you can note it as intercepting at $\infty$
3. Take the reciprocal of each fractional intercept $\left(\frac{1}{\infty}=0\right)$ to yield integer $\mathrm{hkl}$ values which define the plane
Finally, to sketch a plane given $(h k l)$:
1. Find the reciprocal value of $h, k$, and $l\left(\frac{1}{0}=\infty\right)$
2. Mark the fractional values associated with $h, k$, and $l$ on the $\overrightarrow{a_1}, \overrightarrow{a_2}$, and $\overrightarrow{a_3}$ axes
3. Connect the marks on the edge of the unit cell to form the plane
Example $1$
For a) and b), determine the plane shown in the unit cell. Then, for c) and d), sketch the given plane on the axes provided. For c), sketch the (101) plane. For d), sketch the (001) plane.
Solution
For a), the plane intersects $\overrightarrow{a_1}$ at the edge of the unit cell, which is 1. It is entirely parallel to $\overrightarrow{a_2}$ and $\overrightarrow{a_3}$, so we say it intercepts at $\infty$. Taking the reciprocals of these intercepts yields $\left(\frac{1}{1} \frac{1}{\infty} \frac{1}{\infty}\right)$, so this must be the (100) plane. For b), the plane intersects $\overrightarrow{a_1}$ at 1 and $\overrightarrow{a_3}$ at 1 . If we extend the plane along the positive $\overrightarrow{a_2}$ direction, it doesn't intercept the axis. But we have to remember to consider the other half of the crystal too: if we were to extend the plane backwards, it would intercept $\overrightarrow{a_2}$ at $-1$. Taking reciprocals, $\left(\frac{1}{1} \frac{1}{1} \frac{1}{1}\right)$ yields the (1111) plane. Finally, the (101) and (001) planes are shown below:
You might notice that some directions and planes in the crystal look the same: the atoms are spaced the same amount. This is due to symmetry in a crystal: for example, if you look down any of the diagonals in a cubic lattice, the atoms you "see" look the same! These equivalent directions and planes are called families. For example, the $\langle 100\rangle$ family of directions includes $[100],[010],[001],[00 \overline{1}],[0 \overline{1} 0]$, and $[00 \overline{1}]$ in a cubic crystal.
Interplanar spacing
To find the distance between neighboring parallel planes, we calculate the interplanar spacing, $d_{h k l}$.
$d_{h k l}=\dfrac{a}{\sqrt{h^2+k^2+l^2}} \nonumber$
Here, $a$ is the lattice parameter, usually expressed in [nm]. | textbooks/chem/Inorganic_Chemistry/Introduction_to_Solid_State_Chemistry/06%3A_Recitations/6.14%3A_Miller_Indices_and_Interplanar_Spacing.txt |
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