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Infrared Spectroscopy
Infrared spectroscopy measures the frequency of absorption when a sample is irradiated with infrared electromagnetic radiation. It is a technique used to study the vibrations between atoms because atomic vibrational excitations occur in the infrared region of the electromagnetic spectrum. The bonds between atoms can be thought of as a spring connecting two masses. In the spring-mass analogy the moving system can be approximated by a simple harmonic oscillator. The frequency oscillation is proportional to ${\displaystyle {\sqrt {\frac {k}{m}}}}$,where k is the spring constant and m is the mass of the object. In a similar approximation, the frequency of vibration between two atoms is unique and varies depending on the strength of the bond (k) and the size of the atoms (m). When the frequency of electromagnetic radiation matches the natural frequency of vibration between atoms, the atoms are able to absorb this energy and exhibit vibrations. These vibrations can be detected as signals if they produce a change in the dipole moment between two atoms which can interact with the electric field. Representations for normal modes of vibration will be active in the infrared if they transform similar to any one of the cartesian coordinates (x,y,z). This would mean the vibrational motion has shifted the charge distribution in any of the x, y, or z directions resulting in a change in the dipole moment. Selection rules such as these are used to tell us whether such transitions are allowed, and therefore observed, or whether they are forbidden.
Raman Spectroscopy
Unlike IR spectroscopy which measures the energy absorbed, Raman spectroscopy consists of exposing a sample to high energy monochromatic light that interacts with the molecule and causes electronic, vibrational, or translational excitations. Upon interaction, the energy of the light is shifted either up or down and these changes can give information about the molecule’s various vibrational states. A photon of light excites the molecule to an excited state and upon relaxation to a different rotational or vibrational state the molecule emits a photon of a different energy. A vibration will give rise to a Raman shift, due to a shift from the incident light, if it has the same symmetry as the molecular polarizability. Polarizability measures the ability for a molecule’s electron cloud to become distorted. A dense electron cloud is more difficult to change than a more spread out electron density. The representation for one of the normal modes of vibration will be Raman active if it transforms similarly to the direct products of any one of the x, y, or z coordinates. That includes any of the functions: xy, xz, yz, x2, y2, z2, or any combination.
Rule of Mutual Exclusion
A molecule is centrosymmetric if it has a center of inversion and their corresponding point group contains the class for inversion. In such cases, the unit vectors transform as ungerade, or unsymmetric about the center of inversion, and direct products transform as gerade, or symmetric about the center of inversion. As a result, the normal modes of vibrational will show frequencies in either the IR or Raman, but the same frequency will not be observed in both.
Boron Trifluoride or D3d point group
D3d E 2C3(z) 3C'2 σh (xy) 2S3 v Linear Functions/ Rotations Quadratic Functions Cubic Functions
A'1 +1 +1 +1 +1 +1 +1 - x2+y2, z2 x(x2-3y2)
A'2 +1 +1 -1 +1 +1 -1 Rz - y(3x2-y2)
E' +2 -1 0 +2 -1 0 (x,y) (x2-y2, xy) (xz2, yz2) [x(x2+y2), y(x2+y2)]
A"1 +1 +1 +1 -1 -1 -1 - - -
A"2 +1 +1 -1 -1 -1 +1 z - z3, z(x2+y2)
E" +2 -1 0 -2 +1 0 (Rx, Ry) (xz, yz) [xyz, z(x2-y2)
Normal Modes of Vibration for D3d: | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Advanced_Inorganic_Chemistry_(Wikibook)/1.13%3A_Selection_Rules_for_IR_and_Raman_Spectroscopy.txt |
Stretching Modes of Vibration
Often, we are only concerned with the stretching modes of vibrations for certain groups in a molecule. In this case, we can use the bonds to describe the stretching vibration for that group, though we have to consider all the bonds of the same type together. The general strategy is to come up with a reducible representation for only the bond vectors of the groups we are concerned with, then reduce them into their irreducible representations.
Consider the compound fac-Mo(CO)3(NCCH3)3 shown in Figure \(1\). We first determine the point group of the molecule, which is C3v for this molecule (character table shown in Table \(PageIndex{1}\)). We then use the Mo-CO bonds as a basis and generate a reducible representation using the character table. We therefore apply each operation to this basis and if a bond does not move we add a character of 1, if a bond is inverted it gets a character of -1, and if it moves we don't add anything to that character. Under the operation E, all three bonds are left unchanged so we give it a character of 3. Under the C3 operation, all of the bond vectors move, so we have a character of 0. Lastly, for the σv two bond vectors move and one stays in place, so we give it a character of 1. So our reducible representation is:
E C2 σv (xz) σ'v (yz)
Γ 3 1 0 3
This reduces into Γ= A1 + E. We can then determine from the character table that both of these will be IR and Raman active stretches.
Structure Determination
In chemistry, we are often concerned with the exact molecular structure of the compounds we are working with. We can use spectroscopy to further identify the structure of the molecule by considering the number of IR or Raman modes of vibration present. For example, Mo(CO)3(NCCH3)3 is can come in either its facial or meridional form (Figure 1) and therefore have different symmetries present - facial is C3v and meridional is C2v (C2v character table shown in Table \(PageIndex{2}\)). If we generate a reducible representation for the CO stretching vibrations in the meridional form, we get that:
E C2 σv (xz) σ'v (yz)
Γ 3 1 0 3
This can then be reduced to Γmer = 2A1 + E. All three of these stretching vibrations are IR active. If we compare that to the facial version: Γfac= A1 + E as determined above, we can see that the fac-Mo(CO)3(NCCH3)3 will have 2 IR active CO vibrational modes and mer-Mo(CO)3(NCCH3)3 has 3 IR active CO vibrational modes. This information can be used to distinguish between the two structures experimentally using IR spectroscopy. The same technique can be applied to Raman active modes.
Centrosymmetric Point Groups
We can further use Raman and IR spectroscopy to determine if our molecule has an inversion center present. This is because for molecules with an inversion center, all IR active modes are not Raman active and all Raman active modes are not IR active. This is known as the rule of mutual exclusion.
Table \(1\). C3v Character Table
E 2C3 v
A1 1 1 1 z x2+y2, z2
A2 1 1 -1 Rz
E 2 -1 0 (x,y), (Rx, Ry) (x2-y2, xy), (xz, yz)
Table \(2\). C2v Character Table
E C2 σv (xz) σ'v (yz)
A1 1 1 1 1 z x2, y2, z2
A2 1 1 -1 -1 Rz xy
B1 1 -1 1 -1 x, Ry xz
B2 1 -1 -1 1 y, Rx yz | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Advanced_Inorganic_Chemistry_(Wikibook)/1.14%3A_Stretching_Frequencies_and_Structure_Determination.txt |
Groups with axial symmetry are also known as continuous groups due to the infinite amount of rotations and reflections that leave the molecule invariant. Due to the sheer number of symmetry operations, determining irreducible representations is not feasible. There is one aspect of group theory that can be taken advantage of and these are the group, subgroup relationships.
Groups and Subgroups
A subgroup is a set of elements within a group that also form their own group, such that every operation in the subgroup stays in the subgroup. Consider a concrete example. The ${C_{2h}}$ point group contains 4 elements: {$E$, $C_2$, $i$, $σ$}. However, the elements $E$ and $C_2$ form the $C_2$ group. Hence $C_2$ is a point group of the $C_{2h}$ point group, and this is designated as, ${ C_{2}\subseteq C_{2h}}$.
The character tables of subgroups are appropriately reduced down. Consider $D_{4h}$. We know that $C_{4v}$ is a subgroup of $D_{4h}$. On Figure 2, the C4v elements are boxed with their respective characters. Notice that $D_{4h}$ has many more possible irreducible representations. However, there are different irreducible representations in $D_{4h}$ that become identical due to the fact that the characters are the same with the elements of the $C_{4v}$ point group. Conversely, degenerate irreducible reps may become nondegenerate.
One has to figure out properly classify the irreducible representations. For example, $A_{1g}$ and $A_{2u}$ in $D_{4h}$ are both $A_1$ in $C_{4v}$. The complete relationship between different irreducible representations is a correlation diagram and essentially matches identical sets of characters to one another.
The importance of this is that a physical property, i.e. a wavefunction, that serves as a basis for one of the irreducible representations for a point group will transform as its correlated irreducible representation in its subgroup.
Subgroups of Linear Molecules
The electronic energy terms of a linear molecule are classified by the angular momentum that lies along the principal axis of symmetry. The physical reason for this is that only angular momentum about the axis is conserved around the axis. Therefore, only the m quantum number classifies a linear molecule. In analogy with s,p,d.., the new designations for the energy levels are $Σ$, $Π$, $Δ$, etc. A ($±$) subscript is reserved for $Σ$ energies which have no angular momentum about the axis, so it designates between wavefunctions symmetric and antisymmetric with respect to reflection about the principal axis. $g$ and $u$ designate wavefunctions that are symmetric and antisymmetric with respect to inversion. In order to completely determine energy levels, one must Figure out how reflections, inversion, and rotations affect the Hamiltonian of a wavefunction. The various symmetry operations produce physically distinct, but degenerate, energy levels. This method of analysis is free from approximations from molecular orbitals, but molecular orbitals of diatomic molecules must still adhere to these symmetry rules hence a pattern.
The important aspect of this discussion is that these physically realized energy levels help simplify the issue of dealing with an infinite point group. There are an infinite amount of subgroups but the correct subgroup are point groups with two fold rotational symmetry. For a ${ D_{\infty h}}$, the appropriate subgroup is a ${ D_{2h}}$, and for a ${ C_{\infty v}}$, the appropriate subgroup is a ${ C_{2v}}$.
How to Solve Symmetry Problems using Subgroups
The technique for solving the irreducible representations of linear groups is straightforward.
1. Identify the point group of the molecule and then treat the point group as the appropriate subgroup as designated in the above tables.
2. Solve for the reducible as the subgroup and reduce down to the irreducible representations.
3. Use the correlation diagram to convert back to the proper form and then pull out relevant physical information.
Vibrations of Linear Molecules
The continuous rotation of linear molecules is not physically meaningful as a degree of freedom, hence the total number of vibrational degrees of freedom is $3N-5$, subtracting away two rotational degrees of freedom instead of 3. Using the method outlined above, determine the molecule’s irreducible representation in the corresponding two-fold rotation group and then convert back to the infinite point group. Because the correlation diagram links between irreducible representations that transform the same way, one can use the irreducible representations in the two-fold rotation group to Figure out which ones are IR and Raman active.
Example $1$: Carbon Suboxide
Both infrared and electron diffraction studies have argue that carbon suboxide ($\ce{C_{3}O_{2}}$) has a bent structure in the gas phase, however, it exhibits an average linear geometry in the solid phase and would be in the $D_{ꝏh}$ point group.
Solving this as a $D_{2h}$ point group just as we solve a vibrational problem, by appropriately tossing out translational and rotational degrees of freedom, one can arrive at an irreducible representation of
$2A_g + 2B_{2g} + 2B_{3g} + 3B_{1u} +3B_{2u} + 3B_{3u}. \nonumber$
Using the correlation table above, we find that the true irreducible representation is actually
$2\Sigma _{g}^{+} + \Pi _{g} + 2\Sigma _{u}^{+} + 2\Pi _{u}. \nonumber$
• $\Sigma _{g}^{+}$ transforms as ${ A_{g}}$
• ${ \Pi _{g}}$ transforms as both ${ B_{2g}}$ and ${ B_{3g}}$.
Therefore ${ A_{g},B_{2g},}$ and ${ B_{3g}}$ are all Raman active.
• ${ \Sigma _{u}^{+}}$ transforms as ${ B_{1u}}$
• ${ \Pi _{u}}$ transforms as ${ B_{2u}}$ and ${ B_{3u}}$.
Therefore ${ B_{1u},B_{2u}}$, and ${ B_{3u}}$ are IR active.
Making these correlations allow us to effectively treat this molecule as a typical vibrational problem and pull out all the same relevant information. | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Advanced_Inorganic_Chemistry_(Wikibook)/1.15%3A_Vibrational_Spectroscopy_of_Linear_Molecules.txt |
UV-Vis Spectroscopy
UV-Vis spectroscopy is an analytical chemistry technique used to determine the presence of various compounds, such as transition metals/transition metal ions, highly conjugated organic molecules, and more. However, due to the nature of this course, only transition metal complexes will be discussed. UV-Vis spectroscopy works by exciting a metal’s d-electron from the ground state configuration to an excited state using light. In short, when energy in the form of light is directed at a transition metal complex, a d-electron will gain energy and a UV-Vis spectrophotometer measures the abundance of transition metal atoms with excited electrons at a specific wavelength of light, from the visible light region to the UV region.
When using a UV-Vis Spectrophotometer, the solution to be analyzed is prepared by placing the sample in a cuvette then placing the cuvette inside the spectrophotometer. The machine then shines light waves from the visible and ultraviolet wavelengths and measures how much light of each wavelength the sample absorbs and then emits.
Absorbance of the sample can be calculated via Beer’s Law:
$A=εlc \nonumber$
where $A$ is the absorbance, $ε$ is the molar absorptivity of the sample, $l$ is the length of the cuvette used, and $c$ is the concentration of the sample.[1] When the spectrophotometer produces the absorption graph, the molar absorptivity can then be calculated.
To illustrate what this looks like, you will find a sample absorbance spectrum to the right.[2] 404-5</ref> </ref> As can be seen, the y-axis represents absorbance and the x-axis represents the wavelengths of light being scanned. This specific transition metal complex, [CrCl(NH3)5]2+, has the highest absorbance in the UV region of light, right around 250-275 nm, and two slight absorbance peaks near 400 nm and 575 nm respectively. The two latter peaks are much less pronounced than the former peak due to the electron’s transition being partially forbidden—a concept that will be discussed later in this chapter. If a transition is forbidden, not many transition metal electrons will undergo the excitation.
Theory Behind UV-Vis Spectroscopy
Splitting of the D-Orbitals
As is widely known, the d-orbitals contain five types of sub-orbitals: $d_{xy}$, $d_{yz}$, $d_{xz}$, $d_{x^2-y^2}$, and $d_{z^2}$ which are all shown in Figure $2$. When in the absence of a magnetic field—such as when there are no electrons present—all the sub-orbitals combine together to form a degenerate spherical orbital. This singular orbital promptly differentiates back into its sub-orbitals when electrons are introduced or the transition metal is bonded to a set of ligands (Figure $3$).
The Origination of Color in Transition Metal Complexes
When looking at color in transition metal complexes, it is necessary to pay attention to the differentiated d-orbitals. Color in this sense originates from the excitation of d-orbital electrons from one energy level to the next. For instance, an electron in the t2g bonding orbital can be excited by light to the eg* bonding orbital and upon its descent back to the ground state, energy is released in the form of light:
The specific wavelength of light required to excite an electron to the eg* orbital directly correlates to the color given off when the electron moves back down to the ground state. Figure $5$ helps visualize the properties of transition metal color. Whichever color is absorbed, the complimentary color (directly opposite from the color in the Figure) is emitted. For instance, if a metal complex emits green light we can Figure out that the complex absorbed red light with a wavelength between 630 nm-750 nm in length.
Rules of Color Intensity and Forbidden Transitions
The intensity of the emitted color is based on two rules:[4]
1. Spin multiplicity: the spin multiplicity of a complex cannot change when an electron is excited. Multiplicity can be calculated via the equation 2S+1 where S is calculated by (1/2)(number of unpaired d-electrons).
2. If there is a center of symmetry in the molecule (i.e. center of inversion) then a g to g or u to u electron excitation is not allowed.
If a complex breaks one of these rules, we say it is a forbidden transition. If one rule is broken, it is singly forbidden. If two rules are broken, we call it double forbidden and so on. Even though the convention is to call it forbidden, this does not mean it will not happen; rather, the more rules the complex breaks, the more faded its color will be because the more unlikely the chances the transition will happen. Let’s again look at the example (Figure $4$)
If we apply the intensity rules to it:
1. Multiplicity before transition=2(0.5[1 unpaired electron])+1=3, Multiplicity after transition=2(0.5[1 unpaired electron])+1=3. Both multiplicities are the same, so this transition is allowed under rule 1.
2. If we assume this molecule is octahedral in symmetry, this means it has an inversion center and thus the transition of eg* to t2g is forbidden under rule 2 due to both orbitals being gerade (g).
3. We are only exciting one electron and thus it is allowed under rule 3.
Based on these rules, we can see that this transition is only singly forbidden, and thus it will appear only slightly faded and light rather than a deep, rich green.
Ligand Field Theory: How Ligands Affect Color
As it turns out, the atoms bonded to a transition metal affect the wavelength that the complex needs to absorb in order to give off light; we refer to this as Ligand Field Theory. While certain transition metals like to absorb different wavelengths than other transition metals, the ligand(s) plays the most important role in determining the wavelength of light needed for electron excitation.[5]
The terms low spin and high spin are used to describe the difference in energy levels of the t2g and eg* orbitals, which correlates to the wavelength of light needed to excite an electron from t2g to eg*. When a complex is characterized as low spin, the ligands attached to the metal raise the energy of the eg* orbital so much that the ground state configuration of the complex fills the first six electrons in the t2g orbital before the eg* orbital is filled. As a result, high energy wavelengths of light—violet, blue, and green—are needed to successful excite an electron to the eg* bonding orbital. This means that the transition metal complex will emit yellow, orange, and red light, respectively. Conversely, high spin complexes have ligands which lower the energy level of the eg* orbital so that low energy light—red, orange, and yellow—or even high energy light can successfully excite an electron. Thus, a high spin complex can emit any color. High spin complexes can be thought of as all-inclusive while low spin complexes are half-exclusive in terms of the wavelengths needed to excite an electron.
To determine whether a complex is high spin or low spin:
1. Look at the transition metal. First row metals will want to be high spin unless the ligand(s) forces it to be low spin. Second row metals want to be low spin unless the ligand(s) forces it high spin. Third row metals will low spin.
2. Look at the ligand(s) as they will be the ultimate determining factor. If the ligand matches the transition metal in terms of high spin/low spin, then the complex’s spin will be whatever is “agreed” upon. If they differ, follow the ligand’s spin type. If the ligand is classified neither as high nor low spin, follow the transition metal’s spin type. Ligand spin types are enumerated below.
3. If there are multiple ligands with differing spin types, go with whichever spin type is most abundant in the complex.
The ligands below are ranked from low spin (greatest energy difference between t2g and eg*) to high spin (lowest energy difference):[6]
$\ce{CO} > \ce{CN^{-}} > \ce{PR_3} >\text{ethylene diamine (en)} > \ce{NH_3} > \ce{H_2O} > \ce{F^{-}} > \ce{Cl^{-}} > \ce{Br^{-}} > \ce{I^{-}} \nonumber$
To illustrate this concept, let’s take the following complexes: $\ce{[Ni(NH3)6]^{2+}}$ and $\ce{[Ni(CN)4]^{2-}}$.
The nickel complexes all have the same oxidation state on the metal (2+), and thus the same d-electron count. In the first complex, nickel wants to be high spin, while ammonia prefers neither high nor low spin. Therefore the complex will be high spin and emit blue light, which is an absorbance of orange—weak energy—light. For the second complex, nickel again wants to be high spin, but cyanide prefers low spin. As a result, the complex becomes low spin and will emit yellow light, which is an absorbance of violet—strong energy—light. | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Advanced_Inorganic_Chemistry_(Wikibook)/1.16%3A_Fundamentals_of_Electron_Absorption_Spectroscopy.txt |
Introduction to Jahn-Teller Distortion
The Jahn–Teller effect occurs because the unequal occupation of orbitals with identical energies is unfavorable. To avoid these unfavorable electronic configurations, molecules distort (lowering their symmetry) to render these orbitals no longer degenerate.[1] The Jahn–Teller distortion, describes the geometrical distortion of molecules and ions that result from certain electron configurations.[2] This distortion is normally observed among octahedral complexes where the two axial bonds can be compressed or elongated to result in a different bond length from those of the equatorial bonds as shown in Figure \(1\).[3]
A Brief History of Jahn-Teller Distortion
In 1937, Hermann Jahn and Edward Teller postulated a theorem stating that "stability and degeneracy are not possible simultaneously unless the molecule is a linear one," in regards to its electronic state. This leads to a break in degeneracy which stabilizes the molecule and by consequence, reduces its symmetry.[4]
Orbital Analysis of Jahn-Teller Distortion
Elongation Distortion
Elongation as shown in Figure \(2\), is the most common mechanism of distortion. In ideally octahedral complexes that experience Jahn–Teller distortion, the eg* orbitals change more in energy relative to the t2g orbitals.[5]
When the octahedral complex undergoes an elongation, any of the metal d orbitals that carry a z-directional feature become lower in energy, ligands are easier to bond along z-direction. Geometrically, as the z-direction gets elongated, the repulsion energy and sterical hindrance along z-direction are lower to allow ligands easier to bond.
Compression Distortion
Compression as shown in Figure \(3\), happens less frequently than the elongation.
When the octahedral complex undergoes a compression, any of the metal d orbitals that carry a z-directional feature become higher in energy, ligands are harder to bond along z-direction. Geometrically, as the z-direction gets compressed, the repulsion energy and sterical hindrance along z-direction are higher, which makes ligands more difficult to bond.
The Strength of Jahn-Teller Distortion
General Strength Trend of Jahn-Teller Distortion
For octahedral complexes:
• weak Jahn-Teller effect if t2g is unevenly occupied
• strong Jahn-Teller effect if eg is unevenly occupied
Jahn-Teller Distortion and Spectra
Spectra of the First-order Jahn-Teller Distortion
In the first-order Jahn-Teller distortion, two electron transitions can be observed. This phenomenon can lead to a “double-hump” result on the absorption spectra as shown in Figure \(5\) and \(6\).
The Second-order Jahn-Teller Distortion
A brief introduction to the Second-order Jahn-Teller Distortion
The second order Jahn-Teller distortion occurs when the excited state of the transition metal has unequal occupation of d-orbitals with identical energies.[7] The ground state of the transition metal may not have unequal occupation of degenerate orbitals, therefore, the second order Jahn-Teller distortion is also called Pseudo Jahn-Teller distortion as shown in Figure \(7\).[8]
Spectra of the Second-order Jahn-Teller Distortion
Since the second order Jahn-Teller distortion occurs in excited state, the trough on the spectra likely located at higher frequency, thus higher in energy, as shown in Figure \(8\).
Figure \(8\).
1.18: Definition Importance and History of Organometallics
Introduction
Organometallic chemistry, as defined by Dr. Brian W. Pfennig in his book Principles of Inorganic Chemistry, as “the chemistry of compounds that contain at least one metal–carbon bond (other than cyanide).” Though quite simplistic in definition, organometallic chemistry is a rather modern sub-discipline of chemistry when compared to its contemporary chemistry disciplines. Organometallic chemistry joins the of coordination complexes of inorganic chemistry with the synthetic methods of organic chemistry. Today’s interpretation of organometallic chemistry has now interwoven itself into many other disciplines; including bioorganometallic chemistry and catalytic chemistry. The former being a critical component in modern organic chemistry. So much so that without a organometallic catalyst, reaction such as the 2010 Nobel prize winning Heck reaction could not exist. While the latter is critical to life functions within organisms, such as the iron which coordinates the heme group in the blood cells that transfers oxygen to the tissues. Today, organometallic chemistry is used extensively in the modern world, from the construction of polymers, plastics, and petrol, to electronic circuitry and solar panel construction, to advances in medicine such as immunization inoculations and chemotherapy.[1]
History
French chemist Louis Claude Cadet de Gassicourt isolated the first organometallic compound, tetramethyldiarsine a.k.a. cacodyl, in 1757 by accident. He was experimenting with invisible inks by combining arsenic containing cobalt ore with potassium acetate. Arsenic itself is not a true metal, rather it is considered a metalloid, nonetheless it is still considered an organometallic compound.[2]
The first organometallic compound containing a transition metal was formed 67 years later by Danish organic chemist William Christopher Zeise by placing platinum tetrachloride in boiling ethanol. The resulting ion formed was trichloro-(ethene)-platinate (II) anion. When combined with a potassium counter ion, the Zeise salt is formed. The compound drew plenty of criticism in its day from Zeise colleagues over its actual structure. A problem that wasn’t solved until x-ray crystallography became available in the 20th century. Zeise’s salt kickstarted an interest in organometallic compounds even though the 19th century chemist did not exactly know why or how these compounds form. Compounds such as diethyl zinc and the extremely toxic nickel tertracabonyl were formed in the later half of the 19th century. With the former, synthesized by British chemist Ludwig Mond, initiating an entire new class of compounds called metal carbonyl’s.[3]
At the turn of the 20th century French chemist Victor Grignard discovered a new method of coupling carbon to the carbonyl group of a ketone/aldehyde by nucleophilic addition using an alkyl/aryl halide coupled to magnesium metal. Grignard’s groundbreaking organometallic reagent, which now bears his name, swept through the chemistry labs of the early 20th century and was awarded the 1912 Nobel Prize of Chemistry in addition with Paul Sabatier. Over one hundred years later it is still used extensively as a coupling reagent to a variety of carbonyl derivatives.[4]
The following year, in 1913, the Nobel in chemistry went to Swiss inorganic chemist Alfred Werner for his work involving the coordination chemistry of ligands to metals. In particular the structure of hexaminecolbalt (III) chloride. Werner’s work in coordination chemistry proved to be vital in the understanding of organometallic coordination and chemical reactions of compounds and contributed greatly to opening up the organometallic discipline.Though many new organometallic compounds were being created and used, organometallic chemistry was still not recognized as its own independent sub-discipline of chemistry until half way into the 20th century and the discovery of ferrocene in 1951.[5]
Ferrocene was created in 1951 by American chemists Peter Pauson and Tom Kealy by reacting cyclopentadiene magnesium bromide and ferric chloride together, resulting in an orange powder now known as ferrocene. Unfortunately for the two chemists they did not deduce the actual structure of their organometallic salt and erroneously proposed that the iron acted as a bridge in-between the first carbon of two cyclopentadiene molecules. Later on, English chemist Sir Geoffrey Wilkerson, in collaboration with American chemist Robert Woodward, Figured out that the iron in ferrocene was actually being sandwiched in between two cyclopentadiene molecules. In ferrocene, each cyclopentadiene achieves aromaticity and all 12 electrons covalently bond with the iron atoms available sigma and pi orbitals creating a very stable 18 electron containing molecule. Independently, German chemist Ernst Fischer also came to the same conclusions of the sandwich model for ferrocene. Fisher realized that this sandwich compound was not a result of the metal used, but rather how the coordination of the interaction of the cyclopentadiene ligand and metal occurred. Fisher then expanded the metallocene compounds to include other metals. Together, Wilkerson and Fisher shared the 1973 Nobel Prize of Chemistry for their respective work with metallocenes.[6]
The discovery and understanding of metallocenes officially brought forth organometallic chemistry into to its own sub-discipline of chemistry. In doing so, opened up an explosion of new ideas of how to use organometallic compounds. Though one use stood out from the rest, that being to use organometallic compounds as catalyst’s in reactions. One early catalytic organometallic compound, dicyclopentadiene zircon (IV) dichloride was created jointly by chemist Karl Ziegler and Giulio Natta to polymerize terminal olefins. This actual led to two entire classes of organometallic catalyst’s, now known as Ziegler-Natta catalysts and earned each a Nobel Prize of Chemistry in 1963. Armed with new organometallic catalysts, chemists of the late 20th century designed new ways to couple carbons together. These include the famed Heck reaction, the Sharpless epoxidation, and the Grubbs olefin metathesis. Each of which earned a Nobel Prize of chemistry in 2010, 2001, and 2005 respectively.[7] | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Advanced_Inorganic_Chemistry_(Wikibook)/1.17%3A_Jahn-Teller_Distortions.txt |
The 18 Electron Rule is a useful tool to predict the structure and reactivity of organometallic complexes. It describes the tendency of the central metal to achieve the noble gas configuration in its valence shell, and is somewhat analogous to the octet rule in a simplified rationale. Exceptions to this rule exist, depending on the energy and character of atomic and molecular orbitals.[1]
The 18 Electron Rule
The General Rule
Atoms tend to have all its valence orbitals occupied by paired electrons. For transition metals, the valence orbitals consist of ns, 3 np and 5 (n-1)d orbitals, leading to its tendency of being surrounded by 18 electrons. This is somewhat analogous to the octet and Lewis structure rules of main group elements in a simplified rationale.
Structures that satisfy this preferred electron structure are described as electron-precise. Transition metal complexes with 18 electrons are also referred to as saturated, and there will be no other empty low-lying orbitals available for extra ligand coordination. Complexes with less than 18 electron counts are unsaturated and can electronically bind to additional ligands.
Exceptions to The Rule
The 18 electron rule is usually followed in metal complexes with strong field ligands that are good σ donors and π acceptors (for example, CO ligands). The energy difference (Δ0) between t2g and eg* orbitals is very large, and in this case the three t2g orbitals become bonding and are always filled, while the two eg* orbitals are strongly antibonding and are always empty.
However, when Δ0 between t2g and eg* orbitals are small, for example, in the case of first row transition metals with weaker field ligands, the antibonding character of eg* orbitals weakens, and the complex can have up to 22 electrons.
On the other hand, less than 18 electrons may be observed in complexes of 4th and 5th row transition metals with high oxidation states. In this case, Δ0 is relatively large due to increased repulsion between d orbitals of metals and the ligands. The eg* orbitals are strongly antibonding and remains empty, while t2g orbitals are non-bonding, and may be occupied by 0-6 electrons. [2]
Still, generally, the types of ligands in a complex determine if the complex would follow the 18 electron rule or not.
A few common examples of exceptions to 18 electron rules include:[3]
• 16-electron complexes: The metal center is usually low-spin and is in d8 configuration. These complexes adopt square planar structure, such as Rh(I), Ni(II), Pd(II), and Pt(II) complexes. In a lot of catalytic reactions, the organometallic catalysts convert back and forth between 18 and 16 electron configurations, and thus completes a catalytic cycle.
• Bulky ligands may hinder the completion of 18 electron rule. Examples include complexes with agostic interaction. [4] [5]
• Complexes with ligands of strong π-donating characters often violate 18 electron rule. Examples of this kind of ligands include F-, O2-, RO- and RN2-.
Electron Counting Methods
There are two widely used methods for electron counting of complexes - covalent method and ionic ligand method. Both of the two methods are applicable to all organometallic complexes, and should give the same electron count.
Covalent Method
In this method, all metal-ligand bonds are considered covalent. Ligands are considered neutral in charge, and may donate either 2, 1 or zero electrons to the bond. For example, ligands such as CO and NH3 are considered to have filled valence and contribute 2 electrons. Halide and hydroxo groups, however, do not have octet structure in neutral state, and contribute 1 electron to the bonding. Ligands such as BF3 do not have any free electron available, and the two electrons for bonding would come from the metal center.
Steps for covalent counting method:
1. Identify the group number of the metal center.
2. Identify the number of electrons contributed by the ligands.
3. Identify the overall charge of the metal-ligand complex.
4. At the presence of metal-metal bond, one electron is counted towards each metal center in a bond.
5. Add up the group number of the metal center and the e- count of the ligands, then take into consider the overall charge of the complex to obtain the final electron count.
Ionic Method
The ionic method always assigns filled valences to the ligands. For example, H group is now considered H-, as well as other groups such as halide, hydroxyl and methyl groups. These groups now contribute one more electron than they do in covalent method, and oxidize the metal center when a bond is formed. Groups with neutral charge in octet structure, such as CO and NH3, behaves the same as in valence methods.
Steps for ionic counting method:
1. Determine the overall charge of the metal complex.
2. Identify the charges of the ligands, and the numbers of e-s they donate.
3. Determine the number of valence electrons of the metal center, so that the oxidation state of the metal and charges of the ligands balance the overall charge of the complex. (E- count of metal center = Metal atom group number + ∑(charges of ionic ligands) – overall charge of the complex)
4. If metal-metal bond is present, one bond counts for one electron for each metal atom.
5. Add up the electron count of the metal center and the ligands.
Electron Counts of Some Common Ligands [6]
Ligand Covalent Ionic Charge
H 1 2 (H-) -1
Cl, Br, I 1 2 (X-) -1
OH, OR 1 2 (OH-,OR-) -1
CN 1 2 (CN-) -1
CH3, CR3 1 2 (CH3-,CR3-) -1
NO (bent M-N-O) 1 2 (NO-) -1
NO (linear M-N-O) 3 2 (NO+) +1
CO, PR3 2 2 0
NH3, H2O 2 2 0
=CRR' (carbene) 2 2 0
H2C=CH2 (ethylene) 2 2 0
CNR 2 2 0
=O, =S 2 4 (O2-, S2-) -2
η3-C3H5 (π-allyl) 3 2 (C3H5+) +1
≡CR (carbyne) 3 3 0
≡N 3 6 (N3-) -3
en (Ethylenediamine) 4 4 0
bipy (Bipyridine) 4 4 0
butadiene 4 4 0
η5-C5H5 (cyclopentadienyl) 5 6 (C5H5-) -1
η6-C6H6 (benzene) 6 6 0
η7-C7H7 (cycloheptatrienyl) 7 6 (C7H7+) +1
Examples
Examples of Electron Counting of Some Organometallic Complexes[7]
Complexes Covalent Method Ionic Method Total Number of Electrons
5-C5H5)2Fe
• Fe gives 8e-
• 2 η5-C5H5 give 2×5e-
• Complex Charge is 0
• Fe(II) gives 6e-
• 2 η5-C5H5- give 2×6e-
18
[V(CO)7]+
• V gives 5e-
• 7 CO give 7×2e-
• Complex Charge is +1 (-1e-)
• V(I) gives 4e-
• 7 CO give 7×2e-
18
[Re(CO)5(PF3)]+
• Re gives 7e-
• 5 CO give 5×2e-
• PF3 gives 2e-
• Complex Charge is +1 (-1e-)
• Re(I) gives 6e-
• 2 CO give 5×2e-
• PF3 gives 2e-
18 | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Advanced_Inorganic_Chemistry_(Wikibook)/1.19%3A_Electron_Counting_and_the_18_Electron_Rule.txt |
Dative Ligands: CO and Phosphines
Dative ligands represent a class of compounds that form dative covalent bonds, otherwise known as coordinate bonds, in which both electrons come from the same atom. In the case of transition metals, a dative ligand can form a coordinate bond with a transition metal. The dative ligand can form a sigma bond (σ) with the metal center and can donate both of its electrons to the metal, contributing to a transition metal complex. For transition metals, a dative ligand will always first form a sigma bond with the metal center. The pi* (π*) antibonding orbitals of the dative ligand, if energetically available, can also interact with the filled d orbitals of the transition metal. This interaction is known as pi backbonding and results in the transfer of 2 electrons from the d orbital of the metal to the pi* antibonding orbital, thus relieving negative charge from the metal center. This represents a transfer of electrons from an atomic orbital (d orbital of the transition metal) to a molecular orbital (pi* antibonding orbital). The d orbital and the pi* orbital have the requisite symmetry with one another to facilitate this process.
CO Ligands
Backbonding and IR Absorption
CO ligands are carbon monoxide ligands that participate readily in transition metal complexes via coordinate bonds with a transition metal center. CO ligands are neutral 2 electron donors and first form a sigma bond with a metal center. As discussed above, the d orbitals of the transition metal are symmetric about the pi* orbitals of the CO compound and backbonding occurs between the metal and the CO. This process is illustrated in Figure \(1\). Pi backdonation is supported by IR (infrared spectroscopy) data of various metal-carbonyl complexes. Using group theory, one can consider the consequences of backbonding and the affects this would have on the IR absorption of a carbonyl-containing compound. Pi backbonding donates electrons from the transition metal center to the pi* antibonding orbital of CO, the LUMO of this ligand (Lowest Unoccupied Molecular Orbital). This in turn lowers the bond order between the carbon and oxygen, weakening their interaction. This would cause one to suspect that metal carbonyl complexes would absorb light of a lower frequency upon excitation by infrared radiation. This is consistent with experimental data, with few exceptions, for several metal carbonyl complexes when compared to the IR absorption of free carbonyls. This is depicted in Table \(1\).
Table \(1\). IR Absorption Frequencies for Metal Cabonyls
Compound νCO (cm-1)
CO 2143
V(CO)61- 1859
Cr(CO)6 2000
Mn(CO)6+ 2100
Fe(CO)62+ 2204
Fe(CO)5 2022, 2000
Ru(CO)5 2038, 2022
X Ray Crystallography
Homoleptic carbonyl complexes typically form octahedral complexes and the normal modes of vibration transform as a1g, eg, t1u. The character table of the octahedral complex, however, predicts that only the t1u set will be IR active. In this case, the eg set will be active in
Raman spectroscopy. For metal-carbonyl complexes with substituted ligands, the IR absorptions become more numerous due to the lower symmetry of the each complex. A schematic of this is illustrated in Figure \(2\).. Although IR characterization is a powerful tool to confirm the properties of metal-carbonyl complexes, X-Ray crystallography is another useful method of confirming the existence of pi backbonding. X-ray crystallography can be utilized to confirm the presence of a backbonded metal carbon complex. If a carbon is backbonded to a metal complex then the covalent bond between the two atoms will be stronger and this should indicate a smaller bond distance. X-ray crystallography is capable of determining this bond distance by sending X-rays through the material and measuring the diffraction pattern that is produced to measure the bond distance within a crystal structure.
Fluxional Carbonyls
Fluxional carbonyls arise from the ability of carbonyl groups within a transition metal complex to move from one metal to another via bridging. This is easily accomplished due to the fact that the ground state energy for a terminal carbonyl group is similar to that of a bridging carbonyl, allowing carbonyls to interchange between metal centers. This typically occurs within first row transition metal complexes and the presence of fluxional complexes can be confirmed in disubstituted metal carbonyl complexes by way of cis-trans isomers. NMR spectroscopy can then be used to determine the conditions upon which all isomers can be converted to either a cis or trans configuration based on the peaks in the 13C NMR spectrum of the molecule. If the trans configuration produces chemically distinct environments for two carbonyls then this will be demonstrated in the NMR spectrum, and the reverse case is true for the cis configuration in which carbons in chemically equivalent environments will express only one peak.
Phosphine Ligands
Phosphine Backbonding
PR3 ligands are phosphine ligands that behave quite similarly to carbon monoxide ligands when interacting with a transition metal center. Phosphine ligands are also neutral and also contribu te to 2 electrons to the transition metal center. Similar to CO ligands, PR3 ligands first donate a sigma bond to the transition metal center, and then the d orbitals of the metal pi backbond with the phosphine and donate 2 electrons to the sigma* antibonding orbital. A schematic of this is shown in Figure 4. Although a phosphine ligand has the potential to have large steric groups that impede the formation of transition metal complexes or decrease the bond enthalpy of these coordination bonds, the charged nature of some groups attached to the phosphorus can have significant impacts on the stability of the structure. The electronegativity of the groups attached to the phosphine play a huge role in the stability of the transition metal complex. Strongly electronegative groups on the phosphorus atom can drastically lower the energy of the of the sigma* orbital, thereby making the phosphine a significantly better pi-acceptor (more accessible energy level for donation by the metal). In descending sigma* antibonding orbital energy, the following is obtained:
P-C > P-N > P-O > P-F
Phosphine Bonding and IR Absorption
While electronegativity of the groups attached to the phosphorus are important in the stability of the transition metal complex, several other factors can influence this stability. Namely, steric hinderance can play enormous role in the ability of a phosphine group to coordinate with a transition metal center. The following is a comparison for several phosphine complexes in descending order from the ligand with the most pi-acceptor character.
PF3~CO > PCL3 > P(OMe)3 > PPh3 > Py
This demonstrates that the more electronegative phosphine groups have higher stability as a result of having greater pi-acceptor character (reduction in energy of sigma* antibonding orbital). They are also smaller allowing them to interact more closely with the metal center as opposed to phosphine with phenyl groups attached. Additionally, these groups, as a consequence of their higher stability, will absorb higher frequencies of IR radiation.
Cone Angle and Percent Buried Volume
In order to quantify the affects of steric hinderance, cone angle is often used as a metric to define how readily a phosphine ligand or any ligand will dissociate from a transition metal center. For example, a phosphine ligand such as triphenylphosphine will most likely have a larger cone angle than PF3 and will most likely dissociate much more easily than PF3 due to the large size of the three phenyl groups attached to the phosphorus. The cone angle is a rough measure of this property but can be highly inaccurate. This is because a smaller ligand can coordinate more closely to a metal center, thereby exaggerating the size of its relative cone angle. The opposite case is true in that a larger ligand will coordinate further from the metal center, which could underestimate the relative cone angle of this molecule. Therefore, it can be more useful to instead utilize a parameter known as percent buried volume, which is defined as the percent volume occupied by a ligand within a sphere. This is an arbitrary sphere with a defined radius and contains the transition metal at its center. This method is a more useful characterization tool because it does away with the cone angle and considers the spatial volume occupied by the ligand in all dimensions. Inconsistencies between cone angle predictions of dissociation and percent buried volume predictions of dissociation have been documented.
Reference
[1]Organometallic HyperTextBook: Ligand Cone Angles, www.ilpi.com/organomet/coneangle.html.
[2]Symmetry, www.chemedx.org/JCEDLib/QBank/collection/ConcepTests/inorsym.html.
[3]Libretexts. “Carbon Monoxide and Backbonding.” Chemistry LibreTexts, Libretexts, 5 June 2019, chem.libretexts.org/Courses/Johns_Hopkins_University/030.356_Advanced_Inorganic_Laboratory/Lab_BCD:_Four_Coordinate_Nickel_Complexes:_Ligand_Effects_and_Organometallic_Catalysis/3_Carbon_Monoxide.
[4]“X-Ray Diffraction.” Rigaku, www.rigaku.com/en/techniques/xrd.
1. Organometallic HyperTextBook: Ligand Cone Angles, www.ilpi.com/organomet/coneangle.html.
2. Symmetry, www.chemedx.org/JCEDLib/QBank/collection/ConcepTests/inorsym.html.
3. ↑ Libretexts. “Carbon Monoxide and Backbonding.” Chemistry LibreTexts, Libretexts, 5 June 2019, chem.libretexts.org/Courses/Johns_Hopkins_University/030.356_Advanced_Inorganic_Laboratory/Lab_BCD:_Four_Coordinate_Nickel_Complexes:_Ligand_Effects_and_Organometallic_Catalysis/3_Carbon_Monoxide.
4. ↑ “X-Ray Diffraction.” Rigaku, www.rigaku.com/en/techniques/xrd. | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Advanced_Inorganic_Chemistry_(Wikibook)/1.20%3A_Dative_ligands_-_CO_and_phosphines.txt |
Pi Ligands
Pi ligands are a class of organometallic ligand with extended π systems that include linear molecules including ethylene, and allyl, and cyclic molecules such as cyclopentadienyl. As a dative L-type ligand, these molecules have a direct affect on the reactivity of the organometallic complex.
Linear Pi Systems
Properties
Linear π systems include alkenes, alkynes, and other unsaturated compounds containing π bonds. The ligand donates electron density from its π bonding orbitals to the d-σ orbitals on the metal center in a σ fashion; therefore the ligands HOMO is interacting with the metals LUMO. There is also back donation from the metal centers d orbital to the ligands π* orbitals in a π fashion, and in this case the metals HOMO is interacting with the ligands LUMO. With these types of interactions the ligand is a σ donor and π acceptor. In this type of bonding, the C-C bond strength within the ligand is weakened and the bond is lengthened in comparison to its free form. This can be attributed to the donation of electron density removing π-bonding electron density within the ligand, and the donation of electrons to the π* orbitals in the ligand. [1][2]
Synthesis
These types of complexes are typically synthesized through ligand substitution in which an existing ligand is replaced by the π ligand. Ligand substitution can be either associative or dissociative depending on which ligand moves first. In the associative mechanism, the new ligand binds to the metal, followed by the departure of another. In the dissociative mechanism, a ligand is a removed before the incoming ligand can bind to the metal center. [3]
Reactivity
• Nucleophilic addition- Addition of a nucleophile in the trans position with respect to the metal that is also bound to the π ligand.
• Migratory insertion- A nucleophile that is already bound to the metal ion and the π ligand combine to form one ligand. In this case the nucleophile and the metal are cis across the π system.
Cyclic Pi systems
Properties
Cyclic or arene π systems are either actor or spectator ligands that typically bind to metals through more than 2 atoms. The bonding is similar to that of the linear π systems, consisting of a more typical bond formed by the donation of electron density from the π orbitals of the ligand to the dσ of the metal, and back donation from the dπ of the metal to the π* of the ligand. However, donation from the ligand to the metal is much more common since arenes are highly conjugates making them stronger electron donors. De-aromatization occurs in some cases in order to form a more stable structure. This is called ring-slippage or the removal of one π bond from the system, which leaves the atoms bond to the metal coplanar and the remaining atoms are out of plane. In some cases this forms a stable structure, while in others is used as a means to open a coordination site for further reactivity. Typically these ligands are hydrocarbons, and rarely are heterocycles which contain a lone pair of electrons that can react on their own. [1]
Synthesis
One way in which cyclic π ligand-metal complexes are formed is through salt metathesis. This is a type of double replacement reaction where a ligand attached to the metal is exchanged for the π ligand, resulting in the desired organometallic complex and a salt. For example the generic reaction,\({\displaystyle {\ce {MCl2 + 2NaCp-> MCp2 + 2NaCl}}}\), where the chloride ligands are replaced with the more favorable cyclopentadienyl ligand. Another method is applying heat to trigger a retro-Diels-Alder reaction in order to form an ionic form of the ligand that can more easily bond with a metal. Aromatic ligands are chelating ligands, therefore they can easily replace other ligands in a metal complex that are weaker electro donors such as CO. This method of synthesis is entropically favorable, and therefore just requires the application of heat to move forward. [1]
Reactivity
• Coordinating to a metal increases the ligands electrophilicity, therefore increasing its ability to under go nucleophillic addition.
• Electrophilic aromatic addition is also possible as the metal can stabilize both cation and anions.
• Steric hindrance is increased in this type of complex, allowing for a greater selectivity in reactions.
• Bonding to the metal decreases the electron density of the ligand, leaving it vulnerable to nucleophillic aromatic substitution. The new ligand must be as good or better at coordinating the metal and an oxidant can be utilized to release the aromatic ligand. Oxidants decrease the complexes ability to back bond, making the arene less enthalpically favorable.
Sandwich Complexes
Sandwich complexes are organometallic complexes where the metal is bound to two cyclic π systems forming a "sandwich". Typically these complexes follow the 18 electron rule, except for 1st row transition metals that can have electron counts from 15-20 electrons, and lanthanides and actinides that do not follow the rule.
Metallocenes
Metallocenes are a subgroup of sandwich complexes that consist of a metal bonded to two cyclopentadienyl (Cp) ligands. Common configurations include η1-, η3- and η5- bonding modes. If the electron count is higher than 18 electrons there is occupation in antibonding orbitals, increasing the distance between the ligand and the metal and thus decreasing the amount of energy needed to dissociate. The Cp ligands can be eclipsed or staggered as shown in the Figure. Paramagnetic metallocenes can form ions, allowing the complex to form ionic bonds, replace the Cp ligand, or to add to the complex. [2]
Ferrocene
Ferrocene is the most widely studied metallocene with in the field of organometallics. Looking to the molecular orbital diagram, orbitals that are occupied by electrons are stabilized by the interactions with iron, and are typically that of s and p type orbitals. The orbitals of the most interest are that of the HOMO (dz2) and the LUMO (dxz, dyz). In the case of the HOMO the orbital has mostly metal characteristics as it is cone-shaped and has very little overlap with the orbitals of the ligand, making it almost non-bonding. The LUMO however has a large overlap of the d orbital on the metal and p orbitals of the carbons on the Cp ligands, allowing for π bonding. [2]
Following the 18 electron rule also makes for a more stable compound than that of other metallocenes or sandwich compounds. Comparing to sandwich structures containing benzene in the place of Cp, ferrocene performs electrophillic aromatic substitution at a much faster rate. Benzene being much more reactive than Cp also makes it more vulnerable to elimination. Metallocenes that have electron counts greater than 18 are also more vulnerable to elimination in order to achieve the desired electron count. | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Advanced_Inorganic_Chemistry_(Wikibook)/1.21%3A_Pi-Ligands.txt |
Metal-alkyl complexes were initially thought to be unstable because of the weak metal-carbon bonds, making them difficult to synthesize. In fact, the problem pertains to their kinetic stability. Metal alkyls generally undergo decomposition pathways of low activation barriers. In regards to simple alkyls, they are sigma donors in which they can donate 1 or 2 electrons to the metal [1].
Properties
The metal carbon bond (M-C) bond is composed of a positive, charged metal and negative, charged carbon. As shown in Figure 1, as the electronegativity of metals increase, its reactivity decreases in contrast. To elaborate, the alkyl ligand's reactivity is inversely associated with the metal center's electronegativity.
In regards to the carbon's hybridization, sp-hybridized ligands are the least nucleophilic. The Figure to the right elaborates on the nucleophilic order of the ligands.
Synthesis
1. Nucleophilic attacks with frequently used reagents such as R-Li or Pb-(R)4 is common for alkyl complexes (example displayed in Figure $1$). These nucleophiles range from strong to weak, as stronger nucleophiles can sometimes yield an unwanted reduction. The hard-soft principles of Pearson elaborates and explains the interactions of alkylation. Metals with strong leaving groups are more likely to undergo a nucleophilic attack, as the alkyl ligand transfers from a metal to another metal.
$\mathrm{NbCl}_{5} \stackrel{\mathrm{ZnMe}_{2}}{\longrightarrow} \mathrm{NbMe}_{2} \mathrm{Cl}_{3}+\mathrm{ZnCl}_{2} \nonumber$
Figure $2$. Nucleophilic attack
2. Electrophilic attack can also occur when the anionic metal complex is strong enough of a nucleophile to attack an alkyl and acyl halides in an electrophilic attack (Figure $3$). For this occurrence, the metal needs to have a readily available lone pair and open coordination site. The negative charge is shifted to the electrophile's leaving group while the overall charge of the complex increased by 1 [1].
$\mathrm{NbCl}_{5} \stackrel{\mathrm{ZnMe}_{2}}{\longrightarrow} \mathrm{NbMe}_{2} \mathrm{Cl}_{3}+\mathrm{ZnCl}_{2} \nonumber$
Figure $3$. Electrophilic attack
3. Oxidative addition is another method of synthesis, and there are multiple mechanisms that exist (Figure $4$). A representative example of oxidative addition is M + X-Y -> M-X-Y, where X-Y is cleaved and there is a new formation of ligands (-X and -Y) on the metal. The oxidation state of the metal is then increased and the total electron of the complex is increased by 2.
$\mathrm{Cr}^{2+}(\mathrm{aq}) \stackrel{\mathrm{Mel}}{\longrightarrow} \mathrm{CrMe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5}{ }^{2+}+\mathrm{CrI}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5}^{2+} \nonumber$
Figure $4$. Oxidative Addition
4. Migratory insertion is another method to synthesize alkyl complexes, resulting in an addition across pi bonds, as shown in Figure $5$. Certain alkyl complexes can be synthesized this way. For instance, the perfluoroalkyl complex is stable and so, fluoroalkenes insertion is generally favorable compared to elimination (take note of the Figure on the right).
Eliminations
Most metal alkyls generally are vulnerable to decomposition pathways with low activation barriers. The most common pathway is beta-hydride elimination followed by reductive elimination being the second, most common pathway.
Decomposing metal-alkyls:
Beta-hydride elimination
Beta-hydride elimination, as shown in Figure $6$, is the transfer of a hydrogen atom from the β ligand to the metal center of the complex. To prevent it from occurring, alkyls must not contain β-hydrogens, an orientation where the β positioned hydrogen cannot access the metal, and yield an unstable alkene. To make β elimination possible [3]:
• The β-carbon must contain a hydrogen.
• The metal-carbon and carbon-hydrogen bond must be in a syn-coplanar orientation.
• There must exist an open coordination site.
• The metal must have 16 e-'s or less and is at least d2.
Reductive Elimination
Reductive elimination, as shown in Figure $7$, involves eliminating a molecule from a transition metal complex where the metal is reduced via 2 e-'s. Elimination occurs in the cis orientation with groups being eliminated. An unstable oxidative state can encourage the elimination to proceed [1]. In details, the alkyl ligand interacts with a second type of ligand on the metal, in which the metal is consequently reduced by 2 units, a total electron count reduction by 2. Reductive elimination is generally favorable thermodynamically when X = H. However, reductive elimination is not favorable when X = halogen.
Stable Alkyl Complexes
To prevent beta-hydride elimination from occurring and to stabilize desired species, one can avoid this by increasing to 18 e-'s. A transition metal-alkyl complex with 18 e-'s dictates that all the orbitals are in their full capacity, and there is no interaction that can occur with the beta-hydrogen and transition metal center. Another way is to utilize stable alkyl ligands that do not contain beta-hydrogens and cannot undergo elimination[2]. Such alkyls without beta-hydrogens are WMe6, Ti(CH2Ph)4, and C2F5Mn(CO)5 [5]. The orientation of ligands can also be used to block beta-hydrogen elimination by orienting the alkyl in such a way that prevents the metal center from being accessed by the beta-hydrogen due to steric hindrance (Figure $8$). Such examples of bulky alkyls in which the beta-hydrogen cannot access the metal due to the orientation or bulkiness of the ligand are PdPh2L2, Cr(CMe3)4, and Cr(CHMe3)4 [5]. A third method is to use alkyls that would yield an unstable alkene as the product.
There is importance to metal-alkyl studies, as it is significant in the organometallic and catalytic systems and context. Before metal alkyls were fully studied and analyzed, there was speculation between thermodynamics and kinetic instability of metal-C sigma bonds, determining that it was indeed kinetics instability that made it difficult for organometallic chemists to isolate metal alkyls[1]. Regardless, there are various, significant applications to metal alkyl, in which can be elaborated upon in Applications of Metal Carbonyl Anions in the Synthesis of Unusual Organometallic Compounds and used in Strem Chemicals company for chemical vapor deposition, atomic layer deposition, etc (defined in their website). In essence, it is useful to make comparisons between various metal alkyl complexes and draw predictions pertaining to their characteristics and behavior. | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Advanced_Inorganic_Chemistry_(Wikibook)/1.22%3A_Metal-Alkyl_Complexes.txt |
Dissociative substitution mechanism describes one of the common pathways through which a ligand substitution reaction takes place. Found often in octahedral complexes, dissociative mechanisms are distinguished by having an ion X- dissociate from a metal complex, resulting in an intermediate compound with a lower coordination number.[1] This lost ion is then replaced by an ion Y-, as opposed to associative complexes which form a complex of X-, Y-, and the metal before the metal to X- ion bond is broken. This process is analogous to an SN1 reaction. In a dissociative mechanism, the complex is usually has fully saturated coordination, with 18 or more electrons. While most octahedral dissociate, not all do, with key examples being [Cr(NH3)5H20]3+ which follows Id substitution and [Cr(H2O)6]3+ following Ia.[1]
Mechanism
The mechanism is below as follows with Y- substituting in for X- on a general octahedral complex.
$\mathrm{L}_{5} \mathrm{M}-\mathrm{X} \stackrel{-}{\rightleftharpoons}_{+\mathrm{X}, k_{-1}}^{-\mathrm{X}, k_{1}} \mathrm{L}_{5} \mathrm{M}-\square \stackrel{+\mathrm{Y}, k_{2}}{\longrightarrow} \mathrm{L}_{5} \mathrm{M}-\mathrm{Y}\nonumber$
Usually the rate determining step (RDS) of the mechanism is the dissociation of X from the complex, which is dependent upon the strength of the metal to X- bond as well as other factors such as steric hinderance of the species which helps to speed the mechanism along as crowding favors dissociation. [Y] does not affect the rate of reaction, leading to the simple rate equation:
$\text { Rate }=k_{1}\left[\mathrm{L}_{5} \mathrm{M}-\mathrm{X}\right]\nonumber$
When the metal to X- is not the rate determining step, the rate of reaction is what is seen below:[1]
$\text { Rate }=\frac{d\left[\mathrm{L}_{5} \mathrm{M}-\mathrm{Y}\right]}{d t}=k_{2}\left[\mathrm{L}_{5} \mathrm{M}\right]\nonumber$
Lability
A metal complex is either labile or inert depending on how easily the reaction proceeds. A labile compound undergoes reactions with a relatively high rate of substitution. The opposite to labile is inert, a term describing metal complexes whose reactions are slow.[2] There are three main factors that affect the whether a complex is labile or inert.[2]
1. Size: Smaller metal ions tend to be more inert as ligands are held more tightly.[2]
2. Charge on Metal: The greater the charge on a metal ion in a complex, the greater the tendency towards the complex being inert.
3. d Electron Configuration: in octahedral geometries, d electrons have t2g and eg orbitals meaning the 5 d orbitals are not at the same energy level. The number of d electrons can predict if the metal complex behaves as inert or labile as according to the table below. Each electronic configuration is either labile or inert as a result of partial of full occupancy of the corresponding orbitals. [1]
Number of D-Electrons and Configuration Reactivity Notes
d1 Labile N/A
d2 Labile N/A
d3 Inert N/A
d4 Low Spin Inert N/A
d4 High Spin Labile Especially labile as it is structurally distorted by the Jahn-Teller effect.
d5 Low Spin Inert N/A
d5 High Spin Labile N/A
d6 Low Spin Inert N/A
d6 High Spin Labile N/A
d7 High Spin Labile N/A
d8 Square Planar Inert For d8 and above low spin is the same as high spin.
d8 Intermediate This configuration is intermediate, especially with weak field ligands.
d9 Labile Like d4 H.S. this configuration is especially labile as it is distorted by Jahn-Teller effect.
d10 Labile N/A
Traits of Dissociative Mechanisms
• The rate of substitution varies little with the incoming Y-.[1]
• The rate of substitution varies over five orders of magnitude depending on the nature of the leaving group. The weaker the bond between the metal and the leaving X-, the faster the reaction runs, as the loss of X- occurs in the rate determining step.[1]
• Steric crowding around the metal has the ability to increase the rate of substitution because steric crowding favors dissociation. This runs completely contrary to rate of substitution for associative mechanisms as steric crowding is harder to penetrate.[1]
• Dissociative substitutions have a positive entropy (ΔS±) and usually have a positive volume of activation (ΔV±).[1]
The Shift Mechanism
Another way in which a substitution reaction can take place is through use of the shift mechanism, which is most notably present in ring slippage. The shift mechanism consists of the center metal putting some charge back on a ligand either through breaking a double bond and having the ligand hold the charge or through changing the hapacity of the bond.[3] This lowers the overall coordination number on the metal, allowing it to associate ligands for substitution.[4] This phenomena is most commonly found in associative coordinations rather than dissociative mechanism as it lowers the coordination number to one that can accommodate incoming ligands, An example of this is when a η5-Cp ring undergoes a ring slippage, becoming a η3-Cp bond, thus allowing an incoming ligand while maintaining the 18-electron rule.[5] | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Advanced_Inorganic_Chemistry_(Wikibook)/1.23%3A_Dissociative_Mechanism.txt |
Associative substitution mechanism follows a common pathway for ligand substitution in a metal complex. Resembling an SN2 reaction, the intermediate is formed when the incoming ligand momentarily becomes part of the complex; increasing the coordination of the metal [1]. The leaving ligand detaches from the metal complex leaving the complex with a new ligand which was the incoming ligand. Complexes that are coordinatively unsaturated (metals that have less the 18 electrons) or have ligands that can change its bonding to the metal (metals that have 18 electrons) are prime candidates to follow an associative substitution reaction.
In a regular square planar the mechanism follows X becoming part of the complex momentarily and Y as the leaving group.
$ \nonumber$
The rate determining step of the is the formation of the fully coordinated complex
If the solvent is nucleophilic enough then the solvent is part of the associative mechanism. As shown in Figure $2$. when the solvent is part of the mechanism the solvent becomes part of the complex then the leaving ligand leaves and the solvent is substituted by the incoming ligand [3].
The rate of an associative reaction depends if the solvent is involved in the reaction. If the rate of the reaction that only involves the incoming ligand is comparable to the reaction involving the solvent therefore:
kobs= k1[solvent] + k2[Y]
However, if the solvent isn't nucleophilic enough then only the incoming ligand is involved in the overall reaction such as the top reaction in Figure 2.
kobs= k[Y]
Spectator Ligand Influence
The effect of the ligands trans to the leaving ligand from the complex plays an important role in the rate of reaction. This could come about from the stabilization of the transition state (TS) or the destabilization of the ground state (GS).
Trans Effect
Lowers the energy of the TS, starting material needing less energy to go through with the reaction. This is a kinetic effect. This involved back bonding. The more bonding between the pi start orbital with the d orbital, makes the metal have less electron density because density from the d orbital is being removed. The metal becomes more electrophilic which makes it facile for trigonal planar to form overall decreasing the transition state's energy.
Trans Influence
Raises the energy of the ground state which in fact decreases the energy the starting material has to overcome to start the reaction. This is a thermodynamic effect.
Several ligands such as:
R- ~ H- > PR3 > CO ~ C=C ~ Cl- ~NH3
in order of decreasing order of being strong sigma donors have a higher trans influence respectively. The stronger the sigma bond integration of the trans ligand sigma orbital with the metal d orbital, the less contribution of the d orbital available for the leaving group sigma orbital. This leads in an increase bond length from the rest of the ligands making the complex less stable and higher in energy. This makes the position of the ground state closer in energy to the transition state.[1]
Indenyl Effect
Also known as ring slippage. Associative substitutions aren't usually observed in 18 electron compounds. This is due to the fact that the intermediate would be a compound that has more than n18 electrons. This would need more energy and yield an unstable compound. However, it is possible for an 18 electron compound to go through an associative mechanism. As mentioned before the complex needs to have ligands that can change its bonding to the metal. It is usually seen in cyclical compounds because the matal can change its coordination number with the ligand and avoid making a complex that excess the 18 electrons [1][2][3][4].
Reference
1. Pfenning, B.W (2015) Principles of Inorganic Chemistry.Hoboken, New Jersey. John Wiley & Sons Inc.
2. Miessler, G. L. (2014) Inorganic Chemistry. Pearson.
3. https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_(Inorganic_Chemistry)/Organometallic_Chemistry/Fundamentals_of_Organometallic_Chemistry/Associative_Ligand_Substitution
4. Transition metal indenyl complex | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Advanced_Inorganic_Chemistry_(Wikibook)/1.24%3A_Associative_Mechanism.txt |
Electron transfer reaction
Electron transfer reaction is a reaction in which a single electron is transferred from one molecule to another [1]. For example, a reaction that occurs when steel wool (made of iron atoms) is placed in a solution of CuSO4 is given in Figure \(1\) below:
This demonstration shows that the deep blue color of the CuSO4 solution (left picture), which is caused by the Cu2+ ions, becomes light green due to Fe2+ ions (right picture). As a result, a brown solid made of metallic copper forms and the steel wool disintegrates as the Fe atoms disappear. The reaction has thus led Fe to be converted to Fe2+ while the Cu2+ is converted to Cu. This indicates that the oxidation state of copper has changed from +2 in the Cu2+ ions in solution to 0 in the atoms comprising metallic copper. This is accomplished as each Fe atom gives up two electrons, while each Cu2+ gains two electrons, as a result of which two electrons are transferred from Fe atoms to Cu2+ ions in solution. This is an example of an electron transfer reaction. The reaction is given by: Cu2+ + Fe → Cu + Fe2+.
Mechanism of Electron Transfer Reactions
The process of electron transfer from one species to another species leads to the oxidation of the donor and reduction of the acceptor. The mechanism by which the electron transfer occurs between inorganic complexes can be classified in to two types: inner sphere electron transfer mechanism and outer sphere electron transfer mechanism.
Inner Sphere
Inner sphere electron transfer occurs between complexes via a bridging ligand. At least one of the complexes needs to be labile to allow the bridge to form. Bonds are broken and formed.
Outer Sphere
Outer sphere electron transfer occurs between two species that do not undergo substitution and do not involve the incursion of significant covalent bond formation. It occurs when none of the ligands can function as a bridge. It is faster than inner sphere because the energetic demands are less. No new bonds are broken or formed. Interaction between the two coordination spheres exist but is not as pronounced as for the bridge complex in the inner sphere. Outer Adduct is held together by one of the following: Electrostatic interactions, Vander Waals forces, or Hydrogen bonding.
Marcus Theory
In 1956, Rudolph A. Marcus developed the Marcus theory to explain the rates of electron transfer reactions. R.A. Marcus received the Nobel Prize in Chemistry in 1992 for this theory [3]. The rate of transfer refers movement of an electron from the electron donor to the electron acceptor. The theory was formulated to address outer sphere electron transfer reactions, in which the two chemical species (electron donor and acceptor) only change in their charge with an electron jumping but do not undergo large structural changes. Later, the theory included inner sphere electron transfer contributions, in which a change of distances or geometry in the solvation or coordination shells of the two chemical species is taken into account.
When electron transfer reactions involve no making or breaking of bonds, the Marcus theory limits the reaction partners to be weakly coupled and retain their individuality. The reorganization of the surroundings is thermally induced whereby the outer sphere (solvent) and the inner sphere (solvent sheath or ligands) create the geometrically favorable situation prior to and independent of the electron jump. The solvent is very important aspect of the theory and leads the way to the calculation of the Gibbs free energy of activation, using the polarization properties of the solvent, the size of the reactants, the transfer distance and the Gibbs free energy ΔG0 of the redox reaction. The most surprising result of Marcus' theory was the "inverted region": whereas the reaction rates usually become higher with increasing exergonicity of the reaction, electron transfer should, according to Marcus theory, become slower in the very negative ΔG0 domain.
If a redox reaction is assumed, one partner acts as an electron donor D the other as an acceptor A. As these are neutral molecules electrostatic forces may be ignored. In addition, either D or A may be considered to be in a photoexcited state (photoinduced electron transfer, PET). Other than a change in the starting stage energies, the principles of Marcus’ model apply equally well to both ground and excited state electron transfer. For a reaction to take place D and A must diffuse together. They form the precursor complex, usually a kinetic, unstable, solvated encounter complex, which by electron transfer is transformed to the successor complex, and finally this separates by diffusion. For a one electron transfer the reaction is:
k12 k23 k30 D+A<=> [D…A] <=> [D+…A-] D+ + A- k21 k32
Where, k12, k21 and k30 are diffusion constants, k23 and k32 rate constants of activated reactions.
Since redox reactions are preferably run in polar solvents, D and A have a solvent shell and the precursor and successor complexes are solvated also. The closest molecules of the solvent shell, or the ligands in complexes, are tightly bound and constitute the "inner sphere". Reactions in which these participate are called inner sphere redox reactions. The free solvent molecules constitute the "outer sphere". Outer sphere redox reactions do not change the inner sphere, no bonds are made nor broken. Only an electron transfer takes place.
A quite simple example is the Fe2+/Fe3+ redox reaction, the self exchange reaction which is known to be always occurring in an aqueous solution containing both FeSO4 and Fe2(SO4)3 (of course, with equal and measurable rates in both directions and with Gibbs free reaction energy ΔG0 = 0). For outer sphere redox reactions there cannot be such a reaction path, but nevertheless one does observe an activation energy.
Since the electron as an elementary particle cannot be divided, it resides either on the donor or the acceptor and arranges the solvent molecules accordingly in an equilibrium. The "transition state", on the other hand, requires a solvent configuration which would result from the transfer of half an electron, which is impossible. This means that real charge distribution and required solvent polarization are not in an "equilibrium". Yet it is possible that the solvent takes a configuration corresponding to the "transition state", even if the electron sits on the donor or acceptor. This, however, requires energy. This energy may be provided by the thermal energy of the solvent and thermal fluctuations can produce the correct polarization state.
Marcus developed a classical theory with the aim of calculating the polarization energy of the said non-equilibrium state. From thermodynamics it is well known that the energy of such a state can be determined if a reversible path to that state is found. Marcus was successful in finding such a path via two reversible charging steps for the preparation of the "transition state" from the precursor complex. Marcus states that four elements are essential for the model on which the theory is based: 1. A classical, purely electrostatic model must be employed. The charge may be transferred in any portion from one body to another. 2. The fast electron polarization Pe and the slow atom and orientation polarization Pu of the solvent are to be separated on grounds of their time constants differing several orders of magnitude. 3. He further separates the inner sphere (reactant + tightly bound solvent molecules, in complexes + ligands) and the outer sphere (free solvent) 4. In this model Marcus confines himself to calculating the outer sphere energy of the non-equilibrium polarization of the "transition state". The outer sphere energy is often much larger than the inner sphere contribution because of the far reaching electrostatic forces.
In this classical model the transfer of any arbitrary amount of charge Δe is possible. So, the energy of the non-equilibrium state, and consequently of the polarization energy of the solvent, can be probed as a function of Δe. Thus, Marcus has lumped together the coordinates of all solvent molecules into a single coordinate of solvent polarization Δp, which is determined by the amount of transferred charge Δe. So, he reached a simplification of the energy representation to only two dimensions: G = f(Δe). The result for two conducting spheres in a solvent is the formula of Marcus
Where r1 and r2 are the radii of the spheres and R is their separation, εs and εopt are the static and high frequency (optical) dielectric constants of the solvent, Δe the amount of charge transferred. The graph of G vs. Δe is a parabola (Figure \(2\)). In Marcus theory the energy belonging to the transfer of a unit charge (Δe = 1) is called the (outer sphere) reorganization energy λo, i.e. the energy of a state where the polarization would correspond to the transfer of a unit amount of charge, but the real charge distribution is that before the transfer. In terms of exchange direction the system is symmetric.
Marcus theory is used to describe a number of important processes in chemistry and biology, including photosynthesis, corrosion, certain types of chemiluminescence, charge separation in some types of solar cells and more. Besides the inner and outer sphere applications, Marcus theory has been extended to address heterogeneous electron transfer. | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Advanced_Inorganic_Chemistry_(Wikibook)/1.25%3A_Electron_Transfer_Reactions.txt |
Oxidative addition and Reductive elimination are reaction pairs that involve a change in the oxidation state and coordination number of the metal. [1] Oxidative addition is the increase in the oxidation state and coordination number of the metal. Meanwhile, reductive elimination is the decreases in the oxidation state and coordination number of the metal. These two reactions occur through sigma bond and can be described schematically by the following:
Oxidative Addition
Oxidative addition is the reaction where the oxidation state of the metal center increases by two. This reaction cannot occur if the metal center of the complex doesn't have accessible two units higher than the initial oxidation state. Metal center in a complex is known to act as Lewis acid. However, it both behaves as a Lewis acid and a Lewis base under oxidative addition. The metal center of the complexes is a Lewis acid when it takes electrons from the sigma bond (σ) and it's a Lewis base when it donates electrons from the sigma antibonding (σ*).
3 Main Oxidative Addition Mechanisms
Concerted Oxidative Addition Mechanism
• Non-polarized substrates, such as H and C-H and Si-H bond, undergo concerted oxidative addition.
• Under concerted oxidative addition mechanism, ligands end up in cis position although the more stable product is in trans positions due to isomerization.
• Reaction of the Vaska’s comple,trans-IrCl(CO)[P(C6H5)3]2, with dihydrogen is one of the example of the concerted oxidative addition mechanism. In this example, the dihydrogen coordinates with the Iridium. Afterwards, the two hydrogens ends up in cis position with each other and the CO in the complex was pushed towards the trans position with the hydride because of the trans effect.
Non-concerted Oxidative Addition Mechanism
• Non-concerted oxidative addition mechanism is like nucleophilic displacement (SN2) reaction.
• Polarized substrates, such as methyl, allyl, and benzyl halides, undergo non-concerted oxidative addition mechanism.
• Another way to identify that a reaction undergo non-concerted mechanism is by identifying the substrate if it’s optically active.
Radical Mechanism
• Alkyl halides can react with the metal center of a complex through radical reaction.
• Other byproducts can form through the radical reaction.
• Radical reactions are sensitive to dioxygen due to it’s a paramagnetic property.
• There are two types of radical mechanism: Non-chain and Chain radical mechanism.
Reductive Elimination
Reductive Elimination is the opposite of the oxidative addition where the oxidation state of the metal center of the complex decrease by two units. Unlike oxidative addition, reduction elimination only has one mechanism which is the counter part of the concerted oxidative addition mechanism. Reductive elimination is an intramolecular reaction and favored by low electron density of the metal center. Ligands must be in a cis position in order to undergo reductive elimination.
Application
The biggest application of the oxidative addition and reductive elimination is the cross coupling reaction.[2] One of the most important reactions that allow the formation of a new bond (usually carbon) with the help of a metal catalyst.
Catalytic Cycle
Below shows a generic Pd-catalyzed cross-coupling cycle. Aside from Palladium, Nickel, Iron, Cobalt and Copper can also be work as a catalyst. In the given catalytic cycle below, the Pd(0) was generated from a palladium precatalyst. Afterwards it went through oxidative addition followed by transmetallation where metal (M) could be Sn, Zn, B, and Zr. Before reduction elimination could occur, isomerization must conducted first for the ligands to be in cis positions. | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Advanced_Inorganic_Chemistry_(Wikibook)/1.26%3A_Oxidative_Addition_Reductive_Elimination.txt |
C=M complex is an organometallic complex containing metal-carbon double bonds, in which a σ bond and a π bond present. There are two main carbene complex types: Fischer type(carbene) and Schrock type(alkylidene). Fischer type carbene complex, which was developed in 1964, contains a σ donor on carbon and a π-back donate on metal. Schrock-type carbene complexes, which was developed by Schrock several years later, contains more over sharing electrons and commonly known as alkylidenes. Yves Chauvin, Robert Grubbs, Ricahrd Schrock were awarded the 2005 Nobel prize in chemistry. "for the development of the metathesis method in organic synthesis."[1]
Structure
Fisher-Type Carbene Complexes
A carbon on Fisher type is electrophilic because σ donates from the metal to the carbon and has weak back-bonding. Carbon complexes on Fisher have low oxidation state with 18 electron count. For example, Fe(0), Mo(0), Cr(0) (middle to late transition metal) contain good π acceptors ligands in the complex. The stability of Fischer-Type carbene complex can also be enhanced by containing highly electronegative heteroatoms such as O, N, and S. These characteristics of Fischer-Type carbene complex allows to have double bond with Oxygen. The length between these two double bonds in complexes is longer than normal M=C double bond, and shorter than normal M-C single bond. Likewise, the C-X bond distance is shorter than normal C-X bond.
The highly electronegative heteroatoms tend to participate in the π-back bonding with d orbital on the metal, and p orbitals on the carbon. A lone pair is donated from a carbon to an empty d orbital on the metal and another lone pair is donated from π back donate metal to \({\displaystyle P_{z}}\) orbital on carbon. An electron pair corporate in σ donation and there is empty π* orbital which takes π back-donation in MO diagram. Bonding is most likely close to CO bond, and good electrons for back donation with good π acceptor. An example of Fischer-Type carbene complexes is the compound \({\displaystyle Cr(CO)_{5}[C(OCH_{3})C_{6}H_{5}]}\) with a Cr(0).[2]
These resonance form will existing on a temperature-dependent proton NMR (detects the cis and trans separately), and on a carbon NMR. however, π bonding system in complexes of this type will highly more existing on X-ray crystallography(shows double bond character).
Schrock-Type Carbene Complexes
A carbon on Schrock-Type is nucleophilic, because back-bonding metal is strong and contributes σ donation from the ligand to the metal. Carbon complexes on Schrock have high valence oxidation state with 10-18 electron count. For example, Ti(IV), Ta(V), W(VI) (early transition metal) containing good σ or π donor ligands in the complex. An electron help to form σ bond and π bond in MO diagram. It mostly like compose with H or alkyl, which are directly attached to the carbene. These complexes is composed with two covalent bond interactions. one electron donates to the σ bond from each metal and each carbene. An example of Schrock-Type carbene complexes is the compound \({\displaystyle Cp_{2}(Me)Ta=CH_{2}}\) with a Ta(V).[3]
These Schrock-Type carbene complexes have M-C-R linkage and make bond angle of 160-170°. Schrock Carbenes can be displayed by proton NMR, carbon NMR, Infrared Spectroscopy, and Raman Spectroscopy. This complex type displays coupling constant values for typical and agostic interaction between the carbene proton and the metal. These techniques are well used to determine bond angles and structures for Schrock-Type carbene complexes.
Synthesis
Fisher Carbene
Fisher Carbenes' carbon is an electrophilic. Nucleophilic attack at a carbonyl ligand, and most common method.
Alkyl lithium is attaching on metal carbonyl. Zwitterionic resonance is forming by attaching heteroatom to carbene to stabilize Fischer carbenes. Intermediate of Fisher Carbenes is treated as electrophilic to give the Fisher carbene. C-X bond rotation is restricted in syn and anti isomers for alkyl derivatives at low temperature proton NMR because zwitterion resonance forms in Fischer carbenes. These can be observed by X-ray crystallography.
Schrock Carbenes
Schrock Carbenes' carbon is a nucleophilic. There is α-abstraction in the Schrock carbenes synthesis and induced by steric bulk. Schrock carbene is the main key for the both reagents and catalysts. The most famous examples for Schrock Carbenes synthesis are Patasis' Reagent, Tebbe's Reagent, and Grubb's Catalyst. Schrock Carbenes synthesis is used to widespread reactivity such as intermediate of the preparation of organometallic. The most famous of Schrock Carbenes reactivity is Olefin Metathesis
Reactivity
Olefin Metathesis
Olefin metathesis is the main application for carbene. Olefin metathesis reaction utilizes two alkenes to make cyclobutanes and reform the two new types of double bond. A reaction of olefin metathesis works rapidly. d orbitals on the metal alkylidene is present, and it breaks cyclobutanes symmetry and reacts very quickly. Normally, the products of olefin metathesis are statistical, unless the reaction can be driven is some way or the tow alkenes have different reactivities.[4] Titanium, Tungsten, Molybdenum, and Ruthenium are the popular metal for olefin metathesis.
There are five different kinds of reactions process of Olefin Metathesis:
• Cross metathesis CM
• ring-closing metathesis RCM
• ring-opening metathesis ROM
• ring-opening metathesis polymerization ROMP
• acyclic diene metathesis polymerization ADMET | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Advanced_Inorganic_Chemistry_(Wikibook)/1.27%3A_CM_Complexes.txt |
Olefin Metathesis is an organic chemical reaction that uses a metal catalyst for the transfer of substituents between olefins, or alkenes by a 4-membered ring intermediate, also known as a Chauvin Mechanism. This efficient method does not only have a high yield, but it produces less byproducts and hazardous waste compared to other organic reactions. Yves Chauvin, Robert H. Grubbs and Richard R. Schrock won The Nobel Prize in Chemistry of 2005 for their contributions to olefin metathesis.
Historical Overview
Herbert S. Eleuterio, an industrial chemist in the 1950s, found a propylene-ethylene copolymer from a propylene feed on a Mo-Al catalyst as a result of his experiment. He repeated the experiment with a cyclopentene and noticed that “the polymer looked like somebody took a pair of scissors, opened up cyclopentene, and neatly sewed it up again.” Other chemists were also getting similar results of the cleavage and reformation of the olefin double bonds. To his understanding, Grubbs suggested that the rearrangement of substituents occurred through a metallacyclopentane intermediate. Chauvin proposed that olefin metathesis is initiated by a metal carbene. Many scientists agreed with Chauvin that metal carbenes played a major role in the process of olefin metathesis and from there, it was further studied to understand the full mechanism.
Mechanism
The rearrangement of substituents on two olefins occurs through the formation of a 4-membered ring intermediate as in the following Figure.
The initial alkene with substituent R1 forms an intermediate with an olefin metal carbene, allowing the metal element to be attached to the initial alkene. The product then reacts with a second alkene with substituent R2, forming another 4-membered ring intermediate and yields a final combined olefin with both R1 and R2 substituents while producing back the initial metal carbene. This type of 4-membered ring formation is called a [2+2] cycloaddition. A 4-membered ring is not the most stable and is known to cause strain on the molecule. As a result, the formation has a very high activation energy. Interaction with the metal lowers the activation energy of the ring formation, allowing the process to occur at moderate temperatures.
Catalyst
Olefin Metathesis requires the application of a metal catalysts. The most common catalysts used are Grubbs catalyst and Schrock catalyst.
Grubbs catalyst is a commercially available catalyst and is easy to handle as it is pretty stable against water, oxygen, and other small impurities. Although it has its advantages like high functional group tolerance, there is a drawback in that it has a lower reactivity compared to other catalysts like the Mo Imido Alkylidene Catalyst.
Upon interaction with an olefin, a 4-membered ring is formed cis to the carbene and trans to the Cl atoms.
Like the Grubbs catalyst, the Schrock catalyst is also commercially available. It has a higher reactivity than the Grubbs catalyst and is tolerant of multiple functional groups, but has many disadvantages such as needing to be handled under an inert atmosphere using dry solvents and substrates as well as being intolerant to protons on heteroatoms.
Basic Types of Metathesis
• Ring Opening Metathesis Polymerization (ROMP)
• Ring Closing Metathesis (RCM)
• Cross Metathesis (CM)
• Acyclic Diene Metathesis Polymerization (ADMET)
Applications
Olefin metathesis opened up new industrial pathways for petrochemicals, polymers, and so much more. In the petrochemical field, olefin metathesis is used for the Olefins Conversion Technology (OCT) Process and the Shell Higher Olefins Process (SHOP). ROMP has opened up an opportunity to generate useful polymers with special properties in many industrial methods. For future oleochemical advancements, metathesis of natural fats has opened a door of possibilities. Currently, the most important applications are used to make propene, detergent-range olefins, and polymers. | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Advanced_Inorganic_Chemistry_(Wikibook)/1.28%3A_Olefin_Metathesis.txt |
The mechanism of a reaction is a sequence of elementary steps proceeding from the starting materials through a series of intermediates and eventually to the products. Each step has an activation energy barrier. Each intermediate has some measure of stability. The energy changes along a reaction pathway can be tracked with a reaction progress diagram.
The speed of a reaction can be altered by flooding the system with reactant (via the Le Chatelier's Principle) or by adding more (thermal) energy to help overcome the barrier (via the Arrhenius equation), but the reaction follows this same energy pathway. However, this is not the case with a catalyst. A catalyst introduces a completely different pathway/mechanism that was inaccessible before.
Some reactions simply cannot proceed without a catalyst; the energy barrier is too high. This may be true even if the overall reaction is exergonic (\(\Delta{G}\)) and thus thermodynamically favored to proceed.
Consider the reaction progress diagram above. If reactants approach this barrier from the left, they encounter a very large energy barrier. They cannot react, despite the fact that it would be energetically favorable for them to convert to products. The molecules are stuck. The same is true if molecules approach from the right. They cannot overcome the high barrier, though the energy difference between the left and the right is small. The molecules are again stuck.
If the barrier were removed, the molecules could react freely in both directions, eventually coming to a dynamic equilibrium, as in the diagram below:
A catalyst does not fully remove the barrier, but it offers a new pathway with a lower barrier. The reaction is able to proceed back and forth.
The reaction progress diagram plots the changes in energy along one particular coordinate of interest to the molecule. It may track a chemical bond as it lengthens and breaks through the course of a reaction. However, there are always other energetic possibilities unseen in this diagram. This is a single cross-section of a potential energy surface. A potential energy surface is like a landscape, a mountain range in which elevation corresponds to energy. The reaction progress diagram depicts a single pathway from one energetic valley to another. In one of these valleys is the reactant; the product is in the other. The path from one valley to another leads uphill, over a mountain pass, and down into the other valley again.
Suppose a traveler living in a valley must visit a friend in another valley. Each day the traveler takes the same path to the friend's house. He takes the easiest route, over the lowest mountain pass.
One day, instead of visiting the friend on foot, the traveler takes the train. The train takes a completely different pathway than the usual. It does not traverse the same mountain pass; it may take another route that was inaccessible by foot, or it may simply tunnel through the mountain. When the train reaches its destination, it picks up more passengers, repeating the process over and over.
In molecular terms, it is the reactant that takes the train, a low-energy pathway, on its way to the product. The train is a catalyst, and it has several important features:
• A catalyst creates a reaction pathway that has a lower energy barrier than the uncatalyzed pathway.
• A catalyst returns repeatedly to take more molecules through the reaction.
That recycling of the catalyst is sometimes referred to as "turnover". The turnover number of a catalyst is the number of times the catalyst is able to return and carry out the reaction again (eventually something may occur to make the catalyst stop working). The reactant in a catalytic reaction is called the substrate, particularly in cases in which the catalyst is an enzyme of a transition metal catalyst. The speed at which the catalyst is able to carry out the reaction on new substrates is called the turnover frequency. These are important parameters in describing the efficiency of a catalyst.
Problem RK8.1.
Polymerization catalysts take small molecules called monomers and connect them together using the same reaction over and over again to make a long polymer chain. What was the minimum turnover number of the catalyst used to make each of the following polymers? (Mn = number average molecular weight, a statistical estimate of the average size of polymer chain.)
1. polylactide, Mn = 3,000, from lactide, C6H8O4.
2. polystyrene, Mn = 250 thousand, from styrene, C6H5C2H3.
3. high-modulus polyethylene (HMPE), Mn = 6 million, from ethene, C2H4.
Problem RK8.2.
Catalysis often begins with a binding step, followed by one or more subsequent steps needed to carry out the reaction. For the following catalyzed reactions, draw the binding step using curved arrows. | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Catalysis/Catalysis.txt |
Buchwald-Hartwig amination is a palladium-catalyzed cross-coupling reaction of amines and aryl halides that results in formation of C-N bonds. It was first introduced by Kosugi, Kameyama and Migita in 1983[1]. It was a reaction using 1 mol% PdCl2(P(o-Tolyl3)2 with the addition of aryl bromides and N,N-diethylamino-tributyltin in toluene solvent heated at 100°C for three hours. The resulting data showed that only nonsubstitued bromobenzene would give the product with a high yield.
In the following year, Pd(PPh3)4 catalyst mediated -carboline preparation was used to synthesize lavendamycin CDE ring system from 4-aryl pyridines, which was done by Bogen and Panek[2]. After a decade, Hartwig[3] identified and characterized several intermediates in the palladium-catalyzed C-N bond formation. The mechanistic data suggested that the reaction involved oxidation addition and reductive elimination steps.
In the same year, Buchwald[4] published a paper that discussed methods of improving original studies from Migita. One substrate was a volatile amine and the other was an amine with higher boiling point, and the reaction of these two substrates would result in the formation of aminostannes. The transamination was coupled with palladium catalyst to make the reactions available to a broader variety of arylamine substrates.
Mechanism
The catalysis circle is shown in Figure 1. First, Pd(Ⅱ) is reduced to Pd(0) by amines that contain α-H or ligand. Then, Pd kicks one ligand off and undergoes oxidative addition to form Pd(Ⅱ) complex. Next, amines attack Pd, substituting one X under the help of base. The final step is reductive elimination, giving the product and complete the circle. Note that reduction of Pd(Ⅱ) requires amines that contain α-H, otherwise extra ligands should be added to the reaction. An alternative choice is using Pd(0) complex instead of Pd(OAc)2.
Ligands
Buchwald proposed general ligand design strategy shown below. They changed the functional groups to get a library of ligands, for different substrates. Different ligands will be discussed in detail in Scope part. [5]
Electrophile
RX and ArX
Generally, Br, Cl and I can react with amines in certain conditions. ArI is relatively difficult, unlike other C-C coupling reactions. Mechanism studies showed that this results from the unreactive Pd dimers bridged by iodide anions.[5]
Scheme 1
Toluene is favored for this reaction because of poor solubility of Iodine salt in toluene.
ArOTf, ArONf and ArOMs
ArOTf[6]
ArONf[7]
ArOMs[8]
Scheme 2
OTf, ONf, and OMs broaden the range that Buchwald coupling applies, which means that hydroxyl group can react with amine by using OTf.
Nucleophiles
Primary amines
Brettphos[9] is the ligand designed for primary amines.
Scheme 3
Brettphos has selectivity in primary amines toward secondary amines. Using LiHMDS as base combined with Brettphos can get proton tolerance like hydroxyl and carboxyl.
Secondary amines
Everything is almost the same with primary amines, instead of ligand. Ruphos[9] is designed for secondary amines.
Similarly, LiHMDS is utilized to gain proton tolerance. However, it is not easy to have selectivity in secondary amines toward primary amines by steric hindrance control.
Amide[10]
Scheme 4:
Amide is not a good nucleophile, so more reactive ligand tBuBrettpos[10] is designed to solve this problem.
1. NH heterocyclic compounds[11]
Scheme 5
For different types of nucleophile, different ligand is used. For indole, Davephos[11] is a good choice.
Similarly, tBuXphos[12] is for indazole[13].
Scope
It is critical to choose the correct coordinating ligands to the palladium in the Buchwald-Hartwig amination. For intramolecular coupling reaction of aryl bromides with amines having stereocenters at the α-position to the nitrogen atom, the use of Pd(P(o-tolyl)3) would not give racemic mixtures of products. Instead, it would form products with high enantiopurity. [14]
Scheme 6
However, in the case of intermolecular coupling reactions, it is catalyzed by Pd(BINAP) to give the coupled products without loss of enantioselectivity. [14]
Scheme 7
Choice of base and catalysts
For base and catalysts choice, Buchwald gave a good summary in his “user guide”. We cite it here.[5]
Table 1: Base comparison
Base Advantages Disadvantages
NaOT-Bu Permits highest reaction rates and lowest catalysis loadings Incompatible with many electrophilic functional groups
LHMDS
Allows utilization of substrates bearing protic functional groups
Useful for low temperate amination
Solid base is air sensitive
Incompatible with some functional groups at elevated temperature
\(Cs_2CO_3\) Provides excellent functional group tolerance and often highest reaction rate of weak base
Expensive
can be hard to stir on large scale
\(K_3PO_4\) and \(K_3CO_4\)
Excellent functional group tolerance
Often most efficient for arylation of amides
Economically attractive
Can require relatively high catalyst loadings and long reactions times
Advantages and Limitations
Buchwald reactions have many advantages.[5] The catalyst loading is relative low, around 1%-2%, and all ligands are commercially available. It can be done in THF, toluene, t-BuOH and dioxane, and little amount of water is fine. Sometimes water is added intentionally to help Pd(Ⅱ) reduction. Reaction requires argon protected environment, but the reaction system is not very sensitive to oxygen.
As to scope, Buchwald-Hartwig reaction can be applied to various amines, which is discussed above, and most of them have a very good yield. Proton tolerance can be acquired by using LiHMDS. However, functional groups like azo may cause catalyst poisoning, messing the reaction up. Esters and nitro groups are incompatible with KOtBu, but weak base like K2CO3 has a low reaction rate. For more details of this reaction, just turn to Buchwald’s “user guide” published in Chem. Sci. Buchwald-Hartwig reaction is of great significance, which provides a strong tool for C-N coupling.
Reference
1. M. Kosugi, M. Kameyama and T. Migita, Chemistry Letters, 1983, 12, 6, 927-928.
2. D. Bogen and J. Panek, Tetrahedron Lett., 1984, 25, 30, 3175-3178.
3. F. Paul, J. Patt and J. F. Hartwig, J. Am. Chem. Soc., 1994, 116, 5969-5970.
4. A. S. Guram and S. L. Buchwald, J. Am. Chem. Soc., 1994, 116, 17, 7901-7902.
5. D.S.Surry and S.L.Buchwald, Chem.Sci., 2011, 2, 27, 27-50.
6. J. Ahman and S. L. Buchwald, Tetrahedron Lett., 1997, 38, 6363-6366.
7. R. E. Tundel, K. W. Anderson and S. L. Buchwald, J. Org. Chem., 2006, 71, 430-433.
8. B. P. Fors, D. A. Watson, M. R. Biscoe and S. L. Buchwald, J. Am. Chem. Soc., 2008, 130, 41, 13552-13554.
9. D. Maiti, B. P. Fors, J. L. Henderson, Y. Nakamura and S. L. Buchwald, Chem. Sci., 2011, 2, 57-68.
10. B. P. Fors, K. Dooleweerdt, Q. Zeng and S. L. Buchwald, Tetrahedron Lett., 2009, 65, 6576-6583.
11. D. W. Oldm M. C. Harris and S. L. Buchwald, Org. Lett., 2000, 2, 10, 1403-1406.
12. X. Huang, K. W. Anderson, D. Zimi, L. Jiang, A. Klapers and S. L. Buchwald, J. Am. Chem. Soc., 2003, 125, 6653-6655.
13. K. W. Anderson, R. E. Tundel, T. Ikawa, R. A. Altman and S. L. Buchwald, Angew. Chem. Int. Ed., 2006, 45, 6523-6527.
14. S.Wagaw, R.A.Rennel, and S.L.Buchwald, J.Am.Chem.Soc., 1997, 119, 8451-8458.
Contributors and Attributions
• Shiyu Chen (New York University) and Shuhui Chen (New York University) | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Catalysis/Catalyst_Examples/Buchwald-Hartwig_Amination.txt |
Cross-dehydrogenative coupling (CDC) is the class of reaction developed by Chao-Jun Li (McGill U)17 that results in the formation of C-C or C-N bond directly from two unmodified C-H bonds (C-C bond formation) or C-H and N-H bonds (C-N bond formation) (Figure 1). Formally, the reaction occurs with a loss of an equivalent of H2, although hydrogen gas is really the byproduct since the formation of C-C or C-N bond with a loss of H2 is thermodynamically unfavorable and thus requires the use of an oxidant (Cu(ІІ), Ag(І) salts, CAN, benzoquinone, peroxides, O2, hypervalent iodine, persulfates etc.) as the driving force.
Although the same transformations can be achieved by alternative methods, the advantage of CDC approach is that direct tandem oxidation of simple C-H and N-H bonds allows use of simple and readily available reagents and reduces number of steps towards the desired molecule. On the other hand, some challenges are yet to be overcome including low reactivity of some C-H bonds, overoxidation, selective C-H bond functionalization, dimerization etc.
CDC reactions are used to construct bonds between sp3-sp3, sp3-sp2, sp3-sp, sp2-sp2 (including Heck-type reaction), sp2-sp (Sonogashira type) and sp-sp (Glaser reaction) carbon atoms, as well as C(sp3)-N. Because CDC is a group of reaction in which the mechanism and reactivity various dramatically depending on the substrate. These couplings, however, can be roughly divided in four groups (by the proposed mechanism): CDC via Heck-type mechanism, direct arylation, CDC via ionic intermediates and CDC via radical intermediates.1
CDC via Heck-type Mechanism
Heck-type mechanism2 is typical of sp2-sp2 CDCs including arene-alkene, alkene-alkene, arene-benzoquinone coupling; sp3-sp2; and formal sp3-sp2 including enolate-alkene, allyl-arene coupling. As follows from the name, the mechanism is very similar with the Heck Reaction. Palladium3 salts are usually the catalyst of choice, although the same reactivity may be achieved with other transition metals such as Rh(ІІ), Ru(III) and Ir(І) complexes. The mechanism involves the electrophilic palladation of arene ring (ligand assisted or base mediated C-H insertion) with a Pd(II) catalyst to generate arylpalladium intermediate. Subsequent carbopalladation of olefin leads to alkylpalladium complex that undergoes syn-β-H elimination to yield styrenyl product and Pd(0). The final step of the catalytic cycle is oxidation of Pd(0) to Pd(ІІ). In an alternative mechanism, the Pd(ІІ) catalyst is proposed to coordinate to the olefin, which enhances its electrophilicity and propensity to undergo nucleophilic addition with electron-rich aromatic rings (Figure 2).
Catalytic systems are usually not too complex and require no or only a simple ligand such as phosphines, N-protected amino acids or sulfoxides. Inorganic or organic base can also be employed. When more than one product may be formed, the regeoselectivity is normally achieved by installation of a directing group on the aromatic ring, whereas the pyridyl-based groups are the most extensively employed to direct transition-metal-catalyzed C−H activation reactions (both ortho- and meta-).4
Diract Arylation
Direct arylation (also known as 2-fold C-H activation) is used to construct a bond between two arenes.5 Direct arylation strategy is an alternative to such well-known and widely used methods as: Suzuki, Negishi, Stille, Hiyama, Kumada cross-couplings etc, which require independent synthesis and isolation of aryl halide (or pseudohalide) and arylmetal starting materials. Direct arylation can be achieved with electron-rich arenes, arenes bearing pyridine, amide, carbamate, and other oxazoline directing groups, pyridine N-oxides and other electron-deficient arenes. Palladium(ІІ) complexes are almost exclusively used. The method is considered as atom economical approach towards bi- and poly-aryl motifs, which are common structures in medicinal chemistry and material science. The mechanism involves 2-fold C-H activation (either ligand assisted or base mediated) leading to the formation of Pd(ІІ) bis-aryl complex, which undergoes reductive elimination and produces biaryl and Pd(0). The final step of the catalytic cycle is oxidation of Pd(0) to Pd(ІІ) (Figure 4).
The regeoselectivity is usually controlled by thermodynamic factors (e.g. formation of the most stable intermediate through activation of the most acidic C-H bond or through the attack of the most nucleophilic position, Figure 5a)6 and by installation of a directing group (Figure 5b).7
CDC via Ionic Intermediates
Unlike first two mechanisms CDC via ionic intermediates can be promoted by a number of transition metals including Pd, Ru, Zr, Ni, Cu, Fe, Co etc., whereas the last three are the most widely employed. The ionic mechanism predominantly relies on single electron transfer (SET) pathway, although some exceptions are known.8 Mechanistically C-C/N bond formation occurs between an electrophilic carbon species (carbocation) and a carbon- or nitrogen-based nucleophile (amine, amide, carbanion, enamine, heteroaromatic). Carbocations are generated in situ by C-H bond oxidation of a suitable precursor, whereas carbanions can be prepared by deprotonation of relatively acidic C-H or N-H bonds. As a source of carbocations, amines, ethers, allylic and benzylic compounds are usually employed. Oxidant choice depends on a substrate, although organic hydroperoxides were shown to be most efficient in many cases. The catalysis normally performed without ligand additive and simple TM salts are employed (halides, carboxylates, nitrates etc.) While it is hard to derive a single mechanism, the Ionic Intermediates pathway can be demonstrated on the iron(III) catalyzed CDC of ethers with N-nucleophiles, leading to a facile formation of a hemiaminal moiety (which present in a number of bioactive natural products) (Figure 6).
CDC via Radical Intermediates
The radical Intermediates mechanism is very similar to the one described above (Ionic), except that no ionic intermediates are formed. The newly generated carbon-centered radical directly reacts with another active specie (e.g. carbocation, carboanion, C-C multiple bond) forming another radical, which later on gets oxidized either by an oxidant (Figure 7)9 or transition metal (Figure 8)10 realizing the desired product. The radical CDC is exclusively promoted by first-row transition metal complexes such as Cu, Fe, Mn (in some cases vanadium oxo-species) via single-electron transfer type pathway. Selective C-H bond oxidation to a radical can be achieved with phenols, electron-rich arenes, benzylic compounds and C(sp3)-H bond α- to carbonyl or heteroatom (O or N).
The radical CDC is essential in the synthesis of binaphtols11 and related compounds and can be achieved enantioselectively when chiral ligand is used (i.e. Salan, (R)-α-methylbenzylamine etc.).12
Cross-dehydrogenative coupling is a rapidly advancing field and may occur via a wide array of pathways. Therefore some valuable transformations including Au-catalyzed dehydrogenative cyclization between allenes and arenes,13 Cu and Ni-catalyzed cross-coupling of alkynes,14 Cu and Pd-catalyzed CDC of aldehydes and arenes making ketones etc.,15 were not described. In many cases involving C(sp3)-H and sometimes C(sp2)-H bonds CDC may occur without transition metal catalyst. Finally, it is worth noting, that CDC approach is widely used by nature to construct C-C bonds in living organisms, which primarily involves catalysis promoted by cytochrome P450 (CYP) enzymes in the presence of O2 as the terminal oxidant.16
Contributors
• Kirill Korvinson (City University of New York) | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Catalysis/Catalyst_Examples/Cross-Dehydrogenative_Coupling.txt |
The Heck reaction is a famous chemical reaction discovered by Mizoroki and Heck in 1972 through independent research. It involves the cross-coupling reaction between organohalides and alkenes, these two substances react in the presence of a palladium catalyst and a base to form a substituted alkene:
Step A is the oxidative addition of a polar substrate onto a palladium catalyst to form a tetrasubstituted complex. Step B is the migratory insertion of an olefin into the system. Step C is the β-Hydride Elimination of the alkene and Step D is the addition of base to the palladium to regenerate the starting catalyst and close the cycle[1].
Detailed Study of the Mechanism of Heck Reaction
Pre-activation of Palladium Catalyst
It is noteworthy that the first step of the catalytic cycle is actually not the oxidative addition of the substrate, as the palladium catalyst must be activated before the reaction. Therefore, a thorough study of the structure of the palladium catalyst and its properties will be important in understanding Heck reactions [2]. The catalytic precursor Pd(II)(OAc)2, associated with monodentate phosphine ligands such as PPh3 is normally used to catalyze the reaction, but this Pd(II) complex must be reduced to Pd(0) in order to enter the catalytic cycle. There are two different mechanisms involving phosphine-mediated Pd(II) reduction[2]:
The catalytic precursor Pd(II)(OAc)2, associated with monophosphine ligands, is much more efficient in catalyzing Heck reactions when compared to Pd(0)(PPh3)4 catalyst. This is because Pd(0)(PPh3)2(OAc)- can be destabilized by the interaction between OAc- and protons to readily form the unstable Pd(0)(PPh3)2 catalyst, which then enters the catalytic cycle to catalyze the reaction[3]. Since Pd(0)(PPh3)4 is a relatively stable 18-electron complex, it is unlikely that it dissociates two ligands to form an unstable 14-electron structure[3].
Furthermore, Pd(0) catalysts must possess an appropriate coordination number to enter the catalytic cycle[2]. If there are too many monophosphine ligands, it may inhibit the catalyst because a coordinatively saturated metal complex will be formed via ligand association:
The activation of Pd(II) catalysts can also be achieved without the assistance of phosphines. For example, triethylamine is a good reagent to selectively reduce Pd(II):
Oxidative Addition
Oxidative addition is the most difficult step of the entire catalytic cycle. However, the presence of electron-donating groups on the phosphine ligands can activate the Pd(0) catalyst such that the R-X bond can be easily broken along with the formation of Pd-R and Pd-X bond. The rate of oxidative addition also depends on the chemical property of halides, the following trend is generally observed [2]:
Olefin Addition to the Palladium Complex
Before migratory insertion of the olefin to the palladium-R bond, the olefin must first associate onto the palladium complex, which requires the dissociation of the existing ligands. Historically, the Heck reaction was viewed as the functionalization of olefins through aryl halides typically without ligands for aryl iodide or with monodentate phosphines for the other compounds [1]. This cycle produces a strong Pd-X bond while there is a weak Pd-PR3 attachment. The oncoming group would attack and form a net neutral square planar molecule, as is shown in Path A. This reaction mechanism was found to not be ideal as, “chelating diphosphines… in general do not produce useful catalysts [1].” This concern was addressed through the introduction of triflates as leaving groups which would allow the mechanism coordination-insertion step to follow Path B. Catalytic and stoichiometric studies have been utilized to determine these pathways.
• Path A: If a neutral ligand dissociates, then the neutral mechanism predominates the reaction. The neutral mechanism occurs when X=Cl, Br, I (i.e. strong sigma-donor).
• Path B: If an anionic halide ligand dissociates, then the cationic mechanism dominates the reaction. The cationic mechanism is believed to happen when X= OTs- or OAc- (e.g. weakly associated ligand).
The two mechanisms shown above display the possible coordination-insertion paths found in Heck reactions. It has been found that the insertion of ethylene into the Pt-H bond is critical in the reaction characterization [1]. Thorn and Hoffman conducted orbital studies to determine key information from this step. The first conclusion they reached is that there must be a coplanar assembly of the metal, hydride and ethylene for insertion to occur. This indicated that insertion is stereoselective and occurs in a syn manner. Experimental data supported this claim [1]. They also determined that the energy barrier for a pentacoordinated complex is significantly higher than that of a tetracoordinated complex, indicating pentacoordinated complexes are not involved in the mechanism. This observation was supported by kinetic and experimental data [1].
Path B was unknown until 1991, when Ozawa and Hayashi proposed the existence of a cationic form of the square planar complex. To obtain this complex, triflates must be used as the leaving group [1]. The lability of Pd-OTf bonds in the oxidation addition complex aids in this formation. Bidentate phosphorus or nitrogen ligands, along with the triflate leaving group, allow for the reaction to follow this path. It was also determined that this path can yield high asymmetric induction when the diphosphine is chiral. This effect was not seen in Path A [1].
Ligand dissociation mechanisms are also affected by different types of phosphine ligands. Monodentate phosphine ligands lead to the occurrence of both the neutral and cationic mechanisms, whereas bidentate phosphine ligands merely induce a cationic mechanism, but the neutral mechanism is still possible in the presence of a large bite angle [2].
Migratory Insertion
Migratory insertion of the olefin into the Pd-R bond is a crucial step for the catalytic cycle because it can control the stereo-selectivity and regio-selectivity of Heck reactions. For a neutral palladium complex, the regioselectivity is governed by sterics, which means nucleophilic attack happens on the less hindered site of the alkene [4][5]:
For cationic palladium complexes, the regioselectivity is governed by electronics, which implies that nucleophilic attack occurs on the site possessing the least electron density of the alkene [4][5]:
β-Hydride Elimination
β-Hydride Elimination results in the Heck reaction product, which is a new substituted alkene. In this step, the palladium and the hydride attached to it must be syn-coplanar for the initiation of elimination. The product with the Z-conformation is strongly disfavored because of the steric interaction in the transition state [2]:
Afterβ-hydride elimination, the newly formed palladium-alkeneπcomplex is subject to olefin isomerization, resulting in the formation of an undesired Heck product[4]:
These side reactions will occur since this is a reversible reaction. If the olefin dissociation rate is too slow, this problem arises [1]. Fortunately, adding bases or silver salts can significantly reduce the chance of alkene isomerization by facilitating reductive elimination to form an H-X bond [6].
Regeneration of palladium catalyst
The addition of base is necessary to reduce the L2PdHX complex back to the starting L2Pd(0)[1]. Some common bases used are trialjylamines such as Et3N or inorganic salts such as AcONa. A proton sponge or Tl(I) or Ag(I) salts may also be employed to close the cycle[1].
Intramolecular Heck Reaction
Heck reactions can also be performed in a single molecule which is quite useful for macrocyclization. This intramolecular Heck reaction was first reported by Mori and Ban in 1977:
Mori, M; Ban, K.; Tetrahedron 1977, 12, 1037
The Intramolecular Heck reaction has many advantages compared with the intermolecular Heck reaction. First of all, only mono- or disubstituted alkenes can coordinate into the palladium complex in the intermolecular Heck reaction, whereas tri- and tetrasubstituted alkenes are able to participate readily through the intramolecular mechanism. Secondly, the intramolecular Heck reaction is much more efficient than the intermolecular reaction because of entropic considerations [7]. Finally, regioselectivity and stereoselectivity are dramatically improved in the intramolecular Heck reaction. This advantage inspired both Shibasaki and Overman to explore the asymmetric effect in the intramolecular Heck reaction, and they eventually found the first asymmetric intramolecular Heck reactions. This remarkable finding has provided enlightenment for natural product synthesis [7].
Shibasaki, M.J.Org. Chem. 1984,54,4738
Overman, L. E. J. Org. Chem. 1989, 54, 5846
Regioselectivity and Stereoselectivity
Regioselectivity reactions of Path A were conducted on several classes of olefins. Different olefin families reacted with different regioselectivity[1]. Once aryl triflates were introduced as the leaving groups, reactions via Path B were also examined for regioselectivity. These reactions were carried out with aryl triflate leaving groups and aryl halides with Pd(OAc)2 with bidentate phosphorus ligands[1]. These reactions yielded branched products more readily than Path A.
In Path A, regioselectivity is related to the coordination-insertion pathway and steric factors. A migration of the R group to the less substituted carbon with formation of linear products has seen to be favored in Pathway A [1]. Path B differs in that electronic factors dominate in determining regioselectivity. The increase in polarization is determined by the coordination of a pi-system within a cationic complex. This causes selective migration of the aryl moiety onto the carbon with a lower charge density [1].
Hayashi and Ozawa have investigated stereoselective intermolecular reactions using aryl triflate as the leaving group and chiral (R)-BINAP. Several reactions were performed under analogous conditions and their products compared [1]. As is expected with the triflate leaving group, these reactions followed Path B. In these reactions, the chiral BINAP ligand binds tightly to the metal through both of its phosphorus atoms. This chiral ligand is able to transfer its chiral information from the catalyst to the substrates [1]. These reactions have been found to be most selective when using both chiral BINAP and the triflate leaving group. Of note is that this (R)-BINAP determination of selectivity will only occur in systems that are electron rich. This information aligns with the assertion that Path B relies heavily on electronic factors in that electron rich systems react much more efficiently than electron poor [1].
Overman has reported a stereoselective synthesis for quaternary carbons through asymmetric intramolecular Heck reactions. These results were of note, as they did not follow the widely held belief that long reaction times result in isomerization of the double bond and multiple product formation [1]. PMP was used as the base in these reactions without Ag(I) salt present. Good selectivity was observed in these reactions despite the slow reaction time and the difficulty in transferring chirality through Path A [1]. It was determined that with flexible substrates, the products were almost racemic. However, with rigid substrates, the single phosphorus coordination of (R)-BINAP was still able to transfer the chirality via Path A[1].
Limitations of the Heck Reaction
The Heck reaction is widely used in the pharmaceutical, medical and industrial areas because of its ability to efficiently generate large polycyclic structures. However, this synthetically useful reaction has its own disadvantages [8].One of the major disadvantages is that the Pd catalyst will be lost at the end of the catalytic cycle, therefore, it is necessary for researchers to find an effective method to recycle the palladium catalyst. Another significant weakness is that the phosphine ligands attached to the palladium catalyst can be toxic and expensive, so the phosphine-free ligands must be discovered to improve the efficiency of the reaction [8].
Applications of the Heck Reaction
Heck couplings with dehydrocostus lactone and aryl halides can produce guaianolide sesquiterpene lactones derivatives, which have been proved effective in inhibiting resistant acute leukemic cells:
Y. H. Ding, H. X. Fan, J. Long, Q. Zhang, &Y. Chen, Bioorganic and Medicinal Chemistry Letters, vol. 23, 22, 6087–6092, 2013.
The Heck reaction can also be used to synthesize the smoking cessation aid, Chantix ®
Coe, J. W.; Brooks, P.R.; Veteline, M.G.; Bashore, C.G.;Bianco, K.;Flick, A,A.C. Tetrahedron Lett. 2011, 52, 953-954
Contributors and Attributions
• Haley Merritt and Yifan Qi | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Catalysis/Catalyst_Examples/Heck_Reaction.txt |
Hydroamination is a reaction that involves the addition of a hydrogen and an amino group across an unsaturated C-C bond, such as that in alkenes or in alkynes. It is a form of hydrofunctionalization with the functional group being an amine or amino derivative.
Hydroamination is most often performed catalytically. The catalysts for hydroamination are broad and varied, including acids, bases, and transition metals in both homogenous and heterogeneous media. This report will focus primarily on the homogeneous transition metal catalysis of hydroamination.
The general reaction for hydroamination is illustrated in Scheme 1.
Scheme 1: Amine 2 is added across unsaturated C-C bond present in 1 to afford hydroamination product 3.
Regioselectivity for addition across asymmetrically substituted unsaturated C-C bonds is an important consideration in this reaction. Recall from general organic chemistry that regioselectivity is characterized by Markovnikov addition (to form branched products) or anti-Markovnikov addition (to form linear products). Scheme 2 builds upon the example in Scheme 1 to demonstrate the differentiating principle for these two classes of regioselective addition.
Scheme 2: Amine 2 can add across unsymmetrically substituted alkene 4 in two different ways to yield products 5 or 6.
5 is the Markovnikov product, in which the H adds to the carbon bound to the larger number of hydrogen atoms. 6 is the anti-Markovnikov product, in which H adds to the carbon bearing the least number of hydrogen atoms.
Note that hydroamination can proceed intermolecularly or intramolecularly.
This report will be divided into four different sections to discuss hydroamination based on catalyst type:
1. Lanthanide Metal Catalysts
2. Early Transition Metal (Groups 4/5) Catalysts
3. Late Transition Metal Catalysts
4. Other Catalytic Methods
Each of the three sections discussing transition metal chemistry will provide a thorough mechanistic discussion of reported hydroamination reactions, as well as illustrative examples and synthetic applications of the catalysts to total synthesis.
LANTHANIDE METAL CATALYSTS
Lanthanide or rare-earth metal catalysts for hydroamination have been studied extensively. This section will be broken down into the substrate scope of these lanthanide-metal catalysts, looking at hydroamination of aminoalkenes, aminoalkynes, aminoallenes, aminodienes. These discussions will be followed by notes on catalyst design and applications of these catalysts to total synthesis.
Aminoalkenes
We will first consider the lanthanide catalyzed hydroamination of a terminal aminoalkene (Scheme 3).
Scheme 3: Terminal amino alkene 1 is converted into cyclic amine 2 with conversion greater than 95% at a turnover frequency (TOF) of 140 h-1 through the use of lanthanide catalyst. [1]
The lanthanide catalyst used is a lanthanocene catalyst, illustrated in Figure 1.
Lanthanocene catalysts such as A contain a key structural element: the metal is ‘sandwiched’ between two cyclopentadienyl rings, in a manner similar to the famous organometallic compound ferrocene. Hence the ‘-cene’ suffix is added to the metal to describe this general group of organometallic catalysts. Such nomenclature will be used throughout this report.
The reaction in Scheme 3 was found to proceed for numerous terminal alkenes containing methyl substituents on the alkyl chain for lanthanum and neodymium catalysts. The general mechanism in Scheme 4 was proposed based on kinetic experiments and isotopic labeling experiments, which are described below.
Scheme 4: Mechanism for intramolecular hydroamination (Scheme 3) catalyzed by lanthanocene catalyst A.1
This mechanism begins by protonolysis of the lanthanocene precatalyst by one equivalent of the aminoalkene to yield La-amido intermediate B. This intermediate has been characterized by single crystal X-ray diffraction data and spectroscopic data. Intermediate B undergoes b-insertion through proposed transition state C to form the La-alkyl aza-cyclic intermediate D. Note the cis relationship between the La and N atom with respect to the C-C bond, which is a key feature of these lanthanide catalyst mechanisms. This intermediate then undergoes protonolysis with a second equivalent of aminoalkene to eliminate the product cyclic amine and re-generate intermediate B. The electronic structure of the metal is consistently at the (III) oxidation state.
Kinetic studies provided a rate law that was zero-order in amine and first-order in catalyst. As suggested in Scheme 4, this indicates that the b-insertion step is the rate-determining step of the reaction. It is expected to proceed through the four-membered cyclic transition state C denoted in Scheme 4. An isotopic-labeling experiment additionally indicated that cyclization to intermediate D is irreversible.1
Catalyst design was found to influence reaction rate. As the ionic radius of the lanthanide metal increased, the rate of the reaction increased (e.g., La > Sm > Lu).1 Additionally, the less-sterically encumbered the catalyst ligation sphere, the greater the rate of the reaction (Figure 2).
It was found that A1 reacted the fastest, with a bare cyclopentadienyl substituent and a tetramethyl-cyclopentadienyl substituent composing its lanthanocene structure. A2, where both cyclopentadienyl units are methylated at all five carbons, creating greater steric hindrance of the ligation sphere, reacted the slowest.1
One limitation of the above reported lanthanocene catalyst is that it cannot catalyze intramolecular hydroamination across substituted alkenes. This is due to unfavorable, non-bonded steric repulsions between catalyst ligands and the R groups of the alkene in the four-membered transition state C in Scheme 4.
A modified lanthanide catalyst E in Figure 3 was prepared to conduct hydroamination across more hindered alkenes in an effort to reduce these non-bonded steric repulsions (Scheme 5).
Scheme 5: Reported intramolecular hydroamination across hindered alkene 3 with modified catalyst E to afford product 4. Internal alkene 5 interestingly yields no reaction.1
The key feature of catalyst E is a more open ligation environment, meaning more free space around the metal center due to less sterically bulky ligands (i.e., modified cyclopentadienyl ligands). In addition to the less sterically hindered ligation environment, these reactions required high temperature and long reaction time (many days).
Using a metal with larger ionic radius at elevated reaction temperature successfully afforded intramolecular hydroamination of internal alkenes as shown in Scheme 6. These reactions were carried out at modest TOF with excellent conversion.
Scheme 6: Intramolecular hydroamination of internal alkene 6 catalyzed by lanthanocene catalyst to afford product 7.1
Note that the reactions in Schemes 5 and 6 are all purported to follow the mechanism presented in Scheme 4.
Hence alkenes undergo intramolecular hydroamination by lanthanide-metal catalysis. Terminal alkenes are the easiest substrates with which to conduct this reaction. With internal and substituted alkenes, unfavorable steric repulsions in the transition state arise and require modified catalysts with less sterically bulky ligation environments.1
While many organolanthanide complexes facilitate intramolecular hydroamination, intermolecular lanthanide-catalyzed transformations are also possible. With the use of a phenylene-bridged binuclear organolanthanide complex F, intermolecular hydroamination has been reported (Figure 4).[2]
An example intermolecular hydroamination reaction with F is provided in Scheme 7.
Scheme 7: Intermolecular hydroamination between styrene 8 and amine 9 to afford anti-Markovnikov product 10 using catalyst F.2
The substrate scope of F is rather limited, catalyzing reactions with only phenyl-substituted olefins or alkynes. The TOF of these reactions were low, but the yields averaged >95%. The regioselectivity of this reaction is attributed to electronic effects in the intermediate C in Scheme 4. The electronic rich phenyl substituent coordinates to the electrophilic metal center, thus forcing nucleophilic attack of the amine to take place at the terminal end of the alkenyl moiety.
Aminoalkynes
Aminoalkynes are proposed to undergo hydroamination along the same catalytic cycle as aminoalkenes (Scheme 4) because they follow the same rate law (zeroth order in amine, first order in catalyst).1
Alkynes undergo hydroamination much more readily than alkenes, which is owed to important differences in the thermodynamics of the b-insertion step: alkene insertion into the M-N bond is approximately thermoneutral whereas alkyne insertion into the M-N bond is highly exothermic.1 Given this information, we anticipate that alkyne hydroamination will avoid the limitations encountered in alkene hydroamination that had required catalyst optimization as discussed in the previous subsection. Our supposition is true: both terminal and internal alkynes readily undergo hydroamination by the same lanthanide-metal catalyst (Scheme 8). Additionally, alkyne hydroamination is accomplished much more easily and effectively than hydroamination of alkenes.
Alkyne hydroamination yields enamines for secondary amines and imines for primary amines. Examples of each reaction are reported in Scheme 8 below.
Scheme 8: Intramolecular hydroamination of internal alkyne with primary amine 11 yielding imine 12 and terminal alkyne with secondary amine 13 yielding enamine 14 using the same samarium catalyst.1
The rate-determining step in the catalytic cycle for alkyne hydroamination with lanthanide catalysts is b-insertion, as it was for alkenes.[3] In general, alkynes are found to produce higher TOFs than alkenes, which can be explained by easier b-insertion of alkynes into the M-N bond to afford the cyclic transition state H in Scheme 9 (akin to C in Scheme 4).
In lanthanide-metal catalyst design, the rate of alkyne hydroamination interestingly decreases as ionic radius of the metal increases – in stark contrast to the trend observed with alkenes.2 Thus Samarium catalysts are commonly reported metals for these transformations, as seen in Scheme 8.
Lanthanide-metal catalyzed hydroamination of alkynes has been shown to facilitate the construction of very complex azacyclic systems via tandem bicyclization, shown in Scheme 9.
Scheme 9: Tandem bicyclization of C-N followed by C-C bond formation through intramolecular lanthanide-metal catalyzed hydroamination of an internal alkyne.1
As stated previously, the steps for hydroamination in Schemes 9 and 4 are identical. One important difference of note, however, is formation of a Sm-vinyl aza-cyclic intermediate I, in which a pi bond remains after insertion into the alkyne.
G is the catalytically active form of the Sm catalyst which undergoes rate-determining b-insertion as expected to afford I through transition state H. The additional cyclization results from insertion of the pendent terminal alkene into the Sm-C bond in I to afford J. Upon protonolysis of J with another substrate of aminoalkyne, the aza-bicyclic product is released and G is regenerated.
Aminoallenes
Allenes exhibit the same propensity for hydroamination with lanthanide metal catalysts as that of alkenes and of alkynes. Allenes follow the same general catalytic cycle for hydroamination presented in Scheme 4. Density functional theory (DFT) studies indicate that the rate-determining step for allene hydroamination is protonolysis of the Ln-vinyl aza-cyclic intermediate (akin to I in Scheme 9) rather than b-insertion to form said aza-cyclic intermediate (as is the case for alkenes and alkynes).3
Allenes notably react faster than alkenes but slower than alkynes. In terms of catalyst design, the best TOF results from lanthanide-metals of intermediate ionic radius, (Y > Sm > Lu > La).1
Aminoallene hydroamination often occurs stereoselectively which makes this transformation useful for natural product synthesis. An incredibly efficient and elegant example of such stereoselective reactions is presented in Scheme 10.
Scheme 10: Aminoallene 15 undergoes intramolecular hydroamination to afford the Z-stereoisomeric cyclic product 16 regioselectively.1
Aminodienes
A final substrate to consider is the aminodiene, whose reactivity incorporates features of the aminoallene system and the aminoalkene system. The mechanism for diene hydroamination follows that of Scheme 4, but like aminoallenes, the rate-determining step is believed to be protonolysis of the Ln-allyl aza-cyclic intermediate.3
The rate of hydroamination of aminodienes increases as the ligation environment of the catalyst becomes more open/less sterically encumbered. This indicates the same sensitivity to catalyst design as that demonstrated by aminoalkenes. Aminodienes respond to such changes in ligation environment more readily than aminoalkenes. This is best explained by the ability for the diene to datively coordinate to the metal center and form an h3-allyl intermediate L, shown and discussed in Figure 5 below.
Although this allyl intermediate enhances the reaction rate, it requires a less sterically-encumbered ligation environment around the metal center in order for it to form in the first place. Thus the sensitivity of the aminodiene system to catalyst design is not owed to sterics as it was for the aminoalkene system, but rather to the ability to form the allyl intermediate.1,3
An example reaction of aminodiene hydroamination is shown below in Scheme 11.
Scheme 11: Intramolecular hydroamination of aminodiene 17 with lanthanocene catalyst to afford aza-cyclic product 18.1
Additional Notes on Catalyst Design
The substrate scope of lanthanide-metal catalyzed hydroamination reactions is diverse and affords both intramolecular and intermolecular systems. These reactions follow the same general mechanism illustrated in Scheme 4, with slight variations in rate owed to differences in the ability of substrates to b-insert into the Ln-N bond and purported variations in the rate-determining step.
It should be noted that lanthanide metals generally maintain (III) oxidation state during the catalytic cycle. Formation of the catalytically active species involves replacement of a s-bonded ligand with an amine.
Sterically hindered unsaturated C-C bonds react best with more coordinatively open catalysts. This inspired the formation of half-lanthanocene catalysts, such as the samarium catalyst depicted in Figure 6, which provides solutions to the problem of ligated encumbrance retarding the catalytic reaction. This was a very important discovery in hydroamination chemistry and must be highlighted.
As a final note, hydroamination reactions catalyzed by actinide metals have also been reported.3
Application to Total Synthesis
Lanthanide metals are found in a variety of natural product total synthesis schemes. Consider the example in Scheme 12.
Scheme 12: Two intramolecular hydroamination reactions are catalyzed by a ‘half-lanthanocene’ samarium catalyst over an allene and an alkene moiety in 19 to afford bicyclic product 20 with cis stereochemistry. Hydrogenation of the alkene affords natural product (+)-xenovenine 21.1
Note the creation of a stereospecific, fused N-bicyclic system in one step using the remarkable catalytic hydroamination chemistry of lanthanide metals.
EARLY TRANSITION METAL (GROUPS 4/5) CATALYSTS
With Early Transition Metal Catalysts, mechanistic differences arise depending on the metal and the substrate undergoing hydroamination. Titanium and zirconium (group 4), and to a lesser extent vanadium and tantalum (group 5), have demonstrated catalytic activity for hydroamination.3
Like section 1, this section will again be divided into an analysis of the substrates undergoing hydroamination, considering alkynes, allenes, and alkenes and concluding with applications to total synthesis.
Alkynes
Both zirconium and titanium have demonstrated catalytic ability for the hydroamination of alkynes. Each metal is believed to follow approximately the same mechanistic pathway, although the rate-determining step in each catalytic cycle is interestingly proposed to be different.
Zirconium has been reported to catalyze intermolecular hydroamination between internal alkynes and primary arylamines through a bisamide complex to afford both enamine and imine product, depending on the alkyne substituents (Scheme 13).
Scheme 13: Zirconium catalyzed intermolecular hydroamination of internal alkynes 1 and 4 with primary arylamine 2 affording imine 3 and enamine 5, respectively.[4]
Note that 1 initially forms a transient enamine which then rapidly tautomerizes to 3.
The following catalytic cycle (Scheme 14) was proposed for this reaction based on the data which will be presented below.
Scheme 14: Proposed catalytic cycle for zirconium bisamide catalyzed hydroamination of an internal alkyne.4
The precatalyst bisamide undergoes reversible a-elimination to generate the catalytically active Zr imido (=N) species A. Upon coordination of added alkyne 4, A undergoes [2+2] cycloaddition to form the azametallocyclobutene B. This metallocycle is then protonated upon addition of amine 2 to form C which then undergoes a-elimination to release enamine product 5 and to regenerate Zr imido species A. Note that the metal center maintains a (IV) oxidation state throughout the catalytic cycle.
Kinetic studies indicated that the reaction rate is first-order in catalyst, zeroth order in alkyne, and inverse order in amine, which is consistent with formation of Zr imido species A as the rate-determining step. One final note on this reaction is that, for asymmetrically substituted alkynes, it is proposed that [2+2] cycloaddition occurs such that the cyclic species in B contains the larger alkyne substituent a to the Zr metal. This results in an ultimately anti-Markovnikov addition of the amine substrate across the alkyne which is a commonly reported feature of this catalyst system.
Analogous to and slightly more reactive than zirconium catalysts for alkyne hydroamination are the titanium catalysts.3 Various titanocene catalysts have been studied to reach the general conclusion that the regioselectivity is determined by the bulkiness of the incoming amine and not the bulkiness of the titanium ligands. Thus a good, broadly applicable catalyst for this reaction is the bis(indole) titanocene catalyst, D, shown in Figure 7.
Titanocene catalysts are expected to undergo the same catalytic cycle as that depicted for zirconocene catalysts. The difference arises in the kinetic data obtained for the titanocene system, which indicates that the rate is first order in alkyne, first order in catalyst, and inverse first order in amine.3 This is different from the zirconocene data, which indicated zeroth order rate dependence in alkyne.4 This then suggests a different rate-determining step for the titanocene mechanism, which is proposed to be [2+2] cycloaddition to form the azametallocyclobutene B.
This catalyst was found to tolerate primary aryl, tert-alkyl, sec-alkyl, and linear-alkyl amines as well as internal and terminal alkynes. Terminal arylalkynes selectively formed anti-Markovnikov product while terminal alkylalkynes selectively formed Markovnikov product. This regioselectivity increased with the steric bulk of the amine, but such amines needed to be added slowly to the reaction mixture (given the inverse order rate law in amines) for sufficient reaction. Computational studies indicated that the regioselectivity of these titanocene catalysts depended on the stability of the Ti imido alkyne species prior to cyclization to afford the azametallocyclobutene B, which is largely a steric consideration.4
Advancements in catalyst design include the development of bis(amidate) titanium complexes, which are able to perform intramolecular hydroamination of aminoalkynes at room temperature. It was additionally found that mono(cyclopentadienyl) titanium (half-titanocene) catalysts catalyzed alkyne hydroamination much faster than the bis(cyclopentadienyl) titanium (titanocene) catalysts. Catalysts were designed featuring this structural trait, as shown in Figure 8.
Half-titanocenes E and F react much more quickly than titanocene G. E features a pendant ether which donates into the metal, protecting it from unwanted side reactions/ coordination. The pendant ether was found to prevent undesirable dimerization of the Ti imido species A into a four-membered aggregate, which is believed to account for the significantly increased reaction rate.
Half-titanocenes are much more reactive than titanocenes, and they are considered in some reactions to be the catalytically active species (forming a Ti imido amido intermediate) rather than the bis(substituted) species.3
One key feature of this reaction is formation of the azametallocycle intermediate B. With terminal alkynes, selectivity for Markovnikov and for anti-Markovnikov hydroamination can be rationalized via steric interactions present in the formation of this intermediate. Sterically demanding amine substrates, such as those bearing a tert-butyl group, yield anti-Markovnikov products by placement of the terminal alkynyl H (rather than the bulkier alkynyl substituent) near the tert-butyl group to form a lower energy (and thus more stable) azametallocycle intermediate. With aryl amine substrates, this steric interaction can be avoided by strategic rotation of the aryl group to be perpendicular to the plane of the azametallocycle, and thus the Markovnikov product can be obtained.3
Lastly, catalysts containing group 5 metals such as V and Ta are known to catalyze hydroamination of alkynes although their efficiency is less than that of the group 4 metals. The catalytic cycles for these group 5 catalysts are thought to be the same as those for the group 4 metals, although there is current speculation to that regard.3
Allenes
Hydroamination of allenes follows the same catalytic cycle as hydroamination of alkynes (Scheme 14) with early transition metal catalysts. A significant difference in allene hydroamination is the ability to form various regioisomers, such as imines or allylamines, from the ability to form the azametallocycle from one of the two different double bonds. An illustrative example of intramolecular hydroamination of aminoallenes is provided in Scheme 15.
Scheme 15: Intramolecular hydroamination of an amionallene 6 yielding a regioisomeric mixture of products, imine 7 and allyl amine 8.3
Regioselective hydroamination of allenes has been reported with early transition metal catalysts. In general, monosubstituted aminoallenes will form preferentially the imine product and 1,3-disubstituted or trisubstituted aminoallenes will form preferentially the ally amine product. Allene hydroamination has also been performed intramolecularly and intermolecularly.3
Alkenes
Alkenes undergo early transition metal catalyzed hydroamination through two different mechanistic pathways: (1) the cationic catalyst pathway and (2) the neutral catalyst pathway. Secondary aminoalkenes require a cationic catalyst because they cannot form the catalytically active metal imido species. Primary aminoalkenes undergo the neutral catalyst pathway as they can from this metal imido species.
We will first consider the primary aminoalkene hydroamination with an example reaction (Scheme 16) followed by the mechanism (Scheme 17). This mechanism (neutral catalyst) resembles the mechanistic pathway of alkynes introduced in the beginning of this section.
Scheme 16: Intramolecular hydroamination of a primary aminoalkene 9 to afford cyclic amine 10 in 52% yield.3
Scheme 17: The mechanism for the neutral early transition metal catalyst for hydroamination of a primary alkene.3
One feature of the mechanism depicted in Scheme 17 is the formation of the Ti imido species H from the Ti precatalyst. This species undergoes intramolecular [2+2] cycloaddition to form intermediate I, containing the familiar azametallocycle. Upon addition of another equivalent of aminoalkene 9, I undergoes a-elimination to afford product 10 and regenerate species H. It should be noted that other possible metals to perform this transformation include Zr and Hf.3
Hydroamination of secondary aminoalkenes cannot proceed by the mechanism presented above. Cationic metal catalysts, such as J in Figure 9, are employed to facilitate this reaction. An illustrative example of such a reaction is presented in Scheme 18.
Scheme 18: Cationic zirconium catalyzed intramolecular hydroamination of a secondary aminoalkene 11 to form cyclic amine 12.
We will consider catalytic cycle for this hydroamination in Scheme 19 below:
Scheme 19: Cationic metal mechanism for hydroamination of a secondary amine, where the [B(C6F5)4]- anion was removed for the sake of simplicity (but note that it should be present in all steps of the reaction to balance the charge on Zr).
The Zr catalyst undergoes a-elimination to yield the cationic Zr amido intermediate K. A proposed 4-membered transition state L yields the cyclic product bound to the cationic Zr intermediate M which upon a-elimination yields the catalytically active intermediate K and product 12. Note that the above mechanism resembles the mechanism presented for lanthanide-metal catalyzed hydroamination.
Application to Total Synthesis
The first natural product to be synthesized using early transition metal hydroamination was monomorine (15, Scheme 20), a pheromone found in monomorium pharaonis that causes ants to be lure ants.[5] McGrane and Livinghouse synthesized a titanium catalyst to convert a substituted alkyne 13 into a cyclic organotitanium compound 14, which was then used to synthesize the final product monomorine 15. Scheme 20 shows the synthetic pathway undertaken to produce monomorine. The mechanism proceeds through a [2+2] reaction, presented in Scheme 14.
Scheme 20: The synthesis of the natural product monomorine 15, which utilizes early transition metal-catalyzed intramolecular hydroamination to convert alkyne 13 to organotitanium compound 14 in the synthetic sequence.5
LATE TRANSITION METAL CATALYSTS
Hydroamination catalysts derived from late transition metals are a chemically rich group. The d8/d10 metal species associated with these catalysts and their ability to act as Lewis Acids provide a high degree of activity toward hydroamination by varied mechanistic pathways. These robust catalysts generally demonstrate low reactivities with air and water and good functional group tolerance. In addition, these metals catalyze reactions which mainly proceed with Markovnikov regioselectivity.
Metals such as Ru, Rh, Pd, Ir, Pt, Ni, Cu, Zn, and Ag have all been found to catalyze hydroamination. Thus, this section deals with a group of catalysts that is too broad to consider by substrate scope as in the previous sections. Rather, we will consider four essential mechanistic pathways through which late transition metal catalyzed hydroamination is proposed to take place, providing example catalysts and reactions as well as considering substrate scope for each mechanistic pathway. The four mechanistic possibilities are divided into two groups, each of which features one of two key steps: (1) nucleophilic attack by nitrogen on a metal-coordinated, unsaturated C-C bond or (2) metal-hydride or metal-amide bond insertion by unsaturated C-C bond.
Nucleophilic Attack: Nitrogen on Coordinated Alkene/Alkyne
This mechanistic pathway is perhaps the most intuitive. It has been thoroughly studied and proposed for many different catalysts, including Ir, Pd, Pt, Cu, and Au. It requires alkene or alkyne as substrate and can occur intermolecularly or intramolecularly.
Consider the illustrative example reaction in Scheme 21.
Scheme 21: Aminoalkyne 1 undergoes cationic Pd-catalyzed hydroamination to afford enamine 2 which rapidly tautomerizes to imine product 3.[6]
The catalyst used for this reaction is shown in Figure 10 and the proposed mechanistic pathway is presented in Scheme 22.
Scheme 22: Cationic Pd catalyzed intramolecular hydroamination of aminoalkyne via nucleophilic attack of nitrogen on a coordinated alkyne. Note two CF3SO3 anions have been omitted from species B1 – D for simplicity.3
Let us consider the mechanistic pathway in Scheme 22. The precatalyst PdCl2(COD) is treated with Triphos ligand and 2 equivalents of Brønstead base AgCF3SO3 to generate the catalytically active [Pd(Triphos)](CF3SO3)2 abbreviated as [Pd]2+ for simplicity. Notice the metal center maintains a (II) oxidation state throughout the catalytic cycle, characteristic of late transition metal catalyzed hydroamination reactions. Aminoalkyne substrate 1 first coordinates to the cationic, Lewis Acidic Pd species. This can happen by one of two ways: (1) through alkyne coordination, B1, or (2) through amine coordination, B2. In species B1, the coordination through alkyne thus activates this unit for intramolecular nucleophilic attack by the pendant nitrogen.
As a general principle, the nucleophile attacks at the internal position, allowing the bulkier Pd substituent to occupy the terminal end. This affords zwitterionic species C which undergoes protolytic cleavage to afford the enamine product 2 coordinated to the metal center, D. Upon addition of an equivalent of substrate 1, the enamine 2 dissociates. Given that the starting material is a primary amine, this material then rapidly tautomerizes to the imine 3.
The most important feature of this cycle is activation of the unsaturated C-C bond via metal coordination. Thus, the metal in the catalytic cycle must be able to serve as a Lewis Acid for this purpose. Additionally, Markovnikov addition products are expected for this cycle due to amine attack at the internal carbon. The rate-determining step was found by kinetic studies and DFT studies to be the protolytic cleavage of the Pd-C bond in intermediate C.3
We would like to highlight the use of a Brønstead acid co-catalyst in this reaction, which has been shown to improve the rate of the reaction. Although the role of co-catalytic acid is not fully understood, some possible explanations3 include:
(1) inhibition of amine coordination (B2) by forming ammonium salt
(2) assisting in the rate-determining protolytic cleavage step (C to D) as external proton source
(3) assisting in tautomerization of enamine 2 to imine 3 as external proton source, preventing the enamine from coordinating to metal catalyst which would impede catalytic activity.
All of these explanations seem reasonable and likely, yet the mechanism for the rate-determining protolytic cleavage is not fully understood.
The choice of counter ion in co-catalyst also affects the rates of these hydroamination reactions. It was found that bulkier anions tend to increase reaction rates (e.g., OTf- > BF4- > PF6- > NTf2-), likely as a function of their poorer ability to coordinate to the metal center and inhibit catalytic activity.3
From the acidic co-catalyst discussion, it is clear that interaction of amine with the Lewis Acidic metal center must be taken into consideration for these systems. Thus, choice of amine substrate is important in these reactions. More basic amines will poison the catalyst by preventing coordination of unsaturated C-C bonds to initiate the reaction. This problem is often addressed by using protected amines or less nucleophilic anilines, carboxamides, and sulfonamides.3
Choice of ligand also influences these reactions. Strongly binding tridentate ligands (such as Triphos, shown in Figure 10) improves the reaction by preventing b-H elimination in species B (given that the starting material is an alkene and not an alkyne – alkynic starting material will not possess b hydrogens). Therefore, although the effect of the tridentate ligand was not in play during this reaction, it is an important consideration and comment on catalyst design.3
As previously stated, the protolytic step of this reaction is not well understood. We will consider the Pt-mediated intermolecular hydroamination of alkene and arylamine to illustrate the nuances of late transition metal chemistry (Scheme 23).
Scheme 23: Pt-mediated intermolecular hydroamination, where protonolysis of the Pt-C bond and release of product proceeds by two different pathways, Path I or Path II.3
Scheme 23 depicts the familiar first two steps of this reaction, olefin coordination to catalyst E to form F followed by nucleophilic attack on the coordinated olefin by amine to form G. It is here, however, that chemists consider two possible pathways for protonolysis and release of product. Path I, a one-step protonation of the a-carbon, causes release of product and returns the system to catalyst E. Alternatively, protonation of the metal center to form H followed by reductive elimination to release product and return to E is also feasible.
DFT experiments indicate that group 10 metals prefer Path I and group 9 metals prefer Path II.3 Thus variations within the same general catalytic cycle as functions of the metal center family are possible for hydroamination, illustrating the uniqueness of this group of metal catalysts.
Nucleophilic Attack: Nitrogen on Coordinated Allylic Species
The following mechanism is used to describe hydroamination of allenes, dienes, and trienes with catalysts such as Pd and Ni. By extension, it can be used to describe hydroamination of vinyl arenes and other molecules containing unsaturated C-C bonds in conjugation with another unsaturation that can support an h3-allyl structure.
We will begin by considering an illustrative example of this reaction in Scheme 24 followed by its proposed catalytic cycle in Scheme 25.
Scheme 24: Pd catalyzes the hydroamination of styrene 4 with aniline 5 to afford the N-phenyl phenylethyl amino product 6 in high yield.[7]
Scheme 25: Pd-catalyzed hydroamination via nucleophilic attack on an allylic intermediate.7
This reaction begins with formation of precatalyst into the catalytically active Pd(0) species I by introducing the bidentate phosphine ligand, DPPF. Metal-hydride J is formed via oxidative addition of HOTf. Styrene 4 coordinates via its terminal alkene to the Pd metal center, forming a cationic Pd intermediate K which is balanced with the OTf anion. b-Hydride insertion into the olefin leads to two possible regioisomers, one of which places the a-carbon in the allylic position of a double bond in the phenyl ring. This in turn leads to an h3-allyl bond to the Pd metal center, L. Aniline 5 nucleophilically attacks this allylic system opposite the side coordinated to the metal center in a trans-like fashion as shown in transition state M. Coordinated OTf anion deprotonates the ammonium species to afford Markovnikov product 6 which easily de-coordinates from the Pd center, yielding 6 and regenerating catalyst I.
The key feature of this mechanism is the formation of the
h3-allylic species K. These species have been isolated and characterized by single crystal X-ray diffraction, indicating that they are lingering intermediates in the catalytic cycle. This suggests that the rate-determining (slow) step is nucleophilic attack of amine 5 on allylic species K. Kinetic isotope effect studies revealed that the regio-determining step is also the rate-determining step (nucleophilic attack of amine). It had been previously hypothesized that regioselectivity was determined by formation of the h3 intermediate L, but these studies revealed formation of h3 intermediate L is reversible. The rate of this reaction increases as the basicity of the amine increases and decreases with more sterically hindered amines, which more generally states the greater nucleophilicity of the amine, the greater the reaction rate.3
It should be noted for this reaction that use of stereochemically-enriched bidentate ligands such as BINAP affords enantioselective products.7 Thus hydroamination with this catalyst system can proceed enantioselectively.
There is an interesting debate regarding the order of the steps from species I to the allylic species L. The path shown above in Scheme 25 is oxidative addition of the acid across the metal, forming metal hydride J which then coordinates olefin, species K. An alternate path involves coordination of olefin first, followed by addition of acid generating allylic species. This latter pathway is proposed for hydroamination of trienes.3
Notice the above example proceeded with Markovnikov selectivity.3 Anti-Markovnikov selectivity has been reported in a Rh- and Ru-mediated hydroamination of vinyl arenes shown in Scheme 26.
Scheme 26: Brief mechanistic explanation for observed anti-Markovnikov addition of amines across styrene with Ru catalysts.3
Ru is able to coordinate an h6-arene system with a pendant ethyl alkene group. Resonance donation into the ring allows for a partial cationic charge to develop on the terminal carbon, creating an electrophilic carbon and enabling anti-Markovnikov addition of the amine. The Ru metal center can absorb the negative charge that gets donated into the ring as a result of this addition; this forms a stable 18 e- species.3
Olefin Insertion into M-H Bond
We will revisit a previously mentioned reaction under a new set of conditions for a brief discussion of the lesser-known mechanistic pathway of olefin insertion into M-H bond. This mechanistic pathway applies mostly to alkene substrates and Pd catalysts.
Scheme 27: Heterogenous Pd-catalyzed hydroamination of styrene 4 with aniline 5 afforded anti-Markovnikov product 7 at high temperatures.[8]
The mechanism for Markovnikov product 6 was analyzed as part of the discussion of nucleophilic attack on metal coordinated-allylic species (Scheme 25). We will now consider a different mechanistic pathway that will afford anti-Markovnikov product, presented in Scheme 28 below.
Scheme 28: Pd-catalyzed hydroamination of styrene with aniline to afford anti-Markovnikov product 7 via olefin insertion into Pd-H bond.8
The mechanism for Markovnikov and anti-Markovnikov product begin in similar ways. The difference occurs upon coordination of styrene 4 to catalytically active species J to afford the five-coordinate 18 e- Pd species N. b-hydride insertion to afford O followed by nucleophilic attack at the a-carbon of the alkyl chain releases product 7 and regenerates catalyst J.
The rate determining step was found to be the nucleophilic attack of amine, from O to J. Additionally regioselectivity was determined by the insertion step, from N to O, with insertion occurring preferentially at the carbon that would reduce steric interactions of the ligand with the alkenyl substituent (phenyl group). This explains the anti-Markovnikov selectivity for this reaction.8
Olefin Insertion into M-N Bond
This final mechanistic pathway resembles closely the pathway presented in (III), except the initial step is oxidative addition of an N-H bond of the amine substrate. This forms two s-bonds, M-N and M-H, of an hydrido-amido complex. The unsaturated C-C bond inserts itself preferentially into the M-N bond, as will be explained below. This mechanistic pathway has been proposed for many late transition metals, including Pd, Pt, Ir, and Ru. It applies to hydroamination of alkenes, alkynes, and vinyl arenes.
We will consider one example reaction (Scheme 29) and follow its catalytic cycle (Scheme 30).
Scheme 29: 2-vinyl pyridine 8 undergoes hydroamination with piperidine 9 with Rh catalyst to yield anti-Markovnikov product 10 in high yield.9
Scheme 30: The catalytic cycle for oxidative addition of piperidine 9 to Rh catalyst for hydroamination of alkene 8 to afford anti-Markovnikov product 10.[9]
The identity of the catalytically active [Rh] species P is not fully defined but is proposed to be a cationic 14 or 16 electron species that can undergo oxidative addition with amine 9. This yields the Rh-hydrido-amido complex Q which coordinates to alkene 8 and allows for b-insertion into the Rh-N bond, yielding Rh-alkyl-hydrido complex R. This complex readily reductively eliminates product 10 and reforms catalytically active species P.
The alkene could also have b-inserted into the Rh-H bond, but this is generally less favorable because reductive elimination of the resultant C-N bond to form product is more difficult to achieve than reductive elimination of C-H bond to form product.3
This reaction proceeded with anti-Markovnikov addition across the double bond. The explanation for regioselectivity in this mechanism appears to be sterics, but there is little experimental evidence to make this statement conclusive.8
DFT studies indicate that oxidative addition and reductive elimination were the rate-determining steps in a Pt-catalyzed hydroamination reaction that proceeded through this mechanism, although the kinetics for the reaction in Scheme 29 are not definitively known.3
Application to Total Synthesis
Secondary enamides are frequently found in biologically active molecules and are important synthons in natural product synthesis. Traditional organic chemistry affords secondary enamides via harsh conditions without stereoselectivity. Transition metal chemistry allows for an efficient method to stereoselectively synthesize these valuable substrates, thus overcoming a dilemma of traditional organic chemistry.
Gooßen’s Ru catalyst can synthesize the E or Z conformations as a function of added ligands.[10] Scheme 31 illustrates usage of Gooßen’s catalyst to afford the Z stereoisomer of 13, also known as lansiumamide A, a natural product.
Scheme 31: Hydroamidation across alkyne 12 with amide 11 using Gooßen’s ruthenium catalyst with the appropriate ligands selected for the desired stereochemical outcome affords nautral product lansiumamide A 13.[11]
OTHER CATALYTIC METHODS
Although the scope of this paper is to address hydroamination through transition metal catalysts and organometallic chemistry, it would be negligent to study hydroamination without considering catalytic methods outside of organometallic chemistry.
Hydroamination can be accomplished through use of acid catalysts. An example is provided in Scheme 32 below.
Scheme 32: Hydroamination of tosylamine 2 over styrene 1 is catalyzed by HOTf in toluene at elevated temperature and sufficient reaction time to yield Markovnikov product 3.3
Additionally, alkali metals such as Na and Li can catalyze hydroamination. Consider the example reaction below.
Scheme 33: Hydroamination of amine 4 with styrene 1 through assistance of n-BuLi to yield anti-Markovnikov product 5 in good yield.3
These are classic reactions taught in introductory organic chemistry classes. They efficiently produce the products of hydroamination that are attained through transition metal chemistry. They often require functionalization of the starting materials to obtain high yields. There are also many opportunities for functional group incompatibility, usually at the expense of yield.
We will very briefly consider the extensive field of mercury facilitated hydroamination. An example reaction is provided below (Scheme 34).
Scheme 34: Mercury facilitated intramolecular hydroamination of an aminoalkene.3
These generally efficient reactions proceed via a three-membered mercuronium, Hg(II), intermediate. One major drawback to these reactions is the toxicity of mercury.3 For applications in total synthesis and the pharmaceutical industry, hydroamination reactions must be consistent and predictable (in terms of regioselectivity, stereoselectivity), afford good yields, proceed at reasonable reaction times and temperatures, and be suitable for human consumption after product purification. Although mercury facilitated hydroamination is well studied, the latter restriction on the reaction’s usefulness in industry puts it at a clear disadvantage. The various transition metal catalysts discussed in sections 1 to 3 fulfill these criteria most effectively, as shown in the total synthesis examples provided at the end of each section.
SUMMARY
This report dealt with catalytic organometallic hydroamination reactions. The first group of catalysts considered were those composed of the Lanthanide series of metals, which catalyze hydroamination of alkenes, alkynes, allenes, and dienes. The catalytic cycle for these reactions involves a 4-membered transition state that forms an azacylce, that can be either vinylic to or allylic to the metal center. These reactions result in a cis addition of amine and hydrogen across the unsaturated C-C bond. Lanthanide metal catalyzed reactions are usually restricted to intramolecular reactions, which leads to inner-sphere attack of the nitrogen on the unsaturated C-C bond and affords cis product.
The second group of catalysts considered were the Early Transition Metals. These catalysts require formation of an M=N (imido) or M-N (amido) species that undergoes a [2+2] cycloaddition mechanism to afford a azametallocycle. There are both intermolecular and intramolecular examples of hydroamination afforded by these catalysts for alkenes, alkynes, and allenes. Thus outer-sphere attack can occur, opening up the possibility for formation of trans product.
Lastly we considered the Late Transition Metals, which offer a broad range of mechanistic possibilities. Both intermolecular and intramolecular examples of these reactions exist, and they generally occur through use of co-catalytic acid. In general, the mechanism is decided based on whether nitrogen attacks the C-C p-bond or whether the C-C p-bond inserts (“attacks”) the M-N bond. There are also examples of the C-C p-bond inserting into M-H bonds, although this is a less favorable process than inserting into M-N bonds. Late Transition Metals catalyze hydroamination across alkenes, alkynes, dienes, and allenes. Examples of anti-Markovnikov addition are abundant in Early Transition Metal and Late Transition Metal catalyzed reactions, and Markovnikov addition is found often in all catalyst groups. Enantioselectivity has been reported through use of chiral ligands. There is still much to be understood about hydroamination chemistry, but its offerings to synthetic organic chemists thus far render it a field of valuable organometallic research.
Table 1 concludes this report and provides the student of organometallic chemistry with an easy reference guide for the information presented above. The table is organized by catalyst type, which constitutes the three main sections of this report, a general assignment of the oxidation state of the metal center throughout each catalytic cycle, schemes for relevant catalytic cycles for further reference, key features of each catalytic cycle, and factors that have been reported to affect regioselectivity.
Table 1: A general breakdown of the content in this report categorized by the three main sections of transition metal (TM) catalyst types (Lanthanide, Early TM, and Late TM), the general oxidation state of the metal during the catalytic cycle, pertinent mechanistic schemes for each general catalytic cycle, key features of each mechanism, and factors that determine regioselectivity discussed in this report.
Catalyst Type
Metal Center Oxidation State
Catalytic Cycles
Key Features
Regioselectivity (Markovnikov vs. anti-Markovnikov)
Lanthanide
(III)
Scheme 4
*Azametallocycle transition state/transient species
*Azacycle intermediate
*Electronic effects govern addition
Early TM
(IV)
Schemes 14, 19
*Primary amines form metal imido (M=N) species
*Secondary amines form metal amido (M-N) species
*Azametallocycle intermediate
*Steric effects govern addition
Late TM
(II)
Schemes 22, 24, 28, 30
*Varied mechanistic schemes
*Often with co-catalytic acid
*Occur via nucleophilic attack by amine or olefin insertion
* C-C p -bond coordinates to metal first (for all but Scheme 30)
*Oxidative addition of amine first (Scheme 30)
*Steric effects govern addition
REFERENCES
[1] Hong, S.; Marks, T.J. Acc. Chem. Res. 2004, 37, 673-686.
[2] Yuen, H.F.; Marks, T.J. Organometallics, 2009, 28(8). 2423–2440.
[3] Müller, T.E.; Hultzsch, K.C.; Yus, M.; Foubelo, F.; Tada, M. Chem. Rev. 2008, 108, 3795-3892.
[4] Müller, T.E.; Beller, M. Chem. Rev. 1998, 98(2), 675-703.
[5] McGrane, P.L.; Livinghouse, T. J. Org. Chem. 1992, 57(5), 1323-1324.
[6] Penizen, J.; Su, R.Q.; Müller, T.E. J. Mol. Cat. A: Chem. 2002, 182-183, 489-498.
[7] Vo, L.K.; Singleton, D.A. Org. Lett., 2004, 6(14), 2469-2472.
[8] Sievers, C.; Jiménez, O.; Knapp, R.; Lin, X.; Müller, T.E.; Türler, A.; Wierczinski, B.; Lercher, J.A. J. Mol. Cat. A.: Chem. 2008, 279, 187-199.
[9] Beller, M.; Trauthwein, H.; Eichberger, M.; Breindl, C.; Müller, T.E. Eur. J. Inorg. Chem. 1999, 1121-1132.
[10] Gooßen, L.J.; Rauhaus, J.E.; Deng, G. Angew. Chem. Int. Ed. 2005, 44, 4042-4045.
[11] Gooßen, L.J.; Blanchot, M.; Arndt, M.; Kifah, S.M. Synlett. 2010, 11, 1685-168
Contributors and Attributions
• Alec Beaton and Victoria Banas | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Catalysis/Catalyst_Examples/Hydroamination.txt |
Introduction
Hydrosilylation, also known as hydrosilation, is one of the most useful catalytic reactions leading to the formation of organsilanes and organosilicones, which have a variety of applications in industry and as intermediates in organic chemistry. Hydrosilylation occurs via the addition of H-Si to an unsaturated bond such as carbon-carbon bond, carbon-oxygen bond, carbon-nitrogen bond, nitrogen-nitrogen bond and nitrogen-oxygen bond using a metal catalyst, Lewis Acid, or radical initiator. The basic reaction can be described using the following scheme:
Mechanism of Metal Catalyzed Hydrosilanes
Hydrosilylation reaction is often completed with the association of a transition metal catalyst. In the first reported hydrosilylation reaction, the reaction was performed with trichlorosilane and 1-octene using crystalline diacetyl peroxide as the catalyst[3]. Due to the sensitivity and reactivity of diacetyl peroxide, safer metal catalysts are being used, the most common being platinum. Here is a typical catalyst cycle describing the mechanism:[4]
In the first step a Si-H bond undergoes oxidative addition to a metal center. The alkene coordinates and then undergoes insertion. The hydrogen is then added to the alkene through H beta-insertion. In the final step the alkylsilyl Pd(II) complex undergoes reductive elimination to deliver the adduct and return the metal to the original oxidation state.
A well-known mechanism of metal catalyzed hydrosilylation is the Chalk-Harrod mechanism. It proceeds through oxidative addition of the Si-H to the metal center and then alkene insertion to the metal hydride. Reductive elimination of Si-C then occurs and gives the final product. A modified version of the Chalk-Harrod method was also discovered where the reaction proceeds through an ethylene insertion into the metal silyl bond. Then reductive elimination of C-H occurs to give the vinylsilane product.[5]
Reactivity
The reactivity of hydrosilylation is influenced by many factors: substrate, silane, transition metal catalyst, ligand etc. which makes the reaction diverse as well as complicated. Studies have been performed to determine the reactivity influenced by each of these factors. Because there are a large variety of substrates that can be used, we will focus on alkenes. Here are some general rules:
Substrate Group: Alkene
Reaction rate: 1-alkene > 2-alkene > 3-alkene
Generally, the more substituted the alkene group, the slower the reaction would be. This is due to the more substituted alkene group being bulkier, therefore it will be harder for the alkene coordination to happen to the metal. More substituted alkene groups also make it more difficult for the H atom to undergo compete beta-insertion.
Silyl Group
Reaction rate: SiHCl3 > (C6H5)3SiH > (C2H5)3SiH
V0(105): 110 12 1.2
From this set of data, we can derive that the rate of reactivity would increase depending on the substituted groups on the silanes. The order of reactivity is: chlorine > aromatic rings > alkanes. However, this rule is not always true. If we substituted the aromatic groups in (C6H5)3SiH with Cl, the reaction rate would be:
Reaction rate: (C6H5)3SiH > (C6H5)2SiH2 > (C6H5)SiH3
V0(105): 12 1.7 0.69
This shows that the substituted groups do not affect the reactivity individually. The overall regularity for change of silanes is complicated and varies with each group.
Transition Metal
The two-stage hydrosilyation of acetylene by trichlorosilane at 150oC in the gaseous phase using a flow method provided a test reaction for the catalytic activity of different transition metals. This reaction was supported by gamma-Al2O3 and an active carbon. Below is the yield order for the reaction:
Yield for gamma-Al2O3 supported reaction: Pd > Pt > Ru > Rh > Ir
Yield for active carbon supported reaction: Rh > Ir > Pd > Ru > Pt
Such a sequence could be used to help choosing catalyst based on different supporting materials.
Ligand Type
The metal catalyst is usually at an oxidation state of 0 or two oxidation states lower than what is the maximum for that metal. If there is complete saturation of the metal, then a ligand must be lost before the incorporation of the silane in the oxidative addition step. The ligands attached to the metal are usually made up of neutral and anionic ligands. Before coordination with the silane, the metal center must be unsaturated and therefore usually a ligand is lost prior to oxidative addition.
$L_{n}M^{X+}\rightarrow L_{n-1}M^{X+}\frac{\rightarrow}{HSiR_{3}}(H)L_{n-1}M^{(X+2)}SiR_{3} \nonumber$
L = neutral
$L_{n-1}M^{X+}$= oxidative metal
$(H)L_{n-1}M^{(X+2)}SiR_{3}$= oxidative addition product
Common ligands used in TM complexes: CO and phosphines
Regio and Stereoselectivity
The stereo- and regioselectivity of hydrosilylation reactions are highly dependent on the substrate, catalyst, and solvent. There are almost always side products, and cocatalysts or directing groups are often necessary for regiocontrol. In the Chalk-Harrod mechanism the H-Si group is added across an alkene and the anti-markovnikov products are obtained. Hydrosilylation of terminal alkynes have three possible products whereas hydrosilylation of internal alkynes has four possible products:
Terminal Alkyne:
Internal Alkyne:
In terminal alkynes, cis- addition of hydrosilane gives the (E)-product while trans- addition gives the (Z)-product. An example of specific selectivity is in platinum catalized hydrosilylation: the reaction proceeds through cis-addition and produces α and β (E)-alkenylsilanes.
Other Methods of Hydrosilylation
Radical Initiation
Another method of hydrosilylation is through a free radical mechanism. In this case, a radical initiator such as a superoxide is needed to provide a radical source. The mechanism can be described using two steps through the following scheme:[9]
Step 1:
Step 2:
Lewis Acid
Hydrosilylation can also occur through the addition of an alkene or an alkyne to a silyl group and using an acid.[10]
Contributors and Attributions
• Giovanna Romero and Xinyu (Wayne) Wang | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Catalysis/Catalyst_Examples/Hydrosilylation.txt |
As you may have learnt in your general chemistry course, metathesis is the exchange of atoms or functional groups in the substrates and the rearrangement of their matching partners to form new compounds. Simple examples were well-studied in traditional inorganic chemistry, but their analogs in organic reactions were vague and hard to proceed without suitable catalysts.
eg:
Olefin Metathesis:
The Nobel Prize in Chemistry of 2015 was shared by Yves Chauvin, Robert H.Grubbs and Richard R.Schrock for their contributions to the field of Olefin Metathesis.
Olefin Metathesis[1] involves two olefin substrates which form a four-membered ring intermediate (first proposed by Chauvin) and then rearrange the substituents to form two new carbon-carbon double bonds. Late transition metal alkylidene compounds (eg. Grubbs Ru catalyst) greatly broaden the methodology to form carbon-carbon double bonds with high functional group tolerance. Schlock designed and synthesized another group of Mo,W and Re catalysts which exhibit higher efficiency for this reaction[2][3].
From left to right: Yves Chauvin, Robert H.Grubbs and Richard R.Schrock
Catalysts used for this reaction [4] [5a-g]:
1. The earliest heterogenous catalytic system is composed of high-valent transition metal halide, oxide and oxo-halide. Alkyl zinc or alkyl aluminum is used as the co-catalyst. Also, aluminum oxide or silica is added as support.
Characteristics:
• Good Reactivity
• Extremely poor functional group tolerance
(Why? Because it is a lewis acid, it will be poisoned)
• Complicated and ambiguous catalytic mechanisms
2. Titanocene-based catalyst as Tebbe’s reagent is transformed into “Cp2Ti=CH2” in the presence of base. And it acts as an active metathesis catalyst.
Characteristics:
• Mild Reactivity
• Resemble stoichiometric Wittig-Like reactions
• Mechanism resembles Olefin Metathesis except the irreversibility of the last step
(What’s the limitation of this method? It can only perform methylenation, not alkylidenation.)
3. Schrock pioneered the preparation of a series of W, Mo and Re catalysts for olefin metathesis reactions. And the Mo complex with arylimido ligand stands out with impressive activity and high turnover frequency[6a,b].
Characteristics:
• Lower functional group tolerance
• Air and water sensitive
• Multi-substituted and hindered substrates accessible
(Why high reactivity? Because the coordinative and electronic unsaturation along with the bulky ligands make them good electrophilic agents and reduce the bimolecular decomposition)
4. Grubbs later rationalized the design and synthesized another group of Ru-based catalysts with unique properties. His 2nd generation of catalyst employed the nitrogen heterocyclic carbene which facilitated the dissociation of trans-ligands. Hoveyda collaborated with Grubbs to make a system with chelating ligands which was more applicable to sterically demanding and electron-poor substrates [7][8].
Characteristics:
• High functional group tolerance
• Stable in water conditions
• Lower reactivity than Schrock’s catalyst
Comparison Between Grubbs and Schrock Catalysts [9][10]:
Catalyst
Grubbs
Schrock
Reactivity
Low
High
Substrate
Less bulky and strained alkenes
Sterically-demanding and electron-deficient alkenes
Rxn Condition
Stable in bench conditions
Sensitive to air and water
Functional Group Tolerance
High
Poor
General Mechanism:
Chauvin incubated the earliest imaginary mechanism involving a [2+2] cycloaddition reaction between the transition metal alkylidene complex and the olefin substrate to form a four membered ring metallacyclobutane intermediate. Then the metallacycle breaks up in an opposite fashion to yield both the new alkylidene and the olefin product.
According to the traditional organic chemistry using MO theory, such a ring addition reaction is symmetrically forbidden and requires photochemical condition to proceed. However, the introduction of metal alkylidene fragment with d-orbitals breaks the symmetry and facilitates the reaction.
Categories of Olefin Metathesis:
1. Cross Metathesis
The transalkylidenation of two terminal alkenes with release of ethene is catalyzed by the Grubbs catalyst. Both homocoupling and heterocoupling can occur and the E/Z selectivity is hard to control.
Mechanism [4]:
2. Ring-Closing Metathesis
Ring-closing metathesis allows formation of cyclic alkenes ranging from 5 to 30 members, in which the E/Z selectivity is related to the ring strain. The 2nd generation Grubbs catalyst is more versatile for this reaction.
Mechanism:
3. Ring-Opening Polymerization Metathesis (ROMP) [11a-c]
Ru based catalysts can open the strained ring with a second alkene via the cross-metathesis mechanism to form products containing terminal vinyl groups. Further metathesis can occur to form long polymer chains.
Mechanism[12a]:
4. Enyne Metathesis[12b-c]
Alkyne and alkene can have similar reaction to produce 1,3-diene, and this intermolecular process is called cross-enyne metathesis, whereas the intramolecular reactions are referred to as ring-closing anyone metathesis(RCEYM).
5. Acyclic Diene Metathesis Polymerization(ADMET)[12d]
Alpha-omega dienes can be used to produce polymers through ADMET method, and the reverse of this reaction has been studied as a potential means to recycle rubber tires.
(What’s the driving force for these reactions? To form volatile alkenes or to release ring strain)
Detailed mechanistic studies of Grubbs group catalysts in Cross Metathesis[13]
Classification of Olefins:
Type I:Fast homodimerization
Olefins homodimerize rapidly and both the homodimers
and monomers can enter the catalytic CM cycle.
eg. Terminal Olefins
Primary allylic alcohols
Esters
Decreasing in Olefin Reactivity.
Increasing Steric Congestion.
Type II:Slow homodimerization
Slow olefin homodimerization and the homodimers can
be only partly consumed in latter CM catalytic reactions.
eg. Styrene
Secondary allylic alcohols
Increasing Electron Deficiency.
Type III:No homodimerization
Substrates can’t homodimerize but can facilely undergo
CM with other type I or type II olefins.
eg. Vinyl siloxanes
Type IV:Spectators
Don’t participate CM catalysis but also don't block the
catalyst’s activity toward other olefins.
eg. Quaternary allylic carbon-containing olefins
Hint: Some reactants are called bridge type olefins and the reactivity depends on the substituent pattern and catalyst choices. In other words, these substrates can be classified into different groups depending on the reaction conditions.
Reactivity Matrix for Cross Metathesis[13][14][15]:
ST: Statistical, selectivity achieved by excessive amount of one substrate(10 equiv).
S: Selective, selectivity achieved by intrinsic thermodynamics.
SI: Slow reaction.
NR: No reaction.
NS: Non-selective
Reactivity Matrix for Cross Metathesis
I
ST
II
S
NS
III
S
SI
NS
IV
NR
NR
NR
NR
I
II
III
IV
Pros and Cons of Cross Metathesis Reactions:
Pros
Cons
Excellent functional tolerance
Commercially available olefin substrates
Low catalyst loading
Mode of selectivity makes reactivity predictable
Model of E:Z unclear
Intrinsic cross product selectivity remains problematic
For complex molecules, unpredictable
Contributors
• Zhe Xu and Jianing Xu | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Catalysis/Catalyst_Examples/Olefin_Metathesis.txt |
The Ziegler-Natta (ZN) catalyst, named after two chemists: Karl Ziegler and Giulio Natta, is a powerful tool to polymerize α-olefins with high linearity and stereoselectivity (Figure 1). A typical ZN catalyst system usually contains two parts: a transition metal (Group IV metals, like Ti, Zr, Hf) compound and an organoaluminum compound (co-catalyst). The common examples of ZN catalyst systems include TiCl4 + Et3Al and TiCl3 + AlEt2Cl.
In 1953, German chemist Karl Ziegler discovered a catalytic system able to polymerize ethylene into linear, high molecular weight polyethylene which conventional polymerization techniques could not make.1 The system contained a transition metal halide with a main group element alkyl compound (Figure 2).
Following the catalytic design, Italian chemist Giulio Natta found that polymerization of α-olefins resulted in stereoregular structures,2 either syndiotactic or isotactic, depending on the catalyst used (Figure 3). Because of these important discoveries, Karl Ziegler and Giulio Natta shared the Nobel Prize in Chemistry in 1963.
Advantages over traditional polymerization method
Traditionally, polymerization of α-olefins was done by radical polymerization (Figure 4). Problem with this technique was that the formation of undesired allylic radicals leaded to branched polymers.3 For example, radical polymerization of propene gived branched polymers with large molecular weight distribution. Also, radical polymerization had no control over stereochemistry. Linear unbranched polyethylene and stereoregulated polypropylene could not be fabricated by free radical polymerization. This technique largely limited the potential applications of these polymeric materials.
The invention of ZN catalyst successfully addressed these two problems. The catalyst can give linear α-olefin polymers with high and controllable molecular weights. Moreover, it makes the fabrication of polymers with specific tacticity possible. By controlling the stereochemistry of products, either syndiotactic or isotactic polymers can be achieved.
Mechanism of Ziegler-Natta catalytic polymerization
Activation of Ziegler-Natta catalyst
It is necessary to understand the catalyst’s structure before understanding how this catalyst system works. Herein, TiCl4+AlEt3 catalyst system is taken as an example. The titanium chloride compound has a crystal structure in which each Ti atom is coordinated to 6 chlorine atoms. On the crystal surface, a Ti atom is surrounded by 5 chlorine atoms with one empty orbital to be filled. When Et3Al comes in, it donates an ethyl group to Ti atom and the Al atom is coordinated to one of the chlorine atoms. Meanwhile, one chlorine atom from titanium is kicked out during this process. Thus, the catalyst system still has an empty orbital (Figure 5). The catalyst is activated by the coordination of AlEt3 to Ti atom.
Initiation step
The polymerization reaction is initiated by forming alkene-metal complex. When a vinyl monomer like propylene comes to the active metal center, it can be coordinated to Ti atom by overlapping their orbitals. As shown in Figure 6, there is an empty dxy orbital and a filled dx2-y2 orbital in Ti’s outermost shell (the other four orbitals are not shown here). The carbon-carbon double bond of alkene has a pi bond, which consists of a filled pi-bonding orbital and an empty pi-antibonding orbital. So, the alkene's pi-bonding orbital and the Ti's dxy orbital come together and share a pair of electrons. Once they're together, that Ti’s dx2-y2 orbital comes mighty close to the pi-antibonding orbital, sharing another pair of electrons.
The formed alkene-metal complex (1) then goes through electron shuffling, with several pairs of electrons shifting their positions: The pair of electrons from the carbon-carbon pi-bond shifts to form Ti-carbon bond, while the pair of electrons from the bond between Ti and AlEt3’ ethyl group shifts to form a bond between the ethyl group and the methyl-substituted carbon of propylene (Figure 7). After electron shuffling, Ti is back with an empty orbital again, needing electrons to fill it (2).
Propagation step
When other propylene molecules come in, this process starts over and over, giving linear polypropylene (Figure 8).
Termination step
Termination is the final step of a chain-growth polymerization, forming “dead” polymers (desired products). Figure 9 illustrates several termination approaches developed with the aid of co-catalyst AlEt3.4
Mechanistic study: kinetic isotope effect experiments
Unlike the mechanism discussed above, there is also a competing mechanism proposed by Ivin and coworkers.5 They proposed that a 1,2-hydrogen shift occurs prior to monomer association, giving a carbene intermediate (Figure 10).
To determine the actual mechanism, Grubbs6 conducted kinetic isotope effect (KIE) experiments. Due to their different weights, carbon-deuterium bond reacts slower than carbon-hydrogen bond (Figure 11). If such bonds are involved in the rate-determining step, the isotopic species should proceed in lower rate.
In Grubbs’s experiment, deuterated ethylene and normal ethylene were catalyzed by Ziegler-Natta catalyst with 1:1 ratio (Figure 12). The result showed that H/D ratio in the resulting polymers was still 1:1.6 This indicated that no carbon-hydrogen bond cleavage or formation is involved in the rate determining step. Therefore, the mechanism proposed by Ivin was excluded.
Regio- and stero-selectivity
Regio-selectivity
For propene polymerization, most ZN catalysts are highly regioselective, favoring 1,2-primary insertion [M-R + CH2=CH(Me)=M-CH2-CH(Me)(R)] (Figure 13), due to electronic and steric effects.7
Stereo-selectivity
Stereochemistry of polymers made from ZN-catalyst can be well regulated by rational design of ligands. By using different ligand system, either syndiotactic or isotactic polymers can be obtained (Figure 14).
The relative stereochemistry of adjacent chiral centers within a macromolecule is defined as tacticity. Three kinds of stereochemistry are possible: isotactic, syndiotactic and atactic. In isotactic polymers, substituents are located on the same side of the polymer backbone, while substituents on syodiotactic polymers have alternative positions. In atactic polymers, substituents are placed randomly along the chain.
Choice of ZN catalyst regulates the stereochemistry. We use propylene polymerization as an example here. Recall the mechanism section, a monomer approaches the metal center and forms a four-membered ring intermediate. The binding of a monomer to the reactive metal-carbon bond should occur in from the least hindered site.8 As shown in Figure 15, the trans-complex in which methyl on the growing chain is trans to the methyl group on the incoming monomer, should be energetically favored than the cis-complex, as the trans-complex experience less steric effect.
Following the trans- complex, the methyl group on the newly added monomer is trans to that on the previous monomer. The step repeats so that a syndiotactic polypropylene is obtained (Figure 16a). However, if the metal center is coordinated with bulky ligands (e.g. –iBu, -Et2 groups), as denoted by Y in Figure 16b, the incoming monomer will adopt a cis-conformation to avoid serious steric effect with the bulky ligand. Thus, at the presence of bulky ligand, the propylene monomer is cis to the growing chain. This results in a isotactic product.
Applications
ZN catalysts have provided a worldwide profitable industry with production of more than 160 billion pounds and creation of numerous positions.9 These products have been extensively applied in different areas, largely improving the quality of people’s life. They can catalyze α-olefins to produce various commercial polymers, like polyethylene, polypropylene and Polybutene-1. Polyethylene and polypropylene is reported to be the top two widely used synthetic plastic in the word.10
Among the polyethylene fabricated by ZN catalysts, there are three major classes: high density polyethylene (HDPE), linear low density polyethylene (LLDPE), and ultra-high molecular weight polyethylene (UHMWPE). HDPE, a linear homopolymer, is widely applied in garbage containers, detergent bottles and water pipes because of its high tensile strength. Compared with HDPE whose branching degree is quite low, LLDPE has many short branches. Its better toughness, flexibility and stress-cracking resistance makes it suitable for materials like cable coverings, bubble wrap and so on. UHMWPE is polyethylene with a molecular weight between 3.5 and 7.5 million. This material is extremely tough and chemically resistant. Therefore, it is often used to make gears and artificial joints.
Compared to polyethylene, polypropylene has enhanced mechanical properties and thermal resistance because of the additional methyl group. Moreover, isotactic polypropylene is stiffer and more resistant to creep than atactic polypropylene. Polypropylene has a wide range of applications in clothing, medical plastics, food packing, and building construction.
Substrate scope and limitations
ZN catalysts are effective for polymerization of α-Olefins (ethylene, propylene) and some dienes (butadiene, isoprene). However, they don’t work for some other monomers, such as 1.2 disubstituted double bonds. Vinyl chloride cannot be polymerized by ZN catalyst either, because free radical vinyl polymerization is initiated during the reaction. Another situation that ZN catalysts don’t work is when the substrate is acrylate. The reason is that ZN catalysts often initiate anionic vinyl polymerization in those monomers.11
Contributors and Attributions
• Fangyuan Dong and Ru Deng | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Catalysis/Catalyst_Examples/Olefin_Polymerization_with_Ziegler-Natta_Catalyst.txt |
The Sonogashira reaction (also called the Sonogashira-Hagihara reaction) is the cross coupling of aryl or vinyl halides with terminal alkynes to generate conjugated enynes and arylalkynes (Scheme 1). The reaction typically proceeds in the presence of a palladium(0) catalyst, a copper(I) cocatalyst, and an imine base.1 Alternative procedures describe Sonogashira coupling reactions performed without the Cu(I) cocatalyst.
Scheme 1: General reaction for Sonogashira Coupling.
Preceding the Sonogashira reaction, Heck and Diek and Cassar reactions reported the arylation or alkenylation of alkenes via phosphane-palladium complexes.2, 3 However, both procedures required harsh reaction conditions, such as high temperatures (100oC). In 1975, Kenkichi Sonogashira, Yasuo Tohda, and Nobue Hagihara reported the cross coupling of iodobenzene and acetylene in the presence of Pd(Ph3)4Cl2 and CuI.4 The use of the Cu+ cocatalyst, a procedural design that built on the work of Castro and Stephens,5 increased the reactivity of the reagents and allowed the reaction to proceed at room temperature. As a result, the Sonogashira reaction remains one of the most popular cross-coupling reactions for the formation of C(sp2)-C(sp) bonds, with its use extending to applications in pharmaceuticals, nanomaterials, and natural products. In fact, Chinchilla and Najera (2007) reported a significant increase in interest for Sonogashira coupling within the past thirteen years.1 Efforts within the field aim to further improve reaction conditions, catalyst designs, and mechanistic understandings of Sonogashira coupling.
MECHANISM
Palladium and Copper Co-catalyzed Cycle
Palladium and copper co-catalyzed Sonogashira coupling is reported to proceed via two independent catalytic cycles (Scheme 2).6-18
Scheme 2: Catalytic cycle of palladium and copper co-catalyzed Sonogashira reaction.
Prior to oxidative addition, the 14-electron PdL2 complex is generated through a reductive process known as σ-complexation-dehydropalladation-reductive elimination. In this process, palladium(II) is reduced by forming a complex with electron donor compounds that function as either ligands or solvent within the reaction (Scheme 3).
Scheme 3: σ-complexation-dehydropalladation-reductive elimination with amide.
The Pd0 complex then undergoes oxidative addition of R1-X to form a four-coordinated palladium complex. At this point in the reaction, the palladium cycle intersects with the copper cycle. For the co-catalyzed copper cycle, the base typically used is amide. Due to the relatively low basicity of the amide, a π-alkyne-copper complex must be formed to increase the acidity of the alkyne for the alkyne to undergo deprotonation. Following deprotonation in the copper-catalyzed cycle, copper acetylide is formed. It is noteworthy that the formation of the palladium acetylide from the copper acetylide and four-coordinated palladium complex is typically considered the rate determining step. This step is called transmetalation. Trans/cis isomerization occurs next, which is then followed by reductive elimination to afford the final product.1
Copper-free Catalytic Cycle
For copper-free Sonogashira coupling (Scheme 4), the first step is also the oxidative addition to form the four-coordinated palladium complex. Because the basicity of amine is insufficient for the deprotonation of alkyne, the dissociation of a neutral ligand and the formation of a π-alkyne-palladium complex occurs.19 This is followed by deprotonation of the alkyne to form palladium acetylide. Then trans/cis isomerization and reductive elimination occur stepwise to generate the final product and Pd0L2 catalyst.
Scheme 4: Catalytic cycle of copper-free Sonogashira reaction.
Mechanistic Studies
While the exact mechanism of Sonogashira coupling is not known, the catalytic cycles depicted in Scheme 2 are generally accepted. Efforts to isolate and characterize proposed palladium intermediates of the homogeneous Sonogashira reaction have proven difficult, yet some transient species have successfully been identified with multinuclear NMR spectroscopy.1 Alternative methods have turned towards using heterogeneous catalysts and analyzing surface transient organometallic intermediates, as well as performing kinetic measurements using gas chromatography, to corroborate the proposed mechanism.20-22
The real catalyst involved in the catalytic cycle is still under debate. A monoligated palladium catalyst can be observed when the neutral ligand is bulky, suggesting that the catalyst undergoes dissociation prior to oxidative addition.23 It has also been shown that the Pd0L2 will form a corresponding anionic palladium complex if the solution contains anions such as halides. For instance, [L2Pd0Cl]- can be generated from Pd0L2 in the presence of a chloride ion.24 Therefore, it is possible that an anionic palladium complex is the real catalyst in the catalytic cycle.25
As discussed previously, amide is typically the base used in the Sonogashira reaction, and it is proposed that a π-alkyne-copper complex is formed to increase the acidity of the alkyne. However, this proposed π-alkyne-copper complex has not been directly observed.24 In 2005, Berger et al observed a π-alkyne-silver complex using NMR in the palladium and silver co-catalyzed Sonogashira coupling.25 Experimental reports such as this provide evidence for a similar π-alkyne complex forming in the palladium and copper co-catalyzed coupling reaction.
The mechanism for copper-free Sonogashira coupling has been studied with chemical kinetic experiments and density functional theory (DFT) calculations. In particular, DFT calculations have recently proposed an alternative mechanistic pathway, named carbopalladation (Scheme 5). However, the DFT calculations and theoretical mechanistic studies concluded that the activation barrier for this cycle is too high, and subsequently supported Scheme 3 as the observed mechanism.26
Scheme 5: Carbopalladation cycle for copper-free Sonogashira coupling.
Regio- and Stereoselectivity
Disubstituted aryl or vinyl halides have allowed studies to probe at the regioselectivity of Sonogashira coupling. For compounds with two different halides, such as 2-bromo-4-iodo-quinoline, the acetylene adds to the site with the more reactive iodide substituent (Scheme 6). Substrates that have the same halide substituent demonstrate alkynylation at the more electrophilic site (Scheme 7).27
Scheme 6: Sonogashira coupling of disubstituted quinoline.
Scheme 7: Sonogashira coupling of dichloro aryl substrate.
Sonogashira coupling offers a straightforward synthetic route with regards to stereochemistry. The formation of the C(sp2)-C(sp) bonds does not generate new stereocenters in the final product, and no migrations have been reported.28 Additionally, the stereochemical information present in the starting materials is retained in the final product (Scheme 8).29
Scheme 8: Preparation of enyne where stereochemistry is retained.
REACTION CONDITIONS
Substrate Scope
The first step of the catalytic cycle is oxidative addition of the aryl or vinyl halide to the phosphane-palladium complex. The general trend of substrate reactivity towards oxidative addition follows:
vinyl iodide ≥ vinyl triflate > vinyl bromide > vinyl chloride > aryl iodide > aryl triflate ≥ aryl bromide >> aryl chloride
Activation of aryl chlorides is particularly difficult due to the lower electrophilicity of the compound. Despite the favorable reactivity of iodo-substrates, these reagents are expensive and unstable compared to the other halide and triflate compounds. Less reactive substrates have been modified to increase activation towards oxidative addition. Such modifications include placing electron withdrawing groups (EWGs) on R1 to decrease electron density between R1 and X (Scheme 1). The reactivity towards oxidative addition also proceeds more favorably with the use of electron rich phosphane ligands.
The terminal alkyne substrate exhibits a relatively broad range of functional group compatibility. Sonogashira et al originally used acetylene, but subsequent studies have shown that the alkyne can accommodate aryl, hetaryl, alkenyl, alkyl, and trialkylsilyl functional groups. In some cases, the functional group of the alkyne helps facilitate Sonogashira coupling in the absence of the copper cocatalyst. For example, trimethylsilylacetylene has been used to generate the terminal acetylene in a process known as “sila”-Sonogashira coupling.30
Ligand Design and Applications
Changes in catalytic activity and stability of palladium catalysts in Sonogashira reactions can be achieved by modifying the catalyst ligands. Electron-rich phophane ligands, instead of the commonly used triphenylphosphane ligands, can increase the rate of oxidative addition of the aryl halides.1 Additionally, sterically bulky ligands can promote the dissociation of the active palladium catalyst from the resting state.31 Thus, bulky ligands are used to generate corresponding palladium complexes from weakly coordinated palladium complexes, such as Pd(OAc)2, Pd2(dba)3. The combination of bulky and electron rich ligands achieves more efficient Sonogashira couplings. Ligand 1, both bulky and electron rich, is used in the copper-free Sonogashira reaction in the presence of (PhCN)2Cl2 and cesium carbonate as a base. 32 Ligand 2 can catalyze the co-catalyzed Sonogashira reaction of aryl iodides at room temperature when combined with Pd2(dba)3.33 Furthermore, the corresponding palladium catalyst generated from Pd(OAc)2 and bis-(tert-butyl)aminomethylphosphane, ligand 3, is reactive enough that it can catalyze the copper-free Sonogashira reaction of aryl chloride efficiently.34
Classification and properties of palladium catalysts
Palladium-phosphorus complexes.
Typically, Sonogashira reactions proceed under homogeneous reaction conditions in the presence of palladium phosphorus complexes and copper(I) catalysts. The most common catalyst used in Sonogashira reactions is Pd(PPh3)4 and Pd(PPh3)2Cl2, although bidentate ligands have also been used, such as Pd(dppe)Cl2, Pd(dppp)Cl2, or Pd(dppf)Cl2. While Pd(PPh3)2Cl2 is more soluble and stable than Pd(PPh3)4, both catalysts require up to 5% loading to afford good yield.1 Therefore, an important aspect of the field is developing more efficient palladium phosphorus catalysts. For instance, it has been shown that the mixing of Pd2(dba)3/P(t-Bu)3 can catalyze the cross-coupling of terminal alkyne and p-toluenesulfonyl chloride to generate the corresponding alkyne in good yield in the presence of potassium carbonate (Scheme 9).35
Scheme 9: Palladium phosphorus complex catalyzed desulfitative cross-coupling.
Palladium-nitrogen complexes
Palladium-nitrogen catalysts are also effective for Sonogashira coupling, particularly for copper-free reactions.36 Catalyst 3, a bis-imidazolyl-derived palladium complex, can facilitate the coupling of terminal alkynes and aryl iodides with very low catalyst loading (0.02 mol %) when used in the presence of piperidine.37 Similarly, bis-oxazoline palladium catalyst 4 is used for the cross coupling of phenylacetylene and iodobenzene with low catalyst yield (0.055 mol %) in the presence of CuI.38, 39 The dipyrimidyl-palladium catalyst 5 can be used in the cross coupling of aryl diiodides or aryl dibromide to achieve double coupling with phenylacetylene in good yield (Scheme 10). 40
Scheme 10: Dipyrimidyl-palladium complex catalyzed Sonogashira reaction.
N-Heterocyclic Carbene (NHC) Palladium Complexes.
N-Heterocyclic carbenes (NHCs) can efficiently replace phosphane ligands on palladium complexes due to their ability to behave similarly to σ-donor ligands. Catalyst 6 can catalyze the Sonogashira reaction of aryl bromides in the presence of PPH3 and CuI.41 Bis-carbene palladium catalyst 7 has also been reported to catalyze copper co-catalyzed the Sonogashira coupling of aryl bromides with alkynes in boiling pyrrolidine with good yield.42
NHC palladium catalysts can also be used to catalyze copper-free Sonogashira reactions. For example, NHC-derived catalyst 8 is used for the cross-coupling of phenylacetylene and 2-bromoacetopheone in the presence of triethylamine.43 NHC-derived palladacycle 9 is used for the same copper-free Sonogashira reaction with relatively low catalyst loading (0.1 mol %), although the reaction achieves relatively low yield.44 In general, though, palladacycle complexes are stable and demonstrate favorable catalytic activity in coupling reactions.
LIMITATIONS
Despite the popularity of the reaction, Sonogashira coupling exhibits several important limitations. The copper salts used in the co-catalyzed reaction are environmentally unfriendly and difficult to recover from the reaction mixture. Additionally, upon exposure to air, the copper acetylide will undergo a homocoupling side reaction, decreasing the overall efficiency of the reaction. 45
Although copper-free Sonogashira reactions can avoid some problems discussed, this kind of reaction is also environmentally unfriendly because it requires the high loading of bases such as amines, 1 and it is also reported that some commercially available palladium catalysts such as PdCl2 contain trace amount of copper salt. Therefore, it is possible that copper-free Sonogashira reactions are not truly “copper-free,” and a trace amount of copper salt may still have a notable influence on the overall reaction. 46
APPLICATIONS IN SYNTHESIS
The mild reaction conditions and relatively broad range of substrate compatibility allow Sonogashira coupling to find use in a great number of synthetic applications. As a reaction that retains the stereospecific information contained in the starting material, Sonogashira coupling is particularly useful for the generation of enynes and enediynes, an important component for natural products. Many metabolites require the incorporation of enyne moieties. The enyne moiety is generated by coupling a vinylic halide with an alkyne. Sonogashira coupling is used in the synthesis of (-)-isoprelaurefucin, a metabolite that comes from red algae. The intermediate of this total synthesis is achieved with the coupling of an vinyl iodide with trimethylsilylacetylene (Scheme 11).1 The total syntheses of natural products such as benzylisoquinolines, indole alkaloids, and benzofuropyranones also use Sonogashira coupling to form intermediate products. For example, the Sonogashira protocol is used for the coupling of aryl iodide to an aryl acetylene to eventually form bulgaramine, a benzindenoazepine alkaloid (Scheme 12).1
Scheme 11: Synthesis of (-)-isoprelaurefucin.
Scheme 12: Synthesis of bulgaramine.
While an important reaction in the synthesis of natural products, Sonogashira coupling also plays an important role in materials chemistry. The alkyne substrates provides the opportunity to generate extended networks of highly conjugated pi systems, giving rise to eletrooptical and electronic properties in molecules. Poly(phenyleneethynylene)s (PPEs) are synthesized in a one-pot polymerization via the coupling of aryl alkynes with aryl iodides (Scheme 13).1
Scheme 13: Polymerization with Sonogashira Coupling.
REFERENCES
1. Chinchilla, R.; Najera, C. Chem. Rev. 2007, 107, 874.
2. Diek, H. A.; Heck, F. R. J. Organomet. Chem. 1975, 93, 259.
3. Cassar, L. J. Organomet. Chem. 1975, 93, 253.
4. Sonogashira, K.; Tohda, Y.; Hagihara, N. Tetrahedron Lett. 1975, 16, 4467.
5. Stephens, R. D.; Castro, C. E. J. Org. Chem. 1963, 28, 2163.
6. Zapf., A.; Beller, M. Top. Catal. 2002, 19, 101.
7. Tucker, C. E.; de Vries, J. G. Top. Catal. 2002, 19, 111.
8. Brase, S.; Kirchhoff, J. H.; Kobberling, J. Tetrahedron 2003, 59, 885.
9. van de Weghe, P. Lett. Org. Chem. 2005, 2, 113.
10. Brandsma, L. Synthesis of Acetylenes, Allenes and Cumulenes: Methods and Techniques; Elsevier: Oxford, 2004; p 293.
11. Sonogashira, K. In Metal-Catalyzed Cross-Coupling Reactions; Diederich, F., de Meijera, A., Eds.; Wiley-VCH: Weinheim, 2004; Vol. 1, p 319.
12. Tykwinski, R. R. Angew. Chem., Int. Ed. 2003, 42, 1566.
13. Negishi, E.; Anastasia, L. Chem. Rev. 2003, 103, 1979.
14. Sonogashira, K. In Handbook of Organopalladium Chemistry for Organic Synthesis; Negishi, E., de Meijere, A., Eds.; Wiley-Interscience: New York, 2002; p 493.
15. Sonogashira, K. J. Organomet. Chem. 2002, 653, 46.
16. Rossi, R.; Carpita, A.; Bellina, F. Org. Prep. Proced. Int. 1995, 27, 127.
17. Sonogashira, K. InComprehensiVe Organic Synthesis; Trost, B. M., Fleming, I., Eds.;Pergamon: Oxford, 1991; Vol. 3, p 521.
18. Handbook of Organopalladium Chemistry for Organic Synthesis; Negishi, E., de Meijere, A., Eds.; Wiley: New York, 2002.
19. Soheili, A.; Albaneze-Walker, J.; Murry, J. A.; Dormer, P. G.; Hughes, D. L. Org. Lett. 2003, 5, 4191.
20. Choudary, B. M.; Madhi, S.; Kantam, M. L.; Sreedhar, B.; Iwasawa, Y. J. Am. Chem. Soc. 2004, 126, 2292.
21. Posset, T.; Bumel, J. J. Am. Chem. Soc. 2006, 128, 8394.
22. Niemela, E. H.; Lee, A. F.; Fairlamb, I. J. S. Tetrahedron Lett. 2004, 45, 3593.
23. Stambuli, J. P.; Buhl, M.; Hartwig, J. F. J. Am. Chem. Soc. 2002, 124, 9346.
24. Bertus, P.; Fecourt, F.; Bauder, C.; Pale, P. New J. Chem. 2004, 28, 12.
25. Letinois-Halbes, U.; Pale, P.; Berger, S. J. Org. Chem. 2005, 70, 9185.
26. Karak, M.; Barbosa, L. A.; Hargaden, G. C. RSC Adv., 2014, 4, 53442
27. Deng, X.; Mani, N. S. Org. Lett. 2006, 8, 269.
28. Name Reactions for Homologation, Part 1; Li, J. J., Eds.; Wiley: New York, 2009.
29. Tetrahedron, S.; Lopez, J. C.Valverde, A.; Barrio, A.; PedregosaGomez, A. M.; Lett. 2004, 45, 6307.
30. Tykwinski, R. R. Angew. Chem. Int. Ed. 2003, 42, 1566.
31. Barrios-Landeros, F.; Hartwig, J. F. J. Am. Chem. Soc. 2005, 127, 6944.
32. Gelman, D.; Buchwald, S. L. Angew. Chem., Int. Ed. 2003, 42, 5993.
33. Adjabeng, G.; Brenstrum, T.; Frampton, C. S.; Robertson, A. J.; Hillhouse, J.; McNulty, J.; Capretta, A. J. Org. Chem. 2004, 69, 5082.
34. Brenstrum, T.; Clattenburg, J.; Britten, J.; Zavorine, S.; Dyck, J.; Robertson, A. J.; McNully, J.; Capretta, A. Org. Lett. 2006, 8, 103.
35. Amatore, C.; Jutand, A. J. Organomet. Chem. 1999, 576, 254.
36. Buchmeiser, M. R.; Schareina, T.; Kempe, R.; Wurst, K. J. Organomet. Chem. 2001, 634, 39.
37. Park, S. B.; Alper, H. Chem. Commun. 2004, 1306.
38. Gossage, R. A.; Jenkins, H. A.; Yadav, P. N. Tetrahedron Lett. 2004, 45, 7689.
39. Eisnor, C. R.; Gossage, R. A.; Yadav, P. N. Tetrahedron 2006, 62, 3395.
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46. Gil-Moltó, J.; Nájera, C. Adv. Synth. Catal. 2006, 348, 1874
Contributors and Attributions
• Noalle Fellah and Xiaolong Zhu | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Catalysis/Catalyst_Examples/Sonogashira_Coupling.txt |
The Stille reaction, named after the late John Kenneth Stille, is a palladium-catalyzed cross coupling reaction. Heavily used in organic synthesis, it involves the coupling of an organic halide with an organotin compound.1,2 The reaction proceeds according to the following equation11:
General Stille Coupling Reaction
The reaction proceeds using a palladium phosphine catalyst. In this reaction, R1 and R2 represent the organic chains that are to be coupled. These are usually chains containing an sp2 hybridized carbon (aryl, alkenyl, allyl.) X represents a leaving group such as a halide or triflate (Cl, Br, I, OTf, etc.) In some instances, it is possible to use sp3 hybridized alkyls and aromatics as R1. It’s worth noting that, although air stable, tin reagents tend to be highly toxic.2
History
Prior to Stille’s work, palladium-catalyzed cross coupling reactions had been observed. The foundations were laid by Eaborn and Migita in the late 70’s. Eaborn’s process involved formation of diary product using an organotin reagent. This was expanded by Migita who coupled acyl chlorides with organotin reagents to form ketones with yields ranging from 53-87%.3
Later on, Migita was able to couple aryl and acyl halides with allyl tin reagents. He was able to do so at lower temperatures due to allyl’s ability to interact more efficiently with the palladium catalyst. Yields for both aryl and acyl halides were inconsistent ranging from as little as 4% to as much as 100%.4
Migita succeeded in coupling acyl chlorides to organotin reagents under harsh conditions4
Stille was able to build upon these foundations using a variety of alkyl tin reagents. In 1978, he published a report on the successful coupling of many alkyl tin reagents with myriad acyl and aryl halides. He was able to carry out these reactions under much more mild conditions and obtaining much higher yields (76-99%). Subsequent research focused on synthesizing a variety of compounds in this fashion.5,6
Stille's research focused on coupling many organic halides to organotin reagents using palladium under milder conditions.5,6
Scope & Limitations
Stille coupling is a commonly used procedure because of its wide scope. There are ample choices for both nucleophile and electrophile. Furthermore, organotin reagents are air stable, commercially available or readily synthesized, making Stille coupling an easily accessible method for synthesis. A wide variety of aryl, vinyl and acyl halides or pseudo halides can be used as electrophiles. Furthermore, heterocyclic compounds are also viable choices.12The example below shows that the Stille reaction can also work with purines to build more complex molecules.
.
The Stille reaction can add complexity to purines
The major exceptions to this are organic chlorides which are not reactive enough to undergo oxidative addition with the palladium catalyst. As a result, organic bromides or iodides are preferred.
Similarly, the organotin reagent shows great versatility. As with the electrophile, many classes of organotin reagents work well in Stille coupling. Aryl, alkyl, vinyl and even heterocyclic stannanes are known to work. The following example shows N, S containing heterocyclic ring participating in Stille coupling.7
Heterocyclic compounds are capable of participating in Stille coupling; showing the versatility of the reaction.
Limitations include very bulky or heavily substituted reagents. These tend to react very slowly and may require optimization, typically in the form of co-catalytic copper iodide. Nonetheless, the Stille reaction is a versatile one of great importance to organic synthesis.
Stereo & Regioselectivity
Stereochemistry is typically retained in Stille coupling. Regioselectivity, as with most other coupling reactions is hard to control though optimizations have been found to control regioselectivity to some extent. Despite advances by researchers in controlling stereoselectivity, mechanisms have not been fully elucidated and require further study. 7,8,10
Synthetic Applications
The synthetic applications of Stille coupling are immense. A reaction of this versatility and scope readily found its way into total syntheses of natural products. Stille coupling has been of great use towards the synthesis of many such products, many of these antibiotics or anticancer drugs. Among these are manzamine A, ircinal A, oxazolomycin and many others. 9
Stille coupling is an important step in the total synthesis of the marine alkaloid Ircinal A. Ircinal A is a known antitumor substance.9
Mechanism
Catalyst design
The Stille reaction uses a palladium catalyst. It can use an 18- or 16-electron Pd (0) complex as a source of the catalyst, such as Pd(PPh3)4, Pd(dba)2. Then through ligand dissociation, it can be formed into a 14-electron Pd (0), the active catalyst.
Upon oxidative addition, the electrophile will bind to palladium, turning it into a 16-electron square planer intermediate. The organostannane then comes in as a transmetalation agent to introduce the R’ group and also taking away the halide. Finally, the two R groups will be coupled by reductive elimination through various possible pathways, and turning the catalyst back to the 14-electron Pd (0).
Oxidative addition
The alkyl halide would first perform oxidative addition to the palladium in a concerted fashion, resulting in a 16-electron Pd(II) intermediate. The cis square planer product is in a fast equilibrium with the trans product.The bulky ligands used on the catalyst makes the trans product more thermodynamically stable, therefore most of the intermediate will isomerize into the trans product15,16.
A trans- intermediate can also be formed when the catalyst reacts with an sp3 organohalide in a SN2 mechanism.
Transmetalation
The Stille coupling uses organostannane as a trans coupling reagent. The tin is usually bound to allyl, alkenyl, or aryl groups. The tin and the R’ group will form a four-member ring with the palladium center and the halide, forming an 18-electron transition state. Then the tin halide will leave and the R’ group stays bonded to palladium17.
Reductive elimination
The two trans- R groups must first isomerize back into a cis- conformation, then it can undergo a concerted reductive elimination17-19.
There are other two proposed mechanisms which involves the association or dissociation of the ligands. An extra ligand can be bond to palladium, forming an 18-electron trigonal bipyramidal structure and forces the two R groups to the equatorial position, which is a suitable conformation for them to form C-C bonds.
Another proposed mechanism is that one of the ligands will dissociate first to form a T-shape 14-electron intermediate, which is able to speed up the reductive elimination process.
If the ligands are bulky enough, like phosphines with large cone angles, it is possible to push the two R groups closer to each other into an appropriate coordination angle, thus speeding up the reductive elimination process11.
Ligands
Various ligands are used as the 16- or 18- electron Pd0 catalyst, such as Pd(PPh3)4 and Pd(dba)213.
Some Pd(II) complexes can also be used as a source and can be reduced to the active Pd(0) catalyst, such as Pd(OAc)2, PdCl2(MeCN)2, PdCl2(PPh3)2.
Organostannane
Methyl and butyl group are often used as the other three R groups that binds to stannane, although these alkyl groups are not as reactive in the transmetalation process, there is still possibility that the alkyl coupling side product would form. This is called a” Non-transferable” ligand product. Though there are cases that shows benzylstannane are able to couple through Stille catalysis at higher temperatures. Both electron donating and electron withdrawing properties of the R’ group is favorable for transmetalation. The reactivity of the R group on organostannane have the following order20:
Alkynyl > Alkenyl > Aryl > Allyl = Benzyl > α-alkoxyl > Alkyl
Electrophiles
Vinyl halides, aryl halides and heterocyclic halides can all be used as the electrophile for the reaction. Iodine and bromide are used for the halogen because the chlorides are too inert for oxidative addition, and iodides would react even faster than bromide, as can be seen from the reaction below7:
The stereochemistry is often retained during the catalytic reaction, unless it’s carried out under a harsh condition, so the Stille coupling reaction is quite good towards steroselectivity.
Additives
LiCl is often used to enhance the reaction rate by stabilizing the transition state during oxidative addition. It can also improve the rate of transmetalation by increasing the solvent’s polarity2,7. In addition, Cu(I) and Mn(II) salts are also used to increase reaction rate as well as improving selectivity.
Contributors and Attributions
• Kevin Maldonado and Jeffrey Liu | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Catalysis/Catalyst_Examples/Stille_Coupling.txt |
Suzuki-Miyaura coupling (or Suzuki coupling) is a metal catalyzed reaction, typically with Pd, between an alkenyl (vinyl), aryl, or alkynyl organoborane (boronic acid or boronic ester, or special cases with aryl trifluoroborane) and halide or triflate under basic conditions. This reaction is used to create carbon-carbon bonds to produce conjugated systems of alkenes, styrenes, or biaryl compounds (Scheme 1).
Scheme 1: General reaction scheme of Suzuki cross coupling reaction.
Modified conditions have demonstrated reactivity with less reactive substrates such as alkyl boranes (BR3) or aryl or alkenyl chlorides by amending the base and ligands employed.
Historical Methods: Forming carbon-carbon bonds via cross coupling reactions
The Suzuki coupling is a pioneering reaction in cross coupling, and has been thoroughly studied since. The first Suzuki-type cross coupling reaction between phenylboronic acid and haloarenes was published by Suzuki and Miyaura in 1981 (Scheme 1).1 Commonly, Suzuki coupling is compared to Stille coupling seeing that boron has a similar electronegativity to tin, which is used for transmetallation in Stille coupling. Both couplings have a similar reaction scope and proceed via a similar mechanistic cycle.
Scheme 1: Suzuki cross coupling reaction of phenylboronic acid and haloarenes.
Stille cross coupling reactions can form carbon-carbon bonds between alkenyl (vinyl), aryl, or alkynyl halides and an extended scope of organotin alkynes, alkenes, aryl, allyl benzyl, ketones, and alkyl. However, Stille coupling poses several drawbacks seeing that organotin reagents are: (1) highly toxic, (2) costly, and have a (3) lower functional group tolerance, despite Suzuki coupling not being suitable for base sensitive substrates.2 The example below demonstrates that even under conditions for either a Stille or Suzuki coupling to ensue, primarily Suzuki coupling occurs (Scheme 2).3
Scheme 2: Suzuki versus Stille cross coupling reaction.
Negishi coupling also demonstrates similar transformations to Suzuki coupling in a comparable substrate scope. It uses organo-zinc for transmetallation. However, Negishi coupling tends to occur in lower yields, with less functional group tolerance, and is water and oxygen sensitive.1
Reaction Scheme
Suzuki coupling proceeds via three basic steps: oxidative addition, transmetallation, and reductive elimination.
Reaction Mechanism and Mechanistic Studies
The general catalytic cycle for Suzuki cross coupling involves three fundamental steps: oxidative addition, transmetalation, and reductive elimination as demonstrated in Figure 1.1 The oxidative addition of aryl halides to Pd(0) complex is the initial step to give intermediate 1, a Pd(II) species. Under the participation of base, an organoborane compound reacts with intermediate 1 in transmetalation to afford intermediate 2. This is followed by reductive elimination to give the desired product and regenerate the original Pd(0) species. Depending on different catalytic systems with various catalysts, ligands, and solvents, there are additional processes in the catalytic cycle, including ligand or solvent association and dissociation.2 Undoubtedly, these three fundamental steps occur since intermediates 1 and 2 have been characterized by isolation3 or spectroscopic analyses4.
Oxidative addition is the rate determining step in the catalytic cycle and the relative reactivity decreases in the order of I > OTf > Br > Cl.4a The oxidative addition involving Pd(0) complexes is a SN2-type mechanism based on the unambiguous evidence of stereochemical inversion of configuration at alkyl halides carbon.5 Mechanistic studies have revealed that the apparently simple oxidative addition step consists of four concurrent isomerization pathways from the cis isomer, which is initially formed, to the more stable trans isomer (Figure 2).6
Transmetallation is initiated by base to encourage the transfer of the aryl or alkyl group from the organoborane to the Pd complex. The intermediates and stereochemistry of this step were determined by:
1. Low-temperature rapid injection nuclear magnetic resonance (NMR) spectroscopy & kinetic studies have recently revealed three different species containing palladium-oxygen-boron linkages were identified and characterized as the pre-transmetalation intermediates.
1. Deuterium-labeling experiments demonstrated the retention of stereochemistry in the transmetalation step (Figure 4).8 Here, smaller J-values (approximately 0-7 Hz) correspond to a syn configuration, and larger J-values (8-15 Hz) corresponds to the anti configuration.
Reductive elimination occurs from the cis complex, therefore the trans complexes isomerizes to cis isomer to undergo reductive elimination.9 The elimination was confirmed to proceed intramolecularly from the cis isomer9 based on:
Kinetic studies showed that reductive elimination obeys first order kinetics, hence the rate and reaction is dependent only on the concentration of the post-transmetallation Pd-complex.
Deuterium labeling demonstrated no cross-coupling product from crossover between two Pd complexes (Figure 5).
Catalysts
Palladium catalysts are most widely employed in Suzuki coupling. Normally, the active Pd catalyst consist of two parts: precursors (for example: Pd(OAc)2, Pd2(dba)3, or Pd(PPh3)4) and ligands. To enhance the reactivity and stability of the catalyst, they were developed to be electron-rich and spatially bulky, which affords a high turnover number (TON) and low loading.1,2 For example, palladacycles (Figure 4) were developed and exhibit thermal stability, robust reaction times, insensitivity to air and water, low cost, and environmentally friendly.3,4
In terms of separation and recycling of the catalysts, people have developed polymer-supported heterogeneous catalysis systems for Suzuki coupling reaction.5 Polymer-supported heterogeneous catalysis have tremendous values in pharmaceutical and industrial synthesis due to the advantage of preventing contamination from the ligand residue in products, fast recovery, and the simple recycling of the catalysts
Suzuki Cross Coupling Reaction Ligand Design
Since the 1980’s, Suzuki reactions have been significantly improved, which lends to why the Suzuki reaction is so powerful: the elegant ligand design for Pd catalysts. Ligands are designed with electron-rich and spatially bulky features since electron-rich ligands can facilitate the oxidative addition step, and spatially bulky structures increase the orbital overlapping on the metal which enables reductive elimination.1 Based on which atom is coordinated with the metal, ligands are categorized as Phosphine ligands, Carbon ligands, or Nitrogen ligands.
Phosphine ligands are the most popular Pd ligands in both the laboratory and industry. PPh3 was the earliest and most widely use monodentate phosphine ligand2 in Suzuki reactions. Inspired by the structure of PPh3, scientists have manipulated substitutions on the phosphor atom and aromatic ring to tune the different reactivity of the reaction. For example, Fu and other research groups3 replaced aromatic groups with more electron-rich and bulkier alkyl groups (Figure 1), and the resulting ligands gave high catalytic reactivity on less reactive substrates with lower catalyst loading.
In the late 1990’s, Buchwald’s group introduced new electronic rich and bulky phosphine ligands based on the frame skeleton of dialkylbiaryl phosphine ligands4. Understanding the impact of substitutions on dialkylbiarylphosphines on the efficacy of catalytic reactions, the Buchwald group has synthesized a series of dialkylbiaryl ligands facilitating Palladium-catalyzed C-C, C-N, and C-O bond-forming processes, as well as supporting ligands for other reactions.4 There are other monodentate phosphine ligands with unique structures, which can also promote Suzuki reaction effectively, for example, monodentate ferrocene phosphine ligands.5
Many research groups have also designed and synthesized bidentate phosphine ligands (Figure 3) for the Suzuki reaction, and those ligands showed excellent reactivity on substrates that are generally challenging for monodentate ligands.
Carbon & Nitrogen ligands: Carbon ligands mainly comprise carbene8- and olefin9-type ligands for Suzuki couping. Efficient nitrogen ligands include amines and imines.9
Suzuki Cross Coupling Reaction Substrate Scope
Electrophilic Partners1: The electrophilic partners in Suzuki coupling are: alkenyl, alkynyl, allyl, benzyl, aryl, and alkyl halides or triflates. Aryl halides are activated by electron-withdrawing (EWD) groups in the ortho or para positions or 1-alkenyl halides with EWD groups in the alpha or beta positions. This increases their reactivity in oxidative addition compared to those with electron-donating groups. The order of reactivity of the electrophilic partners based on their leaving groups is:
I >> Br > OTf >> Cl > F
Chloride electrophiles are the most nonreactive as they are reluctant to participate in oxidative addition. Reactivity with chloride electrophiles has been established by using bulky, electron-donating (electron rich) phosphine ligands on the Pd catalyst and stronger bases to encourage dissociation of Cl.2
β- alkyl Suzuki coupling is most successful between hindered electron-rich organoboranes and electron-deficient vinyl or aryl halides or triflates. This counters the trends reported above in that the electrophilic partner would be activated by EWD groups.
Bases1: Transmetallation is initiated with base, and it’s reactivity varies with solvent.
Strong bases: NaOH, TlOH, and NaOMe perform optimally in THF or H2O solvent systems
Thallium bases such as TlOH, and its derivatives Tl2CO3 and TlOEt, enable cross coupling between alkyl boranes and alkyl halides. It also allows this reaction to proceed at lower temperatures (20 °C from the usual 50 to 80 °C). Unfortunately, these bases are air and light sensitive, but are still widely used.3
Weak bases: K2CO3 and K3PO4 perform optimally in DMF
Catalysts: Pd*, Ni, Ru, Fe, Cu
Palladium catalysts are the most widely used for Suzuki coupling and perform best with electron-donating (usually phosphine) ligands.
Nickel catalysts have been recently developed and demonstrate reactivity with inert electrophiles, especially chlorides and unreactive bromides. They have also expanded the electrophile scope to include: aryl fluorides (using a directing group, usually o-NO2), carbamates, sulfamates, esters, phosphate esters and ethers. This is enabled by nickel’s variation in oxidation states (Ni0 à Ni2+ and Ni+ à Ni3+) and increased nucleophilicity due to its small size. Use of Ni in catalysis is favored due to its abundance, and thus low cost.4
Boranes: Boronic acids, esters and tri-fluoro derivatives1,2
The reactivity of the borane to conduct transmetallation depends on its Lewis acidity. Therefore, EWD substituents increase the reactivity of the borane. The general trend of borane reactivity is:
ArBF3 > RB(OH)2 > RB(OR)2 >> R3B
Alkyl boranes are the least reactive, and rarely participate in transmetallation under standard conditions; however this has been amended with the use of stronger bases.
Selecting a borane for a reaction depends on its compatibility with its electrophile coupling partner and desired borane R group.
Regio- and Stereoselectivity1,2
Each step of the catalytic cycle of Suzuki coupling can influence the regio- or stereo- configuration of the product. General rules of regio- and stereoselectivity relative to the substrates are:
Oxidative addition of alkyl and alkenyl halides retains the configuration of the electrophilic substrate, however allylic and benzylic halides invert this configuration. This step initially produces the cis complex which isomerizes to the trans complex.
Transmetallation and reductive elimination both retains the regio- and stereochemistry established in oxidative addition.
The borane substrate can also dictate selectivity of the cross coupling product. Often times the reactive borane species is formed in situ via hydroboration with 9-BBN. This occurs as a syn addition, and reductive elimination of the product retains the stereochemistry established by the hydroboration. However, reversing the order in which the reagents are added yields the anti addition product, which is also reflected in reductive elimination.
Chiral centers are formed by using chiral borane reagents. These reagents perform asymmetric cross coupling with retention of configuration i.e. (R) boranes yield (R) products, which follows suit for (S) boranes, in >90% enantiomeric excess.3
Advantages of Suzuki Coupling
Borane Substrates: Easy to synthesize, stabile, nontoxic, and cost effective1,2
The ease of synthesizing boranes provides convenient access to substrates for Suzuki coupling at a low cost and with minimal health risks. The two common methods to synthesize boronic acids are:
1. React lithium or magnesium Grignard reagents with borate esters, followed by hydrolysis.
1. React arylsilanes with boron tribromide, followed by acid hydrolysis.
Boronic esters are synthesized from boronic acids using alcohols or diols, especially pinicol.
Aryl trifluoroborates are synthesized by subjecting the corresponding alkyl boronic acid or ester to excess KHF2. Aryltrifluoroborates possess beneficial characteristics that allow for robust reactivity, easy purified, and reduced protodeboronation, which quenches the borane reagent.3
Reaction Conditions: Milder and Greener1,2,4
Typically, cross coupling reactions are run in organic conditions; however, Suzuki couplings can be performed in heterogeneous or purely aqueous conditions as organoboranes are water soluble and compatible with water soluble, inorganically supported, ligand-free Pd-catalysts. Boranes are exceptionally nucleophilic, and thus do not require extreme conditions for transmetallation which increases its functional group tolerance. Their high reactivity is also demonstrated in high turnover rates of the Pd catalyst used, which allows for a lower loading.
Organoboranes are nontoxic and stable to extreme heating and exposure to oxygen or water. Consequently, these reagents can be easily used at benchtop, and do not require special equipment or techniques, such as gloveboxes or air-sensitive and dry technique.1 This reaction has a high atom economy seeing that the byproducts are typically salts and water soluble borane byproducts.1.4 The lack of interfering byproducts allows for one-pot syntheses to further reduce waste and increase reaction efficiency.4 These attractive features of organoborane reagents increases its utility and synthetic convenience, and makes Suzuki coupling optimal for large scale and industry synthesis.
Limitations of Suzuki Coupling:
Issues with reactants:
• Borane substrates commercially available do not encompass all R groups desired
• Borane substrates are generally preinstalled to more complicated substrates synthesized in lab, which is often difficult
• Some alkyl borates (sp3) and hetero- substrates don’t show reactivity1
• Chloride substrates react slowly, and with lower yields1
• Without base there are many side products1
Side reactions:
Beta-hydride elimination competes with reductive elimination, which affords a side product that greatly reduces the yield. This is observed with reactants that have beta-hydrides, most commonly alkyl substrates, but sometimes can be avoided by using Ni catalysts or ligands with larger bite angles.2
Scheme 1: Β-hydride elimination in Suzuki cross coupling reaction.2
Synthetic Applications:
Large-scale, industry synthesis of medicinal drugs1
Scheme 1: Suzuki cross coupling reaction to synthesize the drug Linifanib.
Linifanib (Scheme 1) is a tyrosine kinase inhibitor that was FDA approved as an anti-cancer drug that is synthesized on kilogram scale in the presence of different functional groups using Suzuki coupling for its industry production.
Natural products synthesis2
Scheme 2: Suzuki cross coupling reaction application: Epothilone total synthesis.
Suzuki coupling in the total synthesis of Epothilone A (Figure 2) was successful in the presence of numerous stereocenters and functional groups.
Polymer synthesis: Suzuki Polycondensation (SPC)3
Suzuki coupling has been applied to create polymers of aromatic groups, which yielded the longest conjugated chains. These polymers are categorized by their properties and applications:
(a) Polymers with alkyl or alkoxy chains and (c) Polyelectrolytes: SPC yields poly(p-phenylene)’s (PPP’s) which are used in LED lights
(b) Amphiphilic PPPs: SPC produces colloidal structures to create rechargeable, solid state lithium batteries
(d) PPP precursors for ladder polymers: currently being developed to apply their thermal, electrical, and optical properties to conducting materials, energy sources, and instrumentation
(e) Polymers with main-chain chirality and (f) Dendronized PPPs: used in organic synthesis of optically pure compounds
(g) Poly(arylene vinylene)’s and poly(arylene ethinylene)’s: Photoluminescence (PL) and electroluminescence (EL) characteristics for materials and instrumentation
(h) Various polymers: functionalization in synthesis via radical mechanism
Contributors and Attributions
• Peiyuan Qu and Amanda Ramdular | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Catalysis/Catalyst_Examples/Suzuki-Miyaura_Coupling.txt |
A catalyst is a substance that speeds up a reaction without being consumed by it. More specifically, a catalyst provides an alternative, lower activation energy pathway between reactants and products. As such, catalysts are vitally important to chemical technology; approximately 95% of industrial chemical processes involve catalysts of various kinds. In addition, most biochemical processes that occur in living organisms are mediated by enzymes, which are catalysts made of proteins.
It is important to understand that a catalyst affects only the kinetics of a reaction; it does not alter the thermodynamic tendency of the reaction to occur. Therefore, ΔHis the same for the two pathways depicted in the plot above.
1. An Introduction to Types of Catalysis
A detailed look at various types of catalytic action
Examples of Catalysis
This page looks at the the different types of catalyst (heterogeneous and homogeneous) with examples of each kind, and explanations of how they work. You will also find a description of one example of autocatalysis - a reaction which is catalysed by one of its products.
Types of catalytic reactions
Catalysts can be divided into two main types - heterogeneous and homogeneous. In a heterogeneous reaction, the catalyst is in a different phase from the reactants. In a homogeneous reaction, the catalyst is in the same phase as the reactants.
What is a phase? If you look at a mixture and can see a boundary between two of the components, those substances are in different phases. A mixture containing a solid and a liquid consists of two phases. A mixture of various chemicals in a single solution consists of only one phase, because you can't see any boundary between them.
You might wonder why phase differs from the term physical state (solid, liquid or gas). It includes solids, liquids and gases, but is actually a bit more general. It can also apply to two liquids (oil and water, for example) which don't dissolve in each other. You could see the boundary between the two liquids.
If you want to be fussy about things, the diagrams actually show more phases than are labelled. Each, for example, also has the glass beaker as a solid phase. All probably have a gas above the liquid - that's another phase. We don't count these extra phases because they aren't a part of the reaction.
Heterogeneous catalysis
This involves the use of a catalyst in a different phase from the reactants. Typical examples involve a solid catalyst with the reactants as either liquids or gases. Most examples of heterogeneous catalysis go through the same stages:
One or more of the reactants are adsorbed on to the surface of the catalyst at active sites.
There is some sort of interaction between the surface of the catalyst and the reactant molecules which makes them more reactive.
The reaction happens.
The product molecules are desorbed.
Adsorption is where something sticks to a surface. It isn't the same as absorption where one substance is taken up within the structure of another. Be careful!
An active site is a part of the surface which is particularly good at adsorbing things and helping them to react.
This might involve an actual reaction with the surface, or some weakening of the bonds in the attached molecules.
At this stage, both of the reactant molecules might be attached to the surface, or one might be attached and hit by the other one moving freely in the gas or liquid.
Desorption simply means that the product molecules break away. This leaves the active site available for a new set of molecules to attach to and react.
A good catalyst needs to adsorb the reactant molecules strongly enough for them to react, but not so strongly that the product molecules stick more or less permanently to the surface.
Silver, for example, isn't a good catalyst because it doesn't form strong enough attachments with reactant molecules. Tungsten, on the other hand, isn't a good catalyst because it adsorbs too strongly.
Metals like platinum and nickel make good catalysts because they adsorb strongly enough to hold and activate the reactants, but not so strongly that the products can't break away.
Examples of heterogeneous catalysis
The hydrogenation of a carbon-carbon double bond
The simplest example of this is the reaction between ethene and hydrogen in the presence of a nickel catalyst.
In practice, this is a pointless reaction, because you are converting the extremely useful ethene into the relatively useless ethane. However, the same reaction will happen with any compound containing a carbon-carbon double bond.
One important industrial use is in the hydrogenation of vegetable oils to make margarine, which also involves reacting a carbon-carbon double bond in the vegetable oil with hydrogen in the presence of a nickel catalyst.
Ethene molecules are adsorbed on the surface of the nickel. The double bond between the carbon atoms breaks and the electrons are used to bond it to the nickel surface.
Hydrogen molecules are also adsorbed on to the surface of the nickel. When this happens, the hydrogen molecules are broken into atoms. These can move around on the surface of the nickel.
If a hydrogen atom diffuses close to one of the bonded carbons, the bond between the carbon and the nickel is replaced by one between the carbon and hydrogen.
That end of the original ethene now breaks free of the surface, and eventually the same thing will happen at the other end.
As before, one of the hydrogen atoms forms a bond with the carbon, and that end also breaks free. There is now space on the surface of the nickel for new reactant molecules to go through the whole process again.
Catalytic converters
Catalytic converters change poisonous molecules like carbon monoxide and various nitrogen oxides in car exhausts into more harmless molecules like carbon dioxide and nitrogen. They use expensive metals like platinum, palladium and rhodium as the heterogeneous catalyst.
The metals are deposited as thin layers onto a ceramic honeycomb. This maximises the surface area and keeps the amount of metal used to a minimum.
Taking the reaction between carbon monoxide and nitrogen monoxide as typical:
Catalytic converters can be affected by catalyst poisoning. This happens when something which isn't a part of the reaction gets very strongly adsorbed onto the surface of the catalyst, preventing the normal reactants from reaching it.
Lead is a familiar catalyst poison for catalytic converters. It coats the honeycomb of expensive metals and stops it working.
In the past, lead compounds were added to petrol (gasoline) to make it burn more smoothly in the engine. But you can't use a catalytic converter if you are using leaded fuel. So catalytic converters have not only helped remove poisonous gases like carbon monoxide and nitrogen oxides, but have also forced the removal of poisonous lead compounds from petrol.
The use of vanadium(V) oxide in the Contact Process
During the Contact Process for manufacturing sulphuric acid, sulphur dioxide has to be converted into sulphur trioxide. This is done by passing sulphur dioxide and oxygen over a solid vanadium(V) oxide catalyst.
This example is slightly different from the previous ones because the gases actually react with the surface of the catalyst, temporarily changing it. It is a good example of the ability of transition metals and their compounds to act as catalysts because of their ability to change their oxidation state.
The sulphur dioxide is oxidised to sulphur trioxide by the vanadium(V) oxide. In the process, the vanadium(V) oxide is reduced to vanadium(IV) oxide.
The vanadium(IV) oxide is then re-oxidised by the oxygen.
This is a good example of the way that a catalyst can be changed during the course of a reaction. At the end of the reaction, though, it will be chemically the same as it started.
Homogeneous catalysis
This has the catalyst in the same phase as the reactants. Typically everything will be present as a gas or contained in a single liquid phase. The examples contain one of each of these . . .
Examples of homogeneous catalysis
The reaction between persulphate ions and iodide ions
This is a solution reaction that you may well only meet in the context of catalysis, but it is a lovely example!
Persulphate ions (peroxodisulphate ions), S2O82-, are very powerful oxidising agents. Iodide ions are very easily oxidised to iodine. And yet the reaction between them in solution in water is very slow.
If you look at the equation, it is easy to see why that is:
The reaction needs a collision between two negative ions. Repulsion is going to get seriously in the way of that!
The catalysed reaction avoids that problem completely. The catalyst can be either iron(II) or iron(III) ions which are added to the same solution. This is another good example of the use of transition metal compounds as catalysts because of their ability to change oxidation state.
For the sake of argument, we'll take the catalyst to be iron(II) ions. As you will see shortly, it doesn't actually matter whether you use iron(II) or iron(III) ions.
The persulphate ions oxidise the iron(II) ions to iron(III) ions. In the process the persulphate ions are reduced to sulfate ions.
The iron(III) ions are strong enough oxidising agents to oxidise iodide ions to iodine. In the process, they are reduced back to iron(II) ions again.
Both of these individual stages in the overall reaction involve collision between positive and negative ions. This will be much more likely to be successful than collision between two negative ions in the uncatalysed reaction.
What happens if you use iron(III) ions as the catalyst instead of iron(II) ions? The reactions simply happen in a different order.
The destruction of atmospheric ozone
This is a good example of homogeneous catalysis where everything is present as a gas.
Ozone, O3, is constantly being formed and broken up again in the high atmosphere by the action of ultraviolet light. Ordinary oxygen molecules absorb ultraviolet light and break into individual oxygen atoms. These have unpaired electrons, and are known as free radicals. They are very reactive.
The oxygen radicals can then combine with ordinary oxygen molecules to make ozone.
Ozone can also be split up again into ordinary oxygen and an oxygen radical by absorbing ultraviolet light.
This formation and breaking up of ozone is going on all the time. Taken together, these reactions stop a lot of harmful ultraviolet radiation penetrating the atmosphere to reach the surface of the Earth.
The catalytic reaction we are interested in destroys the ozone and so stops it absorbing UV in this way.
Chlorofluorocarbons (CFCs) like CF2Cl2, for example, were used extensively in aerosols and as refrigerants. Their slow breakdown in the atmosphere produces chlorine atoms - chlorine free radicals. These catalyse the destruction of the ozone.
This happens in two stages. In the first, the ozone is broken up and a new free radical is produced.
The chlorine radical catalyst is regenerated by a second reaction. This can happen in two ways depending on whether the ClO radical hits an ozone molecule or an oxygen radical.
If it hits an oxygen radical (produced from one of the reactions we've looked at previously):
Or if it hits an ozone molecule:
Because the chlorine radical keeps on being regenerated, each one can destroy thousands of ozone molecules.
Autocatalysis
The oxidation of ethanedioic acid by manganate(VII) ions
In autocatalysis, the reaction is catalysed by one of its products. One of the simplest examples of this is in the oxidation of a solution of ethanedioic acid (oxalic acid) by an acidified solution of potassium manganate(VII) (potassium permanganate).
The reaction is very slow at room temperature. It is used as a titration to find the concentration of potassium manganate(VII) solution and is usually carried out at a temperature of about 60°C. Even so, it is quite slow to start with.
The reaction is catalysed by manganese(II) ions. There obviously aren't any of those present before the reaction starts, and so it starts off extremely slowly at room temperature. However, if you look at the equation, you will find manganese(II) ions amongst the products. More and more catalyst is produced as the reaction proceeds and so the reaction speeds up.
You can measure this effect by plotting the concentration of one of the reactants as time goes on. You get a graph quite unlike the normal rate curve for a reaction.
Most reactions give a rate curve which looks like this:
Concentrations are high at the beginning and so the reaction is fast - shown by a rapid fall in the reactant concentration. As things get used up, the reaction slows down and eventually stops as one or more of the reactants are completely used up.
An example of autocatalysis gives a curve like this:
You can see the slow (uncatalysed) reaction at the beginning. As catalyst begins to be formed in the mixture, the reaction speeds up - getting faster and faster as more and more catalyst is formed. Eventually, of course, the rate falls again as things get used up.
Warning!
Don't assume that a rate curve which looks like this necessarily shows an example of autocatalysis. There are other effects which might produce a similar graph.
For example, if the reaction involved a solid reacting with a liquid, there might be some sort of surface coating on the solid which the liquid has to penetrate before the expected reaction can happen.
A more common possibility is that you have a strongly exothermic reaction and aren't controlling the temperature properly. The heat evolved during the reaction speeds the reaction up. | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Catalysis/Catalysts.txt |
This page takes a brief look at the catalysts used in the Contact Process to manufacture sulphuric acid, in the Haber Process to manufacture ammonia, and in the conversion of ammonia into nitric acid.
The Contact Process for the manufacture of sulphuric acid
At the heart of the Contact Process is a reaction which converts sulphur dioxide into sulphur trioxide. Sulphur dioxide gas is passed together with air (as a source of oxygen) over a solid vanadium(V) oxide catalyst. This is therefore an example of heterogeneous catalysis.
The fact that this is a reversible reaction makes no difference to the operation of the catalyst. It speeds up both the forward reaction and the back reaction by the same amount.
The Haber Process for the manufacture of ammonia
The Haber Process combines hydrogen and nitrogen to make ammonia using an iron catalyst. This is another reversible reaction, and another example of heterogeneous catalysis.
The manufacture of nitric acid from ammonia
This is yet another example of heterogeneous catalysis.
This process involves oxidation of the ammonia from the Haber Process by oxygen in the air in the presence of a platinum-rhodium catalyst. Large sheets of metal gauze are used in order to reduce expense and to maximise the surface area of the catalyst. Although in principle the sheets would last for ever because the metals are acting as a catalyst, in practice they do deteriorate over time and have to be replaced.
The sheets of gauze are held at a temperature of about 900°C. The reaction is very exothermic, and once it starts the temperature is maintained by the heat evolved.
The ammonia is oxidised to nitrogen monoxide gas.
This is cooled. At ordinary temperatures and in the presence of excess air, it is oxidised further to nitrogen dioxide.
The nitrogen dioxide (still in the presence of excess air) is absorbed in water where it reacts to give a concentrated solution of nitric acid.
3. Examples of Catalysis in the Petrochemical Industry
This page looks briefly at some of the basic processes in the petrochemical industry (cracking, isomerisation and reforming) as examples of important catalytic reactions.
Catalytic cracking
Cracking is the name given to breaking up large hydrocarbon molecules into smaller and more useful bits. This is achieved by using high pressures and temperatures without a catalyst, or lower temperatures and pressures in the presence of a catalyst.
The source of the large hydrocarbon molecules is often the naphtha fraction or the gas oil fraction from the fractional distillation of crude oil (petroleum). These fractions are obtained from the distillation process as liquids, but are re-vaporised before cracking.
The hydrocarbons are mixed with a very fine catalyst powder. These days the catalysts are zeolites (complex aluminosilicates) - these are more efficient than the older mixtures of aluminium oxide and silicon dioxide.
The whole mixture is blown rather like a liquid through a reaction chamber at a temperature of about 500°C. Because the mixture behaves like a liquid, this is known as fluid catalytic cracking (or fluidised catalytic cracking).
Although the mixture of gas and fine solid behaves as a liquid, this is nevertheless an example of heterogeneous catalysis - the catalyst is in a different phase from the reactants.
The catalyst is recovered afterwards, and the cracked mixture is separated by cooling and further fractional distillation.
There isn't any single unique reaction happening in the cracker. The hydrocarbon molecules are broken up in a fairly random way to produce mixtures of smaller hydrocarbons, some of which have carbon-carbon double bonds. One possible reaction involving the hydrocarbon C15H32 might be:
Or, showing more clearly what happens to the various atoms and bonds:
This is only one way in which this particular molecule might break up. The ethene and propene are important materials for making plastics or producing other organic chemicals. The octane is one of the molecules found in petrol (gasoline).
Isomerisation
Hydrocarbons used in petrol (gasoline) are given an octane rating which relates to how effectively they perform in the engine. A hydrocarbon with a high octane rating burns more smoothly than one with a low octane rating.
Molecules with "straight chains" have a tendency to pre-ignition. When the petrol / air mixture is compressed they tend to explode, and then explode a second time when the spark is passed through them. This double explosion produces knocking in the engine.
Octane ratings are based on a scale on which heptane is given a rating of 0, and 2,2,4-trimethylpentane (an isomer of octane) a rating of 100.
In order to raise the octane rating of the molecules found in petrol (gasoline) and so make the petrol burn better in modern engines, the oil industry rearranges straight chain molecules into their isomers with branched chains.
One process uses a platinum catalyst on a zeolite base at a temperature of about 250°C and a pressure of 13 - 30 atmospheres. It is used particularly to change straight chains containing 5 or 6 carbon atoms into their branched isomers.
For example:
Reforming
Reforming is another process used to improve the octane rating of hydrocarbons to be used in petrol, but is also a useful source of aromatic compounds for the chemical industry. Aromatic compounds are ones based on a benzene ring.
Reforming uses a platinum catalyst suspended on aluminium oxide together with various promoters to make the catalyst more efficient. The original molecules are passed as vapours over the solid catalyst at a temperature of about 500°C.
Isomerisation reactions occur (as above) but, in addition, chain molecules get converted into rings with the loss of hydrogen. Hexane, for example, gets converted into benzene, and heptane into methylbenzene. | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Catalysis/Examples/2._Examples_of_Catalysis_in_the_Inorganic_Chemical_Industry.txt |
This page looks at the use of acid catalysts in some organic reactions. It covers the nitration of benzene, the hydration of ethene to manufacture ethanol, and the reactions both to produce esters and to hydrolyse them under acidic conditions. You will find links to the full mechanisms for each of these reactions if you want them.
The nitration of benzene
Benzene is treated with a mixture of concentrated nitric acid and concentrated sulphuric acid at a temperature not exceeding 50°C. As the temperature increases there is a greater chance of getting more than one nitro group, -NO2, substituted onto the ring.
Nitrobenzene is formed.
or:
The concentrated sulfuric acid is acting as a catalyst. Because everything is present in the same liquid phase, this is a good example of homogeneous catalysis.
The hydration of ethene to make ethanol
Ethene is mixed with steam and passed over a catalyst consisting of solid silicon dioxide coated with phosphoric(V) acid. The temperature used is 300°C and the pressure is about 60 to 70 atmospheres. Because the catalyst is in a different phase from the reactants, this is an example of heterogeneous catalysis.
This is a reversible reaction and only about 5% of the ethene reacts on each pass over the catalyst. When the reaction mixture is cooled, the ethanol and any excess steam condense, and the gaseous ethene can be recycled through the process. A conversion rate of about 95% is achieved by continual recycling in this way.
Making esters - the esterification reaction
Esters are what is formed when an organic acid reacts with an alcohol in the presence of concentrated sulfuric acid as the catalyst. Everything is present in a single liquid phase, and so this is an example of homogeneous catalysis.
For example, ethanoic acid reacts with ethanol to produce ethyl ethanoate.
The ethyl ethanoate has the lowest boiling point of anything in the mixture, and so is distilled off as soon as it is formed. This helps to reduce the reverse reaction.
The acid catalyzed hydrolysis of esters
In principle, this is the reverse of the esterification reaction but, in practice, it has to be done slightly differently. The ester is heated under reflux with a dilute acid such as dilute hydrochloric acid or dilute sulfuric acid.
The equation for the reaction is simply the esterification equation written backwards.
The dilute acid used as the catalyst also provides the water for the reaction. You need a large excess of water in order to increase the chances of the forward reaction happening and the ester hydrolyzing. You would normally hydrolyse esters quite differently by heating them with sodium hydroxide solution (alkaline hydrolysis). This is not an example of a catalytic reaction because the hydroxide ions are used up during the reaction.
The main advantage of doing it like this is that it is a one-way reaction. The ester can be completely hydrolysed rather than only partially if the reaction is reversible.
Contributors and Attributions
Jim Clark (Chemguide.co.uk)
5. Examples of Other Catalytic Reactions in Organic Chemistry
This page looks a few odds and ends of examples of catalysts used in organic chemistry. It includes the formation of epoxyethane from ethene, and several reactions from benzene chemistry - Friedel-Crafts reactions and halogenation.
The manufacture of epoxyethane from ethene
Epoxyethane is manufactured by reacting ethene with a limited amount of oxygen in the presence of a silver catalyst at a temperature of about 250 - 300°C and a pressure of less than 15 atmospheres. Because the solid silver is catalysing a gas reaction, this is an example of heterogeneous catalysis.
The reaction is exothermic and the temperature has to be carefully controlled to prevent further oxidation of the ethene to carbon dioxide and water.
The halogenation of benzene
Benzene reacts with chlorine or bromine in the presence of a catalyst. The catalyst is either aluminium chloride (or aluminium bromide if you are reacting benzene with bromine) or iron.
Strictly speaking iron isn't a catalyst, because it gets permanently changed during the reaction. It reacts with some of the chlorine or bromine to form iron(III) chloride, FeCl3, or iron(III) bromide, FeBr3.
These compounds act as the catalyst and behave exactly like aluminium chloride in these reactions.
The reaction with chlorine
The reaction between benzene and chlorine in the presence of either aluminium chloride or iron gives chlorobenzene.
or:
The reaction with bromine
The reaction between benzene and bromine in the presence of either aluminium bromide or iron gives bromobenzene. Iron is usually used because it is cheaper and more readily available.
or:
The Friedel-Crafts alkylation of benzene
Alkylation involves replacing a hydrogen atom on a benzene ring by an alkyl group like methyl or ethyl. This is another example of the use of aluminium chloride as a catalyst. Benzene is treated with a chloroalkane (for example, chloromethane or chloroethane) in the presence of aluminium chloride as a catalyst. The equation shows the reaction using a methyl group, but any other alkyl group could be used in the same way.
Substituting a methyl group gives methylbenzene - once known as toluene.
or:
The Friedel-Crafts acylation of benzene
An acyl group is an alkyl group attached to a carbon-oxygen double bond. Acylation means substituting an acyl group into something - in this case, into a benzene ring. The most commonly used acyl group is CH3CO-. This is called the ethanoyl group. In the example which follows we are substituting a CH3CO- group into the ring, but you could equally well use any other alkyl group instead of the CH3.
The most reactive substance containing an acyl group is an acyl chloride (also known as an acid chloride). Benzene is treated with a mixture of ethanoyl chloride, CH3COCl, and aluminium chloride as the catalyst. A ketone called phenylethanone is formed.
or:
Heterogeneous catalysis
Mechanism
According to Surface adsoprtion theory heterogeneous catalysis has five stages:
Stage 1: Diffusion of Reactant(s) to the Surface: The rate at which reactants will diffuse to the surface will be influenced by their bulk concentration and by the thickness of the boundary layer.
Stage 2: Adsorption of reactants:Bonds are formed as the reactant(s) are adsorbed onto the surface of the catalyst. The ability for an atom or molecule to stick to the surface is known, brilliantly, as the Sticking Co-efficient. This is just the ratio or percentage of molecules that end up sticking on the surface.
Stage 3: Reaction: Bonds form between the atoms and molecules on the surface
Stage 4: Desorption of products: Bonds are broken as the product(s) desorb from the surface.
Stage 5: Diffusion of Product(s) away from the Surface: The products are then desorbed from the surface of the catalyst.
Example $1$: Contact Process
The contact process is the current method of producing sulfuric acid in the high concentrations needed for industrial processes. Platinum used to be the catalyst for this reaction; however, as it is susceptible to reacting with arsenic impurities in the sulfur feedstock, vanadium(V) oxide ($V_2O_5$) is now preferred
In the contact process vanadium [V] oxide (V2O2) is solid whereas the reactants SO2 and O2 are gaseous.
$2SO_2 (g) + O_2 (g)\overset{V_2O_5(s)}\rightleftharpoons 2SO_3 (g) \nonumber$
More detailed method:
$2V_2O_5(s) + 2SO_2(g)\rightleftharpoons 2SO_3(g) + 2V_2O_4(s) \nonumber$
$2V_2O_4(s) + O_2 (g) \rightleftharpoons 2V_2O_5 (s) \nonumber$
Therefore:
$2SO_2(g) + O_2(g)\rightleftharpoons 2SO_3(g) \nonumber$
Example $2$: Haber–Bosch Reaction
The original Haber–Bosch reaction chambers used osmium as the catalyst, less expensive iron-based catalyst, which is still used today
$N_2 (g) + 3H_2 (g) \overset{Fe(s)}\rightleftharpoons 2NH_3(g) \nonumber$
The reaction mechanism, involving the heterogeneous catalyst, is believed to involve the following steps:
1. N2 (g) → N2 (adsorbed)
2. N2 (adsorbed) → 2 N (adsorbed)
3. H2 (g) → H2 (adsorbed)
4. H2 (adsorbed) → 2 H (adsorbed)
5. N (adsorbed) + 3 H(adsorbed)→ NH3 (adsorbed)
6. NH3 (adsorbed) → NH3 (g) | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Catalysis/Examples/4._Examples_of_Acid_Catalysis_in_Organic_Chemistry.txt |
Catalysis is the ability of some species to rapidly speed up the rate at which a chemical reaction proceeds. For historical reasons, the discipline is normally split into two sub-catagories; homogeneous (homo = same, geneous = phase) and heterogeneous (hetero = different). Homogeneous catalysis is concerned with catalysts that are in the same phase as the chemical reactions they are speeding up. These reactions are normally in the liquid phase and include all of biology's enzymes. While the majority of homogeneous catalysis is in the liquid phase there are gas phase and solid phase homogeneous catalytic reactions. Heterogeneous catalysts have a catalyst that is in a different phase. This type of catalysis is responsible for the vast majority of 'bulk' chemicals that are produced each year that go into making all the things we take for granted around us such as plastics, and are also extensively used for refining oil in gasoline. This chapter focuses on heterogeneous catalysis and specifically how and why they work at all.
The key concept in catalysis, and indeed all of chemistry, is the tension between the thermodynamics of a particular reaction which tells you whether something should happen, and the kinetics which tells you how fast something will happen. This tension in the world around you is everywhere i.e. thermodynamics says the diamond on an engagement ring should turn into graphite, a more stable allotrope of carbon but the kinetics is such that you would be waiting a very, very long time before it happened! Catalysis works by giving chemical reactions that should happen a way to actually happen.
The standard way to think of a chemical reaction happening is to imagine the energy of a molecule as being represented by a z co-ordinate while the reaction 'proceeds' in the x-y plane. If this sounds confusing, just think of the energy as being like the height of land features like mountains relative to their position. A molecule in this landscape would be a place where you can't go downhill without having to go uphill first, like a crater lake. Even though on a global scale, a crater lake should flow down to the sea, the local conditions around it mean that it would have to first flow uphill, which we could postulate doesn't happen on a regular basis. This analogy helps to visualise what is going on in a chemical reaction: the atoms involved when they're joined in one fashion are in an energy 'lake' and aren't able to change, but given the right conditions, such as temperature or photons, they are able to get over the energy barrier and flow down to another energy 'lake' which has the atoms joined in another fashion. This energy barrier is called the activation energy. In our analogy, imagine drilling a hole from one lake to another lake, since the water doesn't have to go 'up and over', it can flow easily to another position. This is exactly what a catalyst does, it doesn't change the total energy of the system or where the system will eventually end up (the thermodynamics dictate this), it just provides an easier way for the system to get there and hence speeds it up (it increases the rate). This analogy paints a picture that helps you remember the text-book definition of a catalyst: a substance which increases the rate at which a thermodynamic equilibrium is obtained by lowering the activation energy of the reaction pathway.
That is what a catalyst does but it doesn't give us any idea of how. It is only useful in deciding whether you are observing catalysis in action, the mechanism of how heterogeneous catalysts work is far more subtle.
The Effect of a Catalyst on Rate of Reaction
This page explains how adding a catalyst affects the rate of a reaction. It assumes familiarity with basic concepts in the collision theory of reaction rates, and with the Maxwell-Boltzmann distribution of molecular energies in a gas. A catalyst is a substance which speeds up a reaction, but is chemically unchanged at its end. When the reaction has finished, the mass of catalyst is the same as at the beginning. Several examples of catalyzed reactions and their respective catalysts are given below:
reaction catalyst
Decomposition of hydrogen peroxide manganese(IV) oxide, MnO2
Nitration of benzene concentrated sulfuric acid
Manufacture of ammonia by the Haber Process iron
Conversion of SO2 into SO3 during the Contact Process to make sulfuric acid vanadium(V) oxide, V2O5
Hydrogenation of a C=C double bond nickel
The importance of activation energy
Collisions only result in a reaction if the particles collide with a certain minimum energy called the activation energy for the reaction. The position of activation energy can be determined from a on a Maxwell-Boltzmann distribution:
Only those particles represented by the area to the right of the activation energy will react when they collide. The majority do not have enough energy, and will simply bounce apart.
To increase the rate of a reaction, the number of successful collisions must be increased. One possible way of doing this is to provide an alternative way for the reaction to happen which has a lower activation energy. In other words, to move the activation energy to the left on the graph:
Adding a catalyst has this effect on activation energy. A catalyst provides an alternative route for the reaction with a lower activation energy. This is illustrated on the following energy profile:
A word of caution
Care must be taken when discussing how a catalyst operates. A catalyst provides an alternative route for the reaction with a lower activation energy. It does not "lower the activation energy of the reaction". There is a subtle difference between the two statements that is easily illustrated with a simple analogy. Suppose there is a mountain between two valleys such that the only way for people to get from one valley to the other is over the mountain. Only the most active people will manage to get from one valley to the other.
Now suppose a tunnel is cut through the mountain. Many more people will now manage to get from one valley to the other by this easier route. It could be said that the tunnel route has a lower activation energy than going over the mountain, but the mountain itself is not lowered. The tunnel has provided an alternative route but has not lowered the original one. The original mountain is still there, and some people still choose to climb it. In chemical terms, if particles collide with enough energy they can still react in exactly the same way as if the catalyst was not there; it is simply that the majority of particles will react via the easier catalyzed route. | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Catalysis/Introduction_to_Catalysis.txt |
Chemical compounds can generally be classified into two broad groups: molecular compounds and ionic compounds. Molecular compounds involve atoms joined by covalent bonds and can be represented by a variety of formulas. Ionic compounds are composed of ions joined by ionic bonding, and their formulas are generally written using oxidation states.
Chemical Compounds
Stoichiometric coefficients are very convenient with normal chemical equations and their reactions. However, these coefficients are inaccurate in certain compounds due to missing or extra atoms of a specific element. In these compounds, we have a special way of writing the formulas to account for the missing/extra particles.
Contributors and Attributions
• Rachel Feldman, UC Davis
Chemical Compounds
Chemical compounds can generally be classified into two broad groups: molecular compounds and ionic compounds. Molecular compounds involve atoms joined by covalent bonds and can be represented by a variety of formulas. Ionic compounds are composed of ions joined by ionic bonding, and their formulas are generally written using oxidation states.
Molecular compounds are composed of atoms that are held together by covalent bonds. These bonds are formed when electrons are shared between two atoms. The concept of chemical formulas was created to describe many characteristics of molecular compounds in a simple manner. A normal chemical formula encompasses factors such as which elements are in the molecule and how many atoms of each element there are. The number of atoms of each element is denoted by a subscript, a small number that is written to the left of the element.
$CH_3COOH \nonumber$
In the preceding formula, the subscript “3” denotes the fact that there are three hydrogen atoms present in the molecule.
Other types of formulas are used to display more detailed characteristics of molecules.
An empirical formula represents the proportions of atoms in a molecule. It gives important information about a molecule, because it displays the ratios of atoms that are present within the molecule. However, its limitations exist in the sense that it does not represent the exact number of atoms present in the molecule as the molecular formula does. In certain situations, the molecular and the empirical formula can be the same, but in other situations, the molecular formula is a multiple of the ratios of atoms indicated in the empirical formula. Since empirical formulas can be derived from molecular formulas, molecular formulas are generally more useful than empirical formulas.
Empirical vs. molecular compounds
C5H7O is a possible empirical formula, because a ratio of 5:7:1 cannot be simplified any further. In this particular case, the empirical formula could also be the molecular formula, if there are exactly 5 carbon atoms, 7 hydrogen atoms, and 1 oxygen atom per molecule. However, another possible molecular formula for this same molecule is C10H14O2, because while there are 10 carbon atoms, 14 hydrogen atoms, and 2 oxygen atoms present, the ratio 10:14:2 can be simplified to 5:7:1, giving way to the same empirical formula. Additionally, C10H14O2 is not the only possibility of a molecular formula for this molecule; any formula with the same relative proportions of these atoms that can be simplified to a 5:7:1 ratios is a possible molecular formula for this molecule. When given adequate information, the empirical formula and molecular formula can be quantitatively ascertained.
A structural formula is written to denote the details of individual atoms’ bonding. More specifically, it clarifies what types of bonds exist, between which atoms these bonds exist, and the order of the atoms’ bonding within the molecule. Covalent bonds are denoted by lines. A single line represents a single bond, two lines represent a double bond, three lines represent a triple bond, and onwards. A single covalent bond occurs when two electrons are shared between atoms, a double occurs when four electrons are shared between two atoms, etc. In this sense, the higher the number of bonds, the stronger the bond between the two atoms.
A condensed structural formula is a less graphical way of representing the same characteristics displayed by a structural formula. In this type of formula, the molecule is written as a molecular formula with the exception that it indicates where the bonding occurs.
All the representations discussed thus far have not addressed how to show a molecule’s three-dimensional structure. The two ways to illustrate a spatial structure are through the use of the ball-and-stick model as well as the space-filling model.
The ball-and-stick model uses balls to spatially represent a molecule. The balls are the atoms in a molecule and sticks are the bonds between specific atoms.
The space-filling model is also a method of spatially displaying a molecule and its characteristics. A space-filling model shows atoms’ sizes relative sizes to one another.
Ionic Compounds
Ionic compounds are composed of positive and negative ions that are joined by ionic bonds. Ionic bonds are generally formed when electrons are transferred from one atom to another, causing individual atoms to become charged particles, or ions. Ions can be referred to as either monatomic or polyatomic. Monatomic ions such as Cl- are composed of only one ion, while polyatomic ions such as NO3- are defined as polyatomic ions. A combination of these ions that forms a compound whose charge is equal to zero is known as a formula unit of an ionic compound. Ionic compounds generally tend to form crystallized salts. They generally have high boiling/melting points, and are good conductors of electricity. The formulas of ionic compounds are always written with the cation first, followed by the anion. The formula can then be completed with reference to the oxidation states of the elements present.
Coordinative Unsaturation
Coordination unsaturation is effectively the maximum coordination number that a metal ion can adopt.
Coordinative unsaturation is based on ionic radii (ultimately atomic radii from 0.1), where the atomic size increases from right to left (lower nuclear charge) and top to bottom (increasing n level being filled). For any metal, ionization readily occurs to generate a cation, which will yield ionic radii smaller than atomic radii. In this case, the metal cation is considered a Lewis acid with Lewis basic ligands surrounding it through coordinate bonds. The number of Lewis bases surrounding the Lewis acidic metal cation is the coordination number.
The metal cation charge does complicate comparison of say K+ to Ca2+, where Ca2+ is smaller despite being to the right of K+. In this comparison, the coordinative unsaturation is roughly equivalent between K+ and Ca2+ because the additional charge on Ca2+ compensates for the decreased radius to yield both ions tending up to a coordination number of 8.
So the coordinative unsaturation increases as you go down a group: Be2+, 4; Mg2+, 6; Ca2+, 8; Sr2+, 8, Ba2+, 8. | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Chemical_Compounds/Case_Study%3A_Non-Stoichometric_Bonding_in_Materials.txt |
Chemistry is the experimental and theoretical study of materials on their properties at both the macroscopic and microscopic levels. Understanding the relationship between properties and structures/bonding is also a hot pursuit. Chemistry is traditionally divided into organic and inorganic chemistry. The former is the study of compounds containing at least one carbon-hydrogen bonds. By default, the chemical study of all other substances is called inorganic chemistry, a less well defined subject.
However, the boundary between organic and inorganic compounds is not always well defined. For example, oxalic acid, H2C2O4, is a compound formed in plants, and it is generally considered an organic acid, but it does not contain any C-H bond. Inorganic chemistry is also closely related to other disciplines such as materials sciences, physical chemistry, thermodynamics, earth sciences, mineralogy, crystallography, spectroscopy etc.
A chemical formula is a format used to express the structure of atoms. The formula tells which elements and how many of each element are present in a compound. Formulas are written using the elemental symbol of each atom and a subscript to denote the number of elements. This notation can be accredited to Swedish chemist Jons Jakob Berzeliu. The most common elements present in organic compounds are carbon, hydrogen, oxygen, and nitrogen. With carbon and hydrogen present, other elements, such as phosphorous, sulfur, silicon, and the halogens, may exist in organic compounds. Compounds that do not pertain to this rule are called inorganic compounds.
Molecular Geometry and Structural Formula
Understanding how atoms in a molecules are arranged and how they are bonded together is very important in giving the molecule its identity. Isomers are compounds in which two molecules can have the same number of atoms, and thus the same molecular formula, but can have completely different physical and chemical properties because of differences in structural formula.
Methylpropane and butane have the same molecular formula of C4H10, but are structurally different (methylpropane on the left, butane on the right).
Polymers
A polymer is formed when small molecules of identical structure, monomers, combine into a large cluster. The monomers are joined together by covalent bonds. When monomers repeat and bind, they form a polymer. While they can be comprised of natural or synthetic molecules, polymers often include plastics and rubber. When a molecule has more than one of these polymers, square parenthesis are used to show that all the elements within the polymer are multiplied by the subscript outside of the parenthesis. The subscript (shown as n in the example below) denotes the number of monomers present in the macromolecule (or polymer).
Ethylene becomes the polymer polyethylene.
Molecular Formula
The molecular formula is based on the actual makeup of the compound. Although the molecular formula can sometimes be the same as the empirical formula, molecular compounds tend to be more helpful. However, they do not describe how the atoms are put together. Molecular compounds are also misleading when dealing with isomers, which have the same number and types of atoms (see above in molecular geometry and structural formula).
Ex. Molecular Formula for Ethanol: C2H6O.
Empirical Formula
An empirical formula shows the most basic form of a compound. Empirical formulas show the number of atoms of each element in a compound in the most simplified state using whole numbers. Empirical formulas tend to tell us very little about a compound because one cannot determine the structure, shape, or properties of the compound without knowing the molecular formula. Usefulness of the empirical formula is decreased because many chemical compounds can have the same empirical formula.
Ex. Find the empirical formula for C8H16O2.
Answer: C4H8O (divide all subscripts by 2 to get the smallest, whole number ratio).
Structural Formula
A structural formula displays the atoms of the molecule in the order they are bonded. It also depicts how the atoms are bonded to one another, for example single, double, and triple covalent bond. Covalent bonds are shown using lines. The number of dashes indicate whether the bond is a single, double, or triple covalent bond. Structural formulas are helpful because they explain the properties and structure of the compound which empirical and molecular formulas cannot always represent.
Ex. Structural Formula for Ethanol:
Condensed Structural Formula
Condensed structural formulas show the order of atoms like a structural formula but are written in a single line to save space and make it more convenient and faster to write out. Condensed structural formulas are also helpful when showing that a group of atoms is connected to a single atom in a compound. When this happens, parenthesis are used around the group of atoms to show they are together.
Ex. Condensed Structural Formula for Ethanol: CH3CH2OH (Molecular Formula for Ethanol C2H6O).
Line-Angle Formula
Because organic compounds can be complex at times, line-angle formulas are used to write carbon and hydrogen atoms more efficiently by replacing the letters with lines. A carbon atom is present wherever a line intersects another line. Hydrogen atoms are then assumed to complete each of carbon's four bonds. All other atoms that are connected to carbon atoms are written out. Line angle formulas help show structure and order of the atoms in a compound making the advantages and disadvantages similar to structural formulas.
Ex. Line-Angle Formula for Ethanol:
Formulas of Inorganic Compounds
Inorganic compounds are typically not of biological origin. Inorganic compounds are made up of atoms connected using ionic bonds. These inorganic compounds can be binary compounds, binary acids, or polyatomic ions.
Binary compounds
Binary compounds are formed between two elements, either a metal paired with a nonmetal or two nonmetals paired together. When a metal is paired with a nonmetal, they form ionic compounds in which one is a negatively charged ion and the other is positvely charged. The net charge of the compound must then become neutral. Transition metals have different charges; therefore, it is important to specify what type of ion it is during the naming of the compound. When two nonmetals are paired together, the compound is a molecular compound. When writing out the formula, the element with a positive oxidation state is placed first.
Ex. Ionic Compound: BaBr2(Barium Bromide)
Ex. Molecular Compound: N2O4 (Dinitrogen Tetroxide)
Binary acids
Binary acids are binary compounds in which hydrogen bonds with a nonmetal forming an acid. However, there are exceptions such as NH3, which is a base. This is because it shows no tendency to produce a H+. Because hydrogen is positively charged, it is placed first when writing out these binary acids.
Ex. HBr (Hydrobromic Acid)
Polyatomic ions
Polyatomic ions is formed when two or more atoms are connected with covalent bonds. Cations are ions that have are postively charged, while anions are negatively charged ions. The most common polyatomic ions that exists are those of anions. The two main polyatomic cations are Ammonium and Mercury (I). Many polyatomic ions are typically paired with metals using ionic bonds to form chemical compounds.
Ex. MnO4- (Polyatomic ion); NaMnO4 (Chemical Compound)
Oxoacids
Many acids have three different elements to form ternary compounds. When one of those three elements is oxygen, the acid is known as a oxoacid. In other words, oxacids are compounds that contain hydrogen, oxgygen, and one other element.
Ex. HNO3 (Nitric Acid)
Complex Compounds
Certain compounds can appear in multiple forms yet mean the same thing. A common example is hydrates: water molecules bond to another compound or element. When this happens, a dot is shown between H2O and the other part of the compound. Because the H2O molecules are embedded within the compound, the compound is not necessarily "wet". When hydrates are heated, the water in the compound evaporates and the compound becomes anhydrous. These compounds can be used to attract water such as CoCl2. When CoCl2 is dry, CoCl2 is a blue color wherease the hexahydrate (written below) is pink in color.
Ex. CoCl2 ·6 H2O
Formulas of Organic Compounds
Organic compounds contain a combination carbon and hydrogen or carbon and hydrogen with nitrogen and a few other elements, such as phosphorous, sulfur, silicon, and the halogens. Most organic compounds are seen in biological origin, as they are found in nature.
Hydrocarbons
Hydrocarbons are compounds that consist of only carbon and hydrogen atoms. Hydrocarbons that are bonded together with only single bonds are alkanes. The simplest example is methane (shown below). When hydrocarbons have one or more double bonds, they are called alkenes. The simplest alkene is Ethene (C2H4) which contains a double bond between the two carbon atoms.
Ex. Methane on left, Ethene on right
Functional Groups
Functional groups are atoms connected to carbon chains or rings of organic molecules. Compounds that are within a functional group tend to have similar properties and characteristics. Two common functional groups are hydroxyl groups and carboxyl groups. Hydroxyl groups end in -OH and are alcohols. Carboxyl groups end in -COOH, making compounds containing -COOH carboxylic acids. Functional groups also help with nomenclature by using prefixes to help name the compounds that have similar chemical properties.
Ex. Hydroxyl Group on top; Carboxyl Group on bottom
Problems
1. Which of the following formulas are organic?
1. HClO
2. C5H10
3. CO2
2. What is the name of the following formula?
1. Classify the following formulas into their appropriate functional group
1. Acetic acid
2. Butanol
3. Oxalic acid
1. What are the empirical formulas for the following compounds?
1. C12H10O6
2. CH3CH2CH2CH2CH2CH2CH3
3. H3O
1. What is the name of the following figure and what is the molecular formula of the following figure?
Answer Key:
1. b and c. 2. Propane. 3. a. carboxyl group, b. hydroxyl group, c. carboxyl group. 4. a. C6H5O3, b. C7H16, c. H3O. 5. Methylbutane, C5H12
Contributors and Attributions
• Jean Kim (UCD), Kristina Bonnett (UCD) | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Chemical_Compounds/Formulas_of_Inorganic_and_Organic_Compounds.txt |
Chemical bonding is one of the most basic fundamentals of chemistry that explains other concepts such as molecules and reactions. Without it, scientists wouldn't be able to explain why atoms are attracted to each other or how products are formed after a chemical reaction has taken place. To understand the concept of bonding, one must first know the basics behind atomic structure.
Introduction
A common atom contains a nucleus composed of protons and neutrons, with electrons in certain energy levels revolving around the nucleus. In this section, the main focus will be on these electrons. Elements are distinguishable from each other due to their "electron cloud," or the area where electrons move around the nucleus of an atom. Because each element has a distinct electron cloud, this determines their chemical properties as well as the extent of their reactivity (i.e. noble gases are inert/not reactive while alkaline metals are highly reactive). In chemical bonding, only valence electrons, electrons located in the orbitals of the outermost energy level (valence shell) of an element, are involved.
Lewis Diagrams
Lewis diagrams are graphical representations of elements and their valence electrons. Valance electrons are the electrons that form the outermost shell of an atom. In a Lewis diagram of an element, the symbol of the element is written in the center and the valence electrons are drawn around it as dots. The position of the valence electrons drawn is unimportant. However, the general convention is to start from 12o'clock position and go clockwise direction to 3 o'clock, 6 o'clock, 9 o'clock, and back to 12 o'clock positions respectively. Generally the Roman numeral of the group corresponds with the number of valance electrons of the element.
Below is the periodic table representation of the number of valance electrons. The alkali metals of Group IA have one valance electron, the alkaline-earth metals of Group IIA have 2 valance electrons, Group IIIA has 3 valance electrons, and so on. The nonindicated transition metals, lanthanoids, and actinoids are more difficult in terms of distinguishing the number of valance electrons they have; however, this section only introduces bonding, hence they will not be covered in this unit.
Lewis diagrams for Molecular Compounds/Ions
To draw the lewis diagrams for molecular compounds or ions, follow these steps below (we will be using H2O as an example to follow):
1) Count the number of valance electrons of the molecular compound or ion. Remember, if there are two or more of the same element, then you have to double or multiply by however many atoms there are of the number of valance electrons. Follow the roman numeral group number to see the corresponding number of valance electrons there are for that element.
Valance electrons:
Oxygen (O)--Group VIA: therefore, there are 6 valance electrons
Hydrogen (H)--Group IA: therefore, there is 1 valance electron
NOTE: There are TWO hydrogen atoms, so multiply 1 valance electron X 2 atoms
Total: 6 + 2 = 8 valance electrons
2) If the molecule in question is an ion, remember to add or subract the respective number of electrons to the total from step 1.
For ions, if the ion has a negative charge (anion), add the corresponding number of electrons to the total number of electrons (i.e. if NO3- has a negative charge of 1-, then you add 1 extra electron to the total; 5 + 3(6)= 23 +1 = 24 total electrons). A - sign mean the molecule has an overall negative charge, so it must have this extra electron. This is because anions have a higher electron affinity (tendency to gain electrons). Most anions are composed of nonmetals, which have high electronegativity.
If the ion has a positive charge (cation), subtract the corresponding number of electrons to the total number of electrons (i.e. H3O+ has a positive charge of 1+, so you subtract 1 extra electron to the total; 6 + 1(3) = 9 - 1 = 8 total electrons). A + sign means the molecule has an overall positive charge, so it must be missing one electron. Cations are positive and have weaker electron affinity. They are mostly composed of metals; their atomic radii are larger than the nonmetals. This consequently means that shielding is increased, and electrons have less tendency to be attracted to the "shielded" nucleus.
From our example, water is a neutral molecule, therefore no electrons need to be added or subtracted from the total.
3) Write out the symbols of the elements, making sure all atoms are accounted for (i.e. H2O, write out O and 2 H's on either side of the oxygen). Start by adding single bonds (1 pair of electrons) to all possible atoms while making sure they follow the octet rule (with the exceptions of the duet rule and other elements mentioned above).
4) If there are any leftover electrons, then add them to the central atom of the molecule (i.e. XeF4 has 4 extra electrons after being distributed, so the 4 extra electrons are given to Xe: like so. Finally, rearrange the electron pairs into double or triple bonds if possible.
Octet Rule
Most elements follow the octet rule in chemical bonding, which means that an element should have contact to eight valence electrons in a bond or exactly fill up its valence shell. Having eight electrons total ensures that the atom is stable. This is the reason why noble gases, a valence electron shell of 8 electrons, are chemically inert; they are already stable and tend to not need the transfer of electrons when bonding with another atom in order to be stable. On the other hand, alkali metals have a valance electron shell of one electron. Since they want to complete the octet rule they often simply lose one electron. This makes them quite reactive because they can easily donate this electron to other elements. This explains the highly reactive properties of the Group IA elements.
Some elements that are exceptions to the octet rule include Aluminum(Al), Phosphorus(P), Sulfur(S), and Xenon(Xe).
Hydrogen(H) and Helium(He) follow the duet rule since their valence shell only allows two electrons. There are no exceptions to the duet rule; hydrogen and helium will always hold a maximum of two electrons.
Ionic Bonding
Ionic bonding is the process of not sharing electrons between two atoms. It occurs between a nonmetal and a metal. Ionic bonding is also known as the process in which electrons are "transferred" to one another because the two atoms have different levels of electron affinity. In the picture below, a sodium (Na) ion and a chlorine (Cl) ion are being combined through ionic bonding. Na+ has less electronegativity due to a large atomic radius and essentially does not want the electron it has. This will easily allow the more electronegative chlorine atom to gain the electron to complete its 3rd energy level. Throughout this process, the transfer of the electron releases energy to the atmosphere.
Another example of ionic bonding is the crystal lattice structure shown above. The ions are arranged in such a way that shows unifomity and stablity; a physical characteristic in crystals and solids. Moreover, in a concept called "the sea of electrons," it is seen that the molecular structure of metals is composed of stabilized positive ions (cations) and "free-flowing" electrons that weave in-between the cations. This attributes to the metal property of conductivity; the flowing electrons allow the electric current to pass through them. In addition, this explains why strong electrolytes are good conductors. Ionic bonds are easily broken by water because the polarity of the water molecules shield the anions from attracting the cations. Therefore, the ionic compounds dissociate easily in water, and the metallic properties of the compound allow conductivity of the solution.
Covalent Bonding
Covalent bonding is the process of sharing of electrons between two atoms. The bonds are typically between a nonmetal and a nonmetal. Since their electronegativities are all within the high range, the electrons are attracted and pulled by both atom's nuceli. In the case of two identical atoms that are bonded to each other (also known as a nonpolar bond, explained later below), they both emit the same force of pull on the electrons, thus there is equal attraction between the two atoms (i.e. oxygen gas, or O2, have an equal distribution of electron affinity. This makes covalent bonds harder to break.
There are three types of covalent bonds: single, double, and triple bonds. A single bond is composed of 2 bonded electrons. Naturally, a double bond has 4 electrons, and a triple bond has 6 bonded electrons. Because a triple bond will have more strength in electron affinity than a single bond, the attraction to the positively charged nucleus is increased, meaning that the distance from the nucleus to the electrons is less. Simply put, the more bonds or the greater the bond strength, the shorter the bond length will be. In other words:
Bond length: triple bond < double bond < single bond
Polar Covalent Bonding
Polar covalent bonding is the process of unequal sharing of electrons. It is considered the middle ground between ionic bonding and covalent bonding. It happens due to the differing electronegativity values of the two atoms. Because of this, the more electronegative atom will attract and have a stronger pulling force on the electrons. Thus, the electrons will spend more time around this atom.
The symbols above indicate that on the flourine side it is slightly negitive and the hydrogen side is slightly positive.
Polar and Non-polar molecules
Polarity is the competing forces between two atoms for the electrons. It is also known as the polar covalent bond. A molecule is polar when the electrons are attracted to a more electronegative atom due to its greater electron affinity. A nonpolar molecule is a bond between two identical atoms. They are the ideal example of a covalent bond. Some examples are nitrogen gas (N2), oxygen gas (O2), and hydrogen gas (H2).
One way to figure out what type of bond a molecule has is by determining the difference of the electronegativity values of the molecules.
If the difference is between 0.0-0.3, then the molecule has a non-polar bond.
If the difference is between 0.3-1.7, then the molecule has a polar bond.
If the difference is 1.7 or more, then the molecule has an ionic bond. | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Chemical_Compounds/Introduction_to_Chemical_Bonding.txt |
Generally, there are two types of inorganic compounds that can be formed: ionic compounds and molecular compounds. Nomenclature is the process of naming chemical compounds with different names so that they can be easily identified as separate chemicals. Inorganic compounds are compounds that do not deal with the formation of carbohydrates, or simply all other compounds that do not fit into the description of an organic compound. For example, organic compounds include molecules with carbon rings and/or chains with hydrogen atoms (see picture below). Inorganic compounds, the topic of this section, are every other molecule that does not include these distinctive carbon and hydrogen structures.
Compounds between Metals and Nonmetals (Cation and Anion)
Compounds made of a metal and nonmetal are commonly known as Ionic Compounds, where the compound name has an ending of –ide. Cations have positive charges while anions have negative charges. The net charge of any ionic compound must be zero which also means it must be electrically neutral. For example, one Na+ is paired with one Cl-; one Ca2+ is paired with two Br-. There are two rules that must be followed through:
• The cation (metal) is always named first with its name unchanged
• The anion (nonmetal) is written after the cation, modified to end in –ide
Table 1: Cations and Anions:
+1 Charge +2 Charge -1 Charge -2 Charge -3 Charge -4 Charge
Group 1A elements Group 2A elements Group 7A elements Group 6A elements Group 5A elements Group 4A elements
Hydrogen: H+ Beryllium: Be2+ Hydride: H- Oxide: O2- Nitride: N3- Carbide: C4-
Lithium: Li+ Magnesium: Mg2+ Fluoride: F- Sulfide: S2- Phosphide: P3-
Soduim: Na+ Calcium: Ca2+ Chloride: Cl-
Potassium: K+ Strontium: Sr2+ Bromide: Br-
Rubidium: Rb+ Barium: Ba2+ Iodide: I-
Cesium: Cs+
Example 1
Na+ + Cl- = NaCl; Ca2+ + 2Br- = CaBr2
Sodium + Chlorine = Sodium Chloride; Calcium + Bromine = Calcium Bromide
The transition metals may form more than one ion, thus it is needed to be specified which particular ion we are talking about. This is indicated by assigning a Roman numeral after the metal. The Roman numeral denotes the charge and the oxidation state of the transition metal ion. For example, iron can form two common ions, Fe2+ and Fe3+. To distinguish the difference, Fe2+ would be named iron (II) and Fe3+ would be named iron (III).
Table of Transition Metal and Metal Cations:
+1 Charge +2 Charge +3 Charge +4 Charge
Copper(I): Cu+ Copper(II): Cu2+ Aluminum: Al3+ Lead(IV): Pb4+
Silver: Ag+ Iron(II): Fe2+ Iron(III): Fe3+ Tin(IV): Sn4+
Cobalt(II): Co2+ Cobalt(III): Co3+
Tin(II): Sn2+
Lead(II): Pb2+
Nickel: Ni2+
Zinc: Zn2+
Example 2
Ions: Fe2++ 2Cl- Fe3++ 3Cl-
Compound: FeCl2 FeCl3
Nomenclature Iron (II) Chloride Iron (III) Chloride
However, some of the transition metals' charges have specific Latin names. Just like the other nomenclature rules, the ion of the transition metal that has the lower charge has the Latin name ending with -ous and the one with the the higher charge has a Latin name ending with -ic. The most common ones are shown in the table below:
Transition Metal Ion with Roman Numeral Latin name
Copper (I): Cu+ Cuprous
Copper (II): Cu2+ Cupric
Iron (II): Fe2+ Ferrous
Iron (III): Fe3+ Ferric
Lead (II): Pb2+ Plumbous
Lead (IV): Pb4+ Plumbic
Mercury (I): Hg22+ Mercurous
Mercury (II): Hg2+ Mercuric
Tin (II): Sn2+ Stannous
Tin (IV): Sn4+ Stannic
Several exceptions apply to the Roman numeral assignment: Aluminum, Zinc, and Silver. Although they belong to the transition metal category, these metals do not have Roman numerals written after their names because these metals only exist in one ion. Instead of using Roman numerals, the different ions can also be presented in plain words. The metal is changed to end in –ous or –ic.
• -ous ending is used for the lower oxidation state
• -ic ending is used for the higher oxidation state
Example 3
Compound Cu2O CuO FeCl2 FeCl3
Charge Charge of copper is +1 Charge of copper is +2 Charge of iron is +2 Charge of iron is +3
Nomenclature Cuprous Oxide Cupric Oxide Ferrous Chloride Ferric Chloride
However, this -ous/-ic system is inadequate in some cases, so the Roman numeral system is preferred. This system is used commonly in naming acids, where H2SO4 is commonly known as Sulfuric Acid, and H2SO3 is known as Sulfurous Acid.
Compounds between Nonmetals and Nonmetals
Compounds that consist of a nonmetal bonded to a nonmetal are commonly known as Molecular Compounds, where the element with the positive oxidation state is written first. In many cases, nonmetals form more than one binary compound, so prefixes are used to distinguish them.
# of Atoms 1 2 3 4 5 6 7 8 9 10
Prefixes Mono- Di- Tri- Tetra- Penta- Hexa- Hepta- Octa- Nona- Deca-
Example 4
CO = carbon monoxide BCl3 = borontrichloride
CO2 = carbon dioxide N2O5 =dinitrogen pentoxide
The prefix mono- is not used for the first element. If there is not a prefix before the first element, it is assumed that there is only one atom of that element.
Binary Acids
Although HF can be named hydrogen fluoride, it is given a different name for emphasis that it is an acid. An acid is a substance that dissociates into hydrogen ions (H+) and anions in water. A quick way to identify acids is to see if there is an H (denoting hydrogen) in front of the molecular formula of the compound. To name acids, the prefix hydro- is placed in front of the nonmetal modified to end with –ic. The state of acids is aqueous (aq) because acids are found in water.
Some common binary acids include:
HF (g) = hydrogen fluoride -> HF (aq) = hydrofluoric acid
HBr (g) = hydrogen bromide -> HBr (aq) = hydrobromic acid
HCl (g) = hydrogen chloride -> HCl (aq) = hydrochloric acid
H2S (g) = hydrogen sulfide -> H2S (aq) = hydrosulfuricacid
It is important to include (aq) after the acids because the same compounds can be written in gas phase with hydrogen named first followed by the anion ending with –ide.
Example 5
hypo____ite ____ite ____ate per____ate
ClO- ClO2- ClO3- ClO4-
hypochlorite chlorite chlorate perchlorate
---------------->
As indicated by the arrow, moving to the right, the following trends occur:
Increasing number of oxygen atoms
Increasing oxidation state of the nonmetal
(Usage of this example can be seen from the set of compounds containing Cl and O)
This occurs because the number of oxygen atoms are increasing from hypochlorite to perchlorate, yet the overall charge of the polyatomic ion is still -1. To correctly specify how many oxygen atoms are in the ion, prefixes and suffixes are again used.
Polyatomic Ions
In polyatomic ions, polyatomic (meaning two or more atoms) are joined together by covalent bonds. Although there may be a element with positive charge like H+, it is not joined with another element with an ionic bond. This occurs because if the atoms formed an ionic bond, then it would have already become a compound, thus not needing to gain or loose any electrons. Polyatomic anions are more common than polyatomic cations as shown in the chart below. Polyatomic anions have negative charges while polyatomic cations have positive charges. To indicate different polyatomic ions made up of the same elements, the name of the ion is modified according to the example below:
Table: Common Polyatomic ions
Name: Cation/Anion Formula
Ammonium ion NH4+
Hydronium ion
H3O+
Acetate ion
C2H3O2-
Arsenate ion
AsO43-
Carbonate ion
CO32-
Hypochlorite ion
ClO-
Chlorite ion
ClO2-
Chlorate ion
ClO3-
Perchlorate ion
ClO4-
Chromate ion
CrO42-
Dichromate ion
Cr2O72-
Cyanide ion
CN-
Hydroxide ion
OH-
Nitrite ion
NO2-
Nitrate ion
NO3-
Oxalate ion
C2O42-
Permanganate ion
MnO4-
Phosphate ion
PO43-
Sulfite ion
SO32-
Sulfate ion
SO42-
Thiocyanate ion
SCN-
Thiosulfate ion
S2O32-
To combine the topic of acids and polyatomic ions, there is nomenclature of aqueous acids. Such acids include sulfuric acid (H2SO4) or carbonic acid (H2CO3). To name them, follow these quick, simple rules:
1. If the ion ends in -ate and is added with an acid, the acid name will have an -ic ending. Examples: nitrate ion (NO3-) + H+ (denoting formation of acid) = nitric acid (HNO3)
2. If the ion ends in -ite and is added with an acid, then the acid name will have an -ous ending. Example: nitite ion (NO2-) + H+ (denoting formation of acid) = nitrous acid (HNO2)
Problems
1. What is the correct formula for Calcium Carbonate?
a. Ca+ + CO2-
b. CaCO2-
c. CaCO3
d. 2CaCO3
2. What is the correct name for FeO?
a. Iron oxide
b. Iron dioxide
c. Iron(III) oxide
d. Iron(II) oxide
3. What is the correct name for Al(NO3)3?
a. Aluminum nitrate
b. Aluminum(III) nitrate
c. Aluminum nitrite
d. Aluminum nitrogen trioxide
4. What is the correct formula of phosphorus trichloride?
a. P2Cl2
b. PCl3
c. PCl4
d. P4Cl2
5. What is the correct formula of lithium perchlorate?
a. Li2ClO4
b. LiClO2
c. LiClO
d. None of these
6. Write the correct name for these compounds.
a. BeC2O4:
b. NH4MnO4:
c. CoS2O3:
7. What is W(HSO4)5?
8. How do you write diphosphorus trioxide?
9. What is H3P?
10. By adding oxygens to the molecule in number 9, we now have H3PO4? What is the name of this molecule?
Answer
1.C; Calcium + Carbonate --> Ca2+ + CO32- --> CaCO3
2.D; FeO --> Fe + O2- --> Iron must have a charge of +2 to make a neutral compound --> Fe2+ + O2- --> Iron(II) Oxide
3.A; Al(NO3)3 --> Al3+ + (NO3-)3 --> Aluminum nitrate
4.B; Phosphorus trichloride --> P + 3Cl --> PCl3
5.D, LiClO4; Lithium perchlorate --> Li+ + ClO4- --> LiClO4
6. a. Beryllium Oxalate; BeC2O4 --> Be2+ + C2O42- --> Beryllium Oxalate
b. Ammonium Permanganate; NH4MnO4 --> NH4+ + MnO4- --> Ammonium Permanganate
c. Cobalt (II) Thiosulfate; CoS2O3 --> Co + S2O32- --> Cobalt must have +2 charge to make a neutral compund --> Co2+ + S2O32- --> Cobalt(II) Thiosulfate
7. Tungsten (V) hydrogen sulfate
8. P2O3
9. Hydrophosphoric Acid
10. Phosphoric Acid
Contributors and Attributions
• Pui Yan Ho (UCD), Alex Moskaluk (UCD), Emily Nguyen (UCD) | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Chemical_Compounds/Nomenclature_of_Inorganic_Compounds.txt |
A second part is devoted to the subject of conjugation of acids and bases. The relationship between the acidic constant Ka, basic constant Kb, and the constant of autoionization of water, Kw will be discussed. The relationship is useful for weak acids and bases.
Learning Objectives
• Give three definitions for acids.
• Give three definitions for bases.
• Explain conjugate Acid-Base pairs.
• Give the conjugate base of an acid.
• Give the conjugate acid of a base.
Balance Reduction and Oxidation (Redox) Reactions
Skill to Develop
• Balance reduction-oxidation (redox) equations until you have developed a logical method without having to memorize the steps needed to balance redox equations.
Chemical Reactions Overview
Chemical reactions are the processes by which chemicals interact to form new chemicals with different compositions. Simply stated, a chemical reaction is the process where reactants are transformed into products. How chemicals react is dictated by the chemical properties of the element or compound- the ways in which a compound or element undergoes changes in composition.
Describing Reactions Quantitatively
Chemical reactions are constantly occurring in the world around us; everything from the rusting of an iron fence to the metabolic pathways of a human cell are all examples of chemical reactions. Chemistry is an attempt to classify and better understand these reactions.
A chemical reaction is typically represented by a chemical equation, which represents the change from reactants to products. The left hand side of the equation represents the reactants, while the right hand side represents the products. A typical chemical reaction is written with stoichiometric coefficients, which show the relative amounts of products and reactants involved in the reaction. Each compound is followed by a parenthetical note of the compound’s state of 2: (l) for liquid, (s) for solid, (g) for gas. The symbol (aq) is also commonly used in order to represent an aqueous solution, in which compounds are dissolved in water. A reaction might take the following form:
$\ce{A (aq) + B (g) \rightarrow C (s) + D (l)} \nonumber$
In the above example, $A$ and $B$, known as the reactants, reacted to form $C$ and $D$, the products.
To write an accurate chemical equation, two things must occur:
1. Each product and reactant must be written using its chemical formula, e.g., $H_2$
2. The number of atoms of each element must be equal on both sides of the equation. Coefficients are used in front of the chemical formulas in order to help balance the number of atoms, e.g.,
$\ce{2Mg + O_2 \rightarrow 2MgO} \nonumber$
Example $1$: Balancing Reactions
Hydrogen and nitrogen react together in order to produce ammonia gas, write the chemical equation of this reaction.
Solution
Step 1: Write each product and reactant using its chemical formula.
$\ce{H_2 + N_2 \rightarrow NH_3} \nonumber$
Step 2: Ensure the number of atoms of each element are equal on both sides of the equation.
$\ce{3H_2 + N_2 \rightarrow 2NH_3} \nonumber$
In order to balance this equation, coefficients must be used. Since there are only 2 nitrogen atoms present on the left side of the equation, a coefficient of 2 must be added to $NH_3$.
Stoichiometry
The coefficient that used for balancing the equation is called the stoichiometric coefficient. The coefficients tell us the ratio of each element in a chemical equation. For example
$\ce{2Mg + O_2 \rightarrow 2MgO} \nonumber$
means
• 2 moles of MgO is produced for every 2 moles of Mg consumed.
• 2 moles of MgO is produced for every 1 mole of O2 consumed.
When all of the reactants of a reaction are completely consumed, the reaction is in perfect stoichiometric proportions. Often times, however, a reaction is not in perfect stoichiometric proportions, leading to a situation in which the entirety of one reactant is consumed, but there is some of another reactant remaining. The reactant that is entirely consumed is called the limiting reactant, and it determines how much of the products are produced.
Example $2$: Limiting Reagent
4.00 g of hydrogen gas mixed with 20.0g of oxygen gas. How many grams of water are produced?
Solution
$n(H_2)=\dfrac{4g}{(1.008 \times2)g/mol}=1.98mol \nonumber$
So theoretically, it requires 0.99 mol of $O_2$
n(O2)=n(H2)*(1mol O2/2mol H2)=0.99 mol
m(O2)=n(O2)*(16g/mol *2) = 31.7 g O2
Because $O_2$ only has 20.0 g, less than the required mass. It is limiting.
Often, reactants do not react completely, resulting in a smaller amount of product formed than anticipated. The amount of product expected to be formed from the chemical equation is called the theoretical yield. The amount of product that is produced during a reaction is the actual yield. To determine the percent yield:
Percent yield =actual yield/theoretical yield X 100%
Chemical reactions do not only happen in the air, but also exist in solutions. In a solution, the solvent is the compound that is dissolved, and the solute is the compound that the solvent is dissolved in. The molarity of a solution is the number of moles of a solvent divided by the number of liters of solution.
$\ Molarity=\dfrac{ \text{amount of solute (mol)}}{\text{volume of solution (L)}} \nonumber$
$\ M=\dfrac{n}{V} \nonumber$
Example $3$: Concentrations
100.0 g NaCl is dissolved in 50.00 ml water. What is the molarity of the solution?
Solution
a) Find the amount of solute in moles.
100.0g/(22.99 g/mol+35.45 g/mol) =1.711 moles
b) Convert mL to L.
50.00 mL=0.05000 L
c) Find the molarity
1.711 moles/0.05000L=34.22 mol/L
Physical Changes During Chemical Reactions
Physical change is the change in physical properties. Physical changes usually occur during chemical reactions, but does not change the nature of substances. The most common physical changes during reactions are the change of color, scent and evolution of gas. However, when physical changes occur, chemical reactions may not occur.
Types of Chemical Reactions
Precipitation, or double-replacement reaction
A reaction that occurs when aqueous solutions of anions (negatively charged ions) and cations (positively charged ions) combine to form a compound that is insoluble is known as precipitation. The insoluble solid is called the precipitate, and the remaining liquid is called the supernate. See Figure2.1
Real life example: The white precipitate formed by acid rain on a marble statue:
$CaCO_3(aq)+H_2SO_4(aq) \rightarrow CaSO_4(s)+H_2O(l)+CO_2(g) \nonumber$
Example $4$: Precipitation
An example of a precipitation reaction is the reaction between silver nitrate and sodium iodide. This reaction is represented by the chemical equation :
AgNO3 (aq)+ NaI (aq) → AgI (s) + NaNO3 (aq)
Since all of the above species are in aqueous solutions, they are written as ions, in the form:
Ag+ +NO3- (aq)+ Na+ (aq) + I- (aq) → AgI (s) + Na+ (aq) + NO3- (aq)
Ions that appear on both sides of the equation are called spectator ions. These ions do not affect the reaction and are removed from both sides of the equation to reveal the net ionic equation, as written below:
Ag+ (aq) + I- (aq) → AgI (s)
In this reaction, the solid, AgI, is known as the precipitate. The formation of a precipitate is one of the many indicators that a chemical reaction has taken place.
Acid-base, or neutralization reaction
A neutralization reaction occurs when an acid and base are mixed together. An acid is a substance that produces H+ ions in solution, whereas a base is a substance that that produces OH- ions in solution. A typical acid-base reaction will produce an ionic compound called a salt and water. A typical acid-base reaction is the reaction between hydrochloric acid and sodium hydroxide. This reaction is represented by the equation:
$\ce{HCl (aq) + NaOH (aq) \rightarrow NaCl (aq)+ H_2O (l)} \nonumber$
In this reaction, $HCl$ is the acid, $NaOH$ is the base, and $NaCl$ is the salt. Real life example: Baking soda reacts with vinegar is a neutralization reaction.
Video: Vinegar and Baking Soda Reaction with Explanation
Oxidation-Reduction (Redox) Reactions
A redox reaction occurs when the oxidation number of atoms involved in the reaction are changed. Oxidation is the process by which an atom’s oxidation number is increased, and reduction is the process by which an atom’s oxidation number is decreased. If the oxidation states of any elements in a reaction change, the reaction is an oxidation-reduction reaction. An atom that undergoes oxidation is called the reducing agent, and the atom that undergoes reduction is called the oxidizing agent. An example of a redox reaction is the reaction between hydrogen gas and fluorine gas:
$H_2 (g) + F_2 (g) \rightarrow 2HF (g) \label{redox1}$
In this reaction, hydrogen is oxidized from an oxidation state of 0 to +1, and is thus the reducing agent. Fluorine is reduced from 0 to -1, and is thus the oxidizing agent.
Real life example: The cut surface of an apple turns brownish after exposed to the air for a while.
Video: Why Do Apples Turn Brown?
Combustion Reaction
A combustion reaction is a type of redox reaction during which a fuel reacts with an oxidizing agent, resulting in the release of energy as heat. Such reactions are exothermic, meaning that energy is given off during the reaction. An endothermic reaction is one which absorbs heat. A typical combustion reaction has a hydrocarbon as the fuel source, and oxygen gas as the oxidizing agent. The products in such a reaction would be $CO_2$ and $H_2O$.
$C_xH_yO_z+O_2 \rightarrow CO_2+H_2O \;\;\; \text{(unbalanced)} \nonumber$
Such a reaction would be the combustion of glucose in the following equation
$C_6H_{12}O_6 (s) + 6O_2 (g) \rightarrow 6CO_2 (g) + 6H_2O (g) \nonumber$
Real life example: explosion; burning.
Video: Combustion reactions come in many varieties. Here's a collection of various examples, all of which require oxygen, activation energy, and of course, fuel
Synthesis Reactions
A synthesis reaction occurs when one or more compounds combines to form a complex compound. The simplest equation of synthesis reaction is illustrated below.
An example of such a reaction is the reaction of silver with oxygen gas to form silver oxide:
$2Ag (s) +O_2 (g) \rightarrow 2AgO (s) \nonumber$
Real life example: Hydrogen gas is burned in air (reacts with oxygen) to form water:
$2H_2(g) + O_2(g) \rightarrow 2H_2O(l) \nonumber$
Decomposition Reactions
A decomposition reaction is the opposite of a synthesis reaction. During a decomposition reaction, a more complex compound breaks down into multiple simpler compounds. A classic example of this type of reaction is the decomposition of hydrogen peroxide into oxygen and hydrogen gas:
$H_2O_2 (l) \rightarrow H_2 (g) + O_2 (g) \nonumber$
Single Replacement Reactions
A type of oxidation-reduction reaction in which an element in a compound is replaced by another element.
An example of such a reaction is:
$Cu (s) + AgNO_3 (aq) \rightarrow Ag(s) + Cu(NO_3)_2 (aq) \nonumber$
This is also a redox reaction.
Problems
1) C3H6O3 + O2 → CO2 (g) +H2O (g)
a) What type of reaction is this?
b) Is is exothermic or endothermic? Explain.
2) Given the oxidation-reduction reaction :
Fe (s) + CuSO4 (aq)→ FeSO4 (aq)+ Cu (s)
a) Which element is the oxidizing agent and which is the reducing agent?
b) How do the oxidation states of these species change?
3) Given the equation:
AgNO3 (aq) + KBr (aq) → AgBr (s) +KNO3 (aq)
a) What is the net ionic reaction?
b) Which species are spectator ions?
4) 2 HNO3 (aq) + Sr(OH)2 (aq) → Sr(NO3)2 (aq) +2 H2O (l)
a) In this reaction, which species is the acid and which is the base?
b) Which species is the salt?
c) If 2 moles of HNO3 and 1 mole of Sr(OH)2 are used, resulting in 0.85 moles of Sr(NO3)2 , what is the percent yield (with respect to moles) of Sr(NO3)2 ?
5) Identify the type of the following reactions:
a) Al(OH)3 (aq) + HCl (aq) → AlCl3 (aq) + H2O (l)
b) MnO2 + 4H+ + 2Cl- → Mn2+ + 2H2O (l) + Cl2 (g)
c) P4 (s) + Cl2 (g) → PCl3 (l)
d) Ca (s) + 2H2O (l) → Ca(OH)2 (aq) + H2 (g)
e) AgNO3 (aq) + NaCl (aq) → AgCl (s) + NaNO3 (aq)
Solutions
1a) It is a combustion reaction
1b) It is exothermic, because combustion reactions give off heat
2a) Cu is the oxidizing agent and Fe is the reducing agent
2b) Fe changes from 0 to +2, and Cu changes from +2 to 0.
3a) Ag+ (aq) + Br- (aq) → AgBr (s)
3b) The spectator ions are K+ and NO3-
4a) HNO3 is the acid and Sr(OH)2 is the base
4b) Sr(NO3)2 is the salt
4c) According to the stoichiometric coefficients, the theoretical yield of Sr(NO3)2 is one mole. The actual yield was 0.85 moles. Therefore the percent yield is:
(0.85/1.0) * 100% = 85%
5a) Acid-base
5b) Oxidation-reduction
5c) Synthesis
5d) Single-replacement reaction
5e) Double replacement reaction
• Priya Muley | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Chemical_Reactions/Chemical_Reactions_Examples/Acids_and_Bases_-_Conjugate_Pairs.txt |
Part I, Part II
Learning Objectives
• Explain conjugate acids of bases.
• Evaluate Ka of the conjugate acid of a base.
• Treat the conjugate acid of a base as an acid in numerical calculations.
• Reverse the role of acid and base for the previous skills.
Electrolytes
Learning Objectives
• Identify what electrolytes are.
• Distinguish between strong and weak electrolytes.
• Explain what happens when electrolytes dissolve in water.
• Give the equilibrium constant expression for ionizaton.
• Explain ion product of water, autoionization of water, and pH.
• Calculate ionization percentage of weak electrolytes.
• Explain metathesis reactions.
One of the most important properties of water is its ability to dissolve a wide variety of substances. Solutions in which water is the dissolving medium are called aqueous solutions. For electrolytes, water is the most important solvent. Ethanol, ammonia, and acetic acid are some of the non-aqueous solvents that are able to dissolve electrolytes.
Electrolytes
Substances that give ions when dissolved in water are called electrolytes. They can be divided into acids, bases, and salts, because they all give ions when dissolved in water. These solutions conduct electricity due to the mobility of the positive and negative ions, which are called cations and anions respectively. Strong electrolytes completely ionize when dissolved, and no neutral molecules are formed in solution.
For example, $\ce{NaCl}$, $\ce{HNO3}$, $\ce{HClO3}$, $\ce{CaCl2}$ etc. are strong electrolytes. An ionization can be represented by
$\mathrm{NaCl_{\large{(s)}} \rightarrow Na^+_{\large{(aq)}} + Cl^-_{\large{(aq)}}}$
Since $\ce{NaCl}$ is an ionic solid (s), which consists of cations $\ce{Na+}$ and anions $\ce{Cl-}$, no molecules of $\ce{NaCl}$ are present in $\ce{NaCl}$ solid or $\ce{NaCl}$ solution. The ionization is said to be complete. The solute is one hundred percent (100%) ionized. Some other ionic solids are $\ce{CaCl2}$, $\ce{NH4Cl}$, $\ce{KBr}$, $\ce{CuSO4}$, $\ce{NaCH3COO}$ (sodium acetate), $\ce{CaCO3}$, and $\ce{NaHCO3}$ (baking soda).
Small fractions of weak electrolytes' molecules ionize when dissolve in water. Some neutral molecules are present in their solutions. For example, $\ce{NH4OH}$ (ammonia), $\ce{H2CO3}$ (carbonic acid), $\ce{CH3COOH}$ (acetic acid), and most organic acids and bases are weak electrolytes. The following ionization is not complete,
$\mathrm{H_2CO_{3\large{(aq)}} \rightleftharpoons H^+_{\large{(aq)}} + HCO^-_{3\large{(aq)}}}$
In a solution, $\ce{H2CO3}$ molecules are present. The fraction (often expressed as a %) that undergos ionization depends on the concentration of the solution.
On the other hand, ionization can be viewed as an equilibrium established for the above reaction, for which the equilibrium constant is defined as
$\mathrm{\mathit K = \dfrac{[H^+] [HCO_3^-]}{[H_2CO_3]}}$
where we use [ ] to mean the concentration of the species in the [ ]. For carbonic acid, K = 4.2x10-7. You can generalize the definition of K here to give the equilibrium constant expression for any weak electrolyte.
Pure water is a very weak electrolyte.
The ionization or autoionization of pure water can be represented by the ionization equation
$\mathrm{H_2O \rightleftharpoons H^+ + OH^-}$
and the equilibrium constant is
$\mathrm{\mathit K = \dfrac{[H^+] [OH^-]}{[H_2O]}}$
For pure water, $\ce{[H2O]}$ is a constant (1000/18 = 55.6 M), and we often use the ion product, Kw, for water,
$\mathrm{\mathit K_w = \mathit K [H_2O] [H^+] [OH^-]}$
The constant Kw depends on temperature. At 298 K, Kw = 1x10-14. If there is no solute in water, the solution has equal concentrations of $\ce{[H+]}$ and $\ce{[OH-]}$.
$\mathrm{[H^+] = [OH^-] = 1\times10^{-7}}$,
and
$\mathrm{pH = -\log [H^+] = 7}$.
Note that only at 298 K is the pH of water = 7. At higher temperatures, the pH is slightly less than 7, and at lower temperatures, the pH is greater than 7.
Electrolytes in Body Fluids
Our body fluids are solutions of electrolytes and many other things. The combination of blood and the circulatory system is the river of life, because it coordinates all the life functions. When the heart stops pumping in a heart attack, the life ends quickly. Getting the heart restarted as soon as one can is crucial in order to maintain life.
The primary electrolytes required in the body fluid are cations (of calcium, potassium, sodium, and magnesium) and anions (of chloride, carbonates, aminoacetates, phosphates, and iodide). These are nutritionally called macrominerals.
Electrolyte balance is crucial to many body functions. Here's some extreme examples of what can happen with an imbalance of electrolytes: elevated potassium levels may result in cardiac arrhythmias; decreased extracellular potassium produces paralysis; excessive extracellular sodium causes fluid retention; and decreased plasma calcium and magnesium can produce muscle spasms of the extremities.
When a patient is dehydrated, a carefully prepared (commercially available) electrolyte solution is required to maintain health and well being. In terms of child health, oral electrolyte is need when a child is dehydrated due to diarrhea. The use of oral electrolyte maintenance solutions, which is responsible for saving millions of lives worldwide over the last 25 years, is one of the most important medical advances in protecting the health of children in the century, explains Juilus G.K. Goepp, MD, assistant director of the Pediatric Emergency Department of the Children's Center at Johns Hopkins Hospital. If a parent provides an oral electrolyte maintenance solution at the very start of the illness, dehydration can be prevented. The functionality of electrolyte solutions is related to their properties, and interest in electrolyte solutions goes far beyond chemistry.
Electrolytes and Batteries
Solutions of electrolytes are always required in batteries, even in dry cells. The simplest battery consists of two electrodes. The figure here illustrates a copper-zinc battery. The left hand is a zinc electrode. The zinc atoms have a tendency to become ions, leaving the electrons behind.
$\mathrm{Zn_{\large{(s)}} \rightarrow Zn^{2+}_{\large{(aq)}} + 2 e^-}$.
As the zinc ions going into the solution, anions move from the copper cell to the zinc cell to compensate for the charge, and at the same time, electrons go from the $\ce{Zn}$ electrode to the $\ce{Cu}$ electrode to neutralize the copper ions.
$\mathrm{Cu^{2+}_{\large{(aq)}} + 2 e^- \rightarrow Cu_{\large{(s)}}}$
In dry cells, the solution is replaced by a paste so that the solution will not leak out of the package. In this cell, the $\ce{Zn}$ and $\ce{Cu}$ electrode has a voltage of 1.10 V, if the concentrations of the ions are as indicated.
Chemical Reactions of Electrolytes
When solutions of electrolytes are combined, the cations and anions will meet each other. When the ions are indifferent of each other, there is no reaction. However, some cations and anions may form a molecule or solid, and thus the cations and anions change partners. These are called metathesis reactons, which include:
• Solid formation (or precipitation) reactions: the cations and anions form a less soluble solid, resulting in the appearance of a precipitate.
• Neutralization reactions: $\ce{H+}$ of an acid and $\ce{OH-}$ of a base combine to give the neutral water molecule.
• Gas formation reactions: When neutral gaseous molecules are formed in a reaction, they leave the solution forming a gas.
Redox reactions are also possible between the various ions. In fact, the battery operations involve redox reactions.
Confidence Building Problems
1. In a battery consisting of $\ce{Zn}$ and $\ce{Cu}$ electrodes, the reaction is
$\mathrm{Zn + Cu^{2+} \rightarrow Zn^{2+} + Cu}$
and the battery can be represented as
$\mathrm{Zn | Zn^{2+} || Cu^{2+}| Cu}$.
where || means a salt bridge, and $\ce{CuSO4}$ is used to provide $\ce{Cu^2+}$. In the salt bridge, what ions will move toward the $\mathrm{Zn | Zn^{2+}}$ cell?
1. any cation
2. $\ce{Cu^2+}$
3. $\ce{Zn^2+}$
4. $\ce{Na+}$
5. any anion
Hint: e. any anion
Skill:
Explain ion movement in a solution of electrolytes.
1. Do the positive ions move in the salt bridge?
Hint: yes
The two types of ions move in opposite directions.
Skill:
Explain ion movement in a solution of electrolytes.
1. What solution should be used for the $\ce{Cu}$ electrode?
1. any zinc salt
2. any copper salt
3. any chloride
4. any salt
5. an acid
6. a base
Hint: b. any copper salt
Any salt can be used for the $\ce{Zn}$-electrode. But for the $\ce{Cu}$ electrode, $\ce{CuSO4}$ or $\ce{CuCl2}$ is commonly used.
Skill:
Apply chemical knowledge to battery setups.
1. Which of the following will you use as the salt bridge?
1. solid $\ce{NaCl}$
2. concentrated $\ce{NaCl}$ solution
3. $\ce{HNO3}$ solution
4. concentrated $\ce{H2SO4}$ solution
5. deionized water
6. any solid salt
Hint: b. $\ce{NaCl}$ solution
A salt solution is usually used, but solutions of acids and bases will be all right. $\ce{NaCl}$ solution is economical and easy to handle.
1. A solution of which one of the following will best conduct electricity? All solutions have the same concentration in M.
1. alcohol
2. ammonia
3. sugar
4. acetic acid
5. table salt
Hint: e. table salt
Skill:
Distinguish strong and weak electrolytes.
1. Which one of the following solutions has the highest pH?
1. 0.10 M $\ce{NaCl}$ solution
2. 0.10 M $\ce{HCl}$ solution
3. Water at 273 K (freezing point of water)
4. Water at 293 K (room temperature)
5. Water at 373 K (boiling point of water)
Hint: c. water at low temperature
See pH.
Skill:
Define and estimate pH.
1. When solutions of electrolytes are combined, the cations and anions exchange partners. These reactions are called:
1. combustion reactions
2. redox reactions
3. oxidation reactions
4. reduction reactions
5. metathesis reactions
See Redox.
Skill:
Explain metathesis reactions. | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Chemical_Reactions/Chemical_Reactions_Examples/Conjugate_Acids_of_Bases_-_Ka_Kb_and_Kw.txt |
Learning Objectives
• Define some common features such as acid-base reaction (neutralization), combination, combustion, decomposition, displacement, precipitation (exchange or metathesis), and redox reactions.
• Describe features of some common reactions.
• Classify chemical reactions according to their features.
Half Reactions
You may think that the two sulfur atoms in the formula are identical, but they are different. You have to understand the chemistry of these ions and then start to investigate their chemical reaction. The structure of $\ce{HS2O3-}$ may be compared to that of $\ce{HSO4-}$:
S O- O O-
\ / \ /
S S
// \ // \
O OH O OH
Thus, one of the two $\ce{S}$ atoms has an oxidation state of -2, and we represent this $\ce{S}$ atom by ($\ce{=S}$) to indicate that it is attached to another $\ce{S}$ atom by a double bond (=).
1. In this reaction, one $\ce{S}$ atom goes from -2 to 0, whereas the oxidation state of the other $\ce{S}$ atom does not change. You have to assume that the $\ce{S}$ atom is oxidized by a reducing agent, $\ce{H2O}$.
2. Only the key elements are given on the left in the half-reactions:
$\ce{HS(=S)O_3^- \rightarrow S +} \textrm{...}\ce{(HSO_4^- )}$
$\ce{H_2O \rightarrow H_{2\large{(g)}}} + \textrm{...}\ce{(HSO4- )}$
3. Add electrons to compensate for the oxidation changes:
$\ce{HS(=S)O3- \rightarrow S + 2 e- + HSO4-}$
$\ce{H2O + 2 e- \rightarrow H2(g) + HSO4-}$
4. Combining the two half-reactions gives the following balanced chemical equation:
$\ce{HS(=S)O3- + H2O \rightarrow S + H_{2\large{(g)}} + HSO4-}$
Oxidation States
Learning Objectives
• Assign an oxidation state (or oxidation number) to an atom in a molecule or ion.
• Describe oxidation and reduction reactions in terms of oxidation state change.
Revealing the Dates of Buffalo Nickels (Demo)
First, this caveat. Experienced coin collectors will tell you that chemical etching of coins almost never increases their value. It practically always decreases their value, so none of the following will be of any use if you collect coins as an investment. It is simply here to demonstrate a point of interest about revealing hidden dates of buffalo nickels using a gentle etching process.
The U.S. "Buffalo" Nickel was minted between 1913 and 1938. The date was placed on a raised area beneath the image of a native American; the date was usually the first thing to disappear as a Buffalo nickel wore down in people's pockets, purses and being handled in transactions. During the 1950s I found the dates on most circulating buffalo nickels already to be worn beyond recognition. Those minted in the 30s often had readable dates, but those minted between 1913 and 1929 were practically always unreadable. Occasionally I was delighted to find one like that at the right where natural corrosion or some coating process gave a good contrast to a still-existing date.
There were those rare finds of a nickel with a reasonably good date on a bright coin, like that one at the left. A buffalo nickel can be etched with vinegar so as to "develop" a latent image of the date of mint, owing presumably to the variable "etchability" of regions which had experienced high or low die pressures during the minting process.
If a buffalo nickel is dropped into about 10 mL vinegar, date side up, and left for anywhere from one to eight weeks (frequent inspections are advised), a readable date develops in about a third of the cases tried. Here are three from my old coin collection. Most often one obtains marginal results, like the 1916 coin on the right.
Occasionally one gets a clearly identifiable date, as in the case of the 1925 coin at the left.
Here's an enlargement of the date from the coin above.
For some percentage of trials, less than 50% but frustratingly close to it, one gets a result which can be described only as wishful thinking, such as the 1913 coin at the right. However, 1913 nickels were the only minting in which there was "raised ground" under the buffalo. Notice that the "FIVE CENTS" partially emerged, but there was no mint mark under it, so I entered it as a 1913 Philadelphia, even though the date can't be seen.
During a recent discussion on the Internet list of chemical educators, CHEMED-L, someone brought up the topic of etching buffalo nickels to produce latent images of the dates of mint. Local coin stores have large collections of unreadable buffalo nickels because some craftspeople make bracelets and necklaces from them. For 30 cents I bought three unreadable buffalo nickels in December, 1999, and dropped each one into 10 mL standard vinegar, 5% acetic acid.
Here are some "before" (on the left) and "after" (on the right) pictures following an etching period of about eight weeks. The first pair offer the standard marginal success story. Is that a "9" or a "5" as the last digit? Is it 1919 or 1929?
Here's an enlargement to emphasize the 5/9 uncertainty on the last digit and to suggest that it is probably 1925 or 1929 rather than 1915 or 1919 because of the suggested space between the 2nd and fourth digits. For my money (!) I'd say that I can actually see the date 1929. For those old enough to remember it, these pictures are a little reminiscent of the movie "Blowup," aren't they?
The second case is a stunning success. After deciding that I really COULD read the date, I handed this nickel to my wife who is near-sighted and asked her to tell me the date. She removed her glasses, took one look at it and without a pause said, "Oh, that's 1915." Voila!
The image above and to the right appears marginal because the resolution of most screens is less than the image itself. The enlargement to the right shows the date much better.
Finally the third case gives us another example of wishful thinking. Can you even begin to see the first two digits of the date, which you already know to be 19?
Here's an enlargement better to illustrate the problem.
Contributor
• Oliver Seely (California State University, Dominguez Hills).
Solution Stoichiometry
Applying this formula to solve titration problems
• Preparing a solution of prescribed concentration
• Solving any problem involving solution stoichiometry | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Chemical_Reactions/Chemical_Reactions_Examples/Features_of_Chemical_Reaction.txt |
Chemical reactions are the processes by which chemicals interact to form new chemicals with different compositions. Simply stated, a chemical reaction is the process where reactants are transformed into products. How chemicals react is dictated by the chemical properties of the element or compound- the ways in which a compound or element undergoes changes in composition.
• Chemical Reactions Examples
• Limiting Reagents
When there is not enough of one reactant in a chemical reaction, the reaction stops abruptly. To figure out the amount of product produced, it must be determined reactant will limit the chemical reaction (the limiting reagent) and which reactant is in excess (the excess reagent). One way of finding the limiting reagent is by calculating the amount of product that can be formed by each reactant; the one that produces less product is the limiting reagent.
• Physical and Chemical Properties of Matter
We are all surrounded by matter on a daily basis. Anything that we use, touch, eat, etc. is an example of matter. Matter can be defined or described as anything that takes up space, and it is composed of miniscule particles called atoms. It must display the two properties of mass and volume.
• Reactions in Solution
Scientists generally react chemicals in liquid or solution form because reacting chemicals as solids is usually much slower.
• Stoichiometry
The quantitative relationship among reactants and products is called stoichiometry. The term stoichiometry is derived from two Greek words: stoicheion (meaning "element") and metron (meaning "measure"). On this subject, you often are required to calculate quantities of reactants or products. Stoichiometry calculations are based on the fact that atoms are conserved. They cannot be destroyed or created. Numbers and kinds of atoms before and after the reactions are always the same.
• Stoichiometry and Balancing Reactions
Stoichiometry is a section of chemistry that involves using relationships between reactants and/or products in a chemical reaction to determine desired quantitative data. In Greek, stoikhein means element and metron means measure, so stoichiometry literally translated means the measure of elements.
Chemical Reactions
When there is not enough of one reactant in a chemical reaction, the reaction stops abruptly. To figure out the amount of product produced, it must be determined which reactant will limit the chemical reaction (the limiting reagent) and which reactant is in excess (the excess reagent). One way of finding the limiting reagent is by calculating the amount of product that can be formed by each reactant; the one that produces less product is the limiting reagent.
Introduction
The following scenario illustrates the significance of limiting reagents. In order to assemble a car, 4 tires and 2 headlights are needed (among other things). In this example, imagine that the tires and headlights are reactants while the car is the product formed from the reaction of 4 tires and 2 headlights.
$4 \, \text{Tires} + 2 \, \text{Headlights} = 1 \, \text{Car} \nonumber$
If you have 20 tires and 14 headlights, how many cars can be made? With 20 tires, 5 cars can be produced because there are 4 tires to a car. With 14 headlights, 7 cars can be built (each car needs 2 headlights). Although more cars can be made from the headlights available, only 5 full cars are possible because of the limited number of tires available. In this case, the headlights are in excess. Because the number of cars formed by 20 tires is less than number of cars produced by 14 headlights, the tires are the limiting reagent (they limit the full completion of the reaction, in which all of the reactants are used up). This scenario is illustrated below:
The initial condition is that there must be 4 tires to 2 headlights. The reactants must thus occur in that ratio; otherwise, one will limit the reaction. There are 20 tires and 14 headlights, so there are two ways of looking at this problem. For 20 tires, 10 headlights are required, whereas for 14 headlights, 28 tires are required. Because there are not enough tires (20 tires is less than the 28 required), tires are the limiting "reactant."
The limiting reagent is the reactant that is completely used up in a reaction, and thus determines when the reaction stops. From the reaction stoichiometry, the exact amount of reactant needed to react with another element can be calculated. If the reactants are not mixed in the correct stoichiometric proportions (as indicated by the balanced chemical equation), then one of the reactants will be entirely consumed while another will be left over. The limiting reagent is the one that is totally consumed; it limits the reaction from continuing because there is none left to react with the in-excess reactant.
There are two ways to determine the limiting reagent. One method is to find and compare the mole ratio of the reactants used in the reaction (approach 1). Another way is to calculate the grams of products produced from the given quantities of reactants; the reactant that produces the smallest amount of product is the limiting reagent (approach 2).
How to Find the Limiting Reagent: Approach 1
Find the limiting reagent by looking at the number of moles of each reactant.
1. Determine the balanced chemical equation for the chemical reaction.
2. Convert all given information into moles (most likely, through the use of molar mass as a conversion factor).
3. Calculate the mole ratio from the given information. Compare the calculated ratio to the actual ratio.
4. Use the amount of limiting reactant to calculate the amount of product produced.
5. If necessary, calculate how much is left in excess of the non-limiting reagent.
How to Find the Limiting Reagent: Approach 2
Find the limiting reagent by calculating and comparing the amount of product each reactant will produce.
1. Balance the chemical equation for the chemical reaction.
2. Convert the given information into moles.
3. Use stoichiometry for each individual reactant to find the mass of product produced.
4. The reactant that produces a lesser amount of product is the limiting reagent.
5. The reactant that produces a larger amount of product is the excess reagent.
6. To find the amount of remaining excess reactant, subtract the mass of excess reagent consumed from the total mass of excess reagent given.
Example $1$: Photosynthesis
Consider respiration, one of the most common chemical reactions on earth.
$\ce{ C6H_{12}O6 + 6 O_2 \rightarrow 6 CO2 + 6 H2O} + \rm{energy} \nonumber$
What mass of carbon dioxide forms in the reaction of 25 grams of glucose with 40 grams of oxygen?
Solution
When approaching this problem, observe that every 1 mole of glucose ($C_6H_{12}O_6$) requires 6 moles of oxygen to obtain 6 moles of carbon dioxide and 6 moles of water.
Step 1: Determine the balanced chemical equation for the chemical reaction.
The balanced chemical equation is already given.
Step 2: Convert all given information into moles (most likely, through the use of molar mass as a conversion factor).
$\mathrm{25\:g \times \dfrac{1\: mol}{180.06\:g} = 0.1388\: mol\: C_6H_{12}O_6} \nonumber$
$\mathrm{40\:g \times \dfrac{1\: mol}{32\:g} = 1.25\: mol\: O_2} \nonumber$
Step 3: Calculate the mole ratio from the given information. Compare the calculated ratio to the actual ratio.
a. If all of the 1.25 moles of oxygen were to be used up, there would need to be $\mathrm{1.25 \times \dfrac{1}{6}}$ or 0.208 moles of glucose. There is only 0.1388 moles of glucose available which makes it the limiting reactant.
$1.25 \; \rm{mol} \; O_2 \times \dfrac{ 1 \; \rm{mol} \; C_6H_{12}O_6}{6\; \rm{mol} \; O_2}= 0.208 \; \rm{mol} \; C_6H_{12}O_6 \nonumber$
b. If all of the 0.1388 moles of glucose were used up, there would need to be 0.1388 x 6 or 0.8328 moles of oxygen. Because there is an excess of oxygen, the glucose amount is used to calculate the amount of the products in the reaction.
$0.1388\; \rm{ mol}\; C_6H_{12}O_6 \times \dfrac{6 \; \rm{mol} \;O_2}{1 \; \rm{mol} \; C_6H_{12}O_6} = 0.8328 \; \rm{mol}\; O_2 \nonumber$
If more than 6 moles of O2 are available per mole of C6H12O6, the oxygen is in excess and glucose is the limiting reactant. If less than 6 moles of oxygen are available per mole of glucose, oxygen is the limiting reactant. The ratio is 6 mole oxygen per 1 mole glucose, OR 1 mole oxygen per 1/6 mole glucose. This means: 6 mol O2 / 1 mol C6H12O6 .
Therefore, the mole ratio is: (0.8328 mol O2)/(0.208 mol C6H12O6)
This gives a 4.004 ratio of $\ce{O2}$ to $\ce{C6H12O6}$.
Step 4: Use the amount of limiting reactant to calculate the amount of CO2 or H2O produced.
For carbon dioxide produced: $\mathrm{0.1388\: moles\: glucose \times \dfrac{6}{1} = 0.8328\: moles\: carbon\: dioxide}$.
Step 5: If necessary, calculate how much is left in excess.
1.25 mol - 0.8328 mol = 0.4172 moles of oxygen left over
Example $2$: Oxidation of Magnesium
Calculate the mass of magnesium oxide possible if 2.40 g $\ce{Mg}$ reacts with 10.0 g of $\ce{O_2}$
$\ce{ Mg +O_2 \rightarrow MgO} \nonumber$
Solution
Step 1: Balance equation
$\ce{2 Mg + O_2 \rightarrow 2 MgO} \nonumber$
Step 2 and Step 3: Converting mass to moles and stoichiometry
$\mathrm{2.40\:g\: Mg \times \dfrac{1.00\: mol\: Mg}{24.31\:g\: Mg} \times \dfrac{2.00\: mol\: MgO}{2.00\: mol\: Mg} \times \dfrac{40.31\:g\: MgO}{1.00\: mol\: MgO} = 3.98\:g\: MgO} \nonumber$
$\mathrm{10.0\:g\: O_2\times \dfrac{1\: mol\: O_2}{32.0\:g\: O_2} \times \dfrac{2\: mol\: MgO}{1\: mol\: O_2} \times \dfrac{40.31\:g\: MgO}{1\: mol\: MgO} = 25.2\: g\: MgO} \nonumber$
Step 4: The reactant that produces a smaller amount of product is the limiting reagent
$\ce{Mg}$ produces less $\ce{MgO}$ than does $\ce{O2}$ (3.98 g MgO vs. 25.2 g MgO), therefore Mg is the limiting reagent in this reaction.
Step 5: The reactant that produces a larger amount of product is the excess reagent
O2 produces more amount of MgO than Mg (25.2g MgO vs. 3.98 MgO), therefore O2 is the excess reagent in this reaction.
Step 6: Find the amount of remaining excess reactant by subtracting the mass of the excess reagent consumed from the total mass of excess reagent given.
Mass of excess reagent calculated using the limiting reagent:
$\mathrm{2.40\:g\: Mg \times \dfrac{1.00\: mol\: Mg}{24.31\:g\: Mg} \times \dfrac{1.00\: mol\: O_2}{2.00\: mol\: Mg} \times \dfrac{32.0\:g\: O_2}{1.00\: mol\: O_2} = 1.58\:g\: O_2} \nonumber$
OR Mass of excess reagent calculated using the mass of the product:
$\mathrm{3.98\:g\: MgO \times \dfrac{1.00\: mol\: MgO}{40.31\:g\: MgO} \times \dfrac{1.00\: mol\: O_2}{2.00\: mol\: MgO} \times \dfrac{32.0\:g\: O_2}{1.00\: mol\: O_2} = 1.58\:g\: O_2} \nonumber$
Mass of total excess reagent given – mass of excess reagent consumed in the reaction
10.0g – 1.58g = 8.42g O2 is in excess.
Example $3$: Limiting Reagent
What is the limiting reagent if 76.4 grams of $C_2H_3Br_3$ were reacted with 49.1 grams of $O_2$?
$\ce{4 C_2H_3Br_3 + 11 O_2 \rightarrow 8 CO_2 + 6 H_2O + 6 Br_2} \nonumber$
Solution
Using Approach 1:
A.
$\mathrm{76.4\:g \times \dfrac{1\: mole}{266.72\:g} = 0.286\: moles\: of\: C_2H_3Br_3} \nonumber$
$\mathrm{49.1\: g \times \dfrac{1\: mole}{32\:g} = 1.53\: moles\: of\: O_2} \nonumber$
B. Assuming that all of the oxygen is used up, $\mathrm{1.53 \times \dfrac{4}{11}}$ or 0.556 moles of C2H3Br3 are required. Because there are only 0.286 moles of C2H3Br3 available, C2H3Br3 is the limiting reagent.
Using Approach 2:
$\mathrm{76.4\:g\: C_2H_3Br_3 \times \dfrac{1\: mol\: C_2H_3Br_3}{266.72\:g\: C_2H_3Br_3} \times \dfrac{8\: mol\: CO_2}{4\: mol\: C_2H_3Br_3} \times \dfrac{44.01\:g\: CO_2}{1\: mol\: CO_2} = 25.2\:g\: CO_2} \nonumber$
$\mathrm{49.1\:g\: O_2 \times \dfrac{1\: mol\: O_2}{32\:g\: O_2} \times \dfrac{8\: mol\: CO_2}{11\: mol\: O_2} \times \dfrac{44.01\:g\: CO_2}{1\: mol\: CO_2} = 49.1\:g\: CO_2} \nonumber$
Therefore, by either method, C2H3Br3is the limiting reagent.
Example $4$: Limiting Reagent
What is the limiting reagent if 78 grams of $\ce{Na2O2}$ were reacted with 29.4 grams of $\ce{H2O}$?
Solution
Using Approach 1
A.
$\mathrm{78\:g \times \dfrac{1\: mol}{77.96\:g} = 1.001\: moles\: of\: Na_2O_2} \nonumber$
$\mathrm{29.4\:g \times \dfrac{1\: mol}{18\:g}= 1.633\: moles\: of\: H_2O} \nonumber$
B. Assume that all of the water is consumed, $\mathrm{1.633 \times \dfrac{2}{2}}$ or 1.633 moles of Na2O2 are required. Because there are only 1.001 moles of Na2O2, it is the limiting reactant.
Using Approach 2:
\$\mathrm{78\:g\: Na_2O_2 \times \dfrac{1\: mol\: Na_2O_2}{77.96\:g\: Na_2O_2} \times \dfrac{4\: mol\: NaOH}{2\: mol\: Na_2O_2} \times \dfrac{40\:g\: NaOH}{1\: mol\: NaOH} = 80.04\:g\: NaOH} \nonumber$
Using either approach gives Na2O2 as the limiting reagent.
Example $5$: Excess Reagent
How much the excess reagent remains if 24.5 grams of CoO is reacted with 2.58 grams of O2?
$\ce{4 CoO + O_2 \rightarrow 2 Co_2O_3} \nonumber$
Solution
A.
$\mathrm{24.5\:g \times \dfrac{1\: mole}{74.9\:g}= 0.327\: moles\: of\: CoO} \nonumber$
$\mathrm{2.58\:g \times \dfrac{1\: mole}{32\:g}= 0.0806\: moles\: of\: O_2} \nonumber$
B. Assuming that all of the oxygen is used up, $\mathrm{0.0806 \times \dfrac{4}{1}}$ or 0.3225 moles of $CoO$ are required. Because there are 0.327 moles of CoO, CoO is in excess and thus O2 is the limiting reactant.
C. 0.327 mol - 0.3224 mol = 0.0046 moles left in excess.
Example $6$: Identifying the Limiting Reagent
Will 28.7 grams of $SiO_2$ react completely with 22.6 grams of $H_2F_2$? If not, identify the limiting reagent.
$\ce{SiO_2+ 2 H_2F_2 \rightarrow SiF_4+ 2 H_2O} \nonumber$
Solution
A.
$\mathrm{28.7\:g \times \dfrac{1\: mole}{60.08\:g} = 0.478\: moles\: of\: SiO_2} \nonumber$
$\mathrm{22.6\:g \times \dfrac{1\: mole}{39.8\:g} = 0.568\: moles\: of\: H_2F_2} \nonumber$
B. There must be 1 mole of SiO2 for every 2 moles of H2F2 consumed. Because the ratio is 0.478 to 0.568, 28.7 grams of SiO2 do not react with the H2F2.
C. Assuming that all of the silicon dioxide is used up, $\mathrm{0.478 \times \dfrac{2}{1}}$ or 0.956 moles of H2F2 are required. Because there are only 0.568 moles of H2F2, it is the limiting reagent.
Contributors and Attributions
• Sarick Shah (UCD) | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Chemical_Reactions/Limiting_Reagents.txt |
We are all surrounded by matter on a daily basis. Anything that we use, touch, eat, etc. is an example of matter. Matter can be defined or described as anything that takes up space, and it is composed of miniscule particles called atoms. It must display the two properties of mass and volume.
Introduction
The different types of matter can be distinguished through two components: composition and properties. The composition of matter refers to the different components of matter along with their relative proportions. The properties of matter refer to the qualities/attributes that distinguish one sample of matter from another. These properties are generally grouped into two categories: physical or chemical.
Physical Properties and Changes
Physical properties can be observed or measured without changing the composition of matter. Physical properties are used to observe and describe matter. Physical properties of materials and systems are often described as intensive and extensive properties. This classification relates to the dependency of the properties upon the size or extent of the system or object in question.
An intensive property is a bulk property, meaning that it is a physical property of a system that does not depend on the system size or the amount of material in the system. Examples of intensive properties include temperature, refractive index, density, and hardness of an object. When a diamond is cut, the pieces maintain their intrinsic hardness (until their size reaches a few atoms thick). In contrast, an extensive property is additive for independent, non-interacting subsystems. The property is proportional to the amount of material in the system.
Intensive properties: A physical property that will be the same regardless of the amount of matter.
• density: $\rho=\frac{m}{v}$
• color: The pigment or shade
• conductivity: electricity to flow through the substance
• malleability: if a substance can be flattened
• luster: how shiny the substance looks
Extensive Properties: A physical property that will change if the amount of matter changes.
• mass: how much matter in the sample
• volume: How much space the sample takes up
• length: How long the sample is
Physical Change
Change in which the matter's physical appearance is altered, but composition remains unchanged.
A physical change takes place without any changes in molecular composition. The same element or compound is present before and after the change. The same molecule is present through out the changes. Physical changes are related to physical properties since some measurements require that changes be made. The three main states of matter are: Solid, Liquid, Gas
• Solid is distinguished by a fixed structure. Its shape and volume do not change. In a solid, atoms are tightly packed together in a fixed arrangement.
• Liquid is distinguished by its malleable shape (is able to form into the shape of its container), but constant volume. In a liquid, atoms are close together but not in a fixed arrangement.
• Gas is made up of atoms that are separate. However, unlike solid & liquid, a gas has no fixed shape and volume.
Example $1$: Physical Change
When liquid water ($H_2O$) freezes into a solid state (ice), it appears changed; However, this change is only physical as the the composition of the constituent molecules is the same: 11.19% hydrogen and 88.81% oxygen by mass.
Figure $2$: Physical Change: Ice Melting is a physical change. from Wikipedia.
Chemical Properties and Changes
Chemical properties of matter describes its "potential" to undergo some chemical change or reaction by virtue of its composition. What elements, electrons, and bonding are present to give the potential for chemical change. It is quite difficult to define a chemical property without using the word "change". Eventually you should be able to look at the formula of a compound and state some chemical property. At this time this is very difficult to do and you are not expected to be able to do it. For example hydrogen has the potential to ignite and explode given the right conditions. This is a chemical property. Metals in general have they chemical property of reacting with an acid. Zinc reacts with hydrochloric acid to produce hydrogen gas. This is a chemical property.
Chemical change results in one or more substances of entirely different composition from the original substances. The elements and/or compounds at the start of the reaction are rearranged into new product compounds or elements. A CHEMICAL CHANGE alters the composition of the original matter. Different elements or compounds are present at the end of the chemical change. The atoms in compounds are rearranged to make new and different compounds.
Example $1$: Corrosion of Metals
Corrosion is the unwanted oxidation of metals resulting in metal oxides.
$2 Mg + O_2 \rightarrow 2 MgO \nonumber$
Problems
The following questions are multiple choice.
1. Milk turns sour. This is a ________________
• Chemical Change
• Physical Change
• Chemical Property
• Physical Property
• None of the above
2. HCl being a strong acid is a __________, Wood sawed in two is ___________
• Chemical Change, Physical Change
• Physical Change, Chemical Change
• Chemical Property, Physical Change
• Physical Property, Chemical Change
• None of the above
3. CuSO4 is dissolved in water
• Chemical Change
• Physical Change
• Chemical Property
• Physical Property
• None of the above
4. Aluminum Phosphate has a density of 2.566 g/cm3
• Chemical Change
• Physical Change
• Chemical Property
• Physical Property
• None of the above
5. Which of the following are examples of matter?
• A Dog
• Carbon Dioxide
• Ice Cubes
• copper (II) nitrate
• A Moving Car
6. The formation of gas bubbles is a sign of what type of change?
7. True or False: Bread rising is a physical property. 8. True or False: Dicing potatoes is a physical change. 9. Is sunlight matter? 10. The mass of lead is a _____________property.
Solutions
1. chemical change
2. chemical property, physical change
3. physical change
4. physical property
5. All of the above
6. chemical
7. False
8. True
9. No
10. physical property
Contributors and Attributions
• Samantha Ma (UC Davis) | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Chemical_Reactions/Properties_of_Matter.txt |
A solution consists of two or more substances dissolved in a liquid form. Not to get confused with a mixture, which is heterogeneous--multiple substances exist in varying structures-- solutions are homogenous, which means that atoms of the solute are evenly dispersed throughout the solvent (ex. water, ethanol). Think of it as comparing a cup of (dissolved) sugar water and a cup of water with lego blocks in it. The solute is the substance dissolved in the solution, and the solvent is the substance doing the dissolving.
ex.
A solution of NaCl in water A mixture of lego blocks and water
Note: All solutions are mixtures but not all mixtures are solutions.
Solvents
Water (H2O) is the most common solvent, used for dissolving many compounds or brewing coffee. Other common solvents include turpentine (a paint thinner), acetone (nail polish remover), and ethanol (used in some perfumes). Such solvents usually contain carbon and are called organic solvents. Solutions with water as the solvent are called aqueous solutions; they have special properties that are covered here.
Solutes
Different chemical compounds dissolve in solutes in varying degrees. Some compounds, such as the strong acid hydrochloric acid (HCl), dissociate completely in solution into ions. Others, like the weak base ammonia (NH3), only partly dissociate. Yet other compounds like alcohol do not dissociate at all and remain compounds. Laboratory reactions often involve acids and bases, which are covered in more detail here.
Concentration
Concentration is the measure of the amount of solute in a certain amount of solvent. Knowing the concentration of a solution is important determining the strength of an acid or base (pH), among other things. When there is so much solute present in a concentration that it no longer dissolves, the solution is saturated.
Scientists often use molarity to measure concentration.
molarity = moles/Liter
Since reaction stoichiometry relies on molar ratios, molarity is the main measurement for concentration.
A less common unit for concentration is called molality.
Molality = moles/Kg of solvent
Scientists sometimes use molality to measure concentration because liquid volumes change slightly based on the temperature and pressure. Mass, however, stays the same and can be measured accurately using a balance. Commercial concentrated products are usually expressed in mass percent; such as commercial concentrated sulfuric acid, which is 93-98% H2SO4 by mass in water (Hill, Petrucci 116).
Making a Solution
Solutions used in the laboratory are usually made from either solid solutes (often salts) or stock solutions.
To make a solution from solid solutes, first calculate how many moles of solute are in the desired solutions (using the molarity). Calculate the amount of solid you need in grams using the moles needed and the molar mass of the solute and weight out the needed amount. Transfer the solute to a container (preferably a volumetric flask, which most accurately measures volume of solution labeled on the flask) and add a small amount of solvent. Mix thoroughly to dissolve the solute. Once the solute has dissolved, add the remaining solvent to make the solution of the desired volume and mix thoroughly.
For example, to make 0.5 Liters of 0.5 molar NaCl:
1. Multiply the concentration (0.5 mols/Liters) by the volume of solution you want (0.5 Liters) to find the moles of NaCl you need.
0.5 moles/Liter * 0.5 Liters = 0.25 moles of NaCl
2. Multiply the moles of NaCl by its molar mass (58.44 g/mol) to find the grams of solute needed.
(0.25 moles NaCl)*(58.44 grams/mole) = 14.61 grams of NaCl
Making a solution of a certain concentration from a stock solution is called a dilution. When diluting a solution, keep in mind that adding a solvent to a solution changes the concentration of the solution, but not the amount of solute already present.
To dilute a solution with known concentration, first determine the number of moles of solute are in the solution by multiplying the molarity by the volume (in Liters). Then, divide by the desired molarity or volume to find the volume or concentration needed.
The equation to use is simply
M1V1 = M2V2
M1 and V1 are the concentration and volume of the original (stock) solution to dilute; M2 and V2 are the desired concentration and volume of the final solution.
Solution Stoichiometry
For reactions that take place in solutions:
1. Calculate the moles of solute reacting by multiplying the concentration (molarity) by the volume of solution (Liters)
2. Determine the Limiting Reactant, if there is one
3. Follow the stoichiometric process.
4. Convert the resulting moles of solute back to molarity by dividing by the total volume, in liters, of solution used in the reaction.
5. In the case of reactions involving ions (such as in reactions between strong acids and bases), eliminate spectator ions from the net ionic equation. Spectator ions do not react in the equations.
6. If the concentration is not given but the molar mass and volume are, use density (grams/Liter) to find the amount of solute in grams, then convert it to moles.
Problems
1. A solution is prepared by dissolving 44.6 grams of acetone (OC(CH3)2) in water to produce 1.50 Liters of solution. What is the molarity of the resulting solution?
2. A certain laboratory procedure requires 0.025 M H2SO4. How many milliliters of 1.10 M H2SO4 should be diluted in water to prepare 0.500 L of 0.025 M H2SO4?
3. A sample of saturated NaNO3 (aq) is 10.9 M at 25 degrees Celcius. How many grams of NaNO3 are in 230 mL of this solution at the same temperature?
4. A beaker of 175 mL of 0.950 M NaCl is left uncovered for a period of time. If, by the end of the time period, the volume of solution in the beaker has decreased to 137 mL (the volume loss is due to water evaporation), what is the resulting concentration of the solution?
5. A student prepares a solution by dissolving 15.0 mL ethanol (C2H5OH) in water to make a 300.0 mL solution. Calculate the concentration (molarity) of ethanol in the solution. (density = 0.789 g/mL)
1.
2.
3.
4.
5.
Contributors and Attributions
• Stephanie Shieh (UCD) | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Chemical_Reactions/Reactions_in_Solution.txt |
The quantitative relationship among reactants and products is called stoichiometry. The term stoichiometry is derived from two Greek words: stoicheion (meaning "element") and metron (meaning "measure"). On this subject, you often are required to calculate quantities of reactants or products. Stoichiometry calculations are based on the fact that atoms are conserved. They cannot be destroyed or created. Numbers and kinds of atoms before and after the reactions are always the same. This is the basic law of nature.
Stoichiometry
Learning Objectives
• Express amounts of substances in g, kg, or mL, or L.
• Express amounts of substances in moles (mol).
• Convert amounts of substances from mass units (g, kg etc) to the unit mole.
Amounts of Substances in Various Units
Amounts of substances are measured in units of mass (g or kg), volume (L) and mole (mol). Unit interconversions are based on the definitions of the units, and converting amounts from g or kg into mol is based on atomic masses of the elements.
Atomic masses are the masses of one mole of elements. A mole of any element has an Avogadro's number of atoms (= 6.02x1023 atoms per mole).
The natural units of substances are molecules, which are groups of atoms bonded together, except monatomic molecules of inert gases, $\ce{He}$, $\ce{Ne}$, $\ce{Ar}$, $\ce{Kr}$, $\ce{Xe}$, and $\ce{Rn}$. For example, molecules of oxygen, water, and phosphorous are $\ce{O2}$, $\ce{H2O}$, and $\ce{P4}$ respectively. These molecules have 2, 3, and 4 atoms respectively. Masses of one mole of substances are called molecular weights. Atomic and molecular weights are called molar masses.
The above illustrates only a very small number of examples. There are millions of compounds in the world. Please think of some other compounds you know of, and write down their formulas. Then figure out the number of atoms in each mole of your compounds.
Different substances have different molecular masses. Thus, equal masses have different numbers of atoms, molecules, or moles. On the other hand, equal numbers of moles of different substances have different masses. The stoichiometric relationships among reactants and products may be complicated in units of g, but much simpler relationships are seen if we deal with units of moles or natural units of atoms and molecules.
Moles (mol) represent amounts of substances in the unit of Avogadro's number (6.022x1023) of atoms and molecules. Since empirical formulas such as $\ce{Fe^2+}$ ions and $\ce{Fe2O3}$ are used for ionic compounds, a mole represent Avogadro's number of ions or per formula as written. A mole of $\ce{Fe^2+}$ has 6.022x1023 ions, and a mole of $\ce{Fe2O3}$ has 1.204x1024 $\ce{Fe}$ and 1.8066x1024 $\ce{O}$ atoms, a total of 3.0x1024 $\ce{Fe}$ and $\ce{O}$ atoms.
The mole unit is very important for chemical reactions, as is the skill to convert masses in g to mol. The number of moles of a substance in a sample is the mass in g divided by the molar mass, which gives the amount in moles.
$\mathrm{mole = \dfrac{mass\:(g)}{molar\: mass\: (g / mol) }}$
Another common measure of substances is volume. Since density is the mass divided by its volume, conversion between volume and mass is accomplished by the formula:
$\mathrm{density = \dfrac{mass\:(g)}{volume\: (cm^3)}}$
$\mathrm{mass = density\: (g\: cm^{-3}) \times volume\: (cm^3)}$
These fundamental formulas are results of the definition of these terms.
Key Concepts
• Atoms, atomic weights, masses
• Molecules, molecular weights
• Empirical formula, ionic compounds
• Avogadro's number, mole, molar masses
Skill Developing Questions
1. What are the molar masses of the elements hydrogen ($\ce{H}$), oxygen ($\ce{O}$), iron ($\ce{Fe}$) and gold ($\ce{Au}$)?
Know where to find: 1.0, 16.0, 55.9, 197
Skill: know where to find molar masses of elements.
2. The element gold is a precious metal. How many moles of gold are present in a mass of 1.0 kg?
1000/197 = ?
Skill: to convert mass in g into moles.
1. A liter of water has a mass of 1.0 kg. How many moles of water are 1.0 kg?
1000 g / 18 g = 55.6 mol
Skill: calculate molar masses of molecular compounds
1. At standard temperature and pressure, a mole of gas occupies 22.4 L. If 20% of air is oxygen, how many moles of oxygen are contained in 1.0 L?
0.20 mol / 22.4 L = 8.93e-3 mol/L
Skill: express amount in volume.
1. How many grams are there in 8.93e-3 mol of oxygen $\ce{O2}$ (molar mass 32.0)?
32 g/mol * 8.93e-3 mol = 0.286 g
Skill: convert amounts in moles to mass in g or kg.
1. How many moles of $\ce{Fe2O3}$ are present in 1000 kg of the oxide? Atomic weights: $\ce{Fe}$, 55.8, $\ce{O}$, 16.0.
1000000 g/(159.6 g/mol) = 6265 mol
Skill: convert amounts between moles and kg of compounds represented by chemical formulas. | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Chemical_Reactions/Stoichiometry/Amounts_of_Substances.txt |
Learning Objectives
• Explain the meaning of various chemical formulas (empirical, molecular, and structural formulas).
• Calculate weight and mole percentages from a formula.
• Derive chemical formulas from a given set of information.
Key Concepts
• Chemical formulas: empirical formula, molecular formula, and structural formula
• Formula weight, molecular weight and molar mass
• Weight percentage and mole percentage
• Elemental analysis
Chemical formulas such as $\ce{HClO4}$ can be divided into empirical formula, molecular formula, and structural formula. Chemical symbols of elements in the chemical formula represent the elements present, and subscript numbers represent mole proportions of the proceeding elements. Note that no subscript number means a subscript of 1.
From a chemical point of view, an element contained in the substance is a fundamental question, and we represent the elemental composition by a chemical formula, such as $\ce{H2O}$ for water. This formula implies that the water molecules consist of 2 hydrogen and 1 oxygen atoms. The formula $\ce{H2O}$ is also the molecular formula of water. For non-molecular substances such as table salt, we represent the composition with an empirical formula. Sodium chloride is represented by $\ce{NaCl}$, meaning that sodium and chlorine ratio in sodium chloride is 1 to 1. Again, the subscript 1 is omitted. Since table salt is an ionic compound, the formula implies that numbers of $\ce{Na+}$ ions and $\ce{Cl-}$ ions are the same in the solid. The subscript numbers in an empirical formula should have no common divisor.
H H
| |
H-C-C-O-H
| |
H H
Structural of
$\ce{CH3CH2OH}$
A 3-Dimensional
structure of $\ce{C6H12}$
A structural formula reflects the bonding of atoms in a molecule or ion. For example, ethanol can be represented by $\ce{CH3CH2OH}$. This is a simple way of representing a more elaborated structure shown on your left. Molecular structures are often beautiful, but the representation is an artwork. For example, a 3-dimensional structure of cyclohexane is shown on the right. This is a chair form, and another structure has a boat form. You will learn more about it in organic chemistry. The molecular formula of benzene is $\ce{C6H6}$, and its empirical formula is $\ce{CH}$.
You may refer to a substance by its name, and recognize it by its properties. Properties are related to the structure and the composition of the molecules. Knowing the chemical formula is a giant step towards understanding a substance.
Formula Weights, Molecular Weights and Molar Masses
The formula weight is the sum of all the atomic weights in a formula. The evaluation of formula weight is illustrated in this example.
Example 1
What is the formula weight of sufuric acid $\ce{H2SO4}$?
Solution
The formula also indicates a mass as the sum of masses calculated this way
$\mathrm{2\times1.008 + 32.0 + 4\times16.0 = 98.0}$
where 1.008, 32.0 and 16.0 are the atomic weights of $\ce{H}$, $\ce{S}$, and $\ce{O}$ respectively.
Discussion
If the formula is a molecular formula, the mass associated with it is called molecular mass or molecular weight. As an exercise, work out the following problem.
What is the molecular weight of caffeine, $\ce{C8H10N4O2}$?
The diagram shown here is a model of the caffeine molecule.
With the aid of a table of atomic weights, a formula indirectly represents the formula weight. If the formula is a molecular formula, it indirectly represents the molecular weight. For simplicity, we may call these weights molar masses, which can be formula weights or molecular weights.
A chemical formula not only represents what a substance is made of, it provides a great deal of information about the substance. Do you know that chemical formulas are used all over the world, regardless of the language? Chinese, Russian, Japanese, African, and South Americans use the same notations we do. Thus, $\ce{H2S}$ is recognized as a smelly gas all over the world. Chemical formula is an international or universal language.
Weight Percentage and Mole Percentage
A chemical formula not only gives the formula weight, it accurately represents the percentages of elements in a compound. On the other hand, if you know the percentage of a compound, you may figure out its formula. Percentage based on weights is called weight percentage, and percentage based on the numbers of atoms or moles is called mole percentage.
Example 2
What are the weight and mole percentages of $\ce{S}$ in sufuric acid?
Solution
From example 1, we know that there are 32.0 g of $\ce{S}$ in 98.0 g of sulfuric acid. Thus the weight percentage is
$\mathrm{Weight\: percentage = \dfrac{32}{98} = 32.7\%}$
From the formula, there is one $\ce{S}$ atom among 7 atoms in $\ce{H2SO4}$
$\mathrm{Mole\: percentage = \dfrac{1}{7} = 14.3\%}$
Discussion
You have learned what weight and mole percentages are and how to evaluate them in this example. As an exercise, work out the following problem:
What are the weight and mole percentages of $\ce{C}$, $\ce{H}$, $\ce{N}$, and $\ce{O}$ for caffeine, $\ce{C8H10N4O2}$?
Determination of Chemical Formulas
How would you find the chemical formula of a substance? If you know the substance, its formula and other information is usually listed in a handbook. Handbooks such as the CRC Handbook of Chemistry and Physics contain information on millions of substances.
If you are a researcher and you made a new compound that no one has ever made before, then you need to determine its empirical or molecular formula. For an organic compound, you burn it completely to convert all carbon ($\ce{C}$) to $\ce{CO2}$, and all hydrogen ($\ce{H}$) to $\ce{H2O}$.
$\mathrm{C_xH_{2y} \xrightarrow{(burned\: in\: O_2)} x\, CO_2 + y\, H_2O}$
Thus, from the weight of $\ce{CO2}$ and $\ce{H2O}$ produced by burning a definite amount of the substance, you can figure out the percent of $\ce{C}$ and $\ce{H}$ in the compound.
Nitrogen is determined by converting it to $\ce{NH3}$. The amount of $\ce{NH3}$ can be determined by titration, and the percentage can also be determined.
Percentage of $\ce{O}$ is usually obtained by subtracting all percentages of $\ce{C}$, $\ce{H}$, and $\ce{N}$, if the compound does not contain any other element.
Example 3
A compound contains 92.3 weight percent of carbon and 7.7 weight percent of $\ce{H}$. What is the empirical formula?
Solution
Assume that you have 100 g of the compound, then you have 92.3 g of carbon and 7.7 g of hydrogen. Thus the mole ratio of $\ce{C}$ to $\ce{H}$ should be
$\mathrm{\dfrac{92.3}{12}:\dfrac{7.7}{1.008} = 7.7 : 7.7 = 1 : 1}$
Thus, the empirical formula is $\ce{CH}$.
You have learned how to determine a chemical formula if the percentages of various elements present in the compound are known in this example. To test your skill, you may be asked to work out the empirical formula of any compound. Try this problem:
Exercise $1$
Aspartic acid contains 36.09% $\ce{C}$, 5.30% $\ce{H}$, 10.52% $\ce{N}$, and 48.08 $\ce{O}$ by weight. What is the empirical formula for aspartic acid?
Aspartic acid is one of the non-essential amino acids, usually present in young plants. It is obtained by hydrolysis of asparagine, which is abundant in asparagus
Example 4
A compound with an empirical formula of $\ce{CH}$ has a molecular weight of 78 g/mol. What is the molecular formula?
Solution
The formula weight of $\ce{CH}$ is 13.0.
Since 78/13 = 6, the molecular formula is $\ce{C6H6}$, the formula for benzene.
Discussion
This example illustrates the difference between empirical and molecular formula, for which the molecular weight must be known.
Example 5
When 1.00 g of benzene is burned, how much $\ce{CO2}$ and $\ce{H2O}$ should be produced?
Hint
$\mathrm{1\:g\:CH \times \dfrac{1\:mol\:C}{13\:g\:CH}\times\dfrac{1\:mol\:CO_2}{1\:mol\:C}\times\dfrac{44.0\:g\:CO_2}{1\:mol\:CO_2} = 3.38\:g\:CO_2}$
Use the same method to calculate the amount of $\ce{H2O}$ produced (Ans. 0.692 g).
Example 6
When 1.00 g of a compound containing only carbon and hydrogen is burned completely, 3.14 g of $\ce{CO2}$ and 1.29 g of $\ce{H2O}$ is produced. What is the empirical formula?
Hint
Amounts of carbon and $\ce{H}$ in 3.14 g of carbon dioxide,
$\mathrm{3.14\: g\: CO_2\times \dfrac{12\: g\: C}{44\: g\: CO_2}\times\dfrac{1\: mol\: C}{12\: g\: C}= 0.0714\: mol\: C}$
$\mathrm{1.29\: g\: H_2O\times \dfrac{2\: g\: H}{18\: g\: H_2O}\times\dfrac{1\: mol\: H}{1\: g\: H}= 0.143\: mol\: H}$
Thus, mole ratio of $\ce{C}$ : $\ce{H}$ is 0.0714 : 0.143 = 1 : 2. Therefore, the empirical formula is $\ce{CH2}$
Discussion
The molecular formula for ethylene is $\ce{C2H4}$ and cyclohexane is $\ce{C6H12}$. What are their molecular weights?
Example 7
Chloroform is a common solvent used in chemical labs. It has a molecular formula of $\ce{CHCl3}$. What is the weight percentage of chlorine ($\ce{Cl}$)? (Atomic weight, $\ce{Cl}$, 35.453; $\ce{H}$, 1.00794; $\ce{C}$, 12.0110)
Hint:
You should understand the reason for using this formula to calculate it:
$\mathrm{\dfrac{3\times 35.453}{3\times 35.453 + 12.011 + 1.00794}= 89.094\%\: (weight\: percentage)}$
Discussion
What is the weight percentage of $\ce{C}$ in $\ce{CCl4}$?
These examples illustrate some fundamental types of problems related to the understanding of the chemical formula. Their solutions are related to the skills outlined at the start of this page.
Skill Developing Problems
1. What is the molecular weight of sugar $\ce{C12H22O11}$?
~12*12+22+11*16
Skill - Know where to find atomic weights and evaluate molecular weights for:
caffeine, $\ce{C8H10N4O2}$ (194.2);
calcium aspirin, $\ce{Ca(OOCC6H4OCOCH3)2}$ (398.4);
ascorbic acid (vitamin C), $\ce{C6H8O6}$;
quartz (silicon), $\ce{SiO2}$;
ruby, nearly all $\ce{Al2O3}$.
1. What are the weight and mole percentages of $\ce{C}$ in sugar $\ce{C12H22O11}$?
12*12/(12*12+22+11*16) and (12/45)
Skill - Evaluate weight and mole percentages of any element in any substance.
1. The oil of cinnamon contains almost 90% of cinnamaldehyde. By weight, this compound contains 81.79% $\ce{C}$, 6.10% $\ce{H}$ and 12.11% $\ce{O}$. What is the empirical formula for cinnamaldehyde? The empirical formula is also the molecular formula in this case.
$\ce{C9H8O}$
Skill - Find a chemical formula when weight percentages are known.
1. If 1.80 kg glucose $\ce{C6H12O6}$ is burned completely, how many kg of $\ce{CO2}$is produced?
2.64 kg
Skill - Calculate the amount of water produced in this case.
1. Calculate the amount (g) of $\ce{H}$ present in 1.0 L of water. Assume density of water to be 1 g/mL. ($\ce{H}$, 1; $\ce{O}$, 16).
111 g
Skill - Calculate the amount (kg) of $\ce{C}$ in a block of dry ice weighing 88 kg.
1. A sample of a compound containing only $\ce{C}$ and $\ce{H}$ gave 0.090 g of $\ce{H2O}$ and 0.440 g of $\ce{CO2}$. ($\ce{H}$, 1; $\ce{C}$, 12; $\ce{O}$, 16). Find the empirical formula.
$\ce{CH}$
Skill - Describe elemental analysis for the determination of chemical formula.
1. How many moles is 0.130 g of $\ce{C6H6}$?
0.0017 mol
Skill - Determination of chemical formula from elemental analysis.
1. How many hydrogen atoms are present in a water molecule?
Two $\ce{H}$ atoms in an $\ce{H2O}$ molecule.
Skill: Explain the meaning of chemical formula. How many hydrogen atoms are there in an ethanol molecule, $\ce{C2H5OH}$?
1. In an experiment, 12.0 g of carbon gives 44.0 g of carbon dioxide. What is the weight percentage of carbon in $\ce{CO2}$?
27.3%
Skill: The weight percentage and mole percentage can be calculated if you know the chemical formula.
If you reduce 44.0 grams of $\ce{CO2}$ to carbon, how many grams of $\ce{C}$ will you get?
1. The combustion of 1.00 g of carbon gives 2.33 g of carbon oxide. The atomic weight of $\ce{C}$ is 12.011 g/mole and that of $\ce{O}$ is 15.9994 g/mole. What is the mole percent of $\ce{C}$ in the carbon oxide?
50% by mole
Skill: Differentiate mole percentage and weight percentage.
1. The empirical formula for benzene is $\ce{CH}$. Its molecular weight was determined to be 78. What is the molecular formula?
$\ce{C6H6}$
Skill: Differentiate empirical and molecular formulas. | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Chemical_Reactions/Stoichiometry/Chemical_Formulas.txt |
Calculate theoretical and actual yields of chemical reactions.
• What are actual and theoretical yields?
• For certain amounts of reactants, how much products shall be produced? How much actually is produced?
Excess and Limiting Reagents
Key Terms
• Stoichiometric mixture
• Excess reagent
• Limiting reagent
Learning Objectives
• Use stoichiometric calculation to determine excess and limiting reagents in a chemical reaction and explain why.
• Calculate theoretical yields of products formed in reactions that involve limiting reagents.
Chemical reaction equations give the ideal stoichiometric relationship among reactants and products. However, the reactants for a reaction in an experiment are not necessarily a stoichiometric mixture. In a chemical reaction, reactants that are not used up when the reaction is finished are called excess reagents. The reagent that is completely used up or reacted is called the limiting reagent, because its quantity limits the amount of products formed.
Let us consider the reaction between solid sodium and chlorine gas. The reaction can be represented by the equation:
$\mathrm{2 Na_{\large{(s)}} + Cl_{2\large{(g)}} \rightarrow 2 NaCl_{\large{(s)}}} \nonumber$
It represents a reaction of a metal and a diatomic gas chlorine. This balanced reaction equation indicates that two $\ce{Na}$ atoms would react with two $\ce{Cl}$ atoms or one $\ce{Cl2}$ molecule. Thus, if you have 6 $\ce{Na}$ atoms, 3 $\ce{Cl2}$ molecules will be required. If there is an excess number of $\ce{Cl2}$ molecules, they will remain unreacted. We can also state that 6 moles of sodium will require 3 moles of $\ce{Cl2}$ gas. If there are more than 3 moles of $\ce{Cl2}$ gas, some will remain as an excess reagent, and the sodium is a limiting reagent. It limits the amount of the product that can be formed.
Chemical reactions with stoichiometric amounts of reactants have no limiting or excess reagents.
Example $1$
Calculate the number of moles of $\ce{CO2}$ formed in the combustion of ethane $\ce{C2H6}$ in a process when 35.0 mol of $\ce{O2}$ is consumed.
HINT
The reaction is
$\ce{2 C2H6 + 7 O2 \rightarrow 4 CO2 + 6 H2O} \nonumber$
$\mathrm{35.0\: mol\: O_2 \times\dfrac{4\: mol\: CO_2}{7\: mol\: O_2} = 20.0\: mol\: CO_2} \nonumber$
DISCUSSION
A balanced equation for the reaction is a basic requirement for identifying the limiting reagent even if amounts of reactants are known.
Example $2$
Two moles of $\ce{Mg}$ and five moles of $\ce{O2}$ are placed in a reaction vessel, and then the $\ce{Mg}$ is ignited according to the reaction
$\mathrm{Mg + O_2 \rightarrow MgO}$.
Identify the limiting reagent in this experiment.
HINT
Before a limiting reagent is identified, the reaction must be balanced. The balanced reaction is
$\mathrm{2 Mg + O_2 \rightarrow 2 MgO} \nonumber$
Thus, two moles of $\ce{Mg}$ require only ONE mole of $\ce{O2}$. Four moles of oxygen will remain unreacted. Therefore, oxygen is the excess reagent, and $\ce{Mg}$ is the limiting reagent.
DISCUSSION
Answer these questions:
How many moles of $\ce{MgO}$ is formed?
What is the weight of $\ce{MgO}$ formed?
Skill Developing Problems
1. Which of the following will react most vigorously with $\ce{Cl2}$?
$\ce{Li}$, $\ce{Na}$, $\ce{K}$, $\ce{Fe}$, $\ce{Al}$, $\ce{Mg}$, or $\ce{Ar}$?
Hint: Potassium, $\ce{K}$
Skill -
Predict reactivity from location of element in the periodic table.
1. One mole of $\ce{Cl2}$ molecules would react with how many moles of $\ce{Na}$ metal?
Hint: Two moles
Skill -
Write a reaction equation for a chemical reaction.
1. At room temperature (25 °C) what is the state of sodium: solid, gas or liquid?
Hint: Solid
Skill -
Know where to look up properties of elements or substances.
1. Which of the following is the best conductor: sodium metal, solid $\ce{NaCl}$, chlorine gas, dilute $\ce{NaCl}$ solution, or molten $\ce{NaCl}$?
Hint: Sodium metal
1. What positive ions are present in solutions of $\ce{NaCl}$?
Hint: sodium ions
1. If you put equal weights of sodium metal and chlorine gas into a reaction vessel, which is the limiting reagent?
Hint: chlorine is the limiting reagent
Skill -
Identify excess and limiting reagent.
1. Equal weights of $\ce{H2}$ and $\ce{O2}$ are placed in a balloon and then ignited. Assuming reaction goes to completion, which gas is the excess reagent?
Hint: hydrogen
1. Equal weights of $\ce{H2}$ and $\ce{F2}$ are placed in a balloon and then ignited. Assuming reaction goes to completion, which gas is the limiting reagent?
Hint: Fluorine
Skill - Identify the excess and limiting reagent.
2. The reaction:
$\mathrm{2 Al_{\large{(s)}} + Fe_2O_3_{\large{(s)}} \rightarrow 2 Fe_{\large{(liquid)}} + Al_2O_3}$
takes place in the thermite mixture when it is ignited by a magnesium ribbon. A thermite mixture contains a mass ratio of 1 to 2 for $\ce{Al}$ and $\ce{Fe2O3}$. Which one is the limiting reagent?
Hint: Iron oxide
Discussion -
A stoichiometric mixture has a mass ratio of 54:160 (nearly 1:3) for $\ce{Al:Fe2O3}$. | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Chemical_Reactions/Stoichiometry/Chemical_Stoichiometry.txt |
Key Words
• Energy, exothermic reaction, endothermic reaction
• Physical reactions, chemical reactions, phase transitions
• Reactants, products
• Reaction stoichiometry
Learning Objectives
• To distinguish chemical changes from physical changes.
• To write chemical equations to describe a chemical reaction.
• To balance chemical equations.
• To calculate the quantities of reactants required or the quantities produced in a chemical reaction.
Changes in a material or system are called reactions, and they are divided into chemical and physical reactions. Energy is the driving force of all changes, both physical and chemical reactions. Energy is always involved in these reactions. If a system is more stable by losing some energy, a reaction takes place, releasing energy. Such a reaction is said to be exothermic. Supplying energy to a system also causes a reaction. Energy absorbing reactions are called endothermic reactions. Sometimes, the amount of energy involved in a reaction may be so small that the change in energy is not readily noticeable.
An equation can be used to describe a physical reaction, which involves a change of states. For example, melting, sublimation, evaporation, and condensation can be represented as follows. In these equations, (s) stands for solid, (l) for liquid (l), and (g) for gas.
• melting: $\mathrm{H_2O(s) \rightarrow H_2O(l)}$
• sublimation: $\mathrm{H_2O(s) \rightarrow H_2O(g)}$
• evaporation: $\mathrm{C_2H_5OH(l) \rightarrow C_2H_5OH(g)}$
• condensation: $\mathrm{NH_3(g) \rightarrow NH_3(l)}$
In these changes, no chemical bonds are broken or formed, and the molecular identities of the substances have not changed.
Is the phase transition between graphite and diamond a chemical or physical reaction?
$\mathrm{C(graphite) \rightarrow C(diamond)}$.
The crystal structures of diamond and graphite are very different, and bonding between the carbon atoms is also different in the two solid states. Because chemical bonds are broken and new bonds are formed, the phase transition of diamond and graphite is a chemical reaction.
Chemicals or substances change converting to one or more other substances, and these changes are called chemical reactions. At the molecular level, atoms or groups of atoms rearrange resulting in breaking and forming some chemical bonds in a chemical reaction. The substances undergoing changes are called reactants, whereas substances newly formed are called products. Physical appearances of products are often different from reactants. Chemical reactions are often accompanied by the appearance of gas, fire, precipitate, color, light, sound, or odor. These phenomena are related to energy and properties of the reactants and products. For example, the oxidation of propane releases heat and light, and a rapid reaction is an explosion,
$\mathrm{C_3H_8 + 5 O_2 \rightarrow 3 CO_2 + 4 H_2O}$
A balanced equation also shows a macroscopic quantitative relationship. This balanced reaction equation shows that five moles of oxygen reacts with one mole of propane generating three moles of carbon dioxide and four moles of water, a total of 7 moles of products in the combustion reaction.
At the molecular level, this equation shows that for each propane molecule, 5 oxygen molecules are required. The three carbon atoms are converted to three molecules of carbon dioxide, whereas the 8 hydrogen atoms in propane are oxidized to 4 water molecules. The numbers of $\ce{H}$, $\ce{C}$, and $\ce{O}$ atoms are the same on both sides of the equation.
We study properties of substances so that we know how to make use of them. Tendencies of a substance to react, either by itself or with others, are important chemical properties. Via properties, we understand chemical reactions, which are best studied by experimentation and observation. After you have performed many experiments, you may generalize certain rules and facts. Knowing these rules and facts enables you to solve problems that you have not yet encountered.
The most important aspect of a chemical reaction is to know what are the reactants and what are the products. For this, the best description of a reaction is to write an equation for the reaction. A chemical reaction equation gives the reactants and products, and a balanced chemical reaction equation shows the mole relationships of reactants and products. Often, the amount of energy involved in the reaction is given. Dealing with the quantitative aspect of chemical reactions is called reaction stoichiometry.
For example, when clamshells, $\ce{CaCO3}$, are heated, a gas $\ce{CO2}$ will be released, leaving a white powder (solid $\ce{CaO}$) behind. The equation of the reaction is written as:
$\mathrm{CaCO_3 \rightarrow CaO + CO_2} \nonumber$
The equation indicates that one mole of $\ce{CaCO3}$ gives one mole each of $\ce{CaO}$ and $\ce{CO2}$. Amounts of substances represented by chemical formulas have been introduced on the two previous pages, and these concepts should help to figure out the stoichiometry of reactions when a reaction equation is given.
Example 1
When 10.0 g pure calcium carbonate is heated and converted to solid calcium oxide $\ce{CaO}$, how much calcium oxide should be obtained? If only 5.0 grams $\ce{CaO}$ is obtained, what is the actual yield?
HINT
Under ideal conditions, amounts of substance in the reaction equation is as indicated below:
\begin{alignat}{2} \ce{&CaCO_3 \rightarrow &&CaO + &&CO_2}\ &\:100.0 &&\:\:56 &&\:44\:\:\: \mathrm{g/mol\: (formula\: weights)} \end{alignat}
$\mathrm{10.0\: g\: CaCO_3\times\dfrac{1\: mol\: CaCO_3}{100\: g\: CaCO_3}\times\dfrac{1\: mol\: CaO}{1\: mol\: CaCO_3}\times\dfrac{56\: g\: CaO}{1\: mol\: CaO}= 5.6\: g\: CaO} \nonumber$
DISCUSSION
An inefficient conversion is given here, but the method shows the details of consideration. If the amount of $\ce{CaO}$ obtained is not 5.6 g, one can conclude that the sample may not be pure.
Example 2
When 10.0 g pure calcium carbonate is heated and converted to solid calcium oxide $\ce{CaO}$, how much $\ce{CO2}$ at standard condition is released?
HINT
$\mathrm{CaCO_3 \rightarrow CaO + CO_2} \nonumber$
$\mathrm{10.0\: g\: CaCO_3\times\dfrac{1\: mol\: CO_2}{100\: g\: CaCO_3}\times\dfrac{22.4\: L\: CO_2}{1\: mol\: CO_2}= 2.24\: L\: CO_2} \nonumber$
DISCUSSION
We have taken a short cut in this formulation compared to Example 1. Examples 1 and 2 illustrate the evaluation of quantities in g and in L.
Writing Equations for Chemical Reactions
Chemical reaction equations truly represent changes of materials. For many reactions, we may only be able to write equations for the overall reactions. For example, common sense tells us that when sugar is fully oxidized, carbon dioxide and water are the final products. The oxidation reaction is the same as the combustion reaction. Thus we write
$\ce{C12H22O11 + 12 O2 \rightarrow 12 CO2 + 11 H2O} \nonumber$
This illustrates the methods used for writing balanced reaction equations:
1. Determine the reactants and products: In this case, the products are $\ce{CO2}$ and $\ce{H2O}$, determined by common sense. We know that.
2. Apply the fundamental principle of conservation of atoms. Numbers of atoms of each kind must be the same before and after the reactions.
3. Balance one type of atoms at a time: We may use $\ce{H}$ or $\ce{C}$ to begin. Since there are 12 $\ce{C}$ atoms on the left, the coefficient is 12 for $\ce{CO2}$. Similarly, 22 $\ce{H}$ atoms produce 11 $\ce{H2O}$ molecules.
4. Balance the oxygen atoms on both sides: There are a total of 35 $\ce{O}$ atoms on the right hand, and the coefficient for $\ce{O2}$ should be 11.
Example 3
The compound $\ce{N2O5}$ is unstable at room temperature. It decomposes yielding a brown gas $\ce{NO2}$ and oxygen. Write a balanced chemical reaction equation for its decomposition.
HINT
The first step is to write an unbalanced equation indicating only the reactant and products:
$\ce{N2O5 \rightarrow NO2 + O2} \nonumber$
A $\ce{N2O5}$ molecule decomposes into two $\ce{NO2}$ molecule, and half of $\ce{O2}$.
$\ce{N2O5 \rightarrow 2 NO2 + \dfrac{1}{2}O2} \nonumber$
In order to give whole number stoichiometric coefficients to the equation, we multiply all the stoichiometric coefficients by 2.
$\ce{2 N2O5 \rightarrow 4 NO2 + O2} \nonumber$
DISCUSSION
This example illustrates the steps used in writing a balanced equation for a chemical reaction. This balanced equation does not tell us how a $\ce{N2O5}$ molecule decomposes, it only illustrates the overall reaction.
Example 4
When solutions of $\ce{CaCl2}$ and $\ce{AgNO3}$ are mixed, a white precipitate is formed. The same precipitate is also observed when $\ce{NaCl}$ solution is mixed with $\ce{AgCH3CO2}$ solution. Write a balanced equation for the reaction between $\ce{CaCl2}$ and $\ce{AgNO3}$.
HINT
The common ions between $\ce{NaCl}$ and $\ce{CaCl2}$ are $\ce{Cl-}$ ions, and $\ce{Ag+}$ ions are common between the two silver containing compounds. The question illustrates a scientific deduction used in the determination of products. The product is $\ce{AgCl}$, and the balanced reaction is
$\ce{CaCl2 + 2 AgNO3 \rightarrow 2 AgCl + Ca(NO3)2}$
DISCUSSION
In reality, solutions of salts contain ions. In this case, the solutions contain $\ce{Ca^2+}$, $\ce{Cl-}$, $\ce{Ag+}$, and $\ce{NO3-}$ ions. The $\ce{Cl-}$ and $\ce{Ag+}$ ions form an insoluble solid, and a precipitate is formed,
$\ce{Cl- + Ag+ \rightarrow AgCl(s)}$
$\ce{Ca^2+}$ and $\ce{NO3-}$ are spectator ions.
Chemical Reactions
One of the most important topics in chemistry is chemical reaction. In this page, we only concentrate on the stoichiometry conveyed by reaction equations. Other topics related to chemical reactions are:
• Excess and Limiting Reagents or reactants left over or used up
• Features of chemical reactions or classification of reactions
• Chemical kinetics or reaction rates
• Reaction mechanism or how actually reaction proceed
Balancing Redox Reactions
Balancing oxidation and reduction reaction equations is a little more complicated than what we discussed here. You have to have the skills to assign oxidation states, explain oxidation and reduction in terms of oxidation-state change, and write half reaction equations. Then you will be able to balance redox reactions. All these are given in the next module on Chemical Reactions.
Skill Developing Problems
1. What is/are the product(s) containing carbon when methane, $\ce{CH4}$, is burned in the air?
Hint: $\ce{CO2}$
Generalization:
Combustion of $\ce{C}$-containing compounds converts all $\ce{C}$ to $\ce{CO2}$.
1. Use the common sense method to find the molecular formula for hydrogen sulfide, whose molecular weight is 34.1. (Atomic weight, $\ce{H}$, 1.008; $\ce{S}$, 32.066)
Hint: $\ce{H2S}$
Generalization:
Sulfur and oxygen are group 6 elements, and they form $\ce{H2O}$ and $\ce{H2S}$.
1. When 30.0 g of $\ce{Al}$ (atomic weight 27.0) is heated in oxygen (atomic mass 16.0), an aluminum oxide, $\ce{Al2O3}$, is formed. How much oxide should be obtained?
Hint: 56.7 g
A Variation:
How much (in g) oxygen is required?
1. When $\ce{KClO3}$ is heated, it decomposes to give solid $\ce{KCl}$ and oxygen gas. If 0.500 mol $\ce{O2}$ is collected, how many grams of $\ce{KCl}$ should be obtained? (Atomic wt: $\ce{K}$, 39.098; $\ce{Cl}$, 35.453)
Hint: 24.9 g
Method suggestion:
For the reaction: $\ce{2 KClO3 \rightarrow 2 KCl + 3 O2}$
the formulation suggestion is:
$\mathrm{0.50\: mol\: O_2 \times\dfrac{2\: mole\: KCl}{3\: mol\: O_2}\times\dfrac{74.6\: g\: KCl}{1\: mol\: KCl}=\: ??.?\: g\: KCl}$
1. A solution containing pure $\ce{BaCl2}$ is treated with excess amounts of $\ce{H2SO4}$, and the precipitate $\ce{BaSO4}$ is collected and dried. If 13.2 g of $\ce{BaSO4}$ are collected, how many moles of $\ce{Cl-}$ ions are left in the solution?
Atomic wt: $\ce{H}$, 1.008; $\ce{O}$, 16.00; $\ce{S}$, 32.06; $\ce{Cl}$, 35.45; $\ce{Ba}$, 137.33.
Hint: 0.113 mol
Variations:
How much (in g) $\ce{BaCl2}$ is present in the solution?
How much silver nitrate is required to precipitate all the chloride ions?
The reaction is: $\mathrm{BaCl_2 + H_2SO_4 \rightarrow BaSO_4 + 2 H^+ + 2 Cl^-}$.
0.0566 mole of $\ce{Ba}$ correspond to 0.113 mol of $\ce{Cl-}$ in $\ce{BaCl2}$.
Method Suggestion:
$\mathrm{13.2\: g\times\dfrac{1\: mol\: BaSO_4}{233.39\: g}\times\dfrac{2\: mol\: Cl^-}{1\: mol\: BaSO_4}= 0.113\: mol}$
1. A power plant burns coal, and this process is equivalent to burning 999 kg of sulfur a day. How many kg of $\ce{SO2}$ are emitted per day if the power plant does not have pollution control devices to recover the sulfur? Atomic wt: $\ce{C}$, 12.00; $\ce{O}$, 16.00; $\ce{S}$, 32.06.
Hint: 1998 kg
Further consideration:
The molecular weight of $\ce{SO2}$ is about twice the atomic weight of $\ce{S}$. Thus the weight of $\ce{SO2}$ is twice that of $\ce{S}$.
Variations: How much (in mole and L) $\ce{SO2}$ is generated per day?
If all $\ce{SO2}$ is converted to $\ce{H2SO4}$, how much (in mol and kg) sulfuric acid is produced? (3055 kg)
1. How many moles of water will be formed when one mole of propane $\ce{C3H8}$ is burned in an excess amount of air?
Hint: 4 moles; $\ce{C3H8 + 5 O2 \rightarrow 3 CO2 + 4 H2O}$
Skill:
Work out a balanced reaction equation.
Variations:
How many grams of water will be produced?
How many moles of $\ce{CO2}$ will be produced?
1. A mixture containing $\ce{Na2SO4}$, but no other sulfate, is analyzed by precipitation with $\ce{BaCl2}$. A 2.37 g mixture sample gave a 2.57 g $\ce{BaSO4}$ precipitate. What is the percentage of $\ce{Na2SO4}$ in the mixture?
Hint: 66.0%
Skill:
The problem illustrates a strategy for chemical analysis.
1. Suppose 2.33 g of $\ce{CaCl2}$ and $\ce{Ca(NO3)2}$ mixture gives 2.22 g of $\ce{AgCl}$ when $\ce{Ag(NO3)}$ is used as a reagent to precipitate the chloride $\ce{Cl-}$ ions. What is the percentage of $\ce{CaCl2}$ in the mixture?
Hint: 36.9%
Atomic wt: $\ce{N}$, 14.0; $\ce{O}$, 16.0; $\ce{Cl}$, 35.5; $\ce{Ca}$, 40.1; $\ce{Ag}$, 107.9.
Skill:
This problem also illustrates a strategy for chemical analysis. | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Chemical_Reactions/Stoichiometry/Reaction_Equations.txt |
Review Purposes
• To get an overall view of stoichiometry.
• Apply skills learned to perform quantitative chemical analysis.
• Apply theories and rules of chemistry to solve problems.
• Assess areas of strength and weakness for review purposes.
• Improve problem solving strategy and learning efficiency.
Theoretical and Actual Yields
Key Terms
• (Excess reagent, limiting reagent)
• Theoretical and actual yields
• Percentage or actual yield
Learning Objectives
• Use stoichiometric calculation to determine excess and limiting reagents in a chemical reaction and explain why.
• Calculate theoretical yields of products formed in reactions that involve limiting reagents.
• Evaluate percentage or actual yields from known amounts of reactants
Theoretical and Actual Yields
Reactants not completely used up are called excess reagents, and the reactant that completely reacts is called the limiting reagent. This concept has been illustrated for the reaction:
$\mathrm{2 Na + Cl_2 \rightarrow 2 NaCl} \nonumber$
Amounts of products calculated from the complete reaction of the limiting reagent are called theoretical yields, whereas the amount actually produced of a product is the actual yield. The ratio of actual yield to theoretical yield expressed in percentage is called the percentage yield.
$\mathrm{percent\: yield = \dfrac{actual\: yield}{theoretical\: yield}\times100}$
Chemical reaction equations give the ideal stoichiometric relationship among reactants and products. Thus, the theoretical yield can be calculated from reaction stoichiometry. For many chemical reactions, the actual yield is usually less than the theoretical yield, understandably due to loss in the process or inefficiency of the chemical reaction.
Example $1$
Methyl alcohol can be produced in a high-pressure reaction
$\mathrm{CO_{\large{(g)}} + 2 H_{2\large{(g)}} \rightarrow CH_3OH_{\large{(l)}}}$
If 6.1 metric tons of methyl alcohol is obtained from 1.2 metric tons of hydrogen reacting with excess amount of $\ce{CO}$, estimate the theoretical and the percentage yield?
Solution
To calculate the theoretical yield, consider the reaction
\begin{alignat}{2} \ce{&CO_{\large{(g)}} +\, &&2 H_{2\large{(g)}} \rightarrow \, &&CH_3OH_{\large{(l)}}}\ &\:28.0 &&\:4.0 &&\:\:\:32.0 \hspace{45px}\ce{(stoichiometric\: masses\: in\: g,\: kg,\: or\: tons)} \end{alignat}
$\mathrm{1.2\: tons\: H_2 \times\dfrac{32.0\: CH_3OH}{4.0\: H_2}= 9.6\: tons\: CH_3OH}$
Thus, the theoretical yield from 1.2 metric tons (1.2x106 g) of hydrogen gas is 9.6 tons. The actual yield is stated in the problem, 6.1 metric tons. Thus, the percentage yield is
$\mathrm{\%\: yield =\dfrac{6.1\: tons}{9.6\: tons}\times 100 = 64 \%}$
Due to chemical equilibrium or the mass action law, the limiting reagent may not be completely consumed. Thus, a lower yield is expected in some cases. Losses during the recovery process of the product will cause an even lower actual yield.
Example $2$
A solution containing silver ion, $\ce{Ag+}$, has been treated with excess of chloride ions $\ce{Cl-}$. When dried, 0.1234 g of $\ce{AgCl}$ was recovered. Assuming the percentage yield to be 98.7%, how many grams of silver ions were present in the solution?
HINT
The reaction and relative masses of reagents and product are:
\begin{alignat}{2}\ce{ Ag^+_{\large{(aq)}}} &+ \mathrm{Cl^-_{\large{(aq)}}} &&\rightarrow \ce{AgCl_{\large{(s)}}} \ 107.868 &+ 35.453 &&= 143.321 \end{alignat}
The calculation,
$\mathrm{0.1234\: g\: AgCl \times \dfrac{107.868\: g\: Ag^+}{143.321\: g\: AgCl}= 0.09287\: g\: Ag^+}$
shows that 0.1234 g dry $\ce{AgCl}$ comes from 0.09287 g $\ce{Ag+}$ ions. Since the actual yield is only 98.7%, the actual amount of $\ce{Ag+}$ ions present is therefore
$\mathrm{\dfrac{0.09287\: g\: Ag^+}{0.987}= 0.09409\: g\: Ag^+}$
DISCUSSION
One can also calculate the theoretical yield of $\ce{AgCl}$ from the percentage yield of 98.7% to be
$\mathrm{\dfrac{0.1234\: g\: AgCl}{0.987}= 0.1250\: g\: AgCl}$
From 0.1250 g $\ce{AgCl}$, the amount of $\ce{Ag+}$ present is also 0.09409 g.
Skill Developing Problems
1. In an analytical experiment, you are asked to determine the amount of iodide ion $\ce{I+}$ in 10.00 mL of a solution that does not contain any other ions that will form a precipitate with silver ions. You have learned that $\ce{Ag+}$ ions precipitate all the iodide ions in a solution. In performing the experiment, shall you treat $\ce{AgNO3}$ as the excess reagent or limiting reagent? Molar mass or atomic weight: $\ce{Ag}$, 107.868; $\ce{I}$, 126.904 (You should know where to find them).
Hint: $\ce{AgNO3}$ is the excess reagent
Skill -
Apply the concept of excess and limiting reagents for work. You can add $\ce{AgNO3}$ slowly until the clear portion of the solution gives no precipitate when a drop of $\ce{AgNO3}$ solution is added. This indicates that all the $\ce{I-}$ ions are consumed.
1. From the 10.00 mL iodide solution, you have added $\ce{AgNO3}$ solution or solid. How do you know that you have added an excess amount of $\ce{AgNO3}$ to precipitate the iodide ions?
Hint: Test for excess or limiting reagent.
Skill -
Excess reagent can be tested for its presence, and limiting reagent can be tested for its absence.
1. In an analytical experiment, 0.1234 g of $\ce{AgI}$ was obtained from a 10.00-mL solution with excess silver nitrate. How much (in g) iodide ions are present?
Hint: 0.05670 g $\ce{Ag}$ and 0.06670 g $\ce{I}$
Skill -
Calculate the amount of limiting reagent from the amount of products.
1. In an analytical experiment, 0.1234 g of $\ce{AgI}$ was obtained from a 10.00-mL solution with excess silver nitrate. What is the iodide concentration (mol/L or M) in the solution?
Hint: 0.05256 mol/L
Skill -
Calculate the concentration when the amount of solute is known. The concept of concentration will be covered in the unit dealing with solution, but you should be able to convert; see the following relationship.
0.1234 g $\ce{AgI}$ = 0.0005256 mol = 0.5256 mili-mol $\ce{AgI}$ or $\ce{Ag}$ or $\ce{I}$. | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Chemical_Reactions/Stoichiometry/Stoichiometry_-_A_Review.txt |
Stoichiometry is a section of chemistry that involves using relationships between reactants and/or products in a chemical reaction to determine desired quantitative data. In Greek, stoikhein means element and metron means measure, so stoichiometry literally translated means the measure of elements. In order to use stoichiometry to run calculations about chemical reactions, it is important to first understand the relationships that exist between products and reactants and why they exist, which require understanding how to balance reactions.
Balancing
In chemistry, chemical reactions are frequently written as an equation, using chemical symbols. The reactants are displayed on the left side of the equation and the products are shown on the right, with the separation of either a single or double arrow that signifies the direction of the reaction. The significance of single and double arrow is important when discussing solubility constants, but we will not go into detail about it in this module. To balance an equation, it is necessary that there are the same number of atoms on the left side of the equation as the right. One can do this by raising the coefficients.
Reactants to Products
A chemical equation is like a recipe for a reaction so it displays all the ingredients or terms of a chemical reaction. It includes the elements, molecules, or ions in the reactants and in the products as well as their states, and the proportion for how much of each particle reacts or is formed relative to one another, through the stoichiometric coefficient. The following equation demonstrates the typical format of a chemical equation:
$\ce{2 Na(s) + 2HCl(aq) \rightarrow 2NaCl(aq) + H2(g)} \nonumber$
In the above equation, the elements present in the reaction are represented by their chemical symbols. Based on the Law of Conservation of Mass, which states that matter is neither created nor destroyed in a chemical reaction, every chemical reaction has the same elements in its reactants and products, though the elements they are paired up with often change in a reaction. In this reaction, sodium ($Na$), hydrogen ($H$), and chloride ($Cl$) are the elements present in both reactants, so based on the law of conservation of mass, they are also present on the product side of the equations. Displaying each element is important when using the chemical equation to convert between elements.
Stoichiometric Coefficients
In a balanced reaction, both sides of the equation have the same number of elements. The stoichiometric coefficient is the number written in front of atoms, ion and molecules in a chemical reaction to balance the number of each element on both the reactant and product sides of the equation. Though the stoichiometric coefficients can be fractions, whole numbers are frequently used and often preferred. This stoichiometric coefficients are useful since they establish the mole ratio between reactants and products. In the balanced equation:
$\ce{2 Na(s) + 2HCl(aq) \rightarrow 2NaCl(aq) + H2(g)} \nonumber$
we can determine that 2 moles of $HCl$ will react with 2 moles of $Na_{(s)}$ to form 2 moles of $NaCl_{(aq)}$ and 1 mole of $H_{2(g)}$. If we know how many moles of $Na$ reacted, we can use the ratio of 2 moles of $NaCl$ to 2 moles of Na to determine how many moles of $NaCl$ were produced or we can use the ratio of 1 mole of $H_2$ to 2 moles of $Na$ to convert to $NaCl$. This is known as the coefficient factor. The balanced equation makes it possible to convert information about the change in one reactant or product to quantitative data about another reactant or product. Understanding this is essential to solving stoichiometric problems.
Example 1
Lead (IV) hydroxide and sulfuric acid react as shown below. Balance the reaction.
$\ce{Pb(OH)4 + H2SO4 \rightarrow Pb(SO4)2 +H2O} \nonumber$
Solution
Start by counting the number of atoms of each element.
UNBALANCED
Element
Reactant (# of atoms)
Product (# of atoms)
Pb
1
1
O
8
9
H
6
2
S
1
2
The reaction is not balanced; the reaction has 16 reactant atoms and only 14 product atoms and does not obey the conservation of mass principle. Stoichiometric coefficients must be added to make the equation balanced. In this example, there are only one sulfur atom present on the reactant side, so a coefficient of 2 should be added in front of $H_2SO_4$ to have an equal number of sulfur on both sides of the equation. Since there are 12 oxygen on the reactant side and only 9 on the product side, a 4 coefficient should be added in front of $H_2O$ where there is a deficiency of oxygen. Count the number of elements now present on either side of the equation. Since the numbers are the same, the equation is now balanced.
$\ce{ Pb(OH)4 + 2 H2SO4 \rightarrow Pb(SO4)2 + 4H2O} \nonumber$
BALANCED
Element
Reactant (# of atoms)
Product (# of atoms)
Pb
1
1
O
12
12
H
8
8
S
2
2
Balancing reactions involves finding least common multiples between numbers of elements present on both sides of the equation. In general, when applying coefficients, add coefficients to the molecules or unpaired elements last.
A balanced equation ultimately has to satisfy two conditions.
1. The numbers of each element on the left and right side of the equation must be equal.
2. The charge on both sides of the equation must be equal. It is especially important to pay attention to charge when balancing redox reactions.
Stoichiometry and Balanced Equations
In stoichiometry, balanced equations make it possible to compare different elements through the stoichiometric factor discussed earlier. This is the mole ratio between two factors in a chemical reaction found through the ratio of stoichiometric coefficients. Here is a real world example to show how stoichiometric factors are useful.
Example 2
There are 12 party invitations and 20 stamps. Each party invitation needs 2 stamps to be sent. How many party invitations can be sent?
Solution
The equation for this can be written as
$\ce{I + 2S \rightarrow IS2}\nonumber$
where
• $I$ represents invitations,
• $S$ represents stamps, and
• $IS_2$ represents the sent party invitations consisting of one invitation and two stamps.
Based on this, we have the ratio of 2 stamps for 1 sent invite, based on the balanced equation.
Invitations Stamps Party Invitations Sent
In this example are all the reactants (stamps and invitations) used up? No, and this is normally the case with chemical reactions. There is often excess of one of the reactants. The limiting reagent, the one that runs out first, prevents the reaction from continuing and determines the maximum amount of product that can be formed.
Example 3
What is the limiting reagent in this example?
Solution
Stamps, because there was only enough to send out invitations, whereas there were enough invitations for 12 complete party invitations. Aside from just looking at the problem, the problem can be solved using stoichiometric factors.
12 I x (1IS2/1I) = 12 IS2 possible
20 S x (1IS2/2S) = 10 IS2 possible
When there is no limiting reagent because the ratio of all the reactants caused them to run out at the same time, it is known as stoichiometric proportions.
Types of Reactions
There are 6 basic types of reactions.
• Combustion: Combustion is the formation of CO2 and H2O from the reaction of a chemical and O2
• Combination (synthesis): Combination is the addition of 2 or more simple reactants to form a complex product.
• Decomposition: Decomposition is when complex reactants are broken down into simpler products.
• Single Displacement: Single displacement is when an element from on reactant switches with an element of the other to form two new reactants.
• Double Displacement: Double displacement is when two elements from on reactants switched with two elements of the other to form two new reactants.
• Acid-Base: Acid- base reactions are when two reactants form salts and water.
Molar Mass
Before applying stoichiometric factors to chemical equations, you need to understand molar mass. Molar mass is a useful chemical ratio between mass and moles. The atomic mass of each individual element as listed in the periodic table established this relationship for atoms or ions. For compounds or molecules, you have to take the sum of the atomic mass times the number of each atom in order to determine the molar mass
Example 4
What is the molar mass of H2O?
Solution
$\text{Molar mass} = 2 \times (1.00794\; g/mol) + 1 \times (15.9994\; g/mol) = 18.01528\; g/mol \nonumber$
Using molar mass and coefficient factors, it is possible to convert mass of reactants to mass of products or vice versa.
Example 5: Combustion of Propane
Propane ($\ce{C_3H_8}$) burns in this reaction:
$\ce{C_3H_8 + 5O_2 \rightarrow 4H_2O + 3CO_2} \nonumber$
If 200 g of propane is burned, how many g of $H_2O$ is produced?
Solution
Steps to getting this answer: Since you cannot calculate from grams of reactant to grams of products you must convert from grams of $C_3H_8$ to moles of $C_3H_8$ then from moles of $C_3H_8$ to moles of $H_2O$. Then convert from moles of $H_2O$ to grams of $H_2O$.
• Step 1: 200 g $C_3H_8$ is equal to 4.54 mol $C_3H_8$.
• Step 2: Since there is a ratio of 4:1 $H_2O$ to $C_3H_8$, for every 4.54 mol $C_3H_8$ there are 18.18 mol $H_2O$.
• Step 3: Convert 18.18 mol $H_2O$ to g $H_2O$. 18.18 mol $H_2O$ is equal to 327.27 g $H_2O$.
Variation in Stoichiometric Equations
Almost every quantitative relationship can be converted into a ratio that can be useful in data analysis.
Density
Density ($\rho$) is calculated as mass/volume. This ratio can be useful in determining the volume of a solution, given the mass or useful in finding the mass given the volume. In the latter case, the inverse relationship would be used.
Volume x (Mass/Volume) = Mass
Mass x (Volume/Mass) = Volume
Percent Mass
Percents establish a relationship as well. A percent mass states how many grams of a mixture are of a certain element or molecule. The percent X% states that of every 100 grams of a mixture, X grams are of the stated element or compound. This is useful in determining mass of a desired substance in a molecule.
Example 6
A substance is 5% carbon by mass. If the total mass of the substance is 10 grams, what is the mass of carbon in the sample? How many moles of carbon are there?
Solution
10 g sample x (5 g carbon/100 g sample) = 0.5 g carbon
0.5g carbon x (1 mol carbon/12.011g carbon) = 0.0416 mol carbon
Molarity
Molarity (moles/L) establishes a relationship between moles and liters. Given volume and molarity, it is possible to calculate mole or use moles and molarity to calculate volume. This is useful in chemical equations and dilutions.
Example 7
How much 5 M stock solution is needed to prepare 100 mL of 2 M solution?
Solution
100 mL of dilute solution (1 L/1000 mL)(2 mol/1L solution)(1 L stock solution/5 mol solution)(1000 ml stock solution/1L stock solution) = 40 mL stock solution.
These ratios of molarity, density, and mass percent are useful in complex examples ahead.
Determining Empirical Formulas
An empirical formula can be determined through chemical stoichiometry by determining which elements are present in the molecule and in what ratio. The ratio of elements is determined by comparing the number of moles of each element present.
Example 8: Combustion of Organic Molecules
1.000 gram of an organic molecule burns completely in the presence of excess oxygen. It yields 0.0333 mol of CO2 and 0.599 g of H2O. What is the empirical formula of the organic molecule?
Solution
This is a combustion reaction. The problem requires that you know that organic molecules consist of some combination of carbon, hydrogen, and oxygen elements. With that in mind, write the chemical equation out, replacing unknown numbers with variables. Do not worry about coefficients here.
$\ce{C_xH_yO_z(g) + O_2(g) \rightarrow CO_2(g) + H_2O(g)} \nonumber$
Since all the moles of C and H in CO2 and H2O, respectively have to have came from the 1 gram sample of unknown, start by calculating how many moles of each element were present in the unknown sample.
0.0333mol CO2 (1mol C/ 1mol CO2) = 0.0333mol C in unknown
0.599g H2O (1mol H2O/ 18.01528g H2O)(2mol H/ 1mol H2O) = 0.0665 mol H in unknown
Calculate the final moles of oxygen by taking the sum of the moles of oxygen in CO2 and H2O. This will give you the number of moles from both the unknown organic molecule and the O2 so you must subtract the moles of oxygen transferred from the O2.
Moles of oxygen in CO2:
0.0333mol CO2 (2mol O/1mol CO2) = 0.0666 mol O
Moles of oxygen in H2O:
0.599g H2O (1mol H2O/18.01528 g H2O)(1mol O/1mol H2O) = 0.0332 mol O
Using the Law of Conservation, we know that the mass before a reaction must equal the mass after a reaction. With this we can use the difference of the final mass of products and initial mass of the unknown organic molecule to determine the mass of the O2 reactant.
0.333mol CO2(44.0098g CO2/ 1mol CO2) = 1.466g CO2
1.466g CO2 + 0.599g H2O - 1.000g unknown organic = 1.065g O2
Moles of oxygen in O2
1.065g O2(1mol O2/ 31.9988g O2)(2mol O/1mol O2) = 0.0666mol O
Moles of oxygen in unknown
(0.0666mol O + 0.0332 mol O) - 0.0666mol O = 0.0332 mol O
Construct a mole ratio for C, H, and O in the unknown and divide by the smallest number.
(1/0.0332)(0.0333mol C : 0.0665mol H : 0.0332 mol O) => 1mol C: 2 mol H: 1 mol O
From this ratio, the empirical formula is calculated to be CH2O.
Determining Molecular Formulas
To determine a molecular formula, first determine the empirical formula for the compound as shown in the section above and then determine the molecular mass experimentally. Next, divide the molecular mass by the molar mass of the empirical formula (calculated by finding the sum the total atomic masses of all the elements in the empirical formula). Multiply the subscripts of the molecular formula by this answer to get the molecular formula.
Example 9
In the example above, it was determined that the unknown molecule had an empirical formula of CH2O.
1. Find the molar mass of the empircal formula CH2O.
12.011g C + (1.008 g H) * (2 H) + 15.999g O = 30.026 g/mol CH2O
2. Determine the molecular mass experimentally. For our compound, it is 120.056 g/mol.
3. Divide the experimentally determined molecular mass by the mass of the empirical formula.
(120.056 g/mol) / (30.026 g/mol) = 3.9984
4. Since 3.9984 is very close to four, it is possible to safely round up and assume that there was a slight error in the experimentally determined molecular mass. If the answer is not close to a whole number, there was either an error in the calculation of the empirical formula or a large error in the determination of the molecular mass.
5. Multiply the ratio from step 4 by the subscripts of the empirical formula to get the molecular formula.
CH2O * 4 = ?
C: 1 * 4 = 4
H: 2 * 4 = 8
O 1 * 4 = 4
CH2O * 4 = C4H8O4
6. Check your result by calculating the molar mass of the molecular formula and comparing it to the experimentally determined mass.
molar mass of C4H8O4= 120.104 g/mol
experimentally determined mass = 120.056 g/mol
% error = | theoretical - experimental | / theoretical * 100%
% error = | 120.104 g/mol - 120.056 g/mol | / 120.104 g/mol * 100%
% error = 0.040 %
Example 10: Complex Stoichiometry Problem
An amateur welder melts down two metals to make an alloy that is 45% copper by mass and 55% iron(II) by mass. The alloy's density is 3.15 g/L. One liter of alloy completely fills a mold of volume 1000 cm3. He accidentally breaks off a 1.203 cm3 piece of the homogenous mixture and sweeps it outside where it reacts with acid rain over years. Assuming the acid reacts with all the iron(II) and not with the copper, how many grams of H2(g) are released into the atmosphere because of the amateur's carelessness? (Note that the situation is fiction.)
Solution
Step 1: Write a balanced equation after determining the products and reactants. In this situation, since we assume copper does not react, the reactants are only H+(aq) and Fe(s). The given product is H2(g) and based on knowledge of redox reactions, the other product must be Fe2+(aq).
$\ce{Fe(s) + 2H^{+}(aq) \rightarrow H2(g) + Fe^{2+}(aq)} \nonumber$
Step 2: Write down all the given information
Alloy density = (3.15g alloy/ 1L alloy)
x grams of alloy = 45% copper = (45g Cu(s)/100g alloy)
x grams of alloy = 55% iron(II) = (55g Fe(s)/100g alloy)
1 liter alloy = 1000cm3 alloy
alloy sample = 1.203cm3 alloy
Step 3: Answer the question of what is being asked. The question asks how much H2(g) was produced. You are expected to solve for the amount of product formed.
Step 4: Start with the compound you know the most about and use given ratios to convert it to the desired compound.
Convert the given amount of alloy reactant to solve for the moles of Fe(s) reacted.
1.203cm3 alloy(1liter alloy/1000cm3 alloy)(3.15g alloy/1liter alloy)(55g Fe(s)/100g alloy)(1mol Fe(s)/55.8g Fe(s))=3.74 x 10-5 mol Fe(s)
Make sure all the units cancel out to give you moles of $\ce{Fe(s)}$. The above conversion involves using multiple stoichiometric relationships from density, percent mass, and molar mass.
The balanced equation must now be used to convert moles of Fe(s) to moles of H2(g). Remember that the balanced equation's coefficients state the stoichiometric factor or mole ratio of reactants and products.
3.74 x 10-5 mol Fe (s) (1mol H2(g)/1mol Fe(s)) = 3.74 x 10-5 mol H2(g)
Step 5: Check units
The question asks for how many grams of H2(g) were released so the moles of H2(g) must still be converted to grams using the molar mass of H2(g). Since there are two H in each H2, its molar mass is twice that of a single H atom.
molar mass = 2(1.00794g/mol) = 2.01588g/mol
3.74 x 10-5 mol H2(g) (2.01588g H2(g)/1mol H2 (g)) = 7.53 x 10-5 g H2(g) released
Problems
Stoichiometry and balanced equations make it possible to use one piece of information to calculate another. There are countless ways stoichiometry can be used in chemistry and everyday life. Try and see if you can use what you learned to solve the following problems.
1) Why are the following equations not considered balanced?
1. $H_2O_{(l)} \rightarrow H_{2(g)} + O_{2(g)}$
2. $Zn_{(s)} + Au^+_{(aq)} \rightarrow Zn^{2+}_{(aq)} + Ag_{(s)}$
2) Hydrochloric acid reacts with a solid chunk of aluminum to produce hydrogen gas and aluminum ions. Write the balanced chemical equation for this reaction.
3) Given a 10.1M stock solution, how many mL must be added to water to produce 200 mL of 5M solution?
4) If 0.502g of methane gas react with 0.27g of oxygen to produce carbon dioxide and water, what is the limiting reagent and how many moles of water are produced? The unbalanced equation is provided below.
$\ce{CH4(g) + O2(g) \rightarrow CO2(g) + H2O(l)} \nonumber$
5) A 0.777g sample of an organic compound is burned completely. It produces 1.42g CO2 and 0.388g H2O. Knowing that all the carbon and hydrogen atoms in CO2 and H2O came from the 0.777g sample, what is the empirical formula of the organic compound?
Contributors and Attributions
• Joseph Nijmeh (UCD), Mark Tye (DVC) | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Chemical_Reactions/Stoichiometry_and_Balancing_Reactions.txt |
An introduction to the complex ions formed by transition and other metals. Includes bonding, shapes and names, and a simple look at the origin of color. Looks in detail at various ligand exchange reactions and the chemistry of common hexaaqua ions.
• Acidity of the Hexaaqua Ions
This page explains why hexaaqua complex ions are acidic.
• Geometry of Complex Ions
This page describes the shapes of some common complex metal ions. It goes on to look at some simple examples of stereoisomerism (geometric and optical) in complex ions
• Ligand Exchange Reactions (Introduction)
This page describes some common ligand exchange (substitution) reactions involving complex metal ions. It assumes that you are familiar with basic ideas about complex ions. A ligand exchange reaction involves the substitution of one or more ligands in a complex ion with one or more different ligands.
• Ligand Exchange Reactions (Thermodynamics)
This page explains what is meant by a stability constant for a complex ion, and goes on to look at how its size is governed in part by the entropy change during a ligand exchange reaction.
• Nomenclature of Complex Ions
This page explains how to name some common complex metal ions. Although the names of complex ions can look long and worrying, the formulae are simply being coded in much the same way that organic names are coded. Once you have sorted out that code, the names are entirely descriptive.
• Origin of Color in Complex Ions
This page is going to take a simple look at the origin of color in complex ions - in particular, why so many transition metal ions are colored.
• Reactions of the Hexaaqua Ions with Ammonia
Reactions of the hexaaqua ions with ammonia solution are complicated by the fact that the ammonia can have two quite different functions. It can act as a base (in the Brønsted-Lowry sense), but it is also a possible ligand which can replace water molecules around the central metal ion. When it acts as a ligand, it is acting as a Lewis base. We need to look at these two functions separately.
• Reactions of the Hexaaqua Ions with Carbonate Ions
This page describes and explains the reactions between complex ions of the type [M(H₂O)₆]ⁿ⁺ and carbonate ions from, for example, sodium carbonate solution. There is a difference in the reactions depending on whether the metal at the center of the hexaaqua ion carries a 2+ or a 3+ charge. We need to look at each cases separately.
• Reactions of the Hexaaqua Ions with Hydroxide Ions
This page describes and explains the reactions between complex ions of the type [M(H2O)6]n+ and hydroxide ions from, for example, sodium hydroxide solution. It assumes that you know why these ions are acidic, and are happy about the equilibria involved.
• Stereoisomerism in complex ions
Some complex ions can show either optical or geometric isomerism.
Complex Ion Chemistry
This page explains why complex ions of the type [M(H2O)6]n+ are acidic.
The Charge Distribution within Complex Ion
The pH's of solutions containing hexaaqua ions vary a lot from one metal to another (assuming you are comparing solutions of equal concentrations). However, the underlying explanation is the same for all of them. We'll take the hexaaquairon(III) ion, [Fe(H2O)6]3+ as typical. The structure of the ion is:
Each of the six water molecules are attached to the central iron(III) ion via a co-ordinate bond using one of the lone pairs on the oxygen. We'll choose one of these water molecules at random (it doesn't make any difference which one!), and look at the bonding in a bit more detail - showing all the bonds around the oxygen. Imagine for the moment that the 3+ charge is located entirely on the iron.
When the lone pairs on the oxygens form co-ordinate bonds with the iron, there is obviously a movement of electrons towards the iron. That has an effect on the electrons in the O-H bonds. These electrons, in turn, get pulled towards the oxygen even more than usual. That leaves the hydrogen nuclei more exposed than normal. The overall effect is that each of the hydrogen atoms is more positive than it is in ordinary water molecules. The 3+ charge is no longer located entirely on the iron but spread out over the whole ion - much of it on the hydrogen atoms.
The effect of dissolving this ion in water
The hydrogen atoms attached to the water ligands are sufficiently positive that they can be pulled off in a reaction involving water molecules in the solution. The first stage of this process is:
$\ce{ Fe(H2O)6^{3+} + H2O <=> Fe(H2O)5(OH)^{2+} + H3O^{+}} \label{eq1}$
The complex ion is acting as an (Arrhenius) acid by donating a hydrogen ion to water molecules in the solution. The water is, of course, acting as a (Arrhenius) base by accepting the hydrogen ion. Because of the confusing presence of water from two different sources (the ligands and the solution), it is easier to simplify Equation \ref{eq1}:
$\ce{[Fe(H_2O)_6]^{3+} (aq) \rightleftharpoons [Fe(H2O)_5(OH)]^{2+} (aq) + H^{+} (aq)} \label{eq2}$
However, if you write it like this, remember that the hydrogen ion is not just falling off the complex ion. It is being pulled off by a water molecule in the solution. Whenever you write "$H^+(aq)$" what you really mean is a hydroxonium ion, H3O+.
The hexaaquairon(III) ion is quite strongly acidic giving solutions with pH's around 1.5, depending on concentration. You can get further loss of hydrogen ions as well, from a second and a third water molecule since the complex ion is a polyprotic acid.
Losing a second hydrogen ion:
$\ce{ [Fe(H2O)5(OH)]^{2+} (aq) \rightleftharpoons [Fe(H_2O)_4(OH)_2]^{+} (aq) + H^+ (aq)} \label{eq3}$
and a third one:
$\ce{ [ Fe(H_2O)_4(OH)_2]^{+} (aq) \rightleftharpoons [ Fe(H_2O)_3(OH)_3] (s) + H^+ (aq)} \label{eq4}$
This time you end up with a neutral complex and because it has no charge, it does not dissolve to any extent in water, and a precipitate is formed.
In practice
What do you actually get in solution if you dissolve an iron(III) salt in water? In fact you get a mixture of all the complexes that you have seen in the equations above. These reactions are all equilibria, so everything will be present. The proportions depend on how concentrated the solution is.
The color of the solution is very variable and depends in part on the concentration of the solution. Dilute solutions containing iron(III) ions can be pale yellow. More concentrated ones are much more orange, and may even produce some orange precipitate. None of these colors represents the true color of the [Fe(H2O)6]3+ ion - which is a very pale lilac color and is only really easy to see in solids containing the ion.
Looking at the equilibrium showing the loss of the first hydrogen ion:
$\underbrace{\ce{ Fe(H2O)6^{3+}}}_{\text{pale lilac}} + \ce{H2O} \ce{<=>} \underbrace{\ce{Fe(H2O)5(OH)^{2+}}}_{\text{orange}} + \ce{H3O^{+}} \nonumber$
The color of the new complex ion on the right-hand side is so strong that it completely masks the color of the hexaaqua ion. In concentrated solutions, the equilibrium position will be even further to the right-hand side (Le Chatelier's Principle), and so the color darkens. You will also get significant loss of other hydrogen ions (Equation \ref{eq3} and \ref{eq4}) leading to some formation of the neutral complex and thus some precipitate.
The position of this equilibrium can be shifted by adding extra hydrogen ions from a concentrated acid (e.g., by adding concentrated nitric acid to a solution of iron(III) nitrate). The new hydrogen ions push the position of the equilibrium to the left so that you can see the color of the hexaaqua ion. This is slightly easier to follow if you write the simplified version of the equilibrium (Equation \ref{eq2}).
3+ Ions are More Acidic than 2+ ions
Solutions containing 3+ hexaaqua ions tend to have pH's in the range from 1 to 3. Solutions containing 2+ ions have higher pH's - typically around 5 - 6, although they can go down to about 3. Remember that the reason that these ions are acidic is because of the pull of the electrons towards the positive central ion. An ion with 3+ charges on it is going to pull the electrons more strongly than one with only 2+ charges.
In 3+ ions, the electrons in the O-H bonds will be pulled further away from the hydrogens than in 2+ ions. That means that the hydrogen atoms in the ligand water molecules will have a greater positive charge in a 3+ ion, and so will be more attracted to water molecules in the solution. If they are more attracted, they will be more readily lost - and so the 3+ ions are more acidic.
The Effect of ionic radius on acidity
If you have ions of the same charge, it seems reasonable that the smaller the volume this charge is packed into, the greater the distorting effect on the electrons in the O-H bonds. Ions with the same charge but in a smaller volume (a higher charge density) would be expected to be more acidic. You would therefore expect to find that the smaller the radius of the metal ion, the stronger the acid. Unfortunately, it's not that simple!
Unfortunately, almost every data source that you refer to quotes different values for ionic radii. This is because the radius of a metal ion varies depending on what negative ion it is associated with. We are interested in what happens when the metal ion is bonded to water molecules, so haven't got simple ions at all! Whatever values we take are unlikely to represent the real situation. In the graphs which follow I have taken values for ionic radii from two common sources so that you can see what a lot of difference it makes to the argument.
The graphs
If it is true that the smaller the ionic radius, the stronger the acidity of the hexaaqua ion, you would expect some sort of regular increase in pKa (showing weaker acids) as ionic radius increases. The following graphs plot pKa against ionic radii for the 2+ ions of the elements in the first transition series from vanadium to copper. The first graph plots pKa against ionic radii taken from Chemistry Data Book by Stark and Wallace.
You can see that there is a trend for several of the ions, but it is completely broken by vanadium and chromium. The second graph uses ionic radii taken from the Nuffield Advanced Science Book of Data.
There probably is a relationship between ionic radius and acid strength, but that it is nothing like as simple and straightforward as most books at this level pretend. The problem is that there are other more important effects operating as well (quite apart from differences in charge) that can completely swamp the effect of the changes in ionic radius. You have to look in far more detail at the bonding in the hexaaqua ions and the product ions. | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Coordination_Chemistry/Complex_Ion_Chemistry/Acidity_of_the_Hexaaqua_Ions.txt |
This page describes the shapes of some common complex metal ions. These shapes are for complex ions formed using monodentate ligands - ligands which only form one bond to the central metal ion. You will probably be familiar with working out the shapes of simple compounds using the electron pair repulsion theory. Unfortunately that does not work for most complex metal ions involving transition metals.
6-coordinated Complex Ions
These are complex ions in which the central metal ion is forming six bonds. In the simple cases we are talking about, that means that it will be attached to six ligands. These ions have an octahedral shape. Four of the ligands are in one plane, with the fifth one above the plane, and the sixth one below the plane. The diagram shows four fairly random examples of octahedral ions.
It does not matter the nature of the ligands. If you have six of them, this is the shape they will take up. Easy!
4-coordinated Complex Ions
These are far less common, and they can take up one of two different shapes.
Tetrahedral ions
There are two very similar ions which crop up commonly at this level: [CuCl4]2- and [CoCl4]2-. The copper(II) and cobalt(II) ions have four chloride ions bonded to them rather than six, because the chloride ions are too big to fit any more around the central metal ion.
A square planar complex
Occasionally a 4-coordinated complex turns out to be square planar. There is no easy way of predicting that this is going to happen. The only one you might possibly come across at this level is cisplatin which is used as an anti-cancer drug. Cisplatin is a neutral complex, Pt(NH3)2Cl2. It is neutral because the 2+ charge of the original platinum(II) ion is exactly canceled by the two negative charges supplied by the chloride ions.
The platinum, the two chlorines, and the two nitrogens are all in the same plane. We will have more to say about cisplatin immediately below.
Ligand Exchange Reactions (Introduction)
This page describes some common ligand exchange (substitution) reactions involving complex metal ions. It assumes that you are familiar with basic ideas about complex ions. A ligand exchange reaction involves the substitution of one or more ligands in a complex ion with one or more different ligands.
Replacing aqua with chloro ligands
If you add concentrated hydrochloric acid to a solution containing hexaaquacobalt(II) ions (for example, cobalt(II) chloride solution), the solution turns from its original pink color to a dark rich blue. The six aqua molecules are replaced by four chloro ions.
The reaction taking place is reversible.
$\ce{[Co(H2O)6]^{2+} + 4Cl^{-} <=> [CoCl4]^{2-} + 6H2O} \nonumber$
Concentrated hydrochloric acid is used as the source of chloride ions because it provides a very high concentration compared to what is possible with, say, sodium chloride solution. Concentrated hydrochloric acid has a chloride ion concentration of approximately 10 mol dm-3. The high chloride ion concentration pushes the position of the equilibrium to the right according to Le Chatelier's Principle.
Notice the change in the co-ordination of the cobalt. Chloride ions are bigger than water molecules, and there isn't room to fit six of them around the central cobalt ion. This reaction can be easily reversed by adding water to the solution. Adding water to the right-hand side of the equilibrium has the effect of moving the position of equilibrium to the left. The pink color of the hexaaquacobalt(II) ion is produced again (only paler, of course, because it is more dilute).
Replacing the water in the hexaaquacopper(II) ion
In terms of the chemistry, this is exactly the same as the last example - all that differs are the colors. Unfortunately, these are not quite so straightforward. The color of the tetrachlorocuprate(II) ion is almost always seen mixed with that of the original hexaaqua ion. What you normally see is:
The reaction taking place is reversible, and you get a mixture of colors due to both of the complex ions.
$\ce{[Cu(H2O)6]^{2+} <=> [CuCl_4]^{2-} + 6H2O}\nonumber$
You may find the color of the tetrachlorocuprate(II) ion variously described as olive-green or yellow. Adding water to the green solution, replaces the chloro ligands by water molecules again, and returns the solution to blue.
Replacing water molecules by ammonia
Water molecules and ammonia molecules are very similar in size, and so there is no change in co-ordination this time. Unfortunately, the reactions aren't quite so straightforward to describe. Ammonia solution can react with hexaaqua metal ions in two quite distinct ways, because it can act as a base as well as a ligand.
If you add a small amount of ammonia solution you get precipitates of the metal hydroxide - the ammonia is acting as a base. In some cases, these precipitates redissolve when you add more ammonia to give solutions in which a ligand exchange reaction has occurred. In the diagrams below, both steps are shown, but we are only going to consider the chemistry of the overall ligand exchange reaction. The precipitates dissolve because of a complicated series of equilibrium shifts, and we shan't worry about that for the moment.
Replacing the water in the hexaaquacopper(II) ion
This is a slightly untypical case, because only four of the six water molecules get replaced to give the tetraamminediaquacopper(II) ion, [Cu(NH3)4(H2O)2]2+.
Notice that the four ammonias all lie in one plane, with the water molecules above and below. What you see in a test tube is:
The main equilibrium involved in the ligand exchange reaction is:
$\ce{[Cu(H2O)6]^{2+} + 4NH3 <=> [Cu(NH3)4(H2O)2]^{2+} + 4H2O}\nonumber$
The color of the deep blue complex is so strong that this reaction is used as a sensitive test for copper(II) ions in solution. Even if you try to reverse the change by adding large amounts of water to the equilibrium, the strength of the deep blue (even highly diluted) always masks the pale blue of the aqua ion.
Replacing the water in the hexaaquacobalt(II) ion
This time, all the water molecules get replaced.
$\ce{[Co(H2O)6]^{2+} + 6NH3 <=> [Co(NH3)6]^{2+} + 6H2O}\nonumber$
The straw colored solution formed changes color very rapidly on standing to a deep reddish brown. The hexaamminecobalt(II) ions are oxidised by the air to hexaamminecobalt(III) ions. However, that is a quite separate reaction, and is not a part of the ligand exchange reaction.
Replacing the water in the hexaaquachromium(III) ion
Again, all the agua ligands are replaced by ammine ligands. The difference this time is that the reaction is incomplete. The precipitate has to be left to stand in the presence of excess concentrated ammonia solution for some time to get the fully substituted ammine complex. Even so, you still get left with some unreacted precipitate.
$\ce{ [Cr(H2O)6 ]^{3+} + 6NH3 <=> [Cr(NH3)6]^{3+} + 6H2O}\nonumber$
Two more replacements of the water in the hexaaquachromium(III) ion
The color of the hexaaquachromium(III) ion has been shown as a "difficult to describe" violet-blue-grey in all the diagrams above. In practice, when it is produced during a reaction in a test tube, it is often green. A typical example of this is the use of acidified potassium dichromate(VI) as an oxidising agent. Whenever this is used, the orange solution turns green and we nearly always describe the green ion as being Cr3+(aq) - implying the hexaaquachromium(III) ion. That's actually an over-simplification. What happens is that one or more of the ligand water molecules get replaced by a negative ion in the solution - typically sulfate or chloride.
Replacement of the water by sulfate ions
You can do this simply by warming some chromium(III) sulfate solution.
One of the water molecules is replaced by a sulfate ion. Notice the change in the charge on the ion. Two of the positive charges are canceled by the presence of the two negative charges on the sulfate ion.
Replacement of the water by chloride ions
In the presence of chloride ions (for example with chromium(III) chloride), the most commonly observed color is green. This happens when two of the water molecules are replaced by chloride ions to give the tetraaquadichlorochromium(III) ion - [Cr(H2O)4Cl2]+. Once again, notice that replacing water molecules by chloride ions changes the charge on the ion.
A ligand exchange reaction in the test for iron(III) ions
This provides an extremely sensitive test for iron(III) ions in solution. If you add thiocyanate ions, SCN-, (from, say, sodium or potassium or ammonium thiocyanate solution) to a solution containing iron(III) ions, you get an intense blood red solution containing the ion $\ce{[Fe(SCN)(H2O)5]^{2+}}$. | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Coordination_Chemistry/Complex_Ion_Chemistry/Geometry_of_Complex_Ions.txt |
This page explains what is meant by a stability constant for a complex ion, and goes on to look at how its size is governed in part by the entropy change during a ligand exchange reaction.
Replacing aqua ligands with ammine ligands
If you add ammonia solution to a solution containing hexaaquacopper(II) ions, [Cu(H2O)6]2+, four of the water molecules are eventually replaced by ammonia molecules to give [Cu(NH3)4(H2O)2]2+. This can be written as an equilibrium reaction to show the overall effect:
$\ce{[Cu(H2O)6]^{2+} + 4NH3 <=> [Cu(NH3)4(H2O)2]^{2+} + 4H2O} \label{all}$
In fact, the water molecules get replaced one at a time, and so this is made up of a series of part-reactions:
$\ce{[Cu(H2O)6]^{2+} + NH3 <=> [Cu(NH3)(H2O)5]^{2+} + H2O}\label{step1}$
$\ce{ [Cu(NH3)(H2O)5]^{2+} + NH3 <=> [Cu(NH3)2(H2O)4]^{2+} + H2O} \label{step2}$
$\ce{[Cu(NH3)2(H2O)4]^{2+} + NH3 <=> [Cu(NH3)3(H2O)3]^{2+} + H2O} \label{step3}$
$\ce{[Cu(NH3)3(H2O)3]^{2+} + NH3 <=> [Cu(NH3)4(H2O)2]^{2+} + H2O} \label{step4}$
Although this can look a bit daunting at first sight, all that is happening is that first you have one, then two, then three, then four water molecules in total replaced by four ammine ligands.
Individual stability constants
Let's take a closer look at the first of these equilibria (Equation \ref{step1}). Like any other equilibrium, this one has an equilibrium constant, $K_c$ - except that in this case, we call it a stability constant. Because this is the first water molecule to be replaced, we call it $K_1$ and is given by this expression:
$K_1 = \dfrac{[\ce{[Cu(NH3)(H2O)5]^{2+}}]}{\ce{[[Cu(H2O)6]^{2+}}][\ce{NH3]}} \nonumber$
There are two points of possible confusion here - one minor, one more important!
• First, the square brackets have changed their meaning! Square brackets are often used to keep everything in a complex ion together and tidy. Here, they have reverted to their other use, which implies concentrations in mol dm-3.
• In order to avoid complete confusion, the square brackets keeping the complexes together have been removed entirely.
• More importantly, if you compare the equilibrium constant expression with the equation, you will see that the water on the right-hand side hasn't been included. That is normal practice with these expressions.
• Because everything is dissolved in water, the water is present as a huge excess. Generating a little bit more during the reaction is going to make no effective difference to the total concentration of the water in terms of moles of water per dm3.
• The concentration of the water is approximately constant. The equilibrium constant is defined so that you avoid having an extra unnecessary constant in the expression.
With that out of the way, let's go back to where we were - but introduce a value for $K_1$:
$\ce{ [Cu(H2O)6]^{2+} + NH3 <=> [Cu(NH3)(H2O)5]^{2+} + H2O} \nonumber$
The value of the equilibrium constant is fairly large, suggesting that there is a strong tendency to form the ion containing an ammonia molecule. A high value of a stability constant shows that the ion is easily formed. Each of the other equilibria above also has its own stability constant, K2, K3 and K4. For example, K2 is given by:
$\ce{ [Cu(NH3)(H2O)5]^{2+} + NH3 <=> [Cu(NH3)2(H2O)4]^{2+} + H2O} \nonumber$
The ion with two ammonias is even more stable than the ion with one ammonia. You could keep plugging away at this and come up with the following table of stability constants:
ion Kn value (mol-1 dm3) log Kn
[Cu(NH3)(H2O)5]2+ K1 1.78 x 104 4.25
[Cu(NH3)2(H2O)4]2+ K2 4.07 x 103 3.61
[Cu(NH3)3(H2O)3]2+ K3 9.55 x 102 2.98
[Cu(NH3)4(H2O)2]2+ K4 1.74 x 102 2.24
You will often find these values quoted as log K1. All this does is tidy the numbers up so that you can see the patterns more easily. The ions keep on getting more stable as you replace up to 4 water molecules, but notice that the equilibrium constants are gradually getting less big as you replace more and more waters. This is common with individual stability constants.
Overall Formation constants
The overall stability constant is simply the equilibrium constant for the total reaction:
$\ce{ [Cu(H2O)6]^{2+} + 4NH3 <=> [Cu(NH3)4(H2O)2]^{2+} + 4H2O} \nonumber$
It is given by this expression:
You can see that overall this is a very large equilibrium constant, implying a high tendency for the ammonias to replace the waters. The "log" value is 13.1. This overall value is found by multiplying together all the individual values of K1, K2 and so on. To find out why that works, you will need a big bit of paper and some patience! Write down expressions for all the individual values (the first two are done for you above), and then multiply those expressions together. You will find that all the terms for the intermediate ions cancel out to leave you with the expression for the overall stability constant.
Summary
Whether you are looking at the replacement of individual water molecules or an overall reaction producing the final complex ion, a stability constant is simply the equilibrium constant for the reaction you are looking at. The larger the value of the stability constant, the further the reaction lies to the right. That implies that complex ions with large stability constants are more stable than ones with smaller ones.
Stability constants tend to be very large numbers. In order to simplify the numbers a "log" scale is often used. Because of the way this works, a difference of 1 in the log value comes from a 10 times difference in the stability constant. A difference of 2 comes from a 100 (in other words, 102) times difference in stability constant - and so on.
Stability constants and entropy - the chelate effect
The chelate effect is an effect which happens when you replace water (or other simple ligands) around the central metal ion by multidentate ligands like 1,2-diaminoethane (often abbreviated to "en") or EDTA. Compare what happens if you replace two water molecules around a [Cu(H2O)6]2+ ion with either 2 ammonia molecules or one molecule of 1,2-diaminoethane.
This second structure is known as a chelate from a Greek word meaning a crab's claw. You can picture the copper ion as being nipped by the claw of the 1,2-diaminoethane molecule. Chelates are much more stable than complex ions formed from simple monodentate ligands. The overall stability constants for the two ions are:
ion log K
[Cu(NH3)2(H2O)4]2+ 7.86
[Cu(H2O)4(en)]2+ 10.6
The reaction with the 1,2-diaminoethane could eventually go on to produce a complex ion [Cu(en)3]2+. Simplifying the structure of this:
The overall stability constant for this (as log K) is 18.7. Another copper-based chelate comes from the reaction with EDTA.
This also has a high stability constant - log K is 18.8. However many examples you take, you always find that a chelate (a complex ion involving multidentate ligands) is more stable than ions with only monodentate ligands. This is known as the chelate effect.
The reason for the chelate effect
If you compare the two equilibria below, the one with the 1,2-diaminoethane ("en") has the higher equilibrium (stability) constant (for values, see above).
The enthalpy changes of the two reactions are fairly similar. You might expect this because in each case you are breaking two bonds between copper and oxygen atoms and replacing them by two bonds between copper and nitrogen atoms.
If the enthalpy changes are similar, what causes the difference in the extent to which the two reactions happen?
You need to think about the entropy change during each reaction.
Entropy is most easily thought of as a measure of disorder. Any change which increases the amount of disorder increases the tendency of a reaction to happen.
If you look again at the two equiilbria, you might notice that the 1,2-diaminoethane equilibrium does lead to an increase in the entropy. There are only two species on the left-hand side of the equation, but three on the right.
You can obviously get more entropy out of three species than out of only two. Compare that with the other equilibrium. In this case, there is no change in the total number of species before and after reaction, and so no useful contribution to an increase in entropy.
In the case of the complex with EDTA, the increase in entropy is very pronounced.
Here, we are increasing the number of species present from two on the left-hand side to seven on the right. You can get a major amount of increase in disorder by making this change.
Reversing this last change is going to be far more difficult in entropy terms. You would have to move from a highly disordered state to a much more ordered one. That isn't so likely to happen, and so the copper-EDTA complex is very stable.
Summary
Complexes involving multidentate ligands are more stable than those with only monodentate ligands. The underlying reason for this is that each multidentate ligand displaces more than one water molecule. This leads to an increase in the number of species present in the system, and therefore an increase in entropy. An increase in entropy makes the formation of the chelated complex more favorable. | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Coordination_Chemistry/Complex_Ion_Chemistry/Ligand_Exchange_Reactions_%28Thermodynamics%29.txt |
This page explains how to name some common complex metal ions. Although the names of complex ions can look long and worrying, the formulae are simply being coded in much the same way that organic names are coded. Once you have sorted out that code, the names are entirely descriptive.
Naming the ligands
The table shows some common ligands and the code for them in the name of a complex ion. The old names sometimes differ by a letter or so, but never enough for it to be confusing.
ligand coded by (old name)
H2O aqua aquo
NH3 ammine ammino
OH- hydroxo hydroxy
Cl- chloro
F- fluoro
CN- cyano
Take care with the code for ammonia as a ligand - it has 2 "m"s in its name. If you miss one of these out so that you are left with "amine" or "amino", you are refering to the NH2 group in an organic compound. This is probably the only point of confusion with these names.
Coding for the number of ligands
The normal prefixes apply if there is more than one ligand.
no of ligands coded by
2 di
3 tri
4 tetra
5 penta
6 hexa
Putting this together
For a complex ion containing only one type of ligand, there is no problem. For example:
[Cu(H2O)6]2+ is called the hexaaquacopper(II) ion.
(Don't worry about the copper(II) bit for the moment.) The fact that there are two "a"s next to each other in the name is OK.
With more than one type of ligand in an ion, the ligands are named in alphabetical order - ignoring the prefixes. For example:
[Cu(NH3)4(H2O)2]2+ is called the tetraamminediaquacopper(II) ion.
The "ammine" is named before the "aqua" because "am" comes before "aq" in the alphabet. The "tetra" and "di" are ignored.
Naming the metal
You might have thought that this was fairly obvious, but it isn't necessarily. It depends on whether the complex ion ends up as positively or negatively charged.
For positively charged complex ions
A positively charged complex ion is called a cationic complex. A cation is a positively charged ion. The metal in this is named exactly as you would expect, with the addition of its oxidation state. Going back to a previous example, [Cu(H2O)6]2+ is called the hexaaquacopper(II) ion because the copper's oxidation state is +2.
Copper's oxidation is +2 because the original uncomplexed ion was Cu2+ - NOT because the complex carries 2+ charges.
The oxidation state is frequently left out if a metal only ever has one oxidation state. For example, in its compounds aluminium always has an oxidation state of +3. [Al(H2O)6]3+ is usually just called the hexaaquaaluminium ion rather than the hexaaquaaluminium(III) ion.
For negatively charged complex ions
A negatively charged complex ion is called an anionic complex. An anion is a negatively charged ion. In this case the name of the metal is modified to show that it has ended up in a negative ion. This is shown by the ending -ate.
With many metals, the basic name of the metal is changed as well - sometimes drastically! Common examples include:
metal changed to
cobalt cobaltate
aluminium aluminate
chromium chromate
vanadium vanadate
copper cuprate
iron ferrate
So, for example, suppose you bond 4 chloride ions around a Cu2+ ion to give [CuCl4]2-.
The name shows the 4 (tetra) chlorines (chloro) around a copper in an overall negative ion (cuprate). The copper has on oxidation state of +2. This is the tetrachlorocuprate(II) ion.
Similarly, [Al(H2O)2(OH)4]- is called the diaquatetrahydroxoaluminate ion. Take the name to pieces so that you can see exactly what refers to what. Don't forget that the two different ligands are named in alphabetical order - aqua before hydroxo - ignoring the prefixes, di and tetra. The oxidation state of the aluminum could be shown, but is not absolutely necessary because aluminum only has the one oxidation state in its compounds. The full name is the diaquatetrahydroxoaluminate(III) ion.
Contributors and Attributions
Jim Clark (Chemguide.co.uk) | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Coordination_Chemistry/Complex_Ion_Chemistry/Nomenclature_of_Complex_Ions.txt |
This page is going to take a simple look at the origin of color in complex ions - in particular, why so many transition metal ions are colored.
White light and Colors
If you pass white light through a prism it splits into all the colors of the rainbow. Visible light is simply a small part of an electromagnetic spectrum most of which we cannot see - gamma rays, X-rays, infra-red, radio waves and so on. Each of these has a particular wavelength, ranging from 10-16 meters for gamma rays to several hundred metersfor radio waves. Visible light has wavelengths from about 400 to 750 nm. (1 nanometer = 10-9 meters.)
Example 1: Blue Color of Copper (II) Sulfate in Solution
If white light (ordinary sunlight, for example) passes through copper(II) sulfate solution, some wavelengths in the light are absorbed by the solution. Copper(II) ions in solution absorb light in the red region of the spectrum. The light which passes through the solution and out the other side will have all the colors in it except for the red. We see this mixture of wavelengths as pale blue (cyan). The diagram gives an impression of what happens if you pass white light through copper(II) sulfate solution.
Working out what color you will see is not easy if you try to do it by imagining "mixing up" the remaining colors. You wouldn't have thought that all the other colors apart from some red would look cyan, for example. Sometimes what you actually see is quite unexpected. Mixing different wavelengths of light doesn't give you the same result as mixing paints or other pigments. You can, however, sometimes get some estimate of the color you would see using the idea of complementary colors.
If you arrange some colors in a circle, you get a "color wheel". The diagram shows one possible version of this.
colors directly opposite each other on the color wheel are said to be complementary colors. Blue and yellow are complementary colors; red and cyan are complementary; and so are green and magenta. Mixing together two complementary colors of light will give you white light. What this all means is that if a particular color is absorbed from white light, what your eye detects by mixing up all the other wavelengths of light is its complementary color. Copper(II) sulfate solution is pale blue (cyan) because it absorbs light in the red region of the spectrum. Cyan is the complementary color of red.
We often casually talk about the transition metals as being those in the middle of the Periodic Table where d orbitals are being filled, but these should really be called d block elements rather than transition elements (or metals). This shortened version of the Periodic Table shows the first row of the d block, where the 3d orbitals are being filled.
The usual definition of a transition metal is one which forms one or more stable ions which have incompletely filled d orbitals. Zinc with the electronic structure [Ar] 3d104s2 does not count as a transition metal whichever definition you use. In the metal, it has a full 3d level. When it forms an ion, the 4s electrons are lost - again leaving a completely full 3d level. At the other end of the row, scandium ( [Ar] 3d14s2 ) does not really counts as a transition metal either. Although there is a partially filled d level in the metal, when it forms its ion, it loses all three outer electrons. Technically, the Sc3+ ion does not count as a transition metal ion because its 3d level is empty.
Example 2: Hexaaqua Metal Ions
The diagrams show the approximate colors of some typical hexaaqua metal ions, with the formula [ M(H2O)6 ] n+. The charge on these ions is typically 2+ or 3+.
Non-transition metal ions
These ions are all colorless.
Transition metal ions
The corresponding transition metal ions are colored. Some, like the hexaaquamanganese(II) ion (not shown) and the hexaaquairon(II) ion, are quite faintly colored - but they are colored.
So, what causes transition metal ions to absorb wavelengths from visible light (causing color) whereas non-transition metal ions do not? And why does the color vary so much from ion to ion?
The Origin of Color in Complex Ions containing transition metals
Complex ions containing transition metals are usually colored, whereas the similar ions from non-transition metals are not. That suggests that the partly filled d orbitals must be involved in generating the color in some way. Remember that transition metals are defined as having partly filled d orbitals.
Octahedral Complexes
For simplicity we are going to look at the octahedral complexes which have six simple ligands arranged around the central metal ion. The argument is not really any different if you have multidentate ligands. When the ligands bond with the transition metal ion, there is repulsion between the electrons in the ligands and the electrons in the d orbitals of the metal ion. That raises the energy of the d orbitals. However, because of the way the d orbitals are arranged in space, it doesn't raise all their energies by the same amount. Instead, it splits them into two groups. The diagram shows the arrangement of the d electrons in a Cu2+ ion before and after six water molecules bond with it.
Whenever 6 ligands are arranged around a transition metal ion, the d orbitals are always split into 2 groups in this way - 2 with a higher energy than the other 3. The size of the energy gap between them (shown by the blue arrows on the diagram) varies with the nature of the transition metal ion, its oxidation state (whether it is 3+ or 2+, for example), and the nature of the ligands. When white light is passed through a solution of this ion, some of the energy in the light is used to promote an electron from the lower set of orbitals into a space in the upper set.
Each wavelength of light has a particular energy associated with it. Red light has the lowest energy in the visible region. Violet light has the greatest energy. Suppose that the energy gap in the d orbitals of the complex ion corresponded to the energy of yellow light.
The yellow light would be absorbed because its energy would be used in promoting the electron. That leaves the other colors. Your eye would see the light passing through as a dark blue, because blue is the complementary color of yellow.
Tetrahedral Complexes
Simple tetrahedral complexes have four ligands arranged around the central metal ion. Again the ligands have an effect on the energy of the d electrons in the metal ion. This time, of course, the ligands are arranged differently in space relative to the shapes of the d orbitals. The net effect is that when the d orbitals split into two groups, three of them have a greater energy, and the other two a lesser energy (the opposite of the arrangement in an octahedral complex). Apart from this difference of detail, the explanation for the origin of color in terms of the absorption of particular wavelengths of light is exactly the same as for octahedral complexes.
What about non-transition metal complex ions?
Non-transition metals do not have partly filled d orbitals. Visible light is only absorbed if some energy from the light is used to promote an electron over exactly the right energy gap. Non-transition metals do not have any electron transitions which can absorb wavelengths from visible light. For example, although scandium is a member of the d block, its ion (Sc3+) hasn't got any d electrons left to move around. This is no different from an ion based on Mg2+ or Al3+. Scandium(III) complexes are colorless because no visible light is absorbed. In the zinc case, the 3d level is completely full - there are not any gaps to promote an electron in to. Zinc complexes are also colorless.
Factors Affecting the Color of Transition Metal complexes
In each case we are going to choose a particular metal ion for the center of the complex, and change other factors. color changes in a fairly haphazard way from metal to metal across a transition series.
The nature of the ligand
Different ligands have different effects on the energies of the d orbitals of the central ion. Some ligands have strong electrical fields which cause a large energy gap when the d orbitals split into two groups. Others have much weaker fields producing much smaller gaps. Remember that the size of the gap determines what wavelength of light is going to get absorbed. The list shows some common ligands. Those at the top produce the smallest splitting; those at the bottom the largest splitting.
The greater the splitting, the more energy is needed to promote an electron from the lower group of orbitals to the higher ones. In terms of the color of the light absorbed, greater energy corresponds to shorter wavelengths. That means that as the splitting increases, the light absorbed will tend to shift away from the red end of the spectrum towards orange, yellow and so on.
There is a fairly clear-cut case in copper(II) chemistry. If you add an excess of ammonia solution to hexaaquacopper(II) ions in solution, the pale blue (cyan) color is replaced by a dark inky blue as some of the water molecules in the complex ion are replaced by ammonia.
The first complex must be absorbing red light in order to give the complementary color cyan. The second one must be absorbing in the yellow region in order to give the complementary color dark blue. Yellow light has a higher energy than red light. You need that higher energy because ammonia causes more splitting of the d orbitals than water does.
It is not often as simple to see as this, though! Trying to sort out what is being absorbed when you have murky colors not on the simple color wheel further up the page is much more of a problem. The diagrams show some approximate colors of some ions based on chromium(III).
It is obvious that changing the ligand is changing the color, but trying to explain the colors in terms of our simple theory is not easy.
The oxidation state of the metal
As the oxidation state of the metal increases, so also does the amount of splitting of the d orbitals. Changes of oxidation state therefore change the color of the light absorbed, and so the color of the light you see. Taking another example from chromium chemistry involving only a change of oxidation state (from +2 to +3):
The 2+ ion is almost the same color as the hexaaquacopper(II) ion, and the 3+ ion is the hard-to-describe violet-blue-grey color.
The coordination of the Ion
Splitting is greater if the ion is octahedral than if it is tetrahedral, and therefore the color will change with a change of co-ordination. The problem is that an ion will normally only change co-ordination if you change the ligand - and changing the ligand will change the color as well. Hence, you cannot isolate out the effect of the co-ordination change. For example, a commonly quoted case comes from cobalt(II) chemistry, with the ions [Co(H2O)6]2+ and [CoCl4]2-.
The difference in the colors is going to be a combination of the effect of the change of ligand, and the change of the number of ligands. | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Coordination_Chemistry/Complex_Ion_Chemistry/Origin_of_Color_in_Complex_Ions.txt |
This page describes and explains the reactions between complex ions of the type [M(H2O)6]n+ and ammonia solution.Reactions of the hexaaqua ions with ammonia solution are complicated by the fact that the ammonia can have two quite different functions. It can act as a base (in the Brønsted-Lowry sense), but it is also a possible ligand which can replace water molecules around the central metal ion. When it acts as a ligand, it is acting as a Lewis base. We need to look at these two functions separately.
Ammonia acting as a (Brønsted-Lowry) base
This is what happens when you only add small amounts of dilute ammonia solution to any of the hexaaqua ions. The ligand effect only happens with an excess of ammonia or with concentrated ammonia - and with some metals you don't even see it then.
We'll talk through what happens if you add a small amount of dilute ammonia solution to a solution containing a 2+ hexaaqua ion. These have the formula [M(H2O)6]2+, and they are acidic. Their acidity is shown in the reaction of the hexaaqua ions with water molecules from the solution:
$\ce{ [M(H2O)6]^{2+} (aq) + H2O <=> [M(H2O)5(OH)]^{+} (aq) + H3O^{+} (aq) } \nonumber$
They are acting as acids by donating hydrogen ions to water molecules in the solution. Because of the confusing presence of water from two different sources (the ligands and the solution), it is easier to simplify this:
$\ce{ [M(H2O)6]^{2+} (aq) <=> [M(H2O)5(OH)]^{+} (aq) + H^{+} (aq) } \nonumber$
Adding ammonia solution to this equilibrium - stage 1
There are two possible reactions.
1. Reaction of ammonia with the hydroxonium ions (hydrogen ions)
2. Ammonia will react with these to produce ammonium ions.
According to Le Chatelier's Principle, the position of equilibrium will move to the right, producing more of the new complex ion.
Reaction of ammonia with the hexaaqua ion
Statistically, there is far more chance of an ammonia molecule hitting a hexaaqua metal ion than of hitting a hydrogen ion. There are far more hexaaqua ions present. If that happens, you get exactly the same new complex ion formed as above.
$\ce{ [M(H2O)6]^{2+} (aq) + NH3 <=> [M(H2O)5(OH)]^{+} (aq) + NH4^{+} (aq) } \nonumber$
Notice that this is still a reversible change (unlike the corresponding change when you add hydroxide ions). Ammonia is only a weak base.
The second stage of the reaction
Whichever of the above reactions happens, you end up with [M(H2O)5(OH)]+ ions in solution. These are also acidic, and can lose hydrogen ions from another of the water ligands. Taking the easier version of the equilibrium:
$\ce{ [M(H2O)5(OH)]^{+} (aq) <=> [M(H2O)4(OH)2] (s) + H^{+} (aq) } \nonumber$
Adding ammonia again tips the equilibrium to the right - either by reacting with the hydrogen ions, or by reacting directly with the complex on the left-hand side. When this happens, the new complex formed no longer has a charge - this is a "neutral complex". It is insoluble in water - and so a precipitate is formed. This precipitate is often written without including the remaining water ligands. In other words we write it as M(OH)2. A precipitate of the metal hydroxide has been formed.
Summarizing what has happened
You can also usefully write the complete change as an overall equilibrium reaction. This will be important for later on.
$\ce{ [M(H2O)6]^{2+} (aq) + 2NH3 <=> [M(H2O)4(OH)2] (s) + 2NH4^{+} (aq) } \nonumber$
If you did the same reaction with a 3+ ion, the only difference is that you would have to remove a total of 3 hydrogen ions in order to get to the neutral complex. That would give the overall equation:
$\ce{ [M(H2O)6]^{3+} (aq) + 3NH3 <=> [M(H2O)3(OH)3] (s) + 4NH4^{+} (aq) } \nonumber$
Ions of specific metals
Remember that we are concentrating for the moment on the ammonia acting as a base - in other words, on the formation of hydroxide precipitates when you add small amounts of ammonia solution to solutions containing hexaaqua metal The diagrams, however, will show the complete change so I don't have to repeat them later on. Ignore the cases where the precipitate dissolves in excess ammonia for the moment.
2+ ions
hexaaquairon(II)
Iron is very easily oxidized under alkaline conditions. Oxygen in the air oxidizes the iron(II) hydroxide precipitate to iron(III) hydroxide especially around the top of the tube. The darkening of the precipitate comes from the same effect. This is NOT a ligand exchange reaction.
hexaaquamanganese(II)
I have shown the original solution as very pale pink (the palest I can produce!), but in fact it is virtually colorless. The pale brown precipitate is oxidized to darker brown manganese(III) oxide in contact with oxygen from the air. Again, this isn't a ligand exchange reaction.
hexaaquazinc
You start and finish with colourless solutions, producing a white precipitate on the way.
hexaaquaaluminium
Starting from a colorless solution, you get a white precipitate.
Summary of the effect of adding small amounts of ammonia solution
In each case you get a precipitate of the neutral complex - the metal hydroxide. Apart from minor differences in the exact shade of color you get, these are almost all exactly the same as the precipitates you get when you add a little sodium hydroxide solution to the solutions of the hexaaqua ions. The only real difference lies in the color of the cobalt precipitate.
Ammonia acting as a ligand
The ligand exchange reaction
In some cases, ammonia replaces water around the central metal ion to give another soluble complex. This is known as a ligand exchange reaction, and involves an equilibrium such as this one:
$[Cu(H_2O)_6]^{2+} +4NH_3 \rightleftharpoons [Cu(NH_3)_4(H_2O)_2]^{2+} + 4H_2O \nonumber$
The formation of this new soluble complex causes the precipitate to dissolve. The ammonia attaches to the central metal ion using the lone pair of electrons on the nitrogen atom. Because it is a lone pair donor, it is acting as a Lewis base.
Explaining why the precipitate dissolves
Almost all text books leave the argument at this point, assuming that it is obvious why the formation of the complex causes some precipitates to dissolve - it is not! Consider the copper case as typical of any of them. There are two equilibria involved in this. The first is the one in which ammonia is acting as a base and producing the precipitate:
$[Cu(H_2O)_6]^{2+} + 2NH_3 \rightleftharpoons [Cu(H_2O)_4(OH)_2] + 2NH_4^+ \nonumber$
The other one is the ligand exchange reaction:
$[Cu(H_2O)_6]^{2+} + 4NH_3 \rightleftharpoons [Cu(NH_3)_4(H_2O)_2]^{2+} + 4H_2O \nonumber$
Notice that the hexaaqua ion appears in both of these. There is now an interaction between the two equilibria:
Looking at it like this is helpful in explaining why some precipitates dissolve in excess ammonia while others don't. It depends on the positions of the equilibria. To get the precipitate to dissolve, you obviously need the ligand exchange equilibrium to lie well to the right, but you need the acid-base equilibrium to be easy to pull back to the left. | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Coordination_Chemistry/Complex_Ion_Chemistry/Reactions_of_the_Hexaaqua_Ions_with_Ammonia.txt |
This page describes and explains the reactions between complex ions of the type $\ce{[M(H2O)6]^{n+}}$ and carbonate ions from, for example, sodium carbonate solution. There is a difference in the reactions depending on whether the metal at the center of the hexaaqua ion carries a 2+ or a 3+ charge. We need to look at each cases separately.
3+ ions reacting with carbonate ions
The 3+ hexaaqua ions are sufficiently acidic to react with carbonate ions to release carbon dioxide gas and precipitate a metal hydroxide.
The acidity of the 3+ hexaaqua ions
These have the formula $\ce{[M(H2O)6]^{3+}}$, and they are fairly acidic. They react with water molecules from the solution:
$\ce{ [M(H2O)6]^{3+} (aq) + H2O <=> [M(H2O)6]^{2+} (aq) + H3O^{+} (aq) } \nonumber$
They are acting as acids by donating hydrogen ions to water molecules in the solution. Because of the confusing presence of water from two different sources (the ligands and the solution), it is easier to simplify this:
$\ce{ [M(H2O)6]^{3+} (aq) <=> [M(H2O)6]^{2+} (aq) + H^{+} (aq)} \nonumber$
Carbonate ions acting as a base
Carbonate ions combine with hydrogen ions in two stages - first to make hydrogencarbonate ions, and then to give carbon dioxide and water.
$\ce{CO3^{2-} (aq) + H^{+} (aq) <=> HCO3^{-} + H2O (l)} \nonumber$
$\ce{HCO3^{-} (aq) + H^{+} (aq) <=> CO2 (g) + H2O (l)} \nonumber$
Reacting carbonate ions with the 3+ hexaaqua ions
Provided the proportions are right, the 3+ hexaaqua ions are sufficiently acidic for the reactions to go all the way to carbon dioxide. There are two possible reactions.
1. Reaction of carbonate ions with the hydroxonium ions (hydrogen ions). According to Le Chatelier's Principle, as the hydrogen ions are removed, the position of equilibrium will move to the right, producing more of the new complex ion.
1. Reaction of carbonate ions with the hexaaqua ion. Statistically, there is far more chance of a carbonate ion hitting a hexaaqua metal ion than of hitting a hydrogen ion. It removes a hydrogen ion directly from the hexaaqua complex. If that happens, you get exactly the same new complex ion formed as above. But it doesn't stop there. Carbonate ions keep on removing hydrogen ions from the complex until you end up with a neutral complex. You see that as a precipitate of the metal hydroxide.
You can also usefully write the complete change as an overall reaction.
$\ce{ 2 [M(H2O)6]^{3+} (aq) + 3CO3^{2-} -> 2[M(H2O)3(OH)3] (s) + 3CO2 (g) + 3H2O (l)} \nonumber$
2+ ions reacting with carbonate ions
The 2+ hexaaqua ions aren't strongly acidic enough to release carbon dioxide from carbonates. In these cases, you still get a precipitate - but it is a precipitate of what is loosely described as the "metal carbonate".
$\ce{M^{2+} (aq) + CO_3^{2-} (aq) \rightarrow MCO3 (s)} \nonumber$
Looking at the ions of specific metals
We'll look at the reactions of three 3+ ions and three 2+ ions. The important ones are the 3+ ions.
The 3+ ions
hexaaquaaluminium
Starting from a colourless solution, you get a white precipitate - but with bubbles of gas as well. The precipitate is identical to the one you get if you add small amounts of either sodium hydroxide or ammonia solutions to a solution of the hexaaquaaluminium ions.
hexaaquachromium(III)
Again, the precipitate is just the same as if you had added small amounts of either sodium hydroxide or ammonia solution.
hexaaquairon(III)
. . . and again, exactly the same precipitate as if you had added any other base.
Summary
In each case you get a precipitate of the neutral complex - the metal hydroxide. This is exactly the same precipitate that you get if you add small amounts of either sodium hydroxide solution or ammonia solution to solutions of these ions. Bubbles of carbon dioxide are also given off.
2+ ions
hexaaquacobalt(II)
No gas this time - just a precipitate of "cobalt(II) carbonate".
hexaaquacopper(II)
Again, there isn't any carbon dioxide - just a precipitate of the "copper(II) carbonate".
hexaaquairon(II)
You get a precipitate of the "iron(II) carbonate", but no carbon dioxide.
Summary
Hexaaqua ions with a 2+ charge aren't sufficiently acidic to liberate carbon dioxide from carbonate ions. Instead you get a precipitate which you can think of as being the metal carbonate. | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Coordination_Chemistry/Complex_Ion_Chemistry/Reactions_of_the_Hexaaqua_Ions_with_Carbonate_Ions.txt |
This page describes and explains the reactions between complex ions of the type [M(H2O)6]n+ and hydroxide ions from, for example, sodium hydroxide solution. It assumes that you know why these ions are acidic, and are happy about the equilibria involved.
The general case
Although there are only minor differences, for simplicity we will look at 2+ ions and 3+ ions separately.
Adding hydroxide ions to 2+ hexaaqua ions
These have the form [M(H2O)6]2+. Their acidity is shown in the reaction of the hexaaqua ions with water molecules from the solution:
$\ce{[M(H_2O)_6]^{2+} (aq) + H2O (l) \rightleftharpoons [M(H_2O)_5(OH)]^{+} (aq) + H3O^+ (aq)} \nonumber$
They are acting as acids by donating hydrogen ions to water molecules in the solution. Because of the confusing presence of water from two different sources (the ligands and the solution), it is easier to simplify this:
$\ce{[M(H_2O)_6]^{2+} (aq) \rightleftharpoons [M(H_2O)_5(OH)]^{+} (aq) + H^+ (aq)} \nonumber$
Disturbing this equilibrium by adding hydroxide ions - stage 1
What happens if you add hydroxide ions to this equilibrium? There are two possible reactions.
1. Reaction of hydroxide ions with the hydroxonium ions (hydrogen ions). According to Le Chatelier's Principle, the position of equilibrium will move to the right, producing more of the new complex ion.
1. Reaction of hydroxide ions with the hexaaqua ion. Statistically, there is far more chance of a hydroxide ion hitting a hexaaqua metal ion than of hitting a hydrogen ion. There are far more hexaaqua ions present. If that happens, you get exactly the same new complex ion formed as above. $\ce{[M(H_2O)_6]^{2+} (aq) + OH^{-} (aq) \rightleftharpoons [M(H_2O)_5(OH)]^{+} (aq) + H2O (l)} \nonumber$
Notice that this isn't a ligand exchange reaction. The hydroxide ion has removed a hydrogen ion from one of the ligand water molecules. The reaction has also become essentially irreversible.
The second stage of the reaction
Whichever of the above reactions happens, you end up with [M(H2O)5(OH)]+ ions in solution. These are also acidic, and can lose hydrogen ions from another of the water ligands. Taking the easier version of the equilibrium:
$\ce{ [M(H_2O)_5(OH)]^{+} (aq) \rightleftharpoons [M(H_2O)_4(OH)_2](s) + H^{+} (aq) } \nonumber$
Adding hydroxide ions again tips the equilibrium to the right - either by reacting with the hydrogen ions, or by reacting directly with the complex on the left-hand side. When this occurs, the new complex formed no longer has a charge - we describe it as a "neutral complex". In all the cases we are looking at, this neutral complex is insoluble in water - and so a precipitate is formed. This precipitate is often written without including the remaining water ligands. In other words we write it as M(OH)2. A precipitate of the metal hydroxide has been formed.
Going further
There is no logical reason why the removal of hydrogen ions from the complex should stop at this point. Further hydrogen ions can be removed by hydroxide ions to produce anionic complexes - complexes carrying negative charges. Whether this actually happens in the test tube to any extent varies from metal to metal.
In fact, if you do this using sodium hydroxide solution of the usual concentrations, most of the 2+ ions that you will meet at this level don't go beyond the precipitate. The only one you are likely to come across is the zinc case - and that has a complication. The final ion is [Zn(OH)4]2- - a tetrahedral ion which has lost the remaining 2 water ligands.
Adding hydroxide ions to 3+ hexaaqua ions
The argument here is exactly as before - the only difference is the number of hydrogen ions which have to be removed from the original hexaaqua complex to produce the neutral complex. Going beyond the neutral complex is also rather more common with 3+ than with 2+ ions, and may go as far as having a hydrogen ion removed from each of the six water molecules. This is summarized in the same sort of flow scheme as before:
Looking at the ions of specific metals
In each case the formula of the precipitate will be given as if it were the simple neutral complex. In fact, these "hydroxide" precipitates sometimes rearrange by losing water from combinations of the attached OH groups. This produces oxides closely associated with the lost water. These changes are beyond the scope of this site.
2+ ions
hexaaquairon(II)
Iron is very easily oxidized under alkaline conditions. Oxygen in the air oxidizes the iron(II) hydroxide precipitate to iron(III) hydroxide especially around the top of the tube. The darkening of the precipitate comes from the same effect.
hexaaquamanganese(II)
I have shown the original solution as very pale pink (the palest I can produce!), but in fact it is virtually colourless. The pale brown precipitate is oxidized to darker brown manganese(III) oxide in contact with oxygen from the air.
hexaaquazinc
You start and finish with colorless solutions, producing a white precipitate on the way.
3+ ions
hexaaquaaluminium
This looks exactly the same in the test tube as the corresponding zinc reaction above - but beware the different formulae of the precipitate and the final solution.
Reversing the process
If adding hydroxide ions removes hydrogen ions from the hexaaqua complex one at a time, it doesn't seem unreasonable that you could put them back again by adding an acid. That's just what happens!
We'll look in detail at what happens in the chromium(III) case, but exactly the same principle applies to all the other examples we've looked at - whether for 2+ or 3+ ions. As long as you understand what is happening, you can work out the details if you need to
Adding acid to hexahydroxochromate(III) ions
These are the ions formed at the end of the sequence in which you add hydroxide ions to a solution containing hexaaquachromium(III) ions. Their formula is [Cr(OH)6]3-. A reminder of the color changes when you add sodium hydroxide solution to a solution containing hexaaquachromium(III) ions:
If you add an acid (dilute sulphuric acid, for example), the hydrogen ions get put back on one at a time.
You already know the colors of the significant stages (the beginning, the end, and the neutral complex). It isn't a separate bit of learning! You can apply this to any case. If you know the colors as you remove hydrogen ions, you automatically know them as you put the hydrogen ions back on again. It also doesn't matter where you start from either - whether, for example, you add acid to an ionic complex like [Cr(OH)6]3-, or a neutral one like [Fe(H2O)4(OH)2].
You will know that the [Fe(H2O)4(OH)2] is a dirty green precipitate. When you add the hydrogen ions back to it, it will revert to the very pale green solution of the [Fe(H2O)6]2+ ion. None of this is a new bit of learning - you just have to re-arrange what you already know!
Amphoteric hydroxides
An amphoteric substance has both acidic and basic properties. In other words, it will react with both bases and acids. Some of the metal hydroxides we've been looking at are doing exactly that.
Chromium(III) hydroxide as an amphoteric hydroxide
"Chromium(III) hydroxide" is a simple way of naming the neutral complex [Cr(H2O)3(OH)3]. You have seen that it reacts with bases (hydroxide ions) to give [Cr(OH)6]3-. It also reacts with acids (hydrogen ions) to give [Cr(H2O)6]3+. This is a good example of amphoteric behavior.
Other examples of amphoteric hydroxides are zinc hydroxide and aluminium hydroxide.
Copper(II) hydroxide as a basic oxide
Quite a lot of metal hydroxides won't react any further with hydroxide ions if you use sodium hydroxide solution at the sort of concentrations normally used in the lab. That means that they don't have any significant acidic nature.
"Copper(II) hydroxide" is what we would normally call the neutral complex [Cu(H2O)4(OH)2]. This doesn't dissolve in sodium hydroxide solution at any concentration normally used in the lab. It doesn't show any acidic character.
On the other hand, it will react with acids - replacing the lost hydrogen ions on the water ligands. Because it is accepting hydrogen ions, it is acting as a base. Hydroxides like this (which react with acids, but not bases) are not amphoteric - they are just simple bases.
Stereoisomerism in complex ions
Some complex ions can show either optical or geometric isomerism.
Geometric Isomerism
This occurs in planar complexes like the Pt(NH3)2Cl2 we've just looked at. There are two completely different ways in which the ammonias and chloride ions could arrange themselves around the central platinum ion:
The two structures drawn are isomers because there is no way that you can just twist one to turn it into the other. The complexes are both locked into their current forms.
The terms cis and trans are used in the same way as they are in organic chemistry. Trans implies "opposite" - notice that the ammine ligands are arranged opposite each other in that version, and so are the chloro ligands. Cis means "on the same side" - in this instance, that just means that the ammine and chloro ligands are next door to each other.
Optical isomerism
You recognize optical isomers because they have no plane of symmetry. In the organic case, it is fairly easy to recognize the possibility of this by looking for a carbon atom with four different things attached to it. It isn't qute so easy with the complex ions - either to draw or to visualize! The examples you are most likely to need occur in octahedral complexes which contain bidentate ligands - ions like [Ni(NH2CH2CH2NH2)3]2+ or [Cr(C2O4)3]3-.
The diagram below shows a simplified view of one of these ions. Essentially, they all have the same shape - all that differs is the nature of the "headphones". The charges are left off the ion, because obviously they will vary from case to case. The shape shown applies to any ion of this kind.
If your visual imagination will cope, you may be able to see that this ion has no plane of symmetry. If you find this difficult to visualize, the only solution is to make the ion out of a lump of plasticene (or a bit of clay or dough) and three bits of cardboard cut to shape.
A substance with no plane of symmetry is going to have optical isomers - one of which is the mirror image of the other. One of the isomers will rotate the plane of polarization of plane polarized light clockwise; the other rotates it counter-clockwise. In this case, the two isomers are:
If you have a really impressive visual imagination, you may be able to see that there is no way of rotating the second isomer in space so that it looks exactly the same as the first one. | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Coordination_Chemistry/Complex_Ion_Chemistry/Reactions_of_the_Hexaaqua_Ions_with_Hydroxide_Ions.txt |
Complex ions typically exist in complex equilibrium involving its central metal ion and the ligands.
• Chelation
A metal ion in solution does not exist in isolation, but in combination with ligands (such as solvent molecules or simple ions) or chelating groups, giving rise to complex ions or coordination compounds. These complexes contain a central atom or ion, often a transition metal, and a cluster of ions or neutral molecules surrounding it.
• Complex-Ion Equilibria
In general, chemical equilibrium is reached when the forward reaction rate is equal to the reverse reaction rate and can be described using an equilibrium constant, K. Complex ion equilibria are no exception to this and have their own unique equilibrium constant. This formation constant, Kf , describes the formation of a complex ion from its central ion and attached ligands. This constant may be caled a stability constant or association constant.
• Complex Equilibrium
• Hard and Soft Acids and Bases
The thermodynamic stability of a metal complex depends greatly on the properties of the ligand and the metal ion and on the type of bonding. Metal–ligand interaction is an example of a Lewis acid–base interaction. Lewis bases can be divided into two categories: hard bases contain small, relatively nonpolarizable donor atoms (such as N, O, and F), and soft bases contain larger, relatively polarizable donor atoms (such as P, S, and Cl).
• Stability of Metal Complexes and Chelation
Ligands like chloride, water, and ammonia are said to be monodentate (one-toothed, from the Greek mono, meaning “one,” and the Latin dent-, meaning “tooth”): they are attached to the metal via only a single atom. Ligands can, however, be bidentate (two-toothed, from the Greek di, meaning “two”), tridentate (three-toothed, from the Greek tri, meaning “three”), or, in general, polydentate (many-toothed, from the Greek poly, meaning “many”), indicating that they are attached to the metal at two, th
Complex Ion Equilibria
A metal ion in solution does not exist in isolation, but in combination with ligands (such as solvent molecules or simple ions) or chelating groups, giving rise to complex ions or coordination compounds. These complexes contain a central atom or ion, often a transition metal, and a cluster of ions or neutral molecules surrounding it.
Introduction
Many complexes are relatively unreactive species remaining unchanged throughout a sequence of chemical or physical operations and can often be isolated as stable solids or liquid compounds. Other complexes have a much more transient existence and may exist only in solution or be highly reactive and easily converted to other species. All metals form complexes, although the extent of formation and nature of these depend very largely on the electronic structure of the metal. The concept of a metal complex originated in the work of Alfred Werner, who in 1913 was awarded the first Nobel Prize in Inorganic chemistry.
Complexes may be non-ionic (neutral) or cationic or anionic, depending on the charges carried by the central metal ion and the coordinated groups. The total number of points of attachment to the central element is termed the coordination number and this can vary from 2 to greater than 12, but is usually 6. The term ligand come from the latin word ligare, which meaning to bind) was first used by Alfred Stock in 1916 in relation to silicon chemistry. The first use of the term in a British journal was by H. Irving and R.J.P. Williams in Nature, 1948, 162, 746 in their paper describing what is now called the Irving-Williams series.
Ligands can be further characterised as monodentate, bidentate, tridentate etc. where the concept of teeth (dent) is introduced, hence the idea of bite angle etc. The term chelate was first applied in 1920 by Sir Gilbert T. Morgan and H.D.K. Drew [J. Chem. Soc., 1920, 117, 1456], who stated: "The adjective chelate, derived from the great claw or chela (chely- Greek) of the lobster or other crustaceans, is suggested for the caliperlike groups which function as two associating units and fasten to the central atom so as to produce heterocyclic rings."
Metal complexation is of widespread interest. It is studied not only by inorganic chemists, but by physical and organic chemists, biochemists, pharmacologists, molecular biologists and environmentalists.
Thermodynamic Stability
The "stability of a complex in solution" refers to the degree of association between the two species involved in the state of equilibrium. Qualitatively, the greater the association, the greater the stability of the compound. The magnitude of the (stability or formation) equilibrium constant for the association, quantitatively expresses the stability. Thus, if we have a reaction of the type:
$M + 4L \rightarrow ML_4 \nonumber$
then the larger the stability constant, the higher the proportion of $ML_4$ that exists in the solution. Free metal ions rarely exist in solution so that $M$, will usually be surrounded by solvent molecules which will compete with the ligand molecules, $L$, and be successively replaced by them. For example in water:
$M(H_2O)_4 + L \rightleftharpoons M(H_2O)_3L + H_2O \nonumber$
$M(H_2O)_3L + L \rightleftharpoons M(H_2O)_2L_2 + H_2O \nonumber$
However, for simplicity, we generally ignore these solvent molecules and write four stability constants as follows:
$M + L → ML$ with $K_1 = \dfrac{[ML]}{[M] [L]} \nonumber$
$ML + L → ML_2$ with $K_2 = \dfrac{[ML_2]}{[ML] [L]} \nonumber$
$ML_2 + L → ML_3$ with $K_3 = \dfrac{[ML3]}{[ML_2] [L]} \nonumber$
$ML_3 + L → ML_4$ with $K_4 = \dfrac{[ML_4]}{[ML_3] [L]} \nonumber$
where $K_1$, $K_2$ etc. are referred to as "stepwise stability constants." Alternatively, the "Overall Stability Constant" can be constructed
$M + 4L \rightarrow ML_4 \beta_4 = \dfrac{[ML_4]}{[M] [L]^4} \nonumber$
The stepwise and overall stability constants are therefore related as follows:
$\beta_4 =K_1 \times K_2 \times K_3 \times K_4 \nonumber$
or more generally,
$\beta_n =K_1 \times K_2 \times K_3 \times K_4 \times ... K_n \nonumber$
Table E4 outlines the sequential formation constants for a range of metal-ligand complexes.
Example $1$: The Cuprammonium Ion
Consider the four steps involved in the formation of the cuprammonium ion $Cu(NH_3)_4^{2+}$:
STEP 1:
$Cu^{2+} + NH_3 \rightleftharpoons Cu(NH_3)^{2+} \nonumber$
with
$K_1 = \dfrac{[Cu(NH_3)^{2+}]}{[Cu^{2+}] [NH_3]} \nonumber$
STEP 2:
$CuNH_3^{2+} + NH_3 \rightleftharpoons Cu(NH_3)_2^{2+} \nonumber$
with
$K_2 = \dfrac{ [Cu(NH_3)_2^{2+}]}{[Cu(NH_3)^{2+}] [NH_3]} \nonumber$
STEP 3:
$Cu(NH_3)_2^{2+} + NH_3 \rightleftharpoons Cu(NH_3)_3^{2+} \nonumber$
with
$K_3 = \dfrac{ [Cu(NH_3)_3^{2+}]}{[Cu(NH_3)_2^{2+}] [NH_3]} \nonumber$
STEP 4:
$Cu(NH_3)_3^{2+} + NH_3 \rightleftharpoons Cu(NH_3)_4^{2+} \nonumber$
with
$K_4 = \dfrac{ [Cu(NH_3)_4^{2+}]}{[Cu(NH_3)_3^{2+}] [NH_3]} \nonumber$
where the $\{K\}$ constants are the stepwise stability constants. Also:
$\beta_4 = \dfrac{[Cu(NH_3)_4^{2+}]}{[Cu^{2+}] [NH_3]^4} \nonumber$
The addition of the four ammine groups to copper shows a pattern found for most formation constants, in that the successive stability constants decrease. In this case, the four constants are:
$\log K_1 =4.0, \nonumber$
$\log K_2 =3.2, \nonumber$
$\log K_3 =2.7, \nonumber$
$\log K_4 =2.0 \; \text{or} \; \log \beta_4 =11.9 \nonumber$
Pay attention to definitions: a number of texts may refer to $\beta_4$ as the instability constant or the dissociation constant of coordination complexes (e.g., termed $\beta'_4$ below) , which corresponds to the reciprocal of the formation constant ($K_4$), since the reactions referred to are those where fully formed complexes break down to the aqua ion and free ligands. For example,
$\beta'_4 = \dfrac{[Cu^{2+}] [NH_3]^4}{[Cu(NH_3)_4^{2+}]} \nonumber$
However, this is not definition use here (Table E4) and should be compared with the equation for the formation constant given earlier.
It is usual to represent the metal-binding process by a series of stepwise equilibria which lead to stability constants that may vary numerically from hundreds to enormous values such as 1035 and more. That is
100,000,000,000,000,000,000,000,000,000,000,000.0
For this reason, they are commonly reported as logarithms. So log10 (β) = log10 (1035) = 35. It is additionally useful to use logarithms, since log(K) is directly proportional to the free energy of the reaction.
$ΔG^o = -RT \ln(β) \nonumber$
$ΔG^o = -2.303 RT \log_{10}(β) \nonumber$
$ΔG^o = ΔH^o - TΔS^o \nonumber$
Below are three tables of the formation constants and thermodynamics properties of example ligand-metal complexes; a more complete table can be found in Table E4.
Table $1$: Reaction of ammonia and 1,2-diaminoethane with Cd2+.
# of ligands ΔG° (kJmol-1) ΔH° (kJmol-1) ΔS° (JK-1mol-1) log β
2 NH3(1 en) -28.24 (-33.30) -29.79 (-29.41) -5.19 (+13.05) 4.95 (5.84)
4 NH3(2 en) -42.51 (-60.67) -53.14 (-56.48) -35.50 (+13.75) 7.44 (10.62)
Table $2$: Reaction of pyridine and 2,2'-bipyridine with Ni2+.
# of ligands log β ΔG° (kJmol-1)
2 py (1 bipy) 3.5 (6.9) -20 (-39)
4 py (2 bipy) 5.6 (13.6) -32 (-78)
6 py (3 bipy) 9.8 (19.3) -56 (-110)
Table $3$: Reaction of ammonia and 1,2-diaminoethane with Ni2+.
# of ligands log β ΔG° (kJmol-1)
1 NH3 2.8 -16
2 NH3 (1 en) 5.0 (7.51) -28.5 (-42.8)
3 NH3 6.6 -37.7
4 NH3 (2 en) 7.87 (13.86) -44.9 (-79.1)
5 NH3 8.6 -49.1
6 NH3 (3 en) 8.61 (18.28) -49.2 (-104.4)
The Chelate Effect
The chelate effect can be seen by comparing the reaction of a chelating ligand and a metal ion with the corresponding reaction involving comparable monodentate ligands. For example, comparison of the binding of 2,2'-bipyridine with pyridine or 1,2-diaminoethane (ethylenediamine=en) with ammonia. It has been known for many years that a comparison of this type always shows that the complex resulting from coordination with the chelating ligand is much more thermodynamically stable. This can be seen by looking at the values for adding two monodentates compared with adding one bidentate, or adding four monodentates compared to two bidentates, or adding six monodentates compared to three bidentates.
A number of points should be highlighted from the formation constants in Table E4. In Table $1$, it can be seen that the ΔH° values for the formation steps are almost identical, that is, heat is evolved to about the same extent whether forming a complex involving monodentate ligands or bidentate ligands. What is seen to vary significantly is the ΔS° term which changes from negative (unfavorable) to positive (favorable). Note as well that there is a dramatic increase in the size of the ΔS° term for adding two compared to adding four monodentate ligands. (-5 to -35 JK-1mol-1). What does this imply, if we consider ΔS° to give a measure of disorder?
In the case of complex formation of Ni2+ with ammonia or 1,2-diaminoethane, by rewriting the equilibria, the following equations are produced.
Using the equilibrium constant for the reaction (3 above) where the three bidentate ligands replace the six monodentateligands, we find that at a temperature of 25° C:
$\Delta G^° = -2.303 \, RT \, \log_{10} K \nonumber$
$= -2.303 \,RT \,(18.28 - 8.61) \nonumber$
$= -54 \text{ kJ mol}^{-1} \nonumber$
Based on measurements made over a range of temperatures, it is possible to break down the $\Delta G^°$ term into the enthalpy and entropy components.
$\Delta G^° = \Delta H^° - T \Delta S^° \nonumber$
The result is that: $\Delta H^° = -29 kJ mol^{-1}$
- TΔS° = -25 kJ mol-1
and at 25C (298K)
ΔS° = +88 J K-1 mol-1
For many years, these numbers have been incorrectly recorded in textbooks. For example, the third edition of "Basic Inorganic Chemistry" by F.A. Cotton, G. Wilkinson and P.L. Gaus, John Wiley & Sons, Inc, 1995, on page 186 gives the values as:
ΔG° = -67 kJ mol-1
ΔH° = -12 kJ mol-1
-TΔS° = -55 kJ mol-1
The conclusion they drew from these incorrect numbers was that the chelate effect was essentially an entropy effect, since the TΔS° contribution was nearly 5 times bigger than ΔH°.
In fact, the breakdown of the ΔG° into ΔH° and TΔS° shows that the two terms are nearly equal (-29 and -25 kJ mol-1, respectively) with the ΔH° term a little bigger! The entropy term found is still much larger than for reactions involving a non-chelating ligand substitution at a metal ion. How can we explain this enhanced contribution from entropy? One explanation is to count the number of species on the left and right hand side of the equation above.
It will be seen that on the left-hand-side there are 4 species, whereas on the right-hand-side there are 7 species, that is a net gain of 3 species occurs as the reaction proceeds. This can account for the increase in entropy since it represents an increase in the disorder of the system. An alternative view comes from trying to understand how the reactions might proceed. To form a complex with 6 monodentates requires 6 separate favorable collisions between the metal ion and the ligand molecules. To form the tris-bidentate metal complex requires an initial collision for the first ligand to attach by one arm but remember that the other arm is always going to be nearby and only requires a rotation of the other end to enable the ligand to form the chelate ring.
If you consider dissociation steps, then when a monodentate group is displaced, it is lost into the bulk of the solution. On the other hand, if one end of a bidentate group is displaced the other arm is still attached and it is only a matter of the arm rotating around and it can be reattached again. Both conditions favor the formation of the complex with bidentate groups rather than monodentate groups.
Problems
Calculate the entropy changes at 25°C, for the following reactions:
Zn2+ + 2NH3 <=> [Zn(NH3)2]2+
ΔH = -28.03 kJ mol-1 and log10 β2 = 5.01
Zn2+ + en <=> [Zn(en)]2+
ΔH = -27.6 kJ mol-1 and log10 β = 6.15 (NB R=8.314 JK-1mol-1)
Solution
The calculations make use of the equations:
ΔG° = RTlnβ
ΔG° = 2.303RTlog10β
ΔG° = ΔH° - TΔS°
First it is necessary to calculate the ΔG values from the given formation constants and temperature 25C = 298K.
For the ammonia complex, ΔG = -8.314 * 298 * 2.303 * 5.01 kJmol-1, that is -28.6 kJmol-1.
For the 1,2-diaminoethane complex, ΔG= -8.314 * 298 * 2.303 * 6.15 kJmol-1 which corresponds to -35.1 kJmol-1.
This makes use of the relation between Ln and log10 such that ln(x) = 2.303 log10 (x).
Then using the second relationship above, ΔS can be found.
$\Delta S = \dfrac{\Delta H - \Delta G}{T} \nonumber$
For the ammonia complex this gives 1.9 JK-1mol-1 and for the 1,2-diaminoethane complex 25.2 JK-1mol-1, which are the values given as answer 2.
Contributors and Attributions
• Prof. Robert J. Lancashire (The Department of Chemistry, University of the West Indies) | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Coordination_Chemistry/Complex_Ion_Equilibria/Chelation.txt |
When a metal ion reacts with a Lewis base in solution a complex ion is formed. This reaction can be described in terms of chemical equilibria.
Introduction
A complex ion is comprised of two important parts: the central atom and its surrounding ligands. The central atom can be any metallic ion (usually a transition metal). The ligands are any combination of anions that can donate an electron pair, effectively meaning they are all Lewis bases. When combined they form coordinate covalent bonds.
Formation Constant
In general, chemical equilibrium is reached when the forward reaction rate is equal to the reverse reaction rate and can be described using an equilibrium constant, $K$. Complex ion equilibria are no exception to this and have their own unique equilibrium constant. This formation constant, $K_f$, describes the formation of a complex ion from its central ion and attached ligands. This constant may be called a stability constant or association constant; the units depend on the specific reaction it is describing. Common metal/ligand $K_f$ values are tabulated in Table E4.
At its most basic level, $K_f$ can be explained as the following, where $M$ is a metal ion, $L$ is a ligand, and $x$ and $y$ are coefficients:
$K_f = \dfrac{[M_xL_y]} {[M]^x[L]^y} \nonumber$
$K_f$ values are very large in magnitude for complex ion formation reactions heavily favor the product and very small for poorly forming complex ions. For example, observe the formation reaction of the dicyanoargentate(I) complex ion
$\ce{Ag^{+}(aq) + 2CN^{-} (aq) <=> Ag(CN)2^{-}(aq)} \nonumber$
and the resulting formation constant expression:
$K_f = \dfrac{[Ag(CN)_2^-]}{[Ag^{+}][CN^{-}]^{2}} = 5.6 \times 10^{18} \nonumber$
The larger the $K_f$ value of a complex ion, the more stable it is. Due to how large formation constants often are it is not uncommon to see them listed as logarithms in the form $\log K_f$. You may also see them in the form of a dissociation constant, $K_d$ (which should not be confused with pKa). This is merely the inverse value of the formation constant and as such describes the instability of a complex ion. It may help to think of $K_f$ and $K_d$ as stability and instability constants, respectively.
Using the above example, the dissociation constant expression would be:
$K_f = \dfrac{1}{K_d}$
$K_{d} = \dfrac{1}{K_f} = \dfrac{[Ag^{+}][CN^{-}]^{2}}{[Ag(CN)_2]^{-}} = 1.8 \times 10^{-19}$
In logarithm form $K_f$and $K_d$ would be:
$\log K_{f} = \log(5.6 \times 10^{18}) = 18.7$
$\log K_{d} = \log(1.8 \times 10^{-19}) = -18.7$
Example $1$: Stepwise Equilbria
One might assume that the formation of complex ions is a one step process, however this is not always the case. As an example let's look at the formation of tetraamminecopper(II) ion in solution:
$\ce{[Cu(H2O)4]^{2+}(aq) + 4NH3 (aq) <=> [Cu(NH3)4]^{2+}(aq) + 4H2O(l)} \nonumber$
The reason for using the hydrated form of the copper ion is that cations in solution usually exist in their hydrated form. It is because of this that the creation of tetraamminecopper(II) ion is a multi-step process, the ammine ligands have to displace the existing aqua ligands. In calculations however, the concentration of water and the number of aqua ligands are effectively constant meaning that they can be safely ignored. Each step of this process has its own formation constant, and as with other chemical equilibria, the overall reaction constant is the product of these stepwise formation constants.
Step 1:
$[Cu(H_2O)_4]^{2+}_{(aq)} + NH_{3 \; (aq)} \rightleftharpoons [Cu(H_2O)_3(NH_3)]^{2+}_{(aq)} + H_2O_{(l)} \tag{step 1}$
$K_{1} = \dfrac{[[Cu(H_2O)_3(NH_3)]^{2+}]}{[[Cu(H_2O)_4]^{2+}][NH_3]} = 1.9 \times 10^4 \nonumber$
Step 2:
$[Cu(H_2O)_3(NH_3)]^{2+}_{(aq)} + NH_{3 \; (aq)} \rightleftharpoons [Cu(H_2O)_2(NH_3)_2]^{2+}_{(aq)} + H_2O_{(l)} \tag{step 2}$
$K_{2} = \dfrac{[[Cu(H_2O)_2(NH_3)_2]^{2+}]}{[[Cu(H_2O)_3(NH_3)]^{2+}][NH_3]} = 3.9 \times 10^3 \nonumber$
Step 3:
$[Cu(H_2O)_2(NH_3)_2]^{2+}_{(aq)} + NH_{3 \; (aq)} \rightleftharpoons [Cu(H_2O)(NH_3)_3]^{2+}_{(aq)} + H_2O_{(l)} \tag{step 3}$
$K_{3} = \dfrac{[[Cu(H_2O)(NH_3)_3]^{2+}]}{[[Cu(H_2O)_2(NH_3)_2]^{2+}][NH_3]} = 1.0 \times 10^3 \nonumber$
Step 4:
$[Cu(H_2O)(NH_3)_3]^{2+}_{(aq)} + NH_{3 \; (aq)} \rightleftharpoons [Cu(NH_3)_4]^{2+}_{(aq)} + H_2O_{(l)} \tag{step 4}$
$K_{4} = \dfrac{[[Cu(NH_3)_4]^{2+}]}{[[Cu(H_2O)(NH_3)_3]^{2+}][NH_3]} = 1.5 \times 10^2 \nonumber$
Total
\begin{align*} K_f &= K_{1} \times K_{2} \times K_{3} \times K_{4} \[4pt] &= \dfrac{[[Cu(H_2O)_3(NH_3)]^{2+}]}{[[Cu(H_2O)_4]^{2+}][NH_3]} \times \dfrac{[[Cu(H_2O)_2(NH_3)_2]^{2+}]}{[[Cu(H_2O)_3(NH_3)]^{2+}][NH_3]} \times \dfrac{[[Cu(H_2O)(NH_3)_3]^{2+}]}{[[Cu(H_2O)_2(NH_3)_2]^{2+}][NH_3]} \times \dfrac{[[Cu(NH_3)_4]^{2+}]}{[[Cu(H_2O)(NH_3)_3]^{2+}][NH_3]} \[4pt] &= \dfrac{[[Cu(NH_3)_4]^{2+}]}{[[Cu(H_2O)_4]^{2+}][NH_3]^{4}} \[4pt] &= 1.1 \times 10^{13} \end{align*}
If you wanted to find the formation constant of one of the intermediate steps, you would simply take the product of the $K$ values up until that point. For example the formation constant of $[Cu(H_2O)_2(NH_3)_2]^{2+}$ would be $\beta_2 = K_{1} \times K_{2} = 7.4 \times 10^7$. As seen here and above the standard notation for these values is a lowercase beta.
Example $2$: Using Formation Constants as a Metric of Stability
Based on the given values, rank the following complex ions from most to least stable.
• $[Cr(OH)_4]^{-}$ $K_f = 8.0 \times 10^{29}$
• $[Ni(CN)_4]^{2-}$ $\log(K_d) = -31.3$
• $[Co(NH_3)_6]^{3+}$ $K_d = 2.2 \times 10^{-34}$
• $[Cu(CN)_4]^{3-}$ $\log(K_f) = 30.3$
Solution
The key to solving this problem is to convert all of the given values into the same form. In this case we will convert all of them into the form $K_f$ though any other form could be used as long as you are consistent.
• $[Cr(OH)_4]^{-}$ $K_f = 8.0 \times 10^{29}$
• $[Ni(CN)_4]^{2-}$ $\log(K_d) = -31.3 \longrightarrow K_f = 2.0 \times 10^{31}$
• $[Co(NH_3)_6]^{3+}$ $K_d = 2.2 \times 10^{-34} \longrightarrow K_f = 4.5 \times 10^{33}$
• $[Cu(CN)_4]^{3-}$ $\log(K_f) = 30.3 \longrightarrow K_f = 2.0 \times 10^{30}$
After converting it is easy to rate the ions in terms of stability. From most to least stable:
$[Co(NH_3)_6]^{3+} > [Ni(CN)_4]^{2-} > [Cu(CN)_4]^{3-} > [Cr(OH)_4]^{-}. \nonumber$
You may notice that each stepwise formation constant is smaller than the one before it. This decreasing trend is due to the effects of entropy, causing each step to be progressively less likely to occur. You can think of this in the following way, continuing with the previous example:
When the first ammine ligand goes to displace an aqua ligand it has four sites from which to choose from, making it "easier" to displace one. Yet with every step the number of sites decreases making it increasingly more difficult. As always though, there are exceptions to this rule. If the values do not continually decrease then the structure of the complex ion likely changed during one of the steps.
Chelation Effect
Generally, complex ions with polydentate ligands have much higher formation constants than those with monodentate ligands. This additional stability is known as the chelation effect. For example, the formation of the bis(ethylenediamine)copper(II) ion is identical to the tetraamminecopper(II) ion above except the ethylenediamine ligands displace two aqua ligands at a time. Observe the differences in formation constant (Table E4):
Complex Ion Kf
$[Cu(NH_3)_4]^{2+}$ $1.1 \times 10^{13}$
$[Cu(en)_2]^{2+}$ $1.0 \times 10^{20}$
Example $3$
Of $\ce{[Ni(en)3]^{2+}}$, $\ce{[Ni(EDTA)]^{2-}}$, and $\ce{[Ni(NH3)6]^{2+}}$ which would you expect to have the largest $K_f$ value? The smallest? Explain.
Solution
This is a question involving the effects of chelation. When the complex ion with EDTA is formed, the EDTA ligand displaces six aqua ligands at once. Ethylenediamine ligands displace two at a time and ammine ligands displace one at a time. Based on this we would expect $\ce{[Ni(EDTA)]^{2-}}$ to be the most stable and $\ce{[Ni(NH3)6]^{2+}}$ to be the least stable.
Solubility Effects
The formation of complex ions can sometimes explain the solubility of certain compounds in solution. For example, let's say we have a saturated solution of silver bromide ($AgBr$) which also contains additional undissolved solid silver bromide. If high enough concentrations of ammonia are added to the solution, the solid silver bromide will dissolve despite the solution previously having been saturated. This is because of the formation diammineargentate(I) ion, illustrated by the following chemical equation:
$\ce{AgBr(s) + 2NH3 (aq) \rightarrow [Ag(NH3)2]^{+}(aq) + Br^{-}(l)} \nonumber$
References
1. Beck, Mihály T. Chemistry of Complex Equilibria. Revised ed. New York: Halsted Press, 1990.
2. Brown, Theodore, Eugene Lemay, Bruce Bersten, and Catherine Murphey. Chemistry: The Central Science. 10th ed. Andalusia, Alabama: Pearson Publications Company, 2006.
3. Petrucci, Harwood, Herring, and Madura. General Chemistry: Principles and Modern Applications. 9th ed. Upper Saddle River, New Jersey: Pearson Education, 2007.
4. Wilkinson, Geoffrey, Robert D. Gillard, and Jon A. McCleverty, eds. Comprehensive Coordination Chemistry: The Synthesis, Reactions, Properties & Applications of Coordination Compounds. First ed. Vol. 1. Oxford, England: Pergamon Press, 1987. N. pag. 7 vols.
Example $4$
The creation of $\ce{[Ni(en)_3]^{2+}}$ is a three step process with the following stepwise formation constants $K_1 = 3.3 \times 10^{7}$, $K_2 = 1.9 \times 10^{6}$, and $K_3 = 1.8 \times 10^{4}$. Using this information find the overall formation constant and the formation constant for each of the intermediate steps.
Solution
Take the product of the stepwise formation constants up until the step you are trying to find.
$[Ni(H_2O)_4(en)]^{2+}$ $\beta_1 = K_1 = 3.3 \times 10^{7}$
$[Ni(H_2O)_2(en)_2]^{2+}$ $\beta_2 = K_1 \times K_2 = 6.8 \times 10^{13}$
$[Ni(en)_3]^{2+}$ $\beta_3 = K_f = K_1 \times K_2 \times K_3 = 1.1 \times 10^{18}$
Contributors and Attributions
• Katherine Barrett, Gianna Navarro, Christopher May (UCD) | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Coordination_Chemistry/Complex_Ion_Equilibria/Complex-Ion_Equilibria.txt |
The lone pairs of electrons in water molecules make water an excellent ligand. Metal ions in any body of water in the environment usually are coordinated to 6 water molecules, e.g. Ag(H2O)6+, Cu(H2O)62+, and Cr(H2O)62+, etc. In the literature, these ions are represented by Ag+(aq), Cu2+(aq), and Cr3+(aq) and very often notations (aq) is also omitted.
When a stronger coordinating ligand such as $NH_3$ is present, its molecules compete with water for the coordination site.
$Ag(H_2O)_6^+ + NH_3 \rightleftharpoons Ag(NH_3)(H_2O)_5^+ \nonumber$
$Ag(NH_3)(H_2O)_5^+ + NH_3 \rightleftharpoons Ag(NH_3)_2(H_2O)_4^+ \nonumber$
And these are simply represented by the following with the equilibrium constant as indicated.
$Ag^+ + NH_3\rightleftharpoons Ag(NH_3)^+ \;\;\;K_1 = 2.0 \times 10^3 \nonumber$
$Ag(NH_3)^+ + NH_3 \rightleftharpoons Ag(NH_3)_2^+ \;\;\;K_2 = 8.0 \times 10^3 \nonumber$
Where $K_1$ and $K_2$ are called stepwise equilibrium constants as opposed to the overall equilibrium constants, $\beta_1$ and $\beta_2$. Note, however, that
$K_1 = \dfrac{[Ag(NH_3)^+]}{[Ag^+] [NH_3]} \nonumber$
$\beta_1 = K_1 \nonumber$
$K_2 = \dfrac{[Ag(NH_3)^{+2}]}{[Ag(NH_3)^+][NH_3]} \nonumber$
$\beta_2 = \dfrac{[Ag(NH_3)^{+2}]}{[Ag^+][NH_3]^2} = K_1 K_2 \nonumber$
A mole fraction ($X$) distribution diagram for various complexes of silver can be used to show their variations. The total concentration of silver-ion-containing species $C_0$ is,
$C_0 = [Ag^+] + [Ag(NH_3)^+] + [Ag(NH_3)_2^+] \nonumber$
$= [Ag^+] \left(1 + \beta_1 [NH_3] + \beta_2 [NH_3]^2\right) \nonumber$
And the mole fraction, $X$, of a species $A$, $X(A)$ is easily calculated using the following equations.
$X(Ag^+) = \dfrac{[Ag^+]}{C_0} \nonumber$
$= \dfrac{1}{1 + \beta_1 [NH_3] + \beta_2 [NH_3]^2} \nonumber$
$X(Ag(NH_3)+) = \dfrac{[Ag(NH_3)^+]}{C_0} \nonumber$
= \beta_1 [NH3] / (1 + \beta_1 [NH3] + \beta_2 [NH3]2)
= \beta_1 [NH3] X(Ag+)
X(Ag(NH3)2+) = [Ag(NH3)2+] / C0
= \beta_2 [NH3]2 / (1 + \beta_1 [NH3] + \beta_2 [NH3]2)
= \beta_2 [NH3]2 X(Ag+)
Example 1
Evaluate X(Ag+), X(Ag(NH3)+), and X(Ag(NH3)2+) for a solution containing 0.20 M Ag+ ions mixed with an equal volume of 2.00 M NH3 solution.
Solution
Assume the volume doubles when the solutions are mixed. We have,
$\beta_1 = K_1 = 2.0 \times 10^3 \nonumber$
$\beta_2 = K_1 K_2 = 1.6 \times 10^7 \nonumber$
$[NH_3] = 1.00\; M \nonumber$
$C_0 = [Ag^+] = 0.10\; M \nonumber$
$X(Ag^+) = \dfrac{1}{1 + \beta_1 [NH_3] + \beta_2 [NH_3]^2} \nonumber$
$= 1 / 1.6 \times 10^7 = 6.25 \times 10^{-8} \nonumber$
$X(Ag(NH_3)^+) = \beta_1 [NH_3] X(Ag^+) \nonumber$
$= 1.25 \times 10^{-4} \nonumber$
$X(Ag(NH_3)_2^+) = \beta_2 [NH_3]_2 X(Ag^+) \nonumber$
$= 1 \nonumber$
DISCUSSION
More precisely [NH3] = 0.80 M, but the results are not changed. Since [NH3] is >> [Ag+], the complex with the highest number of ligands is the dominating species.
Example 2
What are the concentrations of NH3 when
• [Ag+] = [Ag(NH3)+],
• [Ag(NH3)+] = [Ag(NH3)2+],
These are cross points for the mole fraction distribution lines.
Solution
Early on, we have derived these relationships.
$X(Ag(NH_3)^+) = \beta_1 [NH_3] X(Ag^+) \nonumber$
$X(Ag(NH_3)_2^+) = \beta_2 [NH_3]^2 X(Ag^+) \nonumber$
When the mole fractions are equal, their concentrations are also equal. Thus, when the mole fraction of Ag+ is the same as the mole fraction of Ag(NH3)+,
$X(Ag^+) = X(Ag(NH_3)^+) \nonumber$
We have
$\beta_1[NH_3] = 1 \nonumber$
$[NH_3] = \dfrac{1}{\beta_1} \nonumber$
By the same arguments, the cross point of X(Ag(NH3)+) line and X(Ag(NH3)2+) line occur when
[NH3] = 1 / (\beta_2)1/2
DISCUSSION
Note that [NH3] refers to the concentration of the free ligand, not the total concentration of NH3. For a chemical engineering application, much more details must be revealed in order to understand the complexity of coordination equilibrium. Here is a case for a simulation model either implemented using spread sheet or programming.
The Cupric Complexes
In the case of $Cu^{2+}$ ions, the equilibria are more complicated, because the coordination number of ligands can be as high as 6. Consider the following data:
$Cu^{2+} + NH_3 \rightarrow Cu(NH_3)^{2+} \nonumber$
with $K_1 = 1.1 \times 10^4 \label{1}$
$Cu(NH_3)^{2+} + NH_3 \rightarrow Cu(NH_3)_2^{2+} \nonumber$
with $K_2 = 2.7 \times 10^3 \label{2}$
$Cu(NH_3)_2^{2+} + NH_3 \rightarrow Cu(NH_3)_3^{2+} \nonumber$
with $K_3 = 6.3 \times 10^2 \label{3}$
$Cu(NH_3)_3^{2+} + NH_3 \rightarrow Cu(NH_3)_4^{2+} \nonumber$
with $K_4 = 30 \label{4}$
At high concentration of $NH_3$ more of $Cu(NH_3)_4^{2+}$ complexes are formed. The higher the number of ammonia is coordinated to the cupric ion, the deeper is the color blue. What we have done for the silver ion complexes can be applied to the cupric complexes, and the calculation is a little more complicated. However, the theory and method are the same, and you may applied the discussion on silver ion to that of the cupric ion cases. | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Coordination_Chemistry/Complex_Ion_Equilibria/Complex_Equilibrium.txt |
The thermodynamic stability of a metal complex depends greatly on the properties of the ligand and the metal ion and on the type of bonding. Metal–ligand interaction is an example of a Lewis acid–base interaction. Lewis bases can be divided into two categories:
• hard bases contain small, relatively nonpolarizable donor atoms (such as N, O, and F), and
• soft bases contain larger, relatively polarizable donor atoms (such as P, S, and Cl).
Metal ions with the highest affinities for hard bases are hard acids, whereas metal ions with the highest affinity for soft bases are soft acids. Some examples of hard and soft acids and bases are given in Table \(1\). Notice that hard acids are usually cations of electropositive metals; consequently, they are relatively nonpolarizable and have higher charge-to-radius ratios. Conversely, soft acids tend to be cations of less electropositive metals; consequently, they have lower charge-to-radius ratios and are more polarizable. Chemists can predict the relative stabilities of complexes formed by the d-block metals with a remarkable degree of accuracy by using a simple rule: hard acids prefer to bind to hard bases, and soft acids prefer to bind to soft bases.
Table \(1\): Examples of Hard and Soft Acids and Bases
Acids Bases
hard H+ NH3, RNH2, N2H4
Li+, Na+, K+ H2O, ROH, R2O
Be2+, Mg2+, Ca2+, VO2+ OH, F, Cl, CH3CO2
Al3+, Sc3+, Cr3+ CO32
Ti4+ PO43
soft BF3, Al2Cl6, CO2, SO3
Cu+, Ag+, Au+, Tl+, Hg22+ H
Pd2+, Pt2+, Hg2+ CN, SCN, I, RS
GaCl3, GaBr3, GaI3 CO, R2S
Hard acids prefer to bind to hard bases, and soft acids prefer to bind to soft bases.
Because the interaction between hard acids and hard bases is primarily electrostatic in nature, the stability of complexes involving hard acids and hard bases increases as the positive charge on the metal ion increases and as its radius decreases. For example, the complex of Al3+ (r = 53.5 pm) with four fluoride ligands (AlF4) is about 108 times more stable than InF4, the corresponding fluoride complex of In3+ (r = 80 pm). In general, the stability of complexes of divalent first-row transition metals with a given ligand varies inversely with the radius of the metal ion, as shown in Table \(2\). The inversion in the order at copper is due to the anomalous structure of copper(II) complexes, which will be discussed shortly.
Table \(2\)
complex stability Mn2+ < Fe2+ < Co2+ < Ni2+ < Cu2+ > Zn2+
ionic radius (pm) 83 78 74.5 69 73 74
Because a hard metal interacts with a base in much the same way as a proton, by binding to a lone pair of electrons on the base, the stability of complexes of hard acids with hard bases increases as the ligand becomes more basic. For example, because ammonia is a stronger base than water, metal ions bind preferentially to ammonia. Consequently, adding ammonia to aqueous solutions of many of the first-row transition-metal cations results in the formation of the corresponding ammonia complexes.
In contrast, the interaction between soft metals (such as the second- and third-row transition metals and Cu+) and soft bases is largely covalent in nature. Most soft-metal ions have a filled or nearly filled d subshell, which suggests that metal-to-ligand π bonding is important. Complexes of soft metals with soft bases are therefore much more stable than would be predicted based on electrostatic arguments.
The hard acid–hard base/soft acid–soft base concept also allows us to understand why metals are found in nature in different kinds of ores. Recall that most of the first-row transition metals are isolated from oxide ores but that copper and zinc tend to occur naturally in sulfide ores. This is consistent with the increase in the soft character of the metals across the first row of the transition metals from left to right. Recall also that most of the second- and third-row transition metals occur in nature as sulfide ores, consistent with their greater soft character. | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Coordination_Chemistry/Complex_Ion_Equilibria/Hard_and_Soft_Acids_and_Bases.txt |
Ligands like chloride, water, and ammonia are said to be monodentate (one-toothed, from the Greek mono, meaning “one,” and the Latin dent-, meaning “tooth”): they are attached to the metal via only a single atom. Ligands can, however, be bidentate (two-toothed, from the Greek di, meaning “two”), tridentate (three-toothed, from the Greek tri, meaning “three”), or, in general, polydentate (many-toothed, from the Greek poly, meaning “many”), indicating that they are attached to the metal at two, three, or several sites, respectively. Ethylenediamine (H2NCH2CH2NH2, often abbreviated as en) and diethylenetriamine (H2NCH2CH2NHCH2CH2NH2, often abbreviated as dien) are examples of a bidentate and a tridentate ligand, respectively, because each nitrogen atom has a lone pair that can be shared with a metal ion. When a bidentate ligand such as ethylenediamine binds to a metal such as Ni2+, a five-membered ring is formed. A metal-containing ring like that shown is called a chelate ring (from the Greek chele, meaning “claw”). Correspondingly, a polydentate ligand is a chelating agent, and complexes that contain polydentate ligands are called chelate complexes.
Experimentally, it is observed that metal complexes of polydentate ligands are significantly more stable than the corresponding complexes of chemically similar monodentate ligands; this increase in stability is called the chelate effect. For example, the complex of Ni2+ with three ethylenediamine ligands, [Ni(en)3]2+, should be chemically similar to the Ni2+ complex with six ammonia ligands, [Ni(NH3)6]2+. In fact, the equilibrium constant for the formation of [Ni(en)3]2+ is almost 10 orders of magnitude larger than the equilibrium constant for the formation of [Ni(NH3)6]2+ (Table E4):
\begin{align} & [\mathrm{Ni(H_2O)_6]^{2+}+6NH_3} & \rightleftharpoons \mathrm{[Ni(NH_3)_6]^{2+}+6H_2O(l)} \hspace{3mm} & K_\mathrm f=\mathrm{4\times10^8} \ & \mathrm{[Ni(H_2O)_6]^{2+}+3en} & \rightleftharpoons \mathrm{[Ni(en)_3]^{2+}+6H_2O(l)} \hspace{3mm} & K_\mathrm f=\mathrm{2\times10^{18}}\end{align} \label{23.10} The formation constants are formulated as ligand exchange reactions with aqua ligands being displaced by new ligands ($NH_3$ or $en$) in the examples above.
Chelate complexes are more stable than the analogous complexes with monodentate ligands.
The stability of a chelate complex depends on the size of the chelate rings. For ligands with a flexible organic backbone like ethylenediamine, complexes that contain five-membered chelate rings, which have almost no strain, are significantly more stable than complexes with six-membered chelate rings, which are in turn much more stable than complexes with four- or seven-membered rings. For example, the complex of nickel (II) with three ethylenediamine ligands is about 363,000 times more stable than the corresponding nickel (II) complex with trimethylenediamine (H2NCH2CH2CH2NH2, abbreviated as tn):
\begin{align} & \mathrm{[Ni(H_2O)_6]^{2+}+3en} & & \rightleftharpoons \mathrm{[Ni(en)_3]^{2+}+6H_2O(l)} & K_\mathrm f=6.76 \times 10^{17} \ & \mathrm{[Ni(H_2O)_6]^{2+}+3tn} & & \rightleftharpoons \mathrm{[Ni(tn)_3]^{2+}+6H_2O(l)} & K_\mathrm f=1.86 \times 10^{12}\end{align} \label{23.11}
*The above measurements were done in a solution of ionic strength 0.15 at 25º C.
Example $1$
Arrange [Cr(en)3]3+, [CrCl6]3−, [CrF6]3−, and [Cr(NH3)6]3+ in order of increasing stability.
Given: four Cr(III) complexes
Asked for: relative stabilities
Strategy:
A Determine the relative basicity of the ligands to identify the most stable complexes.
B Decide whether any complexes are further stabilized by a chelate effect and arrange the complexes in order of increasing stability.
Solution
A The metal ion is the same in each case: Cr3+. Consequently, we must focus on the properties of the ligands to determine the stabilities of the complexes. Because the stability of a metal complex increases as the basicity of the ligands increases, we need to determine the relative basicity of the four ligands. Our earlier discussion of acid–base properties suggests that ammonia and ethylenediamine, with nitrogen donor atoms, are the most basic ligands. The fluoride ion is a stronger base (it has a higher charge-to-radius ratio) than chloride, so the order of stability expected due to ligand basicity is
[CrCl6]3− < [CrF6]3− < [Cr(NH3)6]3+ ≈ [Cr(en)3]3+.
B Because of the chelate effect, we expect ethylenediamine to form a stronger complex with Cr3+ than ammonia. Consequently, the likely order of increasing stability is
[CrCl6]3− < [CrF6]3− < [Cr(NH3)6]3+ < [Cr(en)3]3+.
Exercise $1$
Arrange [Co(NH3)6]3+, [CoF6]3−, and [Co(en)3]3+ in order of decreasing stability.
Answer: [Co(en)3]3+ > [Co(NH3)6]3+ > [CoF6]3−
The Chelate Effect
The chelate effect can be seen by comparing the reaction of a chelating ligand and a metal ion with the corresponding reaction involving comparable monodentate ligands. For example, comparison of the binding of 2,2'-bipyridine with pyridine or 1,2-diaminoethane (ethylenediamine=en) with ammonia. It has been known for many years that a comparison of this type always shows that the complex resulting from coordination with the chelating ligand is much more thermodynamically stable. This can be seen by looking at the values for adding two monodentates compared with adding one bidentate, or adding four monodentates compared to two bidentates, or adding six monodentates compared to three bidentates.
The Chelate Effect is that complexes resulting from coordination with the chelating ligand is much more thermodynamically stable than complexes with non-chelating ligands.
A number of points should be highlighted from the formation constants in Table E4. In Table $1$, it can be seen that the ΔH° values for the formation steps are almost identical, that is, heat is evolved to about the same extent whether forming a complex involving monodentate ligands or bidentate ligands. What is seen to vary significantly is the ΔS° term which changes from negative (unfavorable) to positive (favorable). Note as well that there is a dramatic increase in the size of the ΔS° term for adding two compared to adding four monodentate ligands. (-5 to -35 JK-1mol-1). What does this imply, if we consider ΔS° to give a measure of disorder?
In the case of complex formation of Ni2+ with ammonia or 1,2-diaminoethane, by rewriting the equilibria, the following equations are produced.
Using the equilibrium constant for the reaction (3 above) where the three bidentate ligands replace the six monodentateligands, we find that at a temperature of 25° C:
$\Delta G^° = -2.303 \, RT \, \log_{10} K \nonumber$
$= -2.303 \,RT \,(18.28 - 8.61) \nonumber$
$= -54 \text{ kJ mol}^{-1} \nonumber$
Based on measurements made over a range of temperatures, it is possible to break down the $\Delta G^°$ term into the enthalpy and entropy components.
$\Delta G^° = \Delta H^° - T \Delta S^° \nonumber$
The result is that: $\Delta H^° = -29 kJ mol^{-1}$
- TΔS° = -25 kJ mol-1
and at 25C (298K)
ΔS° = +88 J K-1 mol-1
For many years, these numbers have been incorrectly recorded in textbooks. For example, the third edition of "Basic Inorganic Chemistry" by F.A. Cotton, G. Wilkinson and P.L. Gaus, John Wiley & Sons, Inc, 1995, on page 186 gives the values as:
ΔG° = -67 kJ mol-1
ΔH° = -12 kJ mol-1
-TΔS° = -55 kJ mol-1
The conclusion they drew from these incorrect numbers was that the chelate effect was essentially an entropy effect, since the TΔS° contribution was nearly 5 times bigger than ΔH°.
In fact, the breakdown of the ΔG° into ΔH° and TΔS° shows that the two terms are nearly equal (-29 and -25 kJ mol-1, respectively) with the ΔH° term a little bigger! The entropy term found is still much larger than for reactions involving a non-chelating ligand substitution at a metal ion. How can we explain this enhanced contribution from entropy? One explanation is to count the number of species on the left and right hand side of the equation above.
It will be seen that on the left-hand-side there are 4 species, whereas on the right-hand-side there are 7 species, that is a net gain of 3 species occurs as the reaction proceeds. This can account for the increase in entropy since it represents an increase in the disorder of the system. An alternative view comes from trying to understand how the reactions might proceed. To form a complex with 6 monodentates requires 6 separate favorable collisions between the metal ion and the ligand molecules. To form the tris-bidentate metal complex requires an initial collision for the first ligand to attach by one arm but remember that the other arm is always going to be nearby and only requires a rotation of the other end to enable the ligand to form the chelate ring.
If you consider dissociation steps, then when a monodentate group is displaced, it is lost into the bulk of the solution. On the other hand, if one end of a bidentate group is displaced the other arm is still attached and it is only a matter of the arm rotating around and it can be reattached again. Both conditions favor the formation of the complex with bidentate groups rather than monodentate groups. | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Coordination_Chemistry/Complex_Ion_Equilibria/Stability_of_Metal_Complexes_and_Chelation.txt |
Learning Objectives
• To know the most common structures observed for metal complexes.
• To predict the relative stabilities of metal complexes with different ligands
One of the most important properties of metallic elements is their ability to act as Lewis acids that form complexes with a variety of Lewis bases. A metal complex consists of a central metal atom or ion that is bonded to one or more ligands (from the Latin ligare, meaning “to bind”), which are ions or molecules that contain one or more pairs of electrons that can be shared with the metal. Metal complexes can be neutral, such as Co(NH3)3Cl3; positively charged, such as [Nd(H2O)9]3+; or negatively charged, such as [UF8]4−. Electrically charged metal complexes are sometimes called complex ions. A coordination compound contains one or more metal complexes.
Coordination compounds are important for at least three reasons. First, most of the elements in the periodic table are metals, and almost all metals form complexes, so metal complexes are a feature of the chemistry of more than half the elements. Second, many industrial catalysts are metal complexes, and such catalysts are steadily becoming more important as a way to control reactivity. For example, a mixture of a titanium complex and an organometallic compound of aluminum is the catalyst used to produce most of the polyethylene and polypropylene “plastic” items we use every day. Finally, transition-metal complexes are essential in biochemistry. Examples include hemoglobin, an iron complex that transports oxygen in our blood; cytochromes, iron complexes that transfer electrons in our cells; and complexes of Fe, Zn, Cu, and Mo that are crucial components of certain enzymes, the catalysts for all biological reactions.
History of the Coordination Compounds
Coordination compounds have been known and used since antiquity; probably the oldest is the deep blue pigment called Prussian blue: $\ce{KFe2(CN)6}$. The chemical nature of these substances, however, was unclear for a number of reasons. For example, many compounds called “double salts” were known, such as $\ce{AlF3·3KF}$, $\ce{Fe(CN)2·4KCN}$, and $\ce{ZnCl2·2CsCl}$, which were combinations of simple salts in fixed and apparently arbitrary ratios. Why should $\ce{AlF3·3KF}$ exist but not $\ce{AlF3·4KF}$ or $\ce{AlF3·2KF}$? And why should a 3:1 KF:AlF3 mixture have different chemical and physical properties than either of its components? Similarly, adducts of metal salts with neutral molecules such as ammonia were also known—for example, $\ce{CoCl3·6NH3}$, which was first prepared sometime before 1798. Like the double salts, the compositions of these adducts exhibited fixed and apparently arbitrary ratios of the components. For example, $\ce{CoCl3·6NH3}$, $\ce{CoCl3·5NH3}$, $\ce{CoCl3·4NH3}$, and $\ce{CoCl3·3NH3}$ were all known and had very different properties, but despite all attempts, chemists could not prepare $\ce{CoCl3·2NH3}$ or $\ce{CoCl3·NH3}$.
Although the chemical composition of such compounds was readily established by existing analytical methods, their chemical nature was puzzling and highly controversial. The major problem was that what we now call valence (i.e., the oxidation state) and coordination number were thought to be identical. As a result, highly implausible (to modern eyes at least) structures were proposed for such compounds, including the “Chattanooga choo-choo” model for CoCl3·4NH3 shown here.
The modern theory of coordination chemistry is based largely on the work of Alfred Werner (1866–1919; Nobel Prize in Chemistry in 1913). In a series of careful experiments carried out in the late 1880s and early 1890s, he examined the properties of several series of metal halide complexes with ammonia. For example, five different “adducts” of ammonia with PtCl4 were known at the time: PtCl4·nNH3 (n = 2–6). Some of Werner’s original data on these compounds are shown in Table $1$. The electrical conductivity of aqueous solutions of these compounds was roughly proportional to the number of ions formed per mole, while the number of chloride ions that could be precipitated as AgCl after adding Ag+(aq) was a measure of the number of “free” chloride ions present. For example, Werner’s data on PtCl4·6NH3 in Table $1$ showed that all the chloride ions were present as free chloride. In contrast, PtCl4·2NH3 was a neutral molecule that contained no free chloride ions.
Alfred Werner (1866–1919)
Werner, the son of a factory worker, was born in Alsace. He developed an interest in chemistry at an early age, and he did his first independent research experiments at age 18. While doing his military service in southern Germany, he attended a series of chemistry lectures, and he subsequently received his PhD at the University of Zurich in Switzerland, where he was appointed professor of chemistry at age 29. He won the Nobel Prize in Chemistry in 1913 for his work on coordination compounds, which he performed as a graduate student and first presented at age 26. Apparently, Werner was so obsessed with solving the riddle of the structure of coordination compounds that his brain continued to work on the problem even while he was asleep. In 1891, when he was only 25, he woke up in the middle of the night and, in only a few hours, had laid the foundation for modern coordination chemistry.
Table $1$: Werner’s Data on Complexes of Ammonia with $PtCl_4$
Complex Conductivity (ohm−1) Number of Ions per Formula Unit Number of Cl Ions Precipitated by Ag+
PtCl4·6NH3 523 5 4
PtCl4·5NH3 404 4 3
PtCl4·4NH3 299 3 2
PtCl4·3NH3 97 2 1
PtCl4·2NH3 0 0 0
These data led Werner to postulate that metal ions have two different kinds of valence: (1) a primary valence (oxidation state) that corresponds to the positive charge on the metal ion and (2) a secondary valence (coordination number) that is the total number of ligand-metal bonds bound to the metal ion. If $\ce{Pt}$ had a primary valence of 4 and a secondary valence of 6, Werner could explain the properties of the $\ce{PtCl4·NH3}$ adducts by the following reactions, where the metal complex is enclosed in square brackets:
\begin{align*} \mathrm{[Pt(NH_3)_6]Cl_4} &\rightarrow \mathrm{[Pt(NH_3)_6]^{4+}(aq)+4Cl^-(aq)} \[4pt] \mathrm{[Pt(NH_3)_5Cl]Cl_3} &\rightarrow \mathrm{[Pt(NH_3)_5Cl]^{3+}(aq) +3Cl^-(aq)}\[4pt] \mathrm{[Pt(NH_3)_4Cl_2]Cl_2} &\rightarrow \mathrm{[Pt(NH_3)_4Cl_2]^{2+}(aq) +2Cl^-(aq)}\[4pt] \mathrm{[Pt(NH_3)_3Cl_3]Cl} &\rightarrow \mathrm{[Pt(NH_3)_3Cl_3]^+(aq) + Cl^-(aq)}\[4pt] \mathrm{[Pt(NH_3)_2Cl_4]} &\rightarrow \mathrm{[Pt(NH_3)_2Cl_4]^0(aq)} \end{align*} \label{23.9}
Further work showed that the two missing members of the series—[Pt(NH3)Cl5] and [PtCl6]2−—could be prepared as their mono- and dipotassium salts, respectively. Similar studies established coordination numbers of 6 for Co3+ and Cr3+ and 4 for Pt2+ and Pd2+.
Werner’s studies on the analogous Co3+ complexes also allowed him to propose a structural model for metal complexes with a coordination number of 6. Thus he found that [Co(NH3)6]Cl3 (yellow) and [Co(NH3)5Cl]Cl2 (purple) were 1:3 and 1:2 electrolytes. Unexpectedly, however, two different [Co(NH3)4Cl2]Cl compounds were known: one was red, and the other was green (Figure $\PageIndex{1a}$). Because both compounds had the same chemical composition and the same number of groups of the same kind attached to the same metal, there had to be something different about the arrangement of the ligands around the metal ion. Werner’s key insight was that the six ligands in [Co(NH3)4Cl2]Cl had to be arranged at the vertices of an octahedron because that was the only structure consistent with the existence of two, and only two, arrangements of ligands (Figure $\PageIndex{1b}$. His conclusion was corroborated by the existence of only two different forms of the next compound in the series: Co(NH3)3Cl3.
Example $1$
In Werner’s time, many complexes of the general formula MA4B2 were known, but no more than two different compounds with the same composition had been prepared for any metal. To confirm Werner’s reasoning, calculate the maximum number of different structures that are possible for six-coordinate MA4B2 complexes with each of the three most symmetrical possible structures: a hexagon, a trigonal prism, and an octahedron. What does the fact that no more than two forms of any MA4B2 complex were known tell you about the three-dimensional structures of these complexes?
Given: three possible structures and the number of different forms known for MA4B2 complexes
Asked for: number of different arrangements of ligands for MA4B2 complex for each structure
Strategy:
Sketch each structure, place a B ligand at one vertex, and see how many different positions are available for the second B ligand.
Solution
The three regular six-coordinate structures are shown here, with each coordination position numbered so that we can keep track of the different arrangements of ligands. For each structure, all vertices are equivalent. We begin with a symmetrical MA6 complex and simply replace two of the A ligands in each structure to give an MA4B2 complex:
For the hexagon, we place the first B ligand at position 1. There are now three possible places for the second B ligand: at position 2 (or 6), position 3 (or 5), or position 4. These are the only possible arrangements. The (1, 2) and (1, 6) arrangements are chemically identical because the two B ligands are adjacent to each other. The (1, 3) and (1, 5) arrangements are also identical because in both cases the two B ligands are separated by an A ligand.
Turning to the trigonal prism, we place the first B ligand at position 1. Again, there are three possible choices for the second B ligand: at position 2 or 3 on the same triangular face, position 4 (on the other triangular face but adjacent to 1), or position 5 or 6 (on the other triangular face but not adjacent to 1). The (1, 2) and (1, 3) arrangements are chemically identical, as are the (1, 5) and (1, 6) arrangements.
In the octahedron, however, if we place the first B ligand at position 1, then we have only two choices for the second B ligand: at position 2 (or 3 or 4 or 5) or position 6. In the latter, the two B ligands are at opposite vertices of the octahedron, with the metal lying directly between them. Although there are four possible arrangements for the former, they are chemically identical because in all cases the two B ligands are adjacent to each other.
The number of possible MA4B2 arrangements for the three geometries is thus: hexagon, 3; trigonal prism, 3; and octahedron, 2. The fact that only two different forms were known for all MA4B2 complexes that had been prepared suggested that the correct structure was the octahedron but did not prove it. For some reason one of the three arrangements possible for the other two structures could have been less stable or harder to prepare and had simply not yet been synthesized. When combined with analogous results for other types of complexes (e.g., MA3B3), however, the data were best explained by an octahedral structure for six-coordinate metal complexes.
Exercise $1$
Determine the maximum number of structures that are possible for a four-coordinate MA2B2 complex with either a square planar or a tetrahedral symmetrical structure.
Answer
square planar, 2; tetrahedral, 1
Structures of Metal Complexes
The coordination numbers of metal ions in metal complexes can range from 2 to at least 9. In general, the differences in energy between different arrangements of ligands are greatest for complexes with low coordination numbers and decrease as the coordination number increases. Usually only one or two structures are possible for complexes with low coordination numbers, whereas several different energetically equivalent structures are possible for complexes with high coordination numbers (n > 6). The following presents the most commonly encountered structures for coordination numbers 2–9. Many of these structures should be familiar to you from our discussion of the valence-shell electron-pair repulsion (VSEPR) model because they correspond to the lowest-energy arrangements of n electron pairs around a central atom.
Compounds with low coordination numbers exhibit the greatest differences in energy between different arrangements of ligands.
Coordination Number 2
Although it is rare for most metals, this coordination number is surprisingly common for d10 metal ions, especially Cu+, Ag+, Au+, and Hg2+. An example is the [Au(CN)2] ion, which is used to extract gold from its ores. As expected based on VSEPR considerations, these complexes have the linear L–M–L structure shown here.
Coordination Number 3
Although it is also rare, this coordination number is encountered with d10 metal ions such as Cu+ and Hg2+. Among the few known examples is the HgI3 ion. Three-coordinate complexes almost always have the trigonal planar structure expected from the VSEPR model.
Coordination Number 4
Two common structures are observed for four-coordinate metal complexes: tetrahedral and square planar. The tetrahedral structure is observed for all four-coordinate complexes of nontransition metals, such as [BeF4]2−, and d10 ions, such as [ZnCl4]2−. It is also found for four-coordinate complexes of the first-row transition metals, especially those with halide ligands (e.g., [FeCl4] and [FeCl4]2−). In contrast, square planar structures are routinely observed for four-coordinate complexes of second- and third-row transition metals with d8 electron configurations, such as Rh+ and Pd2+, and they are also encountered in some complexes of Ni2+ and Cu2+.
Coordination Number 5
This coordination number is less common than 4 and 6, but it is still found frequently in two different structures: trigonal bipyramidal and square pyramidal. Because the energies of these structures are usually rather similar for most ligands, many five-coordinate complexes have distorted structures that lie somewhere between the two extremes.
Coordination Number 6
This coordination number is by far the most common. The six ligands are almost always at the vertices of an octahedron or a distorted octahedron. The only other six-coordinate structure is the trigonal prism, which is very uncommon in simple metal complexes.
Coordination Number 7
This relatively uncommon coordination number is generally encountered for only large metals (such as the second- and third-row transition metals, lanthanides, and actinides). At least three different structures are known, two of which are derived from an octahedron or a trigonal prism by adding a ligand to one face of the polyhedron to give a “capped” octahedron or trigonal prism. By far the most common, however, is the pentagonal bipyramid.
Coordination Number 8
This coordination number is relatively common for larger metal ions. The simplest structure is the cube, which is rare because it does not minimize interligand repulsive interactions. Common structures are the square antiprism and the dodecahedron, both of which can be generated from the cube.
Coordination Number 9
This coordination number is found in larger metal ions, and the most common structure is the tricapped trigonal prism, as in [Nd(H2O)9]3+.
Key Takeaways
• Coordination compounds are a major feature of the chemistry of over half the elements.
• Coordination compounds have important roles as industrial catalysts in controlling reactivity, and they are essential in biochemical processes.
Summary
Transition metals form metal complexes, polyatomic species in which a metal ion is bound to one or more ligands, which are groups bound to a metal ion. Complex ions are electrically charged metal complexes, and a coordination compound contains one or more metal complexes. Metal complexes with low coordination numbers generally have only one or two possible structures, whereas those with coordination numbers greater than six can have several different structures. Coordination numbers of two and three are common for d10 metal ions. Tetrahedral and square planar complexes have a coordination number of four; trigonal bipyramidal and square pyramidal complexes have a coordination number of five; and octahedral complexes have a coordination number of six. At least three structures are known for a coordination number of seven, which is generally found for only large metal ions. Coordination numbers of eight and nine are also found for larger metal ions. | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Coordination_Chemistry/Introduction_and_History_of_Coordination_Compounds/History_of_Coordination_Compounds.txt |
Discussion Questions
• How did the study of coordination compounds started?
• How can the number of isomers be used to determine the structure of a coordination compound?
Coordination chemistry is the study of the compounds that form between metals and ligands, where a ligand is any molecule or ion that binds to the metal. A metal complex is the unit containing the metal bound to its ligands. For example, [PtCl2(NH3)2] is the neutral metal complex where the Pt+2 metal is bound to two Cl- ligands and two NH3 ligands. If a complex is charged, it is called a complex ion (ex. [Pt(NH3)4]+2 is a complex cation). A complex ion is stabilized by formation of a coordination compound with ions of opposite charge (ex. [Pt(NH3)4]Cl2). It is convention to write the formula of a complex or complex ion inside of square brackets, while counterions are written outside of the brackets. In this convention, it is understood that ligands inside the brackets are bound directly to the metal ion, in the metal's first coordination sphere (a.k.a inner coordination sphere). Ions written outside of the brackets are assumed to be in the second coordination sphere, and they are not directly bound to the metal.
• Neutral Complex: $[CoCl_3(NH_3)_3]$
• Complex Cation: $[CO(NH_3)_6]^{3+}$
• Complex Anion: $[CoCl_4(NH_3)_2]^-$
• Coordination Compound: $K_4[Fe(CN)_6]$
A common metal complex is Ag(NH3)2+, formed when Ag+ ions are mixed with neutral ammonia molecules.
$Ag^+ + 2 NH_3 \rightarrow Ag(NH_3)_2^+ \nonumber$
A complex Ag(S2O3)23- is formed between silver ions and negative thiosulfate ions:
$Ag^+ + 2 S_2O_3^{2-} \rightarrow Ag(S_2O_3)_2^{3-} \nonumber$
The geometry and arrangement of ligands around the metal center affect the properties of a coordination compounds. Compounds with the same molecular formula can appear as isomers with very different properties. Isomers are molecules that have identical chemical formulas, but have different arrangements of atoms in space. Isomers with different geometric arrangements of ligands are called geometric isomers. Isomers whose structures are mirror images of each other are called optical isomers.
How did the study of coordination compounds started?
The coordination chemistry was pioneered by Nobel Prize winner Alfred Werner (1866-1919). He received the Nobel Prize in 1913 for his coordination theory of transition metal-amine complexes. At the start of the 20th century, inorganic chemistry was not a prominant field until Werner studied the metal-amine complexes such as $[Co(NH_3)_6Cl_3]$. Werner recognized the existence of several forms of cobalt-ammonia chloride. These compounds have different color and other characteristics. The chemical formula has three chloride ions per mole, but the number of chloride ions that precipitate with Ag+ ions per formula is not always three. He thought only ionized chloride ions will form precipitate with silver ion. In the following table, the number below the Ionized Cl- is the number of ionized chloride ions per formula. To distinguish ionized chloride from the coordinated chloride, Werner formulated the Complex formula and explained structure of the cobalt complexes.
Figure 1: Proposed Structure of Cobalt Ammonia Complexes from Number of Ionized Chloride
Solid Color Ionized Cl- Complex formula
CoCl36NH3 Yellow 3 [Co(NH3)6]Cl3
CoCl35NH3 Purple 2 [Co(NH3)5Cl]Cl2
CoCl34NH3 Green 1 trans-[Co(NH3)4Cl2]Cl
CoCl34NH3 Violet 1 cis-[Co(NH3)4Cl2]Cl
The structures of the complexes were proposed based on a coordination sphere of 6. The 6 ligands can be ammonia molecules or chloride ions. Two different structures were proposed for the last two compounds, the trans compound has two chloride ions on opposite vertices of an octahedral, whereas the the two chloride ions are adjacent to each other in the cis compound. The cis and trans compounds are known as geometric isomers.
Other cobalt complexes studied by Werner are also interesting. It has been predicted that the complex Co(NH2CH2CH2NH2)2ClNH3]2+ should exist in two forms, which are mirror images of each other. Werner isolated solids of the two forms, and structural studies confirmed his interpretations. The ligand NH2CH2CH2NH2 is ethylenediamine (en) often represented by en.
Example 1
Sketch the structures of isomers Co(en)33+ complex ion to show that they are mirror images of each other.
Solution
The images are shown on page 242 Inorganic Chemistry by Swaddle. If the triangular face of the end-amino group lie on the paper, you can draw lines to represent the en bidentate ligand. These lines will show that the two images are similar to the left-hand and right-hand screws.
From the description above, sketch the structures.
Exercise 1
• How many isomers does the complex Co(NH2CH2CH2NH2)2ClNH3]2+ have? Draw the structures of the isomers.
• How many isomers does Co(en)2Cl2+ have? Sketch the structures of the isomers.
• How many isomers does Co(NH3)4Cl2+ have? Sketch the structures of the isomers.
How are coordination compounds named?
Structures of coordination compounds can be very complicated, and their names long because the ligands may already have long names. Knowing the rules of nomenclature not only enable you to understand what the complex is, but also let you give appropriate names to them.
Often, several groups of the ligand are involved in a complex. The number of ligand molecules per complex is indicated by a Greek prefix: mono-, di- (or bis), tri-, tetra-, penta-, hexa, hepta-, octa-, nona-, (ennea-), deca- etc for 1, 2, 3, ... 10 etc. If the names of ligands already have one of these prefixes, the names are placed in parentheses. The prefices for the number of ligands become bis-, tris-, tetrakis, pentakis- etc.
For neutral ligands, their names are not changed, except the following few:
• $H_2O$: aqua
• $NH_3$: ammine (not two m's, amine is for organic compounds)
• $CO$: carbonyl
• $NO$: nitrosyl
Normal names that will not change
• C5H5N, pyradine
• NH2CH2CH2NH2, ethylenediamine
• C5H4N-C5H4N, dipyridyl
• P(C6H5)3, triphenylphosphine
• NH2CH2CH2NHCH2CH2NH2, diethylenetriamine
The last "e" in names of negative ions are changed to "o" in names of complexes. Sometimes "ide" is changed to "o". Note the following:
• Cl-, chloride -> chloro
• OH-, hydroxide -> hydroxo
• O2-, oxide -> oxo
• O2- peroxide, -> peroxo
• CN-, cyanide -> cyano
• N3-, azide -> axido
• N3-, nitride -> nitrido
• NH2-, amide -> amido
• CO32-, carbonate -> carbonato
• -ONO2-, nitrate -> nitrato (when bonded through O)
• -NO3-, nitrate -> nitro (when bonded through N)
• S2-, sulfide -> sulfido
• SCN-, thiocyanate -> thiocyanato-S
• NCS-, thiocyanate -> thiocyanato-N
• -(CH2-N(CH2COO-)2)2, ethylenediaminetetraacetato (EDTA)
The names of complexes start with the ligands, the anionic ones first, followed with neutral ligands and the metal. If the complex is negative, the name ends with "ate". At the very end are some Roman numerals representing the oxidation state of the metal.
To give and remember all rules of nomenclature are hard to do. Pay attention to the names whenever you encounter any complexes is the way to learn.
• [Co(NH3)5Cl]Cl2, Chloropentaamminecobalt(III) chloride
• [Cr(H2O)4Cl2]Cl, Dichlorotetraaquochromium(III) chloride
• K[PtCl3NH3], Potassiumtrichloroammineplatinate(II)
• PtCl2(NH3)2, Dichlorodiammineplatinum
• Co(en)3Cl3, tris(ethylenediamine)cobalt(III)chloride
• Ni(PF3)4, tetrakis(phosphorus(III)fluoride)nickel(0)
A bridging ligand is indicated by placing a m- before its name. The m- should be repeated for every bridging ligand. For example,
(H3N)3Co(OH)3Co(NH3)3, Triamminecobalt(III)-m-trihydroxotriamminecobalt(III)
Example 2
Give the structural formula for chlorotriphenylphosphinepalladium(II)- m-dichlorochlorotriphenylphosphinepalladium(II).
Solution
The structure is
(C6H5)3P Cl Cl
\ / \ /
Pd Pd
/ \ / \
Cl Cl P(C6H5)3
Exercise 2
• How many ionized chloride ions are there per formula?
• How many chloride ions act as bridges per formula in this complex?
• How many types of chloride ligands are there in this complex?
Example 3
Name the complex:
NH
(en)2Co< >Co(en)2 Cl3
OH
Solution
The name is Bis(ethylenediamine)cobalt(III)-m- imido-m-hydroxobis(ethylenediamine)cobalt(III) ion.
DISCUSSION
When this compound dissolves in water, is the solution a conductor? What are the ions present in the solution of this compound? How many moles of chloride ions are present per mole of the compound?
When potassiumtrichloroammineplatinate(II) dissolves in water, what ions are produced? What about chloropentaamminecobalt(III) chloride?
How the number of isomers be used to determine the structure of a coordination compound?
Before modern structure determination methods were developed, the study of complexes were mostly done by chemical methods and deduction. The fact that CoCl2(NH3)4Cl has only two isomer suggests that | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Coordination_Chemistry/Structure_and_Nomenclature_of_Coordination_Compounds/Intro_to_Coordination_Chemistry.txt |
There are several types of this isomerism frequently encountered in coordination chemistry and the following represents some of them. Isomers that contain the same number of atoms of each kind but differ in which atoms are bonded to one another are called structural isomers, which differ in structure or bond type. For inorganic complexes, there are three types of structural isomers: ionization, coordination, and linkage and two types of stereoisomers: geometric and optical.
Structural isomers, as their name implies, differ in their structure or bonding, which are separate from stereoisomers that differ in the spatial arrangement of the ligands are attached, but still have the bonding properties. The different chemical formulas in structural isomers are caused either by a difference in what ligands are bonded to the central atoms or how the individual ligands are bonded to the central atoms. When determining a structural isomer, you look at: (1) the ligands that are bonded to the central metal, and (2) which atom of the ligands attach to the central metal. Below is a quick look at the different types of structural isomers. The highlighted ions are the ions that switch or change somehow to make the type of structural isomer it is.
• Optical Isomers in Inorganic Complexes
Optical isomers are related as non-superimposable mirror images and differ in the direction with which they rotate plane-polarised light. These isomers are referred to as enantiomers or enantiomorphs of each other and their non-superimposable structures are described as being asymmetric.
• Stereoisomers: Geometric Isomers in cis-platin
Two compounds that have the same formula and the same connectivity do not always have the same shape. If the molecules appear to be connected the same way on paper, but differ in three dimensional space, then they are called stereoisomers. One simple example of stereoisomers from inorganic chemistry is diammine platinum dichloride. This important compound is sometimes called "platin" for short.
• Stereoisomers: Geometric Isomers in Transition Metal Complexes
The existence of coordination compounds with the same formula but different arrangements of the ligands was crucial in the development of coordination chemistry. Two or more compounds with the same formula but different arrangements of the atoms are called isomers. Because isomers usually have different physical and chemical properties, it is important to know which isomer we are dealing with if more than one isomer is possible.
• Stereoisomers: Geometric Isomers in Transition Metal Complexes II
Geometric Isomers are isomers that differ in the arrangement of the ligands around the metal or the central atom. In other words, these isomers differ from each other based on where the ligands are placed in the coordinate compound. This will be much easier to understand as examples will be considered. There are 2 main types of geometric isomers:
• Stereoisomers of Complex Metal Complexes
Stereoisomers are isomers that have the same molecular formula and ligands, but differ in the arrangement of those ligands in 3D space.
• Structural Isomers: Coordination Isomerism in Transition Metal Complexes
Coordination isomerism occurs compounds containing complex anionic and cationic parts can be thought of as occurring by interchange of some ligands from the cationic part to the anionic part. Hence, there are two complex compounds bound together, one with a negative charge and the other with a positive charge. In coordination isomers, the anion and cation complexes of a coordination compound exchange one or more ligands.
• Structural Isomers: Linkage Isomerism in Transition Metal Complexes
Linkage isomerism occurs with ambidentate ligands that are capable of coordinating in more than one way. The best known cases involve the monodentate ligands: SCN- / NCS- and NO2- / ONO-. The only difference is what atoms the molecular ligands use to attach to the central ion.
• Structural Isomers - Ionization Isomerism in Transition Metal Complexes
Ionization isomers are identical except for a ligand has exchanging places with an anion or neutral molecule that was originally outside the coordination complex. The central ion and the other ligands are identical. For example, an octahedral isomer will have five ligands that are identical, but the sixth will differ. The non-matching ligand in one compound will be outside of the coordination sphere of the other compound.
Isomers
Optical activity refers to whether or not a compound has optical isomers. A coordinate compound that is optically active has optical isomers and a coordinate compound that is not optically active does not have optical isomers. As we will discuss later, optical isomers have the unique property of rotating light. When light is shot through a polarimeter, optical isomers can rotate the light so it comes out in a different direction on the other end. Armed with the knowledge of symmetry and mirror images, optical isomers should not be very difficult. There are two ways optical isomers can be determined: using mirror images or using planes of symmetry.
Optical isomers do not exhibit symmetry and do not have identical mirror images. Let's go through a quick review of symmetry and mirror images. A mirror image of an object is that object flipped or the way the object would look in front of a mirror. For example, the mirror image of your left hand would be your right hand. Symmetry on the other hand refers to when an object looks exactly the same when sliced in a certain direction with a plane. For example imagine the shape of a square. No matter in what direction it is sliced, the two resulting images will be the same.
Method 1: The "Mirror Image Method"
The mirror images method uses a mirror image of the molecule to determined whether optical isomers exist or not. If the mirror image can be rotated in such a way that it looks identical to the original molecule, then the molecule is said to be superimposable and has no optical isomers. On the other hand, if the mirror image cannot be rotated in any way such that it looks identical to the original molecule, then the molecule is said to be non-superimposable and the molecule has optical isomers. Once again, if the mirror image is superimposable, then no optical isomers but if the mirror image is non-superimposable, then optical isomers exist.
Definition: Non-superimposable
Non-superimposable means the structure cannot be rotated in a way that one can be put on top of another. This means that no matter how the structure is rotated, it cannot be put on top of another with all points matching. An example of this is your hands. Both left and right hands are identical, but they cannot be put on top of each other with all points matching.
The examples you are most likely to need occur in octahedral complexes which contain bidentate ligands - ions like \([Ni(NH_2CH_2CH_2NH_2)_3]^{2+}\) or \([Cr(C_2O_4)_3]^{3-}\). The diagram below shows a simplified view of one of these ions. Essentially, they all have the same shape - all that differs is the nature of the "headphones".
A substance with no plane of symmetry is going to have optical isomers - one of which is the mirror image of the other. One of the isomers will rotate the plane of polarization of plane polarised light clockwise; the other rotates it counter-clockwise. In this case, the two isomers are:
You may be able to see that there is no way of rotating the second isomer in space so that it looks exactly the same as the first one. As long as you draw the isomers carefully, with the second one a true reflection of the first, the two structures will be different.
Method 2: The "Plane of Symmetry Method"
The plane of symmetry method uses symmetry, as it's name indicates, to identify optical isomers. In this method, one tries to see if such a plane exists which when cut through the coordinate compound produces two exact images. In other words, one looks for the existence of a plane of symmetry within the coordinate compound. If a plane of symmetry exists, then no optical isomers exist. On the other hand, if there is no plane of symmetry, the coordinate compound has optical isomers. Furthermore, if a plane of symmetry exists around the central atom, then that molecule is called achiral but if a plane of symmetry does not exist around the central molecule, then that molecule has chiral center.
Example \(1\): CHBrClF
Consider the tetrahedral molecule, CHBrClF (note the color scheme: grey=carbon, white=hydrogen, green=chlorine, blue=fluorine, red=bromine)
Is this molecule optically active? In other words, does this molecule have optical isomers?
Solution
First take the Mirror-image method. The mirror image of the molecule is:
Note that this mirror image is not superimposable. In other words, the mirror image above cannot be rotated in any such way that it looks identical to the original molecule. Remember, if the mirror image is not superimposable, then optical isomers exist. Thus we know that this molecule has optical isomers.
Let's try approaching this problem using the symmetry method. If we take the original molecule and draw an axis or plane of symmetry down the middle, this is what we get:
Since the left side is not identical to the right, this molecule does not have a symmetrical center and thus can be called chiral.Additionally, because it does not have a symmetrical center, we can conclude that this molecule has optical isomers. In general, when dealing with a tetrahedral molecule that has 4 different ligands, optical isomers will exist most of the time.
No matter which method you use, the answer will end up being the same.
Optical isomers because they have no plane of symmetry. In the organic case, for tetrahedral complexes, this is fairly easy to recognize the possibility of this by looking for a center atom with four different things attached to it. Unfortunately, this is not quite so easy with more complicated geometries!
Example \(1\): \(\ce{PFeCl3F3}\)
This time we will be analyzing the octahedral compound FeCl3F3. Is this molecule optically active?
(note the color scheme: orange=iron, blue=fluorine, green=chlorine):
Solution
If we try to attempt this problem using the mirror image method, we notice that the mirror image is essentially identical to the original molecule. In other words, the mirror image can be placed on top of the original molecule and is thus superimposable. Since the mirror image is superimposable, this molecule does not have any optical isomers. Let's attempt this same problem using the symmetry method. If we draw an axis or plane of symmetry, this is what we get:
Since the left side is identical to the right side, this molecule has a symmetrical center and is an achiral molecule. Thus, it has no optical isomers.
What is a Polarimeter?
A polarimeter is a scientific instrument used to measure the angle of rotation caused by passing polarized light through an optically active substance. Some chemical substances are optically active, and polarized (uni-directional) light will rotate either to the left (counter-clockwise) or right (clockwise) when passed through these substances. The amount by which the light is rotated is known as the angle of rotation. The angle of rotation is basically known as observed angle.
The polarimeter is made up of a polarizer (#3 on Figure \(1\)) and an analyzer (#7 on Figure \(1\)). The polarizer allows only those light waves which move in a single plane. This causes the light to become plane polarized. When the analyzer is also placed in a similar position it allows the light waves coming from the polarizer to pass through it. When it is rotated through the right angle no waves can pass through the right angle and the field appears to be dark. If now a glass tube containing an optically active solution is placed between the polarizer and analyzer the light now rotates through the plane of polarization through a certain angle, the analyzer will have to be rotated in same angle.
Nomenclature of Optical Isomers
Various methods have been used to denote the absolute configuration of optical isomers such as R or S, Λ or Δ, or C and A. The IUPAC rules suggest that for general octahedral complexes C/A scheme is convenient to use and that for bis and tris bidentate complexes the absolute configuration be designated Lambda Λ (left-handed) and Delta Δ (right-handed).
Priorities are assigned for mononuclear coordination systems based on the standard sequence rules developed for enantiomeric carbon compounds by Cahn, Ingold and Prelog (CIP rules). These rules use the coordinating atom to arrange the ligands into a priority order such that the highest atomic number gives the highest priority number (smallest CIP number). For example the hypothetical complex [Co Cl Br I NH3 NO2 SCN]2- would assign the I- as 1, Br as 2, Cl as 3, SCN as 4, NO2 as 5 and NH3 as 6.
Figure \(2\): Here is one isomer where the I and Cl, and Br and NO2 were found to be trans- to each other.
The reference axis for an octahedral center is that axis containing the ligating atom of CIP priority 1 and the trans ligating atom of lowest possible priority (highest numerical value). The atoms in the coordination plane perpendicular to the reference axis are viewed from the ligand having that highest priority (CIP priority 1) and the clockwise and anticlockwise sequences of priority numbers are compared. The structure is assigned the symbol C or A, according to whether the clockwise (C) or anticlockwise (A) sequence is lower at the first point of difference. In the example shown above this would be C.
The two optical isomers of [Co(en)3]3+ have identical chemical properties and just denoting their absolute configuration does NOT give any information regarding the direction in which they rotate plane-polarised light. This can ONLY be determined from measurement and then the isomers are further distinguished by using the prefixes (-) and (+) depending on whether they rotate left or right.
To add to the confusion, when measured at the sodium D line (589 nm), the tris(1,2-diaminoethane)M(III) complexes (M= Rh(III) and Co(III)) with IDENTICAL absolute configuration, rotate plane polarized light in OPPOSITE directions! The left-handed (Λ)-[Co(en)3]3+ isomer gives a rotation to the right and therefore corresponds to the (+) isomer. Since the successful resolution of an entirely inorganic ion (containing no C atoms) (hexol) only a handful of truly inorganic complexes have been isolated as their optical isomers e.g. (NH4)2Pt(S5)3.2H2O.
For tetrahedral complexes, R and S would be used in a similar method to tetrahedral Carbon species and although it is predicted that tetrahedral complexes with 4 different ligands should be able to give rise to optical isomers, in general they are too labile and can not be isolated.
Contributors and Attributions
• The Department of Chemistry, University of the West Indies)
• Jim Clark (Chemguide.co.uk)
Optical Isomers in Inorganic Complexes
Enantiomers are another kind of isomer that occur in octahedral metal complexes. Like the square planar platinum complexes seen before, these compounds consist of metal ions with other atoms or groups bound to them. More information about the binding of ligands to metals is found in the section on Lewis acids and bases. These groups that bind to metal ions are called ligands. One example of a type of octahedral compounds that can form enantiomers is bidentate complexes. In bidentate complexes (from the Greek two teeth) or chelating complexes (from the Greek crab), a ligand binds very tightly to the metal because it holds onto the metal via more than one atom. Ethylenediamine is one example of a bidentate ligand.
Bidentate ligands bind very tightly to a metal because they form two bonds with it, rather than just one.
In an octahedral complex, the two donor atoms in a bidentate ligand bind cis to each other. They cannot reach all the way around the molecule to bind trans to each other.
The spatial relationship between the metal and the two atoms connected to it from the same ligand forms a plane. If more than one bidentate ligand is connected to the metal, the relative orientation of one plane to another creates the possibility of mirror images. A complex containing three bidentate ligands can take on the shape of a left-handed propeller or a right-handed propeller.
These shapes are alternatively described as a left-handed screw and a right handed screw. If you can picture turning the shape so that it screws into the page behind it, which direction would you turn the screwdriver? If you would twist the screwdriver clockwise, then you have a right handed screw. If you would twist the screwdriver counter-clockwise, then you have a left handed screw.
What do we know about enantiomers?
• enantiomers have identical physical properties, except...
• enantiomers have opposite optical rotation.
These kinds of complexes were historically important in demonstrating how small molecules and ions bound to metal cations. By showing that some metal complexes were chiral and displayed optical activity, early 20th century workers such as Alfred Werner were able to rule out some competing ideas about the structures of metal compounds. Today, we know that metal complexes play important roles in enzymes in biology, and Werner's work on metal complexes laid the groundwork for how we think about these complexes.
In addition, stereochemistry in metal complexes became very important in the late 20th century, especially as pharmaceutical companies looked for catalysts that could aid in the production of one enantiomer of a drug, and not the other, in order to maximize pharmaceutical effectiveness and minimize side effects.
Problem SC18.1.
Identify the relationships between the following pairs of iron(III) oxalate ions.
Problem SC18.2.
Indicate whether each of the following compounds is the Δ or Λ isomer and draw its enantiomer.
Exercise: How many stereoisomers does an octahedral complex have? (contribution from B.J. Johnson) | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Coordination_Chemistry/Structure_and_Nomenclature_of_Coordination_Compounds/Isomers/Optical_Isomers_in_Inorganic_Complexes/Enantiomers_in_Octahedr.txt |
The symmetry of a molecule is determined by the existence of symmetry operations performed with respect to symmetry elements. A symmetry element is a line, a plane or a point in or through an object, about which a rotation or reflection leaves the object in an orientation indistinguishable from the original. A plane of symmetry and the reflection operation is the coincidence of atoms on one side of the plane with corresponding atoms on the other side, as though reflected in a mirror.
Nine Planes of Potential Symmetry
Identifying planes of symmetry in an octahedral geometry can be daunting in evaluating if a molecule is chiral or not. There is a simple approach, which requires identying and testing six possible planes of symmetry that may exist for this geometry. To demonstrated this, we will consider the \(\ce{[Cr(CO)6]}\) complex in Figure \(1\).
While you can no doubt recognize at least one plane of symmetry to confirm that this is an achiral complex, let's identify all six of them. First we have the three planes that include the \(Cr\) metal center (as they all do) and four of the attached ligands (Figure \(2\)). These are often the easy planes to identify.
The next six planes are harder to identify and can be viewed as rotations of those in Figure \(2\) by 45°. These planes bisect one angle between ligands, but still have two ligands on the plane (instead of four in the planes of Figure \(2\)).
The next three planes are even harder to view on the same perspective but are also rotations of those in Figure \(2\) by 45° in the perpendicular direction. These planes bisect an angle between ligands, but still have two ligands on the plane.
All nine planes, in each of planes are combined in the images below.
Now that can you identify these nine planes that may be mirror planes in an octahedral complex, you have to evaluate if they truly are mirror planes. That is simply asking if the ligand fields above the plane is IDENTICAL to the ligand field below. For the \(\ce{[Cr(CO)]}\) complex discussed above, all nine planes are indeed mirror planes. However, in other complexes that may not be the case. Fortunately, to determine if a complex is chiral (optically active), then only one plane needs to be identified.
Polarimetry
In measuring optical rotation, plane-polarized light travels down a long tube containing the sample. If it is a liquid, the sample may be placed in the tube as a pure liquid (its is sometimes called a neat sample). Usually, the sample is dissolved in a solvent and the resulting solution is placed in the tube. There are important factors affecting the outcome of the experiment.
• Optical rotation depends on the number of molecules encountered by the light during the experiment.
• Two factors can be controlled in the experiment and must be accounted for when comparing an experimental result to a reported value.
• The more concentrated the sample (the more molecules per unit volume), the more molecules will be encountered.
• Concentrated solutions and neat samples will have higher optical rotations than dilute solutions.
• The value of the optical rotation must be corrected for concentration.
• The longer the path of light through a solution of molecules, the more molecules will be encountered by the light, and the greater the optical rotation.
• The value of the optical rotation must be corrected for the length of the cell used to hold the sample.
Summary
$[\alpha] = \dfrac{\alpha}{c l} \nonumber$
• $\alpha$ is the measured optical rotation.
• $c$ is the sample concentration in grams per deciliter (1 dL = 10 mL), that is, c = m / V (m = mass in g, V = volume in dL).
• $l$ is the cell length in decimeters (1 dm = 10 cm = 100 mm)
• The square brackets mean the optical rotation has been corrected for these variables.
Exercise $1$
Problem SC7.1.
A pure sample of the naturally-occurring, chiral compound A (0.250 g) is dissolved in acetone (2.0 mL) and the solution is placed in a 0.5 dm cell. Three polarimetry readings are recorded with the sample: 0.775o, 0.806o, 0.682o.
1. What is [a]?
2. What would be the [a] value of the opposite enantiomer?
Answer
TBA
Exercise $2$
Problem SC7.2.
A pure sample of the (+) enantiomer of compound B shows [a] = 32o. What would be the observed a if a solution of the sample was made by dissolving 0.150 g in 1.0 mL of dichloromethane and was then placed in a 0.5 dm cell?
Answer
TBA | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Coordination_Chemistry/Structure_and_Nomenclature_of_Coordination_Compounds/Isomers/Optical_Isomers_in_Inorganic_Complexes/Identifying_Planes_of_S.txt |
The existence of coordination compounds with the same formula but different arrangements of the ligands was crucial in the development of coordination chemistry. Two or more compounds with the same formula but different arrangements of the atoms are called isomers. Because isomers usually have different physical and chemical properties, it is important to know which isomer we are dealing with if more than one isomer is possible. Recall that in many cases more than one structure is possible for organic compounds with the same molecular formula; examples discussed previously include n-butane versus isobutane and cis-2-butene versus trans-2-butene. As we will see, coordination compounds exhibit the same types of isomers as organic compounds, as well as several kinds of isomers that are unique.
Planar Isomers
Metal complexes that differ only in which ligands are adjacent to one another (cis) or directly across from one another (trans) in the coordination sphere of the metal are called geometrical isomers. They are most important for square planar and octahedral complexes.
Because all vertices of a square are equivalent, it does not matter which vertex is occupied by the ligand B in a square planar MA3B complex; hence only a single geometrical isomer is possible in this case (and in the analogous MAB3 case). All four structures shown here are chemically identical because they can be superimposed simply by rotating the complex in space:
For an MA2B2 complex, there are two possible isomers: either the A ligands can be adjacent to one another (cis), in which case the B ligands must also be cis, or the A ligands can be across from one another (trans), in which case the B ligands must also be trans. Even though it is possible to draw the cis isomer in four different ways and the trans isomer in two different ways, all members of each set are chemically equivalent:
Because there is no way to convert the cis structure to the trans by rotating or flipping the molecule in space, they are fundamentally different arrangements of atoms in space. Probably the best-known examples of cis and trans isomers of an MA2B2 square planar complex are cis-Pt(NH3)2Cl2, also known as cisplatin, and trans-Pt(NH3)2Cl2, which is actually toxic rather than therapeutic.
The anticancer drug cisplatin and its inactive trans isomer. Cisplatin is especially effective against tumors of the reproductive organs, which primarily affect individuals in their 20s and were notoriously difficult to cure. For example, after being diagnosed with metastasized testicular cancer in 1991 and given only a 50% chance of survival, Lance Armstrong was cured by treatment with cisplatin.
Square planar complexes that contain symmetrical bidentate ligands, such as [Pt(en)2]2+, have only one possible structure, in which curved lines linking the two N atoms indicate the ethylenediamine ligands:
Octahedral Isomers
Octahedral complexes also exhibit cis and trans isomers. Like square planar complexes, only one structure is possible for octahedral complexes in which only one ligand is different from the other five (MA5B). Even though we usually draw an octahedron in a way that suggests that the four “in-plane” ligands are different from the two “axial” ligands, in fact all six vertices of an octahedron are equivalent. Consequently, no matter how we draw an MA5B structure, it can be superimposed on any other representation simply by rotating the molecule in space. Two of the many possible orientations of an MA5B structure are as follows:
If two ligands in an octahedral complex are different from the other four, giving an MA4B2 complex, two isomers are possible. The two B ligands can be cis or trans. Cis- and trans-[Co(NH3)4Cl2]Cl are examples of this type of system:
Replacing another A ligand by B gives an MA3B3 complex for which there are also two possible isomers. In one, the three ligands of each kind occupy opposite triangular faces of the octahedron; this is called the fac isomer (for facial). In the other, the three ligands of each kind lie on what would be the meridian if the complex were viewed as a sphere; this is called the mer isomer (for meridional):
Example \(1\)
Draw all the possible geometrical isomers for the complex [Co(H2O)2(ox)BrCl], where ox is O2CCO2, which stands for oxalate.
Given: formula of complex
Asked for: structures of geometrical isomers
Solution
This complex contains one bidentate ligand (oxalate), which can occupy only adjacent (cis) positions, and four monodentate ligands, two of which are identical (H2O). The easiest way to attack the problem is to go through the various combinations of ligands systematically to determine which ligands can be trans. Thus either the water ligands can be trans to one another or the two halide ligands can be trans to one another, giving the two geometrical isomers shown here:
In addition, two structures are possible in which one of the halides is trans to a water ligand. In the first, the chloride ligand is in the same plane as the oxalate ligand and trans to one of the oxalate oxygens. Exchanging the chloride and bromide ligands gives the other, in which the bromide ligand is in the same plane as the oxalate ligand and trans to one of the oxalate oxygens:
This complex can therefore exist as four different geometrical isomers.
Exercise \(1\)
Draw all the possible geometrical isomers for the complex [Cr(en)2(CN)2]+.
Answer
Two geometrical isomers are possible: trans and cis.
Summary
Many metal complexes form isomers, which are two or more compounds with the same formula but different arrangements of atoms. Structural isomers differ in which atoms are bonded to one another, while geometrical isomers differ only in the arrangement of ligands around the metal ion. Ligands adjacent to one another are cis, while ligands across from one another are trans. | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Coordination_Chemistry/Structure_and_Nomenclature_of_Coordination_Compounds/Isomers/Stereoisomers%3A_Geometric_Isomers_in_Transition_Metal_Complex.txt |
Learning Objectives
• To understand that there may be more than one way to arrange the same groups around the same atom with the same geometry (stereochemistry).
Two compounds that have the same formula and the same connectivity do not always have the same shape. There are two reasons why this may happen. In one case, the molecule may be flexible, so that it can twist into different shapes via rotation around individual sigma bonds. This phenomenon is called conformation, and it is covered in a different chapter. The second case occurs when two molecules appear to be connected the same way on paper, but are connected in two different ways in three dimensional space. These two, different molecules are called stereoisomers.
One simple example of stereoisomers from inorganic chemistry is diammine platinum dichloride, (NH3)2PtCl2. This important compound is sometimes called "platin" for short. As the formula implies, it contains a platinum ion that is coordinated to two ammonia ligands and two chloride ligands (remember, a ligand in inorganic chemistry is an electron donor that is attached to a metal atom, donating a pair of electrons to form a bond).
Platin is an example of a coordination compound. The way the different pieces of coordination compounds bond together is discussed in the chapter of Lewis acids and bases. For reasons arising from molecular orbital interactions, platin has a square planar geometry at the platinum atom. That arrangement results in two possible ways the ligands could be connected. The two sets of like ligands could be connected on the same side of the square or on opposite corners.
These two arrangements result in two different compounds; they are isomers that differ only in three-dimensional space.
• The one with the two amines beside each other is called cis-platin.
• These two ligands are 90 degrees from each other.
• The one with the amines across from each other is trans-platin.
• These two ligands are 180 degrees from each other.
CIS/TRANS isomers have different physical properties
Although these two compounds are very similar, they have slightly different physical properties. Both are yellow compounds that decompose when heated to 270 degrees C, but trans-platin forms pale yellow crystals and is more soluble than cis-platin in water.
CIS/TRANS isomers have different biological properties
Cis-platin has clinical importance in the treatment of ovarian and testicular cancers. The biological mechanism of the drug's action was long suspected to involve binding of the platinum by DNA. Further details were worked out by MIT chemist Steve Lippard and graduate student Amy Rosenzweig in the 1990's. Inside the cell nucleus, the two ammines in cis-platin can be replaced by nitrogen donors from a DNA strand. To donate to the Lewis acidic platinum, the DNA molecule must bend slightly. Normally that bend is detected and repaired by proteins in the cell. However, ovarian and testicular cells happen to contain a protein that is just the right shape to fit around this slightly bent DNA strand. The DNA strand becomes lodged in the protein and can't be displaced, and so it is unable to bind with other proteins used in DNA replication. The cell becomes unable to replicate, and so cancerous growth is stopped.
Exercise SC2.1
Draw the cis and trans isomers of the following compounds:
1. (NH3)2IrCl(CO)
2. (H3P)2PtHBr
3. (AsH3)2PtH(CO)
Exercise SC2.2
Only one isomer of (tmeda)PtCl2 is possible [tmeda = (CH3)2NCH2CH2N(CH3)2; both nitrogens connect to the platinum]. Draw this isomer and explain why the other isomer is not possible. | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Coordination_Chemistry/Structure_and_Nomenclature_of_Coordination_Compounds/Isomers/Stereoisomers%3A_Geometric_Isomers_in_cis-platin.txt |
Stereoisomers are isomers that have the same molecular formula and ligands, but differ in the arrangement of those ligands in 3D space.
Contributors and Attributions
• The Department of Chemistry, University of the West Indies)
• Angad Oberoi (UCD)
Structural Isomers: Coordination Isomerism in Transition Met
Coordination isomerism occurs in compounds containing complex anionic and cationic parts and can be viewed as the interchange of one or more ligands between the cationic complex ion and the anionic complex ion. For example, \(\ce{[Co(NH3)6][Cr(CN)6]}\) is a coordination isomer with \(\ce{[Cr(NH3)6][Co(CN)6]}\). Alternatively, coordination isomers may be formed by switching the metals between the two complex ions like \(\ce{[Zn(NH3)4][CuCl4]}\) and \(\ce{[Cu(NH3)4][ZnCl4]}\).
Exercise \(1\)
Are \(\ce{[Cu(NH3)4][PtCl4]}\) and \(\ce{[Pt(NH3)4][CuCl4]}\) coordination isomers?
Solution
Here, both the cation and anion are complex ions. In the first isomer, \(\ce{NH3}\) is attached to the copper and the \(\ce{Cl^{-}}\) are attached to the platinum. In the second isomer, they have swapped.
Yes, they are coordination isomers.
Exercise \(2\)
What is one coordination isomer of \(\ce{[Co(NH3)6] [Cr(C2O4)3]}\)?
Solution
Coordination isomers involve swapping the species from the inner coordination sphere to one metal (e.g, cation) to inner coordination sphere of a different metal (e.g., the anion) in the compound. One isomer is completely swapping the ligand sphere, e.g, \(\ce{[Co(C2O4)3] [Cr(NH3)6]}\).
Alternative coordination isomers are \(\ce{ [Co(NH3)4(C2O4)] [Cr(NH3)2(C2O4)2]}\) and \(\ce{ [Co(NH3)2(C2O4)2] [Cr(NH3)4(C2O4)]}\).
Contributors and Attributions
• The Department of Chemistry, University of the West Indies)
Structural Isomers: Linkage Isomerism in Transition Metal Co
Linkage isomerism occurs with ambidentate ligands that are capable of coordinating in more than one way. The best known cases involve the monodentate ligands: \(SCN^- / NCS^-\) and \(NO_2^- / ONO^-\). The only difference is what atoms the molecular ligands bind to the central ion. The ligand(s) must have more than one donor atom, but bind to ion in only one place. For example, the (NO2-) ion is a ligand can bind to the central atom through the nitrogen or the oxygen atom, but cannot bind to the central atom with both oxygen and nitrogen at once, in which case it would be called a polydentate rather than an ambidentate ligand.
The names used to specify the changed ligands are changed as well. For example, the (NO2-) ion is called nitro when it binds with the N atom and is called nitrito when it binds with the O atom.
Example \(1\): Nitro- vs. Nitrito- Linkage Isomers
The cationic cobalt complex [Co(NH3)5(NO2)]Cl2 exists in two separable linkage isomers of the complex ion: (NH3)5(NO2)]2+.
(left) The nitro isomer (Co-NO2) and (right) the nitrito isomer (Co-ONO)
When donation is from nitrogen to a metal center, the complex is known as a nitro- complex and when donation is from one oxygen to a metal center, the complex is known as a nitrito- complex. An alternative formula structure to emphasize the different coordinate covalent bond for the two isomers
• \([Co(ONO)(NH_3)_5]Cl\): the nitrito isomer -O attached
• \([Co(NO_2)(NH_3)_5]Cl\): the nitro isomer - N attached.
The formula of the complex is unchanged, but the properties of the complex may differ.
Another example of an ambidentate ligans is thiocyanate, SCN, which can attach at either the sulfur atom or the nitrogen atom. Such compounds give rise to linkage isomerism. Polyfunctional ligands can bond to a metal center through different ligand atoms to form various isomers. Other ligands that give rise to linkage isomers include selenocyanate, SeCN – isoselenocyanate, NCSeand sulfite, SO32.
Exercise \(1\)
Are [FeCl5(NO2)]3– and [FeCl5(ONO)]3– linkage isomers?
Solution
Here, the difference is in how the ligand bonds to the metal. In the first isomer, the ligand bonds to the metal through an electron pair on the nitrogen. In the second isomer, the ligand bonds to the metal through an electron pair on one of the oxygen atoms. It's easier to see it:
Contributors and Attributions
• The Department of Chemistry, University of the West Indies) | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Coordination_Chemistry/Structure_and_Nomenclature_of_Coordination_Compounds/Isomers/Stereoisomers_of_Complex_Metal_Complexes.txt |
Coordination isomerism is a form of structural isomerism in which the composition of the complex ion varies. In a coordination isomer the total ratio of ligand to metal remains the same, but the ligands attached to a specific metal ion change. Ionization isomers can be thought of as occurring because of the formation of different ions in solution.
Introduction
Ionization isomers are identical except for a ligand has exchanged places with an anion or neutral molecule that was originally outside the coordination complex. The central ion and the other ligands are identical. For example, an octahedral isomer will have five ligands that are identical, but the sixth will differ. The non-matching ligand in one compound will be outside of the coordination sphere of the other compound. Because the anion or molecule outside the coordination sphere is different, the chemical properties of these isomers is different.
The difference between the ionization isomers can be viewed within the context of the ions generated when each are dissolved in solution.
For example, when pentaaquabromocobaltate(II)chloride is dissolved in water, $Cl^-$ ions are generated:
$CoBr(H_2O)_5Cl {(s)} \rightarrow CoBr(H_2O)^+_{5} (aq) + Cl^+ (aq) \nonumber$
whereas when pentaaquachlorocobaltate(II)bromide is dissolved, $Br^-$ ions are generated:
$CoCl(H_2O)_5Br {(s)} \rightarrow CoCl(H_2O)^+_{5} (aq) + Br^+ (aq). \nonumber$
Note
If one dissolved $[PtBr(NH_3)_3]NO_2$ and $[Pt(NO_2)(NH_3)_3]Br$ into solution, then two different set of ions will be general.
• Dissolving $[Pt(NO_2)(NH_3)_3]Br$ in aqueous solution would have the following reaction
$[PtBr(NH_3)_3]NO_2 (s) \rightarrow [PtBr(NH_3)_3]^+ (aq) + NO_2^- (aq) \label{R1}$
• Dissolving of $[Pt(NO_2)(NH_3)_3]Br$ in aqueous solution would be
$[Pt(NO_2)(NH_3)_3]Br (s) \rightarrow [Pt(NO_2)(NH_3)_3]^+ (aq) + Br^- (aq) \label{R2}$
Notice that these two ionization isomers differ in that one ion is directly attached to the central metal, but the other is not.
Equations $\ref{R1}$ and $\ref{R2}$ are valid under the assumption that the platinum-ligand bonds of the complexes are stable (i.e., not labile). Otherwise, they may break and other ligands (e.g., water) may bind.
Example $1$
Are $\ce{[Cr(NH3)5(OSO3)]Br}$ and $\ce{[Cr(NH3)5Br]SO4}$ coordination isomers?
Solution
First, we need confirm that each compound has the same number of atoms of the respective elements (this requires viewing both cations and anions of each compound).
Element number of atoms in $\ce{[Cr(NH3)5(OSO3)]Br}$ number of atoms in $\ce{[Cr(NH3)5Br]SO4}$
$\ce{Cr}$ 1 1
$\ce{N}$ 5 5
$\ce{H}$ 15 15
$\ce{O}$ 4 4
$\ce{S}$ 1 1
$\ce{Br}$ 1 1
Now, let's look at what these two compounds look like (Figure $2$). The sulfate group is a ligand with a dative bond to the chromium atom and the bromide counter ion ($\ce{[Cr(NH3)5(OSO3)]Br}$). For $\ce{[Cr(NH3)5Br]SO4}$, this is the the reverse.
Yes, $\ce{[Cr(NH3)5(OSO3)]Br}$ and $\ce{[Cr(NH3)5Br]SO4}$ are coordination isomers.
Exercise $1$
Are [Co(NH3)5(SO4)]Br and [Co(NH3)5Br]SO4 ionization isomers?
Solution
In the first isomer, SO4 is attached to the Cobalt and is part of the complex ion (the cation), with Br as the anion. In the second isomer, Br is attached to the cobalt as part of the complex and SO4 is acting as the anion.
A hydrate isomer is a specific kind of ionization isomer where a water molecule is one of the molecules that exchanges places.
Solvate or Hydrate Isomerization: A Special kind of Ionization Isomer
A very similar type of isomerism results from replacement of a coordinated group by a solvent molecule (Solvate Isomerism). In the case of water, this is called Hydrate isomerism. The best known example of this occurs for chromium chloride "CrCl3.6H2O" which may contain 4, 5, or 6 coordinated water molecules.
• $[CrCl_2(H_2O)_4]Cl \cdot 2H_2O$: bright-green colored
• $[CrCl(H_2O)_5]Cl_2 \cdot H_2O$: grey-green colored
• $[Cr(H_2O)_6]Cl_3$: violet colored
These isomers have very different chemical properties and on reaction with $AgNO_3$ to test for $Cl^-$ ions, would find 1, 2, and 3 $Cl^-$ ions in solution respectively.
Contributors and Attributions
• The Department of Chemistry, University of the West Indies) | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Coordination_Chemistry/Structure_and_Nomenclature_of_Coordination_Compounds/Isomers/Structural_Isomers_-_Ionization_Isomerism_in_Transition_Metal_.txt |
A metal ion in solution does not exist in isolation, but in combination with ligands (such as solvent molecules or simple ions) or chelating groups, giving rise to complex ions or coordination compounds. These complexes contain a central atom or ion, often a transition metal, and a cluster of ions or neutral molecules surrounding it. Ligands are ions or neutral molecules that bond to a central metal atom or ion. Ligands act as Lewis bases (electron pair donors), and the central atom acts as a Lewis acid (electron pair acceptor). Ligands have at least one donor atom with an electron pair used to form covalent bonds with the central atom.
The term ligand come from the latin word ligare (which meaning to bind) was first used by Alfred Stock in 1916 in relation to silicon chemistry. Ligands can be anions, cations, or neutral molecules. Ligands can be further characterized as monodentate, bidentate, tridentate etc. where the concept of teeth (dent) is introduced, hence the idea of bite angle etc. A monodentate ligand has only one donor atom used to bond to the central metal atom or ion.
Monodentate Ligands
The term "monodentate" can be translated as "one tooth," referring to the ligand binding to the center through only one atom. Some examples of monodentate ligands are: chloride ions (referred to as chloro when it is a ligand), water (referred to as aqua when it is a ligand), hydroxide ions (referred to as hydroxo when it is a ligand), and ammonia (referred to as ammine when it is a ligand).
Bidentate Ligands
Bidentate ligands have two donor atoms which allow them to bind to a central metal atom or ion at two points. Common examples of bidentate ligands are ethylenediamine (en), and the oxalate ion (ox). Shown below is a diagram of ethylenediamine: the nitrogen (blue) atoms on the edges each have two free electrons that can be used to bond to a central metal atom or ion.
Polydentate Ligands
Polydentate ligands range in the number of atoms used to bond to a central metal atom or ion. EDTA, a hexadentate ligand, is an example of a polydentate ligand that has six donor atoms with electron pairs that can be used to bond to a central metal atom or ion.
Unlike polydentate ligands, ambidentate ligands can attach to the central atom in two places. A good example of this is thiocyanate, \(SCN^−\), which can attach at either the sulfur atom or the nitrogen atom.
Example \(1\)
Draw metal complexes using the ligands below and metal ions of your choice.
Chelation
Chelation is a process in which a polydentate ligand bonds to a metal ion, forming a ring. The complex produced by this process is called a chelate, and the polydentate ligand is referred to as a chelating agent. The term chelate was first applied in 1920 by Sir Gilbert T. Morgan and H.D.K. Drew, who stated: "The adjective chelate, derived from the great claw or chela (chely- Greek) of the lobster or other crustaceans, is suggested for the caliperlike groups which function as two associating units and fasten to the central atom so as to produce heterocyclic rings." As the name implies, chelating ligands have high affinity for metal ions relative to ligands with only one binding group (which are called monodentate = "single tooth") ligands. Both Ethylenediamine (Figure \(2\)) and Ethylenediaminetetraaceticacid acid (Figure \(3\)) are examples of chelating agents, but many others are commonly found in the inorganic laboratory.
The Chelate Effect
The chelate effect is the enhanced affinity of chelating ligands for a metal ion compared to the affinity of a collection of similar nonchelating (monodentate) ligands for the same metal.
The macrocyclic effect follows the same principle as the chelate effect, but the effect is further enhanced by the cyclic conformation of the ligand. Macrocyclic ligands are not only multi-dentate, but because they are covalently constrained to their cyclic form, they allow less conformational freedom. The ligand is said to be "pre-organized" for binding, and there is little entropy penalty for wrapping it around the metal ion. For example heme b is a tetradentate cyclic ligand that strongly complexes transition metal ions, including Fe+2 in hemogloben (Figure \(4\)).
Some other common chelating and cyclic ligands are shown below:
• Acetylacetonate (acac-, top) is an anionic bidentate ligand that coordinates metal ions through two oxygen atoms. Acac- is a hard base so it prefers hard acid cations. With divalent metal ions, acac-forms neutral, volatile complexes such as Cu(acac)2 and Mo(acac)2 that are useful for chemical vapor deposition (CVD) of metal thin films.
• 2,2'-Bipyridine 2,2'-Bipyridine (Figure \(5\): left) and related bidentate ligands such as 1,10-phenanthroline form propeller-shaped complexes with metals such as Ru2+. The [Ru(bpy)3]2+ complex is photoluminescent and can also undergo photoredox reactions, making it an interesting compound for both photocatalysis and artificial photosynthesis.
• Crown ethers such as 18-crown-6 2,2'-Bipyridine (Figure \(5\): center) are cyclic hard bases that can complex alkali metal cations. Crowns can selectively bind Li+, Na+, or K+ depending on the number of ethylene oxide units in the ring.
• The chelating properties of crown ethers are mimetic of the natural antibiotic valinomycin ( 2,2'-Bipyridine (Figure \(5\): right), which selectively transports K+ ions across bacterial cell membranes, killing the bacterium by dissipating its membrane potential. Like crown ethers, valinomycin is a cyclic hard base.
Ligands
Contributors and Attributions
• Allen Zeng, UC Davis, SSReno, Parul Jandir, UC Davis | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Coordination_Chemistry/Structure_and_Nomenclature_of_Coordination_Compounds/Ligands/EDTA.txt |
The general stability sequence of high spin octahedral metal complexes for the replacement of water by other ligands is:
$\ce{Mn(II) < Fe(II) < Co(II) < Ni(II) < Cu(II) > Zn(II)} \nonumber$
This trend is essentially independent of the ligand. In the case of 1,2-diaminoethane (en), the first step-wise stability constants (logK1) for M(II) ions are shown below.
The Irving-Williams sequence is generally quoted ONLY for Mn(II) to Zn(II) since there is little data available for the other first row transition metal ions because their M(II) oxidation states are not very stable. The position of Cu(II) is considered out-of-line with predictions based on Crystal Field Theory and is probably a consequence of the fact that Cu(II) often forms Jahn-Teller distorted octahedral complexes.
One explanation
Crystal Field Theory is based on the idea that a purely electrostatic interaction exists between the central metal ion and the ligands. This suggests that the stability of the complexes should be related to the ionic potential; that is, the charge to radius ratio. In the Irving-Williams series, the trend is based on high-spin M(II) ions, so what needs to be considered is how the ionic radii vary across the d-block.
For free metal ions in the gaseous phase it might be expected that the ionic radius of each ion on progressing across the d-block should show a gradual decrease in size. This would come about due to the incomplete screening of the additional positive charge by the additional electron, as is observed in the Lanthanide Contraction.
For high-spin octahedral complexes it is essential to consider the effect of the removal of the degeneracy of the d-orbitals by the crystal field. Here the d-electrons will initially add to the lower t2g orbitals before filling the eg orbitals since for octahedral complexes, the t2g subset are directed in between the incoming ligands whilst the eg subset are directed towards the incoming ligands and cause maximum repulsion.
For d1-d3 (and d6-d8) the addition of the electrons to the t2g orbitals will mean that the screening of the increasing attractive nuclear charge is not very effective and the radius should be smaller than for the free ion.
The position of d4 and d9 on the plot is difficult to ascertain with certainty since six-coordinate complexes are expected to be distorted due to the Jahn-Teller Theorem. Cr(II) is not very stable so few measurements are available. For Cu(II) however, most complexes are found to have 4 short bonds and 2 long bonds although 2 short and 4 long bonds is feasible. The radii are expected to show an increase over the d3 and d8 situation since electrons are being added to the eg subset. The reported values have been found to lie on both sides of the predicted value.
For d0, d5 and d10 the screening expected is essentially that of a spherical arrangement equivalent to the absence of a crystal field. The plot above shows that these points return to the line drawn showing a gradual decrease of the radius on moving across the d-block.
Once the decrease in radius with Z pattern is understood, it is a small step to move to a pattern for q/r since this only involves taking the reciprocal of the radius and holding the charge constant. The radius essentially decreases with increasing $Z$, therefore 1/r must increase with increasing $Z$.
For the sequence Mn(II) to Zn(II), the crystal field (q/r) trend expected would be
$\ce{Mn(II) < Fe(II) < Co(II) < Ni(II) > Cu(II) > Zn(II)} \nonumber$
Apart from the position of Cu(II), this corresponds to the Irving-Williams series (Eqution 1). The discrepancy is once again accounted for by the fact that copper(II) complexes are often distorted or not octahedral at all. When this is taken into consideration, it is seen that the Irving-Williams series can be explained quite well using Crystal Field Theory.
Contributors and Attributions
• Prof. Robert J. Lancashire (The Department of Chemistry, University of the West Indies) | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Coordination_Chemistry/Structure_and_Nomenclature_of_Coordination_Compounds/Ligands/The_Irving-Williams_Series.txt |
Learning Objectives
• To learn the basis for complex ion and compound nomenclature
Coordination complexes have their own classes of isomers, different magnetic properties and colors, and various applications (photography, cancer treatment, etc), so it makes sense that they would have a naming system as well. Consisting of a metal and ligands, their formulas follow the pattern [Metal ligands]±Charge, while names are written Prefix Ligands Metal (Oxidation State).
Introduction
According to the Lewis base theory, ligands are Lewis bases since they can donate electrons to the central metal atom. The metals, in turn, are Lewis acids since they accept electrons. Coordination complexes consist of a ligand and a metal center cation. The overall charge can be positive, negative, or neutral. Coordination compounds are complex or contain complex ions, for example:
• Complex Cation: \(\ce{[Co(NH3)6]^{3+}}\)
• Complex Anion: \(\ce{[CoCl4(NH3)2]^{-}}\)
• Neutral Complex: \(\ce{[CoCl3(NH3)3]}\)
• Coordination Compound: \(\ce{K4[Fe(CN)6]}\)
A ligand can be an anion or a neutral molecule that donates an electron pair to the complex (NH3, H2O, Cl-). The number of ligands that attach to a metal depends on whether the ligand is monodentate or polydentate. To begin naming coordination complexes, here are some things to keep in mind.
1. Ligands are named first in alphabetical order.
2. The name of the metal comes next.
3. The oxidation state of the metal follows, noted by a Roman numeral in parentheses (II, IV).
Rule 1: Anionic Ligands
Ligands that act as anions which end in "-ide" are replaced with an ending "-o" (e.g., Chloride → Chloro). Anions ending with "-ite" and "-ate" are replaced with endings "-ito" and "-ato" respectively (e.g., Nitrite → Nitrito, Nitrate → Nitrato).
Table \(1\): Anionic Monodentate Ligands
Molecular Formula Ligand Name Molecular Formula Ligand Name
F- Fluoro OH- Hydroxo
Cl- Chloro SO42- Sulfato
Br- Bromo S2O32- Thiosulfato
I- Iodo NO2- Nitrito-N-; Nitro
O2- Oxo ONO- Nitrito-O-; Nitrito
CN- Cyano SCN- Thiocyanato-S-; Thiocyanato
NC- Isocyano NCS- Thiocyanato-N-; Isothiocyanato
Rule 2: Neutral Ligands
Most neutral molecules that are ligands carry their normal name. The few exceptions are the first four on the chart: ammine, aqua, carbonyl, and nitrosyl.
Table \(2\): Select Neutral Monodentate Ligands. Note: Ammine is spelled with two m's when referring to a ligand. Amines are a class of organic nitrogen-containing compounds.
Molecular Formula of Ligand Ligand Name
NH3 Ammine
H2O Aqua
CO Carbonyl
NO Nitrosyl
CH3NH2 Methylamine
C5H5N Pyridine
Polydentate ligands follow the same rules for anions and neutral molecules.
Table \(3\): Select Polydentate ligands
Short name Extended name
en Ethylenediamine
ox2- Oxalato
EDTA4- Ethylenediaminetetraacetato
Rule 3: Ligand Multiplicity
The number of ligands present in the complex is indicated with the prefixes di, tri, etc. The exceptions are polydentates that have a prefix already in their name (en and EDTA4- are the most common). When indicating how many of these are present in a coordination complex, put the ligand's name in parentheses and use bis (for two ligands), tris (for three ligands), and tetrakis (for four ligands).
Table \(4\): Prefixes for indicating number of ligands in a complex.
Number of Ligands Monodentate Ligands Polydentate Ligands
1 mono -
2 di bis
3 tri tris
4 tetra tetrakis
5 penta pentakis
6 hexa hexakis
Prefixes always go before the ligand name; they are not taken into account when putting ligands in alphabetical order. Note that "mono" often is not used. For example, \(\ce{[FeCl(CO)2(NH3)3]^{2+}}\) would be called triamminedicarbonylchloroiron(III) ion. Remember that ligands are always named first, before the metal is.
Example \(1\)
What is the name of this complex ion: \(\ce{[CrCl2(H2O)4]^{+}}\)?
Solution
Let's start by identifying the ligands. The ligands here are Cl and H2O. Therefore, we will use the monodentate ligand names of "chloro" and "aqua". Alphabetically, aqua comes before chloro, so this will be their order in the complex's name. There are 4 aqua's and 2 chloro's, so we will add the number prefixes before the names. Since both are monodentate ligands, we will say "tetra[aqua]di[chloro]".
Now that the ligands are named, we will name the metal itself. The metal is Cr, which is chromium. Therefore, this coordination complex is called tetraaquadichlorochromium(III) ion. See the next section for an explanation of the (III).
Exercise \(1\)
What is the name of this complex ion: \(\ce{[CoCl_2(en)_2]^{+}}\)?
Answer
We take the same approach. There are two chloro and ethylenediamine ligands. The metal is Co, cobalt. We follow the same steps, except that \(en\) is a polydentate ligand with a prefix in its name (ethylenediamine), so "bis" is used instead of "di", and parentheses are added. Therefore, this coordination complex is called dichlorobis(ethylenediamine)cobalt(III) ion.
Rule 4: The Metals
When naming the metal center, you must know the formal metal name and the oxidation state. To show the oxidation state, we use Roman numerals inside parenthesis. For example, in the problems above, chromium and cobalt have the oxidation state of +3, so that is why they have (III) after them. Copper, with an oxidation state of +2, is denoted as copper(II). If the overall coordination complex is an anion, the ending "-ate" is attached to the metal center. Some metals also change to their Latin names in this situation. Copper +2 will change into cuprate(II). The following change to their Latin names when part of an anion complex:
Table \(5\): Latin terms for Select Metal Ion
Transition Metal Latin
Iron Ferrate
Copper Cuprate
Tin Stannate
Silver Argentate
Lead Plumbate
Gold Aurate
The rest of the metals simply have -ate added to the end (cobaltate, nickelate, zincate, osmate, cadmate, platinate, mercurate, etc. Note that the -ate tends to replace -um or -ium, if present).
Finally, when a complex has an overall charge, "ion" is written after it. This is not necessary if it is neutral or part of a coordination compound (Example \(3\)). Here are some examples with determining oxidation states, naming a metal in an anion complex, and naming coordination compounds.
Example \(2\)
What is the name of [Cr(OH)4]- ?
Solution
Immediately we know that this complex is an anion. There is only one monodentate ligand, hydroxide. There are four of them, so we will use the name "tetrahydroxo". The metal is chromium, but since the complex is an anion, we will have to use the "-ate" ending, yielding "chromate". The oxidation state of the metal is 3 (x+(-1)4=-1). Write this with Roman numerals and parentheses (III) and place it after the metal to get tetrahydroxochromate(III) ion.
Exercise \(2\)
What is the name of \(\ce{[CuCl4]^{2-}}\)?
Answer
tetrachlorocuprate(II) ion
A last little side note: when naming a coordination compound, it is important that you name the cation first, then the anion. You base this on the charge of the ligand. Think of NaCl. Na, the positive cation, comes first and Cl, the negative anion, follows.
Example \(3\)
What is the name of \([\ce{Pt(NH3)4}][\ce{Pt(Cl)4}]\)?
Solution
NH3 is neutral, making the first complex positively charged overall. Cl has a -1 charge, making the second complex the anion. Therefore, you will write the complex with NH3 first, followed by the one with Cl (the same order as the formula). This coordination compound is called tetraammineplatinum(II) tetrachloroplatinate(II).
Exercise \(3\): The Nitro/Nitrito Ambidentate Ligand
What is the name of \(\ce{[CoCl(NO2)(NH3)4]^{+}}\) ?
Answer
This coordination complex is called tetraamminechloronitrito-N-cobalt(III). N comes before the O in the symbol for the nitrite ligand, so it is called nitrito-N. If an O came first, as in [CoCl(ONO)(NH3)4]+, the ligand would be called nitrito-O, yielding the name tetraamminechloronitrito-O-cobalt(III).
Nitro (for NO2) and nitrito (for ONO) can also be used to describe the nitrite ligand, yielding the names tetraamminechloronitrocobalt(III) and tetraamminechloronitritocobalt(III).
Writing Formulas of Coordination Complexes
While chemistry typically follow the nomenclature rules for naming complexes and compounds, there is disagreement with the rules for constructing formulas of inorganic complex. The order of ligand names in their formula has been ambiguous with different conventions being used (charged vs neutral, number of each ligand, etc.). In 2005, IUPAC adopted the recommendation that all ligand names in formulas be listed alphabetically (in the same way as in the naming convention) irrespective of the charge or number of each ligand type.
However, this rule is not adhered to in many chemistry laboratories. For practice, the order of the ligands in chemical formulas does not matter as long as you write the transition metal first, which is the stance taken here.
Examples \(4\)
Write the chemical formulas for:
1. Amminetetraaquachromium(II) ion
2. Amminesulfatochromium(II)
Solution
1. Amminetetraaquachromium(II) ion could be written as \(\ce{[Cr(H2O)4(NH3)]^{+2}}\) or \(\ce{[Cr(NH3)(H2O)4]^{+2}}\).
2. Amminesulfatochromium (II) could be written as \(\ce{[Cr(SO4)(NH3)]}\) or \(\ce{[Cr(NH3)(SO4)]}\).
Exercise \(4\)
Write the chemical formulas for
1. Amminetetraaquachromium (II) sulfate
2. Potassium hexacyanoferrate (III)
Answer
1. Amminetetraaquachromium (II) sulfate can be written as \(\ce{[Cr(H2O)4(NH3)]SO4}\). Although \(\ce{[Cr(NH3)(H2O)4]SO4}\) is also acceptable.
2. Potassium hexacyanoferrate (III) is be written as \(\ce{K3[Fe(CN)6]}\)
Contributors and Attributions
• Justin Hosung Lee (UCD), Sophia Muller (UCD) | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Coordination_Chemistry/Structure_and_Nomenclature_of_Coordination_Compounds/Nomenclature_of_Coordination_Complexes.txt |
One of the most striking characteristics of transition-metal complexes is the wide range of colors they exhibit. Crystal field theory (CFT) is a bonding model that explains many important properties of transition-metal complexes, including their colors, magnetism, structures, stability, and reactivity. The central assumption of CFT is that metal–ligand interactions are purely electrostatic in nature.
• Colors of Coordination Complexes
The color for a coordination complex can be predicted using the Crystal Field Theory (CFT). Knowing the color can have a number of useful applications, such as the creation of pigments for dyes in the textile industry. The tendency for coordination complexes to display such a wide array of colors is merely coincidental; their absorption energies happen to fall within range of the visible light spectrum.
• Crystal Field Stabilization Energy
A consequence of Crystal Field Theory is that the distribution of electrons in the d orbitals may lead to net stabilization (decrease in energy) of some complexes depending on the specific ligand field geometry and metal d-electron configurations. It is a simple matter to calculate this stabilization since all that is needed is the electron configuration and knowledge of the splitting patterns.
• Crystal Field Theory
Crystal field theory (CFT) describes the breaking of orbital degeneracy in transition metal complexes due to the presence of ligands. CFT qualitatively describes the strength of the metal-ligand bonds. Based on the strength of the metal-ligand bonds, the energy of the system is altered. This may lead to a change in magnetic properties as well as color. This theory was developed by Hans Bethe and John Hasbrouck van Vleck.
• Introduction to Crystal Field Theory
One of the most striking characteristics of transition-metal complexes is the wide range of colors they exhibit. In this section, we describe crystal field theory (CFT), a bonding model that explains many important properties of transition-metal complexes, including their colors, magnetism, structures, stability, and reactivity. The central assumption of CFT is that metal–ligand interactions are purely electrostatic in nature.
• Magnetic Moments of Transition Metals
Magnetic moments are often used in conjunction with electronic spectra to gain information about the oxidation number and stereochemistry of the central metal ion in coordination complexes. A common laboratory procedure for the determination of the magnetic moment for a complex is the Gouy method which involves weighing a sample of the complex in the presence and absence of a magnetic field and observing the difference in weight. A template is provided for the calculations involved.
• Magnetism
Movement of an electrical charge (which is the basis of electric currents) generates a magnetic field in a material. Magnetism is therefore a characteristic property of all materials that contain electrically charged particles and for most purposes can be considered to be entirely of electronic origin.
• Metals, Tetrahedral and Octahedral
• Non-octahedral Complexes
• Octahedral vs. Tetrahedral Geometries
A consequence of Crystal Field Theory is that the distribution of electrons in the d orbitals can lead to stabilization for some electron configurations. It is a simple matter to calculate this stabilization since all that is needed is the electron configuration.
• Orgel Diagrams
Orgel diagrams are useful for showing the energy levels of both high spin octahedral and tetrahedral transition metal ions.
• Tanabe-Sugano Diagrams
Tanabe Sugano diagrams are used to predict the transition energies for both spin-allowed and spin-forbidden transitions, as well as for both strong field (low spin), and weak field (high spin) complexes. In this method the energy of the electronic states are given on the vertical axis and the ligand field strength increases on the horizontal axis from left to right.
• Tetrahedral vs. Square Planar Complexes
High spin and low spin are two possible classifications of spin states that occur in coordination compounds. These classifications come from either the ligand field theory, which accounts for the energy differences between the orbitals for each respective geometry, or the crystal field theory, which accounts for the breaking of degenerate orbital states, compared to the pairing energy.
• Thermodynamics and Structural Consequences of d-Orbital Splitting
The energy level splitting of the d-orbitals due to their interaction with the ligands in a complex has important structural and thermodynamic effects on the chemistry of transition-metal complexes. Although these two types of effects are interrelated, they are considered separately here.
Crystal Field Theory
The color for a coordination complex can be predicted using the Crystal Field Theory (CFT). Knowing the color can have a number of useful applications, such as the creation of pigments for dyes in the textile industry. The tendency for coordination complexes to display such a wide array of colors is merely coincidental; their absorption energies happen to fall within range of the visible light spectrum. Chemists and physicists often study the color of a substance not to understand its sheer appearance, but because color is an indicator of a chemical's physical proprieties on the atomic level.
The Electromagnetic Spectrum
The electromagnetic spectrum (EM) spectrum is made up of photons of different wavelengths. Photons, unique in displaying the properties of both waves and particles, create visible light and colors in a small portion of the EM spectrum. This visible light portion has wavelengths in approximately the 400-700 nanometer range (a nanometer, “nm,” is 10-9 meters). Each specific wavelength corresponds to a different color (Figure $1$), and when all the wavelengths are present, it appears as white light.
The wavelength and frequency of a wave are inversely proportional: as one increases, the other decreases; this is a consequence of all light traveling at the same speed.
$\lambda \propto \nu^{-1} \nonumber$
Because of this relationship, blue light has a much higher frequency and more energy than red light.
Perceiving Color
Color is perceived in two ways, through additive mixing, where different colors are made by combining different colors of light, and through subtractive mixing, where different wavelengths of light are taken out so that the light is no longer pure white. For colors of coordination complexes, subtractive mixing is considered. As shown in Figure $2$, the idea behind subtractive mixing is that white light (which is made from all the colors mixed together) interacts with an object. The object absorbs some of the light, and then reflects or transmits (or both, depending on the object) the rest of the light, which contacts the eye. The object is perceived as whichever color is not absorbed. In Figure $2$, white light (simplified as green, red, and blue bands) is shone through a solution. The solution absorbs the red and green wavelengths; however, the blue light is reflected and passes through, so the solution appears blue. This procedure takes place whenever an object displays visible color. If none of the light is absorbed, and all is reflected back off, the object appears white; if all of the light is absorbed, and there is none left to reflect or transmit through, the object appears black.
Colors of Coordination Complexes: Crystal Field Splitting
When ligands attach to a transition metal to form a coordination complex, electrons in the d orbital split into high energy and low energy orbitals. The difference in energy of the two levels is denoted as ∆, and it is a characteristic a property both of the metal and the ligands. This is illustrated in Figure $3$; the "o" subscript on the ∆ indicates that the complex has octahedral geometry.
If ∆o is large, and much energy is required to promote electrons into the high energy orbitals, the electrons will instead pair in the lower energy orbitals, resulting in a "low spin" complex (Figure $\PageIndex{4A}$); however, if ∆o is small, and it takes little energy to occupy the higher orbitals, the electrons will do so, and remain unpaired (until there are more than five electrons), resulting in a “high spin” complex (Figure $\PageIndex{4B}$). Different ligands are associated with either high or low spina "strong field" ligand results in a large o and a low spin configuration, while a "weak field" ligand results in a small ∆o and a high spin configuration. For more details, see the
A photon equal to the energy difference o can be absorbed, promoting an electron to the higher energy level. As certain wavelengths are absorbed in this process, subtractive color mixing occurs and the coordination complex solution becomes colored. If the ions have a noble gas configuration, and have no unpaired electrons, the solutions appear colorless; in reality, they still have a measured energy and absorb certain wavelengths of light, but these wavelengths are not in the visible portion of the EM spectrum and no color is perceived by the eye.
In general, a larger $∆_o$ indicates that higher energy photons are absorbed, and the solution appears further to the left on the EM spectrum shown in Figure $1$. This relationship is described in the equation
$∆_o=hc/λ \nonumber$
where $h$ and $c$ are constants, and $λ$ is the wavelength of light absorbed.
Using a color wheel can be useful for determining what color a solution will appear based on what wavelengths it absorbs (Figure $6$). If a complex absorbs a particular color, it will have the appearance of whatever color is directly opposite it on the wheel. For example, if a complex is known to absorb photons in the orange range, it can be concluded that the solution will look blue. This concept can be used in reverse to determine ∆ for a complex from the color of its solution.
Relating the Colors of Coordination Complexes to the Spectrochemical Series
According to the Crystal Field Theory, ligands that have smaller $\Delta$) values are considered "weak field" and will absorb lower-energy light with longer $\lambda$ values (ie a "red shift"). Ligands that have larger $\Delta$) values are considered "strong field" and will absorb higher-energy light with shorter $\lambda$ values (ie a "blue shift"). This relates to the colors seen in a coordination complex. Weaker-field ligands induce the absorption of longer wavelength (lower frequency=lower energy) light than stronger-field ligands since their respective $\Delta_o$ values are smaller than the electron pairing energy.
The energy difference, $\Delta_o$, determines the color of the coordination complex. According to the spectrochemical series, the high spin ligands are considered "weak field," and absorb longer wavelengths of light (weak $\Delta_o$), while complexes with low spin ligands absorb light of greater frequency (high $\Delta_o$). The color seen is the complementary color of the color associated with the absorbed wavelength. To predict which possible colors and their corresponding wavelengths are absorbed, the spectrochemical series can be used:
(Strong field/large Δ0/low spin) (weak field/ small Δ0/high spin)
CO-, NO-, CN->NO2->en>py≈NH3>EDTA4->NCS->H2O>ONO->ox2->OH->F->SCN->Cl->Br->I-
Example $1$
If a solution with a dissolved octahedral complex appears yellow to the eye, what wavelength of light does it absorb? Is this complex expected to be low spin or high spin?
Solution
A solution that looks yellow absorbs light that is violet, which is roughly 410 nm from the color wheel. Since it absorbs high energy, the electrons must be raised to a higher level, and $\Delta_o$ is high, so the complex is likely to be low spin.
Example $2$
An octahedral metal complex absorbs light with wavelength 535 nm. What is the crystal field splitting $\Delta_o$ for the complex? What color is it to the eye?
Solution
To solve this question, we need to use the equation
$\Delta_o =\dfrac{hc}{\lambda} \nonumber$
with
• $h$ is Planck’s constant and is $6.625 \times 10^{-34} J \cdot s$ and
• $c$ is the speed of light and is $2.998 \times 10^8\, m/s$.
It is also important to remember that 1 nm is equal to $1 \times 10^{-9}$ meters. With all this information, the final equation looks like this:
$\Delta_o =\dfrac{(6.625 \times 10^{-34}\, J \cdot s)(2.998 \times 10^8\, m/s)}{(535nm) \left(\dfrac{1\,m}{1 \times 10^9\, nm}\right)}= 3.712 \,J/molecule \nonumber$
It is not necessary to use any equations to solve the second part of the problem. Light that is 535 nm is green, and because green light is absorbed, the complex appears red (refer to Figures $1$ and $6$ for this information).
Note: the fact that the complex is octahedral makes no impact when solving this problem. Although the splitting is different for complexes of different structures, the mechanics of solving the problem are identical.
Example $3$
There are two solutions, one orange and one blue. Both solutions are known to be made up of a cobalt complex; however, one has chloride ions as ligands, while the other has ammonia ligands. Which solution is expected to be orange?
Solution
In order to solve this problem, it is necessary to know the relative strengths of the ligands involved. A sample ligand strength list is given here, but see Crystal Field Splitting for a more complete list:
CN- > en > NH3 > H2O > F- >SCN- > Cl-
From this information, it is clear that NH3 is a stronger ligand than Cl-, which means that the complex involving NH3 has a greater ∆, and the complex will be low spin. Because of the larger ∆, the electrons absorb higher energy photons, and the solution will have the appearance of a lower energy color. Since orange light is less energetic than blue light, the NH3 containing solution is predicted to be orange
Problems
1. What color will a complex be that absorbs light that is 600 nm be?
2. What color will a complex an octahedral complex appear if it has a $\Delta_o$ of $3.75 \times 10^{-19}\, J$?
3. Would you expect a violet solution to be high spin or low spin? What about a red solution?
4. There are two solutions, one which is yellow and another which is violet. The solutions are [Co(H2O)6]3+and [Co(CN-)6]3-. What are the colors of each solution?
Answers
1. Blue. The color absorbed is orange.
2. It is red. Using Δ=hc/λ, h=6.626*10-34J*s, c=2.998*108m/s, wavelength would equal 530 nm. So green is absorbed, and the complementary color of green is red, so red is the color of the complex.
3. It would be high spin. The complementary color of violet is yellow, which has a wavelength of 570 nm. For a red solution, the complementary color absorbed is green, with a wavelength of 530 nm, so it would be considered low spin.
4. [Co(H2O)6]3+ is violet and [Co(CN-)6]3- is yellow. Looking at the spectrochemical series, H2O is a weak field ligand, so it absorbs colors of long wavelengths—in this case, the longer wavelength is yellow, so the color reflected is violet. CN- is a strong field ligand, so it absorbs colors of shorter wavelengths-in this case, the shorter wavelength is violet, so the color reflected is yellow.
Contributors and Attributions
• Deyu Wang (UCD) | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Crystal_Field_Theory/Colors_of_Coordination_Complexes.txt |
A consequence of Crystal Field Theory is that the distribution of electrons in the d orbitals may lead to net stabilization (decrease in energy) of some complexes depending on the specific ligand field geometry and metal d-electron configurations. It is a simple matter to calculate this stabilization since all that is needed is the electron configuration and knowledge of the splitting patterns.
Definition: Crystal Field Stabilization Energy
The Crystal Field Stabilization Energy is defined as the energy of the electron configuration in the ligand field minus the energy of the electronic configuration in the isotropic field.
$CFSE=\Delta{E}=E_{\text{ligand field}} - E_{\text{isotropic field}} \label{1}$
The CSFE will depend on multiple factors including:
For an octahedral complex, an electron in the more stable $t_{2g}$ subset is treated as contributing $-2/5\Delta_o$ whereas an electron in the higher energy $e_g$ subset contributes to a destabilization of $+3/5\Delta_o$. The final answer is then expressed as a multiple of the crystal field splitting parameter $\Delta_o$. If any electrons are paired within a single orbital, then the term $P$ is used to represent the spin pairing energy.
Example $1$: CFSE for a high Spin $d^7$ complex
What is the Crystal Field Stabilization Energy for a high spin $d^7$ octahedral complex?
Solution
The splitting pattern and electron configuration for both isotropic and octahedral ligand fields are compared below.
The energy of the isotropic field $(E_{\text{isotropic field}}$) is
$E_{\text{isotropic field}}= 7 \times 0 + 2P = 2P \nonumber$
The energy of the octahedral ligand field $E_{\text{ligand field}}$ is
$E_{\text{ligand field}} = (5 \times -2/5 \Delta_o ) + (2 \times 3/5 \Delta_o) + 2P = -4/5 \Delta_o + 2P \nonumber$
So via Equation \ref{1}, the CFSE is
\begin{align} CFSE &=E_{\text{ligand field}} - E_{\text{isotropic field}} \nonumber \[4pt] &=( -4/5\Delta_o + 2P ) - 2P \nonumber \[4pt] &=-4/5 \Delta_o \nonumber \end{align} \nonumber
Notice that the Spin pairing Energy falls out in this case (and will when calculating the CFSE of high spin complexes) since the number of paired electrons in the ligand field is the same as that in isotropic field of the free metal ion.
Example $2$: CFSE for a Low Spin $d^7$ complex
What is the Crystal Field Stabilization Energy for a low spin $d^7$ octahedral complex?
Solution
The splitting pattern and electron configuration for both isotropic and octahedral ligand fields are compared below.
The energy of the isotropic field is the same as calculated for the high spin configuration in Example 1:
$E_{\text{isotropic field}}= 7 \times 0 + 2P = 2P \nonumber$
The energy of the octahedral ligand\) field $E_{\text{ligand field}}$ is
\begin{align} E_{\text{ligand field}} &= (6 \times -2/5 \Delta_o ) + (1 \times 3/5 \Delta_o) + 3P \nonumber \[4pt] &= -9/5 \Delta_o + 3P \nonumber \end{align} \nonumber
So via Equation \ref{1}, the CFSE is
\begin{align} CFSE&=E_{\text{ligand field}} - E_{\text{isotropic field}} \nonumber \[4pt] &=( -9/5 \Delta_o + 3P ) - 2P \nonumber \[4pt] &=-9/5 \Delta_o + P \nonumber \end{align} \nonumber
Adding in the pairing energy since it will require extra energy to pair up one extra group of electrons. This appears more a more stable configuration than the high spin $d^7$ configuration in Example $1$, but we have then to take into consideration the Pairing energy $P$ to know definitely, which varies between $200-400\; kJ\; mol^{-1}$ depending on the metal.
Table $1$: Crystal Field Stabilization Energies (CFSE) for high and low spin octahedral complexes
Total d-electrons Isotropic Field Octahedral Complex Crystal Field Stabilization Energy
High Spin Low Spin
$E_{\text{isotropic field}}$ Configuration $E_{\text{ligand field}}$ Configuration $E_{\text{ligand field}}$ High Spin Low Spin
d0 0 $t_{2g}$0$e_g$0 0 $t_{2g}$0$e_g$0 0 0 0
d1 0 $t_{2g}$1$e_g$0 -2/5 $\Delta_o$ $t_{2g}$1$e_g$0 -2/5 $\Delta_o$ -2/5 $\Delta_o$ -2/5 $\Delta_o$
d2 0 $t_{2g}$2$e_g$0 -4/5 $\Delta_o$ $t_{2g}$2$e_g$0 -4/5 $\Delta_o$ -4/5 $\Delta_o$ -4/5 $\Delta_o$
d3 0 $t_{2g}$3$e_g$0 -6/5 $\Delta_o$ $t_{2g}$3$e_g$0 -6/5 $\Delta_o$ -6/5 $\Delta_o$ -6/5 $\Delta_o$
d4 0 $t_{2g}$3$e_g$1 -3/5 $\Delta_o$ $t_{2g}$4$e_g$0 -8/5 $\Delta_o$ + P -3/5 $\Delta_o$ -8/5 $\Delta_o$ + P
d5 0 $t_{2g}$3$e_g$2 0 $\Delta_o$ $t_{2g}$5$e_g$0 -10/5 $\Delta_o$ + 2P 0 $\Delta_o$ -10/5 $\Delta_o$ + 2P
d6 P $t_{2g}$4$e_g$2 -2/5 $\Delta_o$ + P $t_{2g}$6$e_g$0 -12/5 $\Delta_o$ + 3P -2/5 $\Delta_o$ -12/5 $\Delta_o$ + P
d7 2P $t_{2g}$5$e_g$2 -4/5 $\Delta_o$ + 2P $t_{2g}$6$e_g$1 -9/5 $\Delta_o$ + 3P -4/5 $\Delta_o$ -9/5 $\Delta_o$ + P
d8 3P $t_{2g}$6$e_g$2 -6/5 $\Delta_o$ + 3P $t_{2g}$6$e_g$2 -6/5 $\Delta_o$ + 3P -6/5 $\Delta_o$ -6/5 $\Delta_o$
d9 4P $t_{2g}$6$e_g$3 -3/5 $\Delta_o$ + 4P $t_{2g}$6$e_g$3 -3/5 $\Delta_o$ + 4P -3/5 $\Delta_o$ -3/5 $\Delta_o$
d10 5P $t_{2g}$6$e_g$4 0 $\Delta_o$ + 5P $t_{2g}$6$e_g$4 0 $\Delta_o$ + 5P 0 0
$P$ is the spin pairing energy and represents the energy required to pair up electrons within the same orbital. For a given metal ion P (pairing energy) is constant, but it does not vary with ligand and oxidation state of the metal ion).
Octahedral Preference
Similar CFSE values can be constructed for non-octahedral ligand field geometries once the knowledge of the d-orbital splitting is known and the electron configuration within those orbitals known, e.g., the tetrahedral complexes in Table $2$. These energies geoemtries can then be contrasted to the octahedral CFSE to calculate a thermodynamic preference (Enthalpy-wise) for a metal-ligand combination to favor the octahedral geometry. This is quantified via a Octahedral Site Preference Energy defined below.
Definition: Octahedral Site Preference Energies
The Octahedral Site Preference Energy (OSPE) is defined as the difference of CFSE energies for a non-octahedral complex and the octahedral complex. For comparing the preference of forming an octahedral ligand field vs. a tetrahedral ligand field, the OSPE is thus:
$OSPE = CFSE_{(oct)} - CFSE_{(tet)} \label{2}$
The OSPE quantifies the preference of a complex to exhibit an octahedral geometry vs. a tetrahedral geometry.
Note: the conversion between $\Delta_o$ and $\Delta_t$ used for these calculations is:
$\Delta_t \approx \dfrac{4}{9} \Delta_o \label{3}$
which is applicable for comparing octahedral and tetrahedral complexes that involve same ligands only.
Table $2$: Octahedral Site Preference Energies (OSPE)
Total d-electrons CFSE(Octahedral) CFSE(Tetrahedral) OSPE (for high spin complexes)**
High Spin Low Spin Configuration Always High Spin*
d0 0 $\Delta_o$ 0 $\Delta_o$ e0 0 $\Delta_t$ 0 $\Delta_o$
d1 -2/5 $\Delta_o$ -2/5 $\Delta_o$ e1 -3/5 $\Delta_t$ -6/45 $\Delta_o$
d2 -4/5 $\Delta_o$ -4/5 $\Delta_o$ e2 -6/5 $\Delta_t$ -12/45 $\Delta_o$
d3 -6/5 $\Delta_o$ -6/5 $\Delta_o$ e2t21 -4/5 $\Delta_t$ -38/45 $\Delta_o$
d4 -3/5 $\Delta_o$ -8/5 $\Delta_o$ + P e2t22 -2/5 $\Delta_t$ -19/45 $\Delta_o$
d5 0 $\Delta_o$ -10/5 $\Delta_o$ + 2P e2t23 0 $\Delta_t$ 0 $\Delta_o$
d6 -2/5 $\Delta_o$ -12/5 $\Delta_o$ + P e3t23 -3/5 $\Delta_t$ -6/45 $\Delta_o$
d7 -4/5 $\Delta_o$ -9/5 $\Delta_o$ + P e4t23 -6/5 $\Delta_t$ -12/45 $\Delta_o$
d8 -6/5 $\Delta_o$ -6/5 $\Delta_o$ e4t24 -4/5 $\Delta_t$ -38/45 $\Delta_o$
d9 -3/5 $\Delta_o$ -3/5 $\Delta_o$ e4t25 -2/5 $\Delta_t$ -19/45 $\Delta_o$
d10 0 0 e4t26 0 $\Delta_t$ 0 $\Delta_o$
$P$ is the spin pairing energy and represents the energy required to pair up electrons within the same orbital.
Tetrahedral complexes are always high spin since the splitting is appreciably smaller than $P$ (Equation \ref{3}).
After conversion with Equation \ref{3}. The data in Tables $1$ and $2$ are represented graphically by the curves in Figure $1$ below for the high spin complexes only. The low spin complexes require knowledge of $P$ to graph.
Figure $1$: Crystal Field Stabilization Energies for both octahedral fields ($CFSE_{oct}$) and tetrahedral fields ($CFSE_{tet}$). Octahedral Site Preference Energies (OSPE) are in yellow. This is for high spin complexes.
From a simple inspection of Figure $1$, the following observations can be made:
• The OSPE is small in $d^1$, $d^2$, $d^5$, $d^6$, $d^7$ complexes and other factors influence the stability of the complexes including steric factors
• The OSPE is large in $d^3$ and $d^8$ complexes which strongly favor octahedral geometries
Applications
The "double-humped" curve in Figure $1$ is found for various properties of the first-row transition metals, including Hydration and Lattice energies of the M(II) ions, ionic radii as well as the stability of M(II) complexes. This suggests that these properties are somehow related to Crystal Field effects.
In the case of Hydration Energies describing the complexation of water ligands to a bare metal ion:
$M^{2+} (g) + H_2O \rightarrow [M(OH_2)_6]^{2+} (aq) \nonumber$
Table $3$ and Figure $1$ shows this type of curve. Note that in any series of this type not all the data are available since a number of ions are not very stable in the M(II) state.
Table $3$: Hydration energies of $M^{2+}$ ions
M ΔH°/kJmol-1 M ΔH°/kJmol-1
Ca -2469 Fe -2840
Sc no stable 2+ ion Co -2910
Ti -2729 Ni -2993
>V -2777 Cu -2996
Cr -2792 Zn -2928
Mn -2733
Graphically the data in Table 2 can be represented by:
Figure $2$: hydration energies of $M^{2+}$ ions
Contributors and Attributions
• The Department of Chemistry, University of the West Indies) | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Crystal_Field_Theory/Crystal_Field_Stabilization_Energy.txt |
Crystal field theory (CFT) describes the breaking of orbital degeneracy in transition metal complexes due to the presence of ligands. CFT qualitatively describes the strength of the metal-ligand bonds. Based on the strength of the metal-ligand bonds, the energy of the system is altered. This may lead to a change in magnetic properties as well as color. This theory was developed by Hans Bethe and John Hasbrouck van Vleck.
Basic Concept
In Crystal Field Theory, it is assumed that the ions are simple point charges (a simplification). When applied to alkali metal ions containing a symmetric sphere of charge, calculations of bond energies are generally quite successful. The approach taken uses classical potential energy equations that take into account the attractive and repulsive interactions between charged particles (that is, Coulomb's Law interactions).
$E \propto \dfrac{q_1 q_2}{r} \nonumber$
with
• $E$ the bond energy between the charges and
• $q_1$ and $q_2$ are the charges of the interacting ions and
• $r$ is the distance separating them.
This approach leads to the correct prediction that large cations of low charge, such as $K^+$ and $Na^+$, should form few coordination compounds. For transition metal cations that contain varying numbers of d electrons in orbitals that are NOT spherically symmetric, however, the situation is quite different. The shapes and occupations of these d-orbitals then become important in building an accurate description of the bond energy and properties of the transition metal compound.
When examining a single transition metal ion, the five d-orbitals have the same energy (Figure $1$). When ligands approach the metal ion, some experience more opposition from the d-orbital electrons than others based on the geometric structure of the molecule. Since ligands approach from different directions, not all d-orbitals interact directly. These interactions, however, create a splitting due to the electrostatic environment.
For example, consider a molecule with octahedral geometry. Ligands approach the metal ion along the $x$, $y$, and $z$ axes. Therefore, the electrons in the $d_{z^2}$ and $d_{x^2-y^2}$ orbitals (which lie along these axes) experience greater repulsion. It requires more energy to have an electron in these orbitals than it would to put an electron in one of the other orbitals. This causes a splitting in the energy levels of the d-orbitals. This is known as crystal field splitting. For octahedral complexes, crystal field splitting is denoted by $\Delta_o$ (or $\Delta_{oct}$). The energies of the $d_{z^2}$ and $d_{x^2-y^2}$ orbitals increase due to greater interactions with the ligands. The $d_{xy}$, $d_{xz}$, and $d_{yz}$ orbitals decrease with respect to this normal energy level and become more stable.
Electrons in Orbitals
According to the Aufbau principle, electrons are filled from lower to higher energy orbitals (Figure $1$). For the octahedral case above, this corresponds to the dxy, dxz, and dyz orbitals. Following Hund's rule, electrons are filled in order to have the highest number of unpaired electrons. For example, if one had a d3 complex, there would be three unpaired electrons. If one were to add an electron, however, it has the ability to fill a higher energy orbital ( dor dx²-y²) or pair with an electron residing in the dxy, dxz, or dyz orbitals. This pairing of the electrons requires energy (spin pairing energy). If the pairing energy is less than the crystal field splitting energy, ∆₀, then the next electron will go into the dxy, dxz, or dyz orbitals due to stability. This situation allows for the least amount of unpaired electrons, and is known as low spin. If the pairing energy is greater than ∆₀, then the next electron will go into the dz² or dx²-y² orbitals as an unpaired electron. This situation allows for the most number of unpaired electrons, and is known as high spin. Ligands that cause a transition metal to have a small crystal field splitting, which leads to high spin, are called weak-field ligands. Ligands that produce a large crystal field splitting, which leads to low spin, are called strong field ligands.
As mentioned above, CFT is based primarily on symmetry of ligands around a central metal/ion and how this anisotropic (properties depending on direction) ligand field affects the metal's atomic orbitals; the energies of which may increase, decrease or not be affected at all. Once the ligands' electrons interact with the electrons of the d-orbitals, the electrostatic interactions cause the energy levels of the d-orbital to fluctuate depending on the orientation and the nature of the ligands. For example, the oxidation state and the strength of the ligands determine splitting; the higher the oxidation state or the stronger the ligand, the larger the splitting. Ligands are classified as strong or weak based on the spectrochemical series:
I- < Br- < Cl- < SCN- < F- < OH- < ox2-< ONO- < H2O < SCN- < EDTA4- < NH3 < en < NO2- < CN-
Note that SCN- and NO2- ligands are represented twice in the above spectrochemical series since there are two different Lewis base sites (e.g., free electron pairs to share) on each ligand (e.g., for the SCN- ligand, the electron pair on the sulfur or the nitrogen can form the coordinate covalent bond to a metal). The specific atom that binds in such ligands is underlined.
In addition to octahedral complexes, two common geometries observed are that of tetrahedral and square planar. These complexes differ from the octahedral complexes in that the orbital levels are raised in energy due to the interference with electrons from ligands. For the tetrahedral complex, the dxy, dxz, and dyz orbitals are raised in energy while the d, dx²-y² orbitals are lowered. For the square planar complexes, there is greatest interaction with the dx²-y² orbital and therefore it has higher energy. The next orbital with the greatest interaction is dxy, followed below by d. The orbitals with the lowest energy are the dxz and dyz orbitals. There is a large energy separation between the d orbital and the dxz and dyz orbitals, meaning that the crystal field splitting energy is large. We find that the square planar complexes have the greatest crystal field splitting energy compared to all the other complexes. This means that most square planar complexes are low spin, strong field ligands.
Description of d-Orbitals
To understand CFT, one must understand the description of the lobes:
• dxy: lobes lie in-between the x and the y axes.
• dxz: lobes lie in-between the x and the z axes.
• dyz: lobes lie in-between the y and the z axes.
• dx2-y2: lobes lie on the x and y axes.
• dz2: there are two lobes on the z axes and there is a donut shape ring that lies on the xy plane around the other two lobes.
Octahedral Complexes
In an octahedral complex, there are six ligands attached to the central transition metal. The d-orbital splits into two different levels (Figure $4$). The bottom three energy levels are named $d_{xy}$, $d_{xz}$, and $d_{yz}$ (collectively referred to as $t_{2g}$). The two upper energy levels are named $d_{x^²-y^²}$, and $d_{z^²}$ (collectively referred to as $e_g$).
The reason they split is because of the electrostatic interactions between the electrons of the ligand and the lobes of the d-orbital. In an octahedral, the electrons are attracted to the axes. Any orbital that has a lobe on the axes moves to a higher energy level. This means that in an octahedral, the energy levels of $e_g$ are higher (0.6∆o) while $t_{2g}$ is lower (0.4∆o). The distance that the electrons have to move from $t_{2g}$ from $e_g$ and it dictates the energy that the complex will absorb from white light, which will determine the color. Whether the complex is paramagnetic or diamagnetic will be determined by the spin state. If there are unpaired electrons, the complex is paramagnetic; if all electrons are paired, the complex is diamagnetic.
Tetrahedral Complexes
In a tetrahedral complex, there are four ligands attached to the central metal. The d orbitals also split into two different energy levels. The top three consist of the $d_{xy}$, $d_{xz}$, and $d_{yz}$ orbitals. The bottom two consist of the $d_{x^2-y^2}$ and $d_{z^2}$ orbitals. The reason for this is due to poor orbital overlap between the metal and the ligand orbitals. The orbitals are directed on the axes, while the ligands are not.
The difference in the splitting energy is tetrahedral splitting constant ($\Delta_{t}$), which less than ($\Delta_{o}$) for the same ligands:
$\Delta_{t} = 0.44\,\Delta_o \label{1}$
Consequentially, $\Delta_{t}$ is typically smaller than the spin pairing energy, so tetrahedral complexes are usually high spin.
Square Planar Complexes
In a square planar, there are four ligands as well. However, the difference is that the electrons of the ligands are only attracted to the $xy$ plane. Any orbital in the xy plane has a higher energy level (Figure $6$). There are four different energy levels for the square planar (from the highest energy level to the lowest energy level): dx2-y2, dxy, dz2, and both dxz and dyz.
The splitting energy (from highest orbital to lowest orbital) is $\Delta_{sp}$ and tends to be larger then $\Delta_{o}$
$\Delta_{sp} = 1.74\,\Delta_o \label{2}$
Moreover, $\Delta_{sp}$ is also larger than the pairing energy, so the square planar complexes are usually low spin complexes.
Example $1$
For the complex ion [Fe(Cl)6]3- determine the number of d electrons for Fe, sketch the d-orbital energy levels and the distribution of d electrons among them, list the number of lone electrons, and label whether the complex is paramagnetic or diamagnetic.
Solution
• Step 1: Determine the oxidation state of Fe. Here it is Fe3+. Based on its electron configuration, Fe3+ has 5 d-electrons.
• Step 2: Determine the geometry of the ion. Here it is an octahedral which means the energy splitting should look like:
• Step 3: Determine whether the ligand induces is a strong or weak field spin by looking at the spectrochemical series. Cl- is a weak field ligand (i.e., it induces high spin complexes). Therefore, electrons fill all orbitals before being paired.
• Step four: Count the number of lone electrons. Here, there are 5 electrons.
• Step five: The five unpaired electrons means this complex ion is paramagnetic (and strongly so).
Example $2$
A tetrahedral complex absorbs at 545 nm. What is the respective octahedral crystal field splitting ($\Delta_o$)? What is the color of the complex?
Solution
\begin{align*} \Delta_t &= \dfrac{hc}{\lambda}\[4pt] &= \dfrac{ (6.626 \times 10^{-34} J \cdot s)(3 \times 10^8 m/s)}{545 \times 10^{-9} m} \[4pt] &=3.65 \times 10^{-19}\; J \end{align*}
However, the tetrahedral splitting ($\Delta_t$) is ~4/9 that of the octahedral splitting ($\Delta_o$).
\begin{align*} \Delta_t &= 0.44\Delta_o \[4pt] \Delta_o &= \dfrac{\Delta_t}{0.44} \[4pt] &= \dfrac{3.65 \times 10^{-19} J}{0.44} \[4pt] &= 8.30 \times 10^{-18}J \end{align*}
This is the energy needed to promote one electron in one complex. Often the crystal field splitting is given per mole, which requires this number to be multiplied by Avogadro's Number ($6.022 \times 10^{23}$).
This complex appears red, since it absorbs in the complementary green color (determined via the color wheel).
Video:
Problems
For each of the following, sketch the d-orbital energy levels and the distribution of d electrons among them, state the geometry, list the number of d-electrons, list the number of lone electrons, and label whether they are paramagnetic or dimagnetic:
1. [Ti(H2O)6]2+
2. [NiCl4]2-
3. [CoF6]3- (also state whether this is low or high spin)
4. [Co(NH3)6]3+ (also state whether this is low or high spin)
5. True or False: Square Planer complex compounds are usually low spin.
Answers
1. octahedral, 2, 2, paramagnetic
2. tetrahedral, 8, 2, paramagnetic (see
3. octahedral, 6, 4, paramagnetic, high spin
4. octahedral, 6, 0, diamagnetic, low spin
5. True
Contributors and Attributions
• Asadullah Awan (UCD), Hong Truong (UCD)
• The Department of Chemistry, University of the West Indies) | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Crystal_Field_Theory/Crystal_Field_Theory.txt |
Learning Objectives
• To understand how crystal field theory explains the electronic structures and colors of metal complexes.
One of the most striking characteristics of transition-metal complexes is the wide range of colors they exhibit. In this section, we describe crystal field theory (CFT), a bonding model that explains many important properties of transition-metal complexes, including their colors, magnetism, structures, stability, and reactivity. The central assumption of CFT is that metal–ligand interactions are purely electrostatic in nature. Even though this assumption is clearly not valid for many complexes, such as those that contain neutral ligands like CO, CFT enables chemists to explain many of the properties of transition-metal complexes with a reasonable degree of accuracy. The Learning Objective of this Module is to understand how crystal field theory explains the electronic structures and colors of metal complexes.
d-Orbital Splittings
CFT focuses on the interaction of the five (n − 1)d orbitals with ligands arranged in a regular array around a transition-metal ion. We will focus on the application of CFT to octahedral complexes, which are by far the most common and the easiest to visualize. Other common structures, such as square planar complexes, can be treated as a distortion of the octahedral model. According to CFT, an octahedral metal complex forms because of the electrostatic interaction of a positively charged metal ion with six negatively charged ligands or with the negative ends of dipoles associated with the six ligands. In addition, the ligands interact with one other electrostatically. As you learned in our discussion of the valence-shell electron-pair repulsion (VSEPR) model, the lowest-energy arrangement of six identical negative charges is an octahedron, which minimizes repulsive interactions between the ligands.
We begin by considering how the energies of the d orbitals of a transition-metal ion are affected by an octahedral arrangement of six negative charges. Recall that the five d orbitals are initially degenerate (have the same energy). If we distribute six negative charges uniformly over the surface of a sphere, the d orbitals remain degenerate, but their energy will be higher due to repulsive electrostatic interactions between the spherical shell of negative charge and electrons in the d orbitals (Figure $\PageIndex{1a}$). Placing the six negative charges at the vertices of an octahedron does not change the average energy of the d orbitals, but it does remove their degeneracy: the five d orbitals split into two groups whose energies depend on their orientations. As shown in Figure $\PageIndex{1b}$, the dz2 and dx2−y2 orbitals point directly at the six negative charges located on the x, y, and z axes. Consequently, the energy of an electron in these two orbitals (collectively labeled the eg orbitals) will be greater than it will be for a spherical distribution of negative charge because of increased electrostatic repulsions. In contrast, the other three d orbitals (dxy, dxz, and dyz, collectively called the t2g orbitals) are all oriented at a 45° angle to the coordinate axes, so they point between the six negative charges. The energy of an electron in any of these three orbitals is lower than the energy for a spherical distribution of negative charge.
The difference in energy between the two sets of d orbitals is called the crystal field splitting energy (Δo), where the subscript o stands for octahedral. As we shall see, the magnitude of the splitting depends on the charge on the metal ion, the position of the metal in the periodic table, and the nature of the ligands. (Crystal field splitting energy also applies to tetrahedral complexes: Δt.) It is important to note that the splitting of the d orbitals in a crystal field does not change the total energy of the five d orbitals: the two eg orbitals increase in energy by 0.6Δo, whereas the three t2g orbitals decrease in energy by 0.4Δo. Thus the total change in energy is
$2(0.6Δ_o) + 3(−0.4Δ_o) = 0.$
Crystal field splitting does not change the total energy of the d orbitals.
Thus far, we have considered only the effect of repulsive electrostatic interactions between electrons in the d orbitals and the six negatively charged ligands, which increases the total energy of the system and splits the d orbitals. Interactions between the positively charged metal ion and the ligands results in a net stabilization of the system, which decreases the energy of all five d orbitals without affecting their splitting (as shown at the far right in Figure $\PageIndex{1a}$).
Electronic Structures of Metal Complexes
We can use the d-orbital energy-level diagram in Figure $1$ to predict electronic structures and some of the properties of transition-metal complexes. We start with the Ti3+ ion, which contains a single d electron, and proceed across the first row of the transition metals by adding a single electron at a time. We place additional electrons in the lowest-energy orbital available, while keeping their spins parallel as required by Hund’s rule. As shown in Figure $2$, for d1–d3 systems—such as [Ti(H2O)6]3+, [V(H2O)6]3+, and [Cr(H2O)6]3+, respectively—the electrons successively occupy the three degenerate t2g orbitals with their spins parallel, giving one, two, and three unpaired electrons, respectively. We can summarize this for the complex [Cr(H2O)6]3+, for example, by saying that the chromium ion has a d3 electron configuration or, more succinctly, Cr3+ is a d3 ion.
When we reach the d4 configuration, there are two possible choices for the fourth electron: it can occupy either one of the empty eg orbitals or one of the singly occupied t2g orbitals. Recall that placing an electron in an already occupied orbital results in electrostatic repulsions that increase the energy of the system; this increase in energy is called the spin-pairing energy (P). If Δo is less than P, then the lowest-energy arrangement has the fourth electron in one of the empty eg orbitals. Because this arrangement results in four unpaired electrons, it is called a high-spin configuration, and a complex with this electron configuration, such as the [Cr(H2O)6]2+ ion, is called a high-spin complex. Conversely, if Δo is greater than P, then the lowest-energy arrangement has the fourth electron in one of the occupied t2g orbitals. Because this arrangement results in only two unpaired electrons, it is called a low-spin configuration, and a complex with this electron configuration, such as the [Mn(CN)6]3− ion, is called a low-spin complex. Similarly, metal ions with the d5, d6, or d7 electron configurations can be either high spin or low spin, depending on the magnitude of Δo.
In contrast, only one arrangement of d electrons is possible for metal ions with d8–d10 electron configurations. For example, the [Ni(H2O)6]2+ ion is d8 with two unpaired electrons, the [Cu(H2O)6]2+ ion is d9 with one unpaired electron, and the [Zn(H2O)6]2+ ion is d10 with no unpaired electrons.
If Δo is less than the spin-pairing energy, a high-spin configuration results. Conversely, if Δo is greater, a low-spin configuration forms.
Factors That Affect the Magnitude of Δo
The magnitude of Δo dictates whether a complex with four, five, six, or seven d electrons is high spin or low spin, which affects its magnetic properties, structure, and reactivity. Large values of Δo (i.e., Δo > P) yield a low-spin complex, whereas small values of Δo (i.e., Δo < P) produce a high-spin complex. As we noted, the magnitude of Δo depends on three factors: the charge on the metal ion, the principal quantum number of the metal (and thus its location in the periodic table), and the nature of the ligand. Values of Δo for some representative transition-metal complexes are given in Table $1$.
Table $1$: Crystal Field Splitting Energies for Some Octahedral (Δo)* and Tetrahedral (Δt) Transition-Metal Complexes
Octahedral Complexes Δo (cm−1) Octahedral Complexes Δo (cm−1) Tetrahedral Complexes Δt (cm−1)
*Energies obtained by spectroscopic measurements are often given in units of wave numbers (cm−1); the wave number is the reciprocal of the wavelength of the corresponding electromagnetic radiation expressed in centimeters: 1 cm−1 = 11.96 J/mol.
[Ti(H2O)6]3+ 20,300 [Fe(CN)6]4− 32,800 VCl4 9010
[V(H2O)6]2+ 12,600 [Fe(CN)6]3− 35,000 [CoCl4]2− 3300
[V(H2O)6]3+ 18,900 [CoF6]3− 13,000 [CoBr4]2− 2900
[CrCl6]3− 13,000 [Co(H2O)6]2+ 9300 [CoI4]2− 2700
[Cr(H2O)6]2+ 13,900 [Co(H2O)6]3+ 27,000
[Cr(H2O)6]3+ 17,400 [Co(NH3)6]3+ 22,900
[Cr(NH3)6]3+ 21,500 [Co(CN)6]3− 34,800
[Cr(CN)6]3− 26,600 [Ni(H2O)6]2+ 8500
Cr(CO)6 34,150 [Ni(NH3)6]2+ 10,800
[MnCl6]4− 7500 [RhCl6]3− 20,400
[Mn(H2O)6]2+ 8500 [Rh(H2O)6]3+ 27,000
[MnCl6]3− 20,000 [Rh(NH3)6]3+ 34,000
[Mn(H2O)6]3+ 21,000 [Rh(CN)6]3− 45,500
[Fe(H2O)6]2+ 10,400 [IrCl6]3− 25,000
[Fe(H2O)6]3+ 14,300 [Ir(NH3)6]3+ 41,000
Source of data: Duward F. Shriver, Peter W. Atkins, and Cooper H. Langford, Inorganic Chemistry, 2nd ed. (New York: W. H. Freeman and Company, 1994).
Factor 1: Charge on the Metal Ion
Increasing the charge on a metal ion has two effects: the radius of the metal ion decreases, and negatively charged ligands are more strongly attracted to it. Both factors decrease the metal–ligand distance, which in turn causes the negatively charged ligands to interact more strongly with the d orbitals. Consequently, the magnitude of Δo increases as the charge on the metal ion increases. Typically, Δo for a tripositive ion is about 50% greater than for the dipositive ion of the same metal; for example, for [V(H2O)6]2+, Δo = 11,800 cm−1; for [V(H2O)6]3+, Δo = 17,850 cm−1.
Factor 2: Principal Quantum Number of the Metal
For a series of complexes of metals from the same group in the periodic table with the same charge and the same ligands, the magnitude of Δo increases with increasing principal quantum number: Δo (3d) < Δo (4d) < Δo (5d). The data for hexaammine complexes of the trivalent Group 9 metals illustrate this point:
[Co(NH3)6]3+: Δo = 22,900 cm−1
[Rh(NH3)6]3+: Δo = 34,100 cm−1
[Ir(NH3)6]3+: Δo = 40,000 cm−1
The increase in Δo with increasing principal quantum number is due to the larger radius of valence orbitals down a column. In addition, repulsive ligand–ligand interactions are most important for smaller metal ions. Relatively speaking, this results in shorter M–L distances and stronger d orbital–ligand interactions.
Factor 3: The Nature of the Ligands
Experimentally, it is found that the Δo observed for a series of complexes of the same metal ion depends strongly on the nature of the ligands. For a series of chemically similar ligands, the magnitude of Δo decreases as the size of the donor atom increases. For example, Δo values for halide complexes generally decrease in the order F > Cl > Br > I− because smaller, more localized charges, such as we see for F, interact more strongly with the d orbitals of the metal ion. In addition, a small neutral ligand with a highly localized lone pair, such as NH3, results in significantly larger Δo values than might be expected. Because the lone pair points directly at the metal ion, the electron density along the M–L axis is greater than for a spherical anion such as F. The experimentally observed order of the crystal field splitting energies produced by different ligands is called the spectrochemical series, shown here in order of decreasing Δo:
$\mathrm{\underset{\textrm{strong-field ligands}}{CO\approx CN^->}NO_2^->en>NH_3>\underset{\textrm{intermediate-field ligands}}{SCN^->H_2O>oxalate^{2-}}>OH^->F>acetate^->\underset{\textrm{weak-field ligands}}{Cl^->Br^->I^-}}$
The values of Δo listed in Table $1$ illustrate the effects of the charge on the metal ion, the principal quantum number of the metal, and the nature of the ligand.
The largest Δo splittings are found in complexes of metal ions from the third row of the transition metals with charges of at least +3 and ligands with localized lone pairs of electrons.
Example $1$
For each complex, predict its structure, whether it is high spin or low spin, and the number of unpaired electrons present.
1. [CoF6]3−
2. [Rh(CO)2Cl2]
Given: complexes
Asked for: structure, high spin versus low spin, and the number of unpaired electrons
Strategy:
1. From the number of ligands, determine the coordination number of the compound.
2. Classify the ligands as either strong field or weak field and determine the electron configuration of the metal ion.
3. Predict the relative magnitude of Δo and decide whether the compound is high spin or low spin.
4. Place the appropriate number of electrons in the d orbitals and determine the number of unpaired electrons.
Solution
1. A With six ligands, we expect this complex to be octahedral.
B The fluoride ion is a small anion with a concentrated negative charge, but compared with ligands with localized lone pairs of electrons, it is weak field. The charge on the metal ion is +3, giving a d6 electron configuration.
C Because of the weak-field ligands, we expect a relatively small Δo, making the compound high spin.
D In a high-spin octahedral d6 complex, the first five electrons are placed individually in each of the d orbitals with their spins parallel, and the sixth electron is paired in one of the t2g orbitals, giving four unpaired electrons.
1. A This complex has four ligands, so it is either square planar or tetrahedral.
B C Because rhodium is a second-row transition metal ion with a d8 electron configuration and CO is a strong-field ligand, the complex is likely to be square planar with a large Δo, making it low spin. Because the strongest d-orbital interactions are along the x and y axes, the orbital energies increase in the order dz2dyz, and dxz (these are degenerate); dxy; and dx2−y2.
D The eight electrons occupy the first four of these orbitals, leaving the dx2−y2. orbital empty. Thus there are no unpaired electrons.
Exercise $1$
For each complex, predict its structure, whether it is high spin or low spin, and the number of unpaired electrons present.
1. [Mn(H2O)6]2+
2. [PtCl4]2−
Answer a
octahedral; high spin; five
Answer b
square planar; low spin; no unpaired electrons
Crystal Field Stabilization Energies
Recall that stable molecules contain more electrons in the lower-energy (bonding) molecular orbitals in a molecular orbital diagram than in the higher-energy (antibonding) molecular orbitals. If the lower-energy set of d orbitals (the t2g orbitals) is selectively populated by electrons, then the stability of the complex increases. For example, the single d electron in a d1 complex such as [Ti(H2O)6]3+ is located in one of the t2g orbitals. Consequently, this complex will be more stable than expected on purely electrostatic grounds by 0.4Δo. The additional stabilization of a metal complex by selective population of the lower-energy d orbitals is called its crystal field stabilization energy (CFSE). The CFSE of a complex can be calculated by multiplying the number of electrons in t2g orbitals by the energy of those orbitals (−0.4Δo), multiplying the number of electrons in eg orbitals by the energy of those orbitals (+0.6Δo), and summing the two. Table $2$ gives CFSE values for octahedral complexes with different d electron configurations. The CFSE is highest for low-spin d6 complexes, which accounts in part for the extraordinarily large number of Co(III) complexes known. The other low-spin configurations also have high CFSEs, as does the d3 configuration.
Table $2$: CFSEs for Octahedral Complexes with Different Electron Configurations (in Units of Δo)
High Spin CFSE (Δo) Low Spin CFSE (Δo)
t2g eg t2g eg
d 0 0
d 1 0.4
d 2 ↿ ↿ 0.8
d 3 ↿ ↿ ↿ 1.2
d 4 ↿ ↿ ↿ 0.6 ↿⇂ ↿ ↿ 1.6
d 5 ↿ ↿ ↿ ↿ ↿ 0.0 ↿⇂ ↿⇂ ↿ 2.0
d 6 ↿⇂ ↿ ↿ ↿ ↿ 0.4 ↿⇂ ↿⇂ ↿⇂ 2.4
d 7 ↿⇂ ↿⇂ ↿ ↿ ↿ 0.8 ↿⇂ ↿⇂ ↿⇂ 1.8
d 8 ↿⇂ ↿⇂ ↿⇂ ↿ ↿ 1.2
d 9 ↿⇂ ↿⇂ ↿⇂ ↿⇂ ↿ 0.6
d 10 ↿⇂ ↿⇂ ↿⇂ ↿⇂ ↿⇂ 0.0
CFSEs are important for two reasons. First, the existence of CFSE nicely accounts for the difference between experimentally measured values for bond energies in metal complexes and values calculated based solely on electrostatic interactions. Second, CFSEs represent relatively large amounts of energy (up to several hundred kilojoules per mole), which has important chemical consequences.
Octahedral d3 and d8 complexes and low-spin d6, d5, d7, and d4 complexes exhibit large CFSEs.
Summary
Crystal field theory, which assumes that metal–ligand interactions are only electrostatic in nature, explains many important properties of transition-metal complexes, including their colors, magnetism, structures, stability, and reactivity. Crystal field theory (CFT) is a bonding model that explains many properties of transition metals that cannot be explained using valence bond theory. In CFT, complex formation is assumed to be due to electrostatic interactions between a central metal ion and a set of negatively charged ligands or ligand dipoles arranged around the metal ion. Depending on the arrangement of the ligands, the d orbitals split into sets of orbitals with different energies. The difference between the energy levels in an octahedral complex is called the crystal field splitting energy (Δo), whose magnitude depends on the charge on the metal ion, the position of the metal in the periodic table, and the nature of the ligands. The spin-pairing energy (P) is the increase in energy that occurs when an electron is added to an already occupied orbital. A high-spin configuration occurs when the Δo is less than P, which produces complexes with the maximum number of unpaired electrons possible. Conversely, a low-spin configuration occurs when the Δo is greater than P, which produces complexes with the minimum number of unpaired electrons possible. Strong-field ligands interact strongly with the d orbitals of the metal ions and give a large Δo, whereas weak-field ligands interact more weakly and give a smaller Δo. The colors of transition-metal complexes depend on the environment of the metal ion and can be explained by CFT. | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Crystal_Field_Theory/Introduction_to_Crystal_Field_Theory.txt |
Magnetic moments are often used in conjunction with electronic spectra to gain information about the oxidation number and stereochemistry of the central metal ion in coordination complexes. A common laboratory procedure for the determination of the magnetic moment for a complex is the Gouy method which involves weighing a sample of the complex in the presence and absence of a magnetic field and observing the difference in weight. A template is provided for the calculations involved.
For first row transition metal ions in the free ion state, i.e. isolated ions in a vacuum, all 5 of the 3d orbitals are degenerate.
A simple crystal field theory approach to the bonding in these ions assumes that when they form octahedral complexes, the energy of the d orbitals are no longer degenerate but are split such that two orbitals, the dx2-y2 and the dz2 (eg subset) are at higher energy than the dxy, dxz, dyz orbitals (the t2g subset).
For octahedral ions with between 4 and 7 d electrons, this gives rise to 2 possible arrangements called either high spin/weak field or low spin/strong field respectively. The energy gap is dependent on the position of the coordinated ligands in the SPECTROCHEMICAL SERIES.
Note
A good starting point is to assume that all Co(III), d6 complexes are octahedral and LOW spin, i.e. t2g6.
In tetrahedral complexes, the energy levels of the orbitals are again split, such that the energy of two orbitals, the $d_{x^2-y^2}$ and the $d_{z^2}$ (e subset) are now at lower energy (more favored) than the remaining three $d_{xy}$, $d_{xz}$, $d_{yz}$ (the $t_2$ subset) which are destabilized.
Tetrahedral complexes are ALL high spin since the difference between the 2 subsets of energies of the orbitals is much smaller than is found in octahedral complexes.
The usual relationship quoted between them is:
$\Delta_{tet} \approx \dfrac{4}{9} \Delta_{oct} \nonumber$
Square planar complexes are less common than tetrahedral and d8 e.g. Ni(II), Pd(II), Pt(II), etc, have a strong propensity to form square planar complexes. As with octahedral complexes, the energy gap between the $d_{xy}$ and $d_{x^2-y^2}$ is $\Delta_{oct}$ and these are considered strong field / low spin hence they are all diamagnetic, μ=0 Bohr Magneton (B.M.)
The formula used to calculate the spin-only magnetic moment can be written in two forms; the first based on the number of unpaired electrons, n, and the second based on the electron spin quantum number, $S$. Since for each unpaired electron, $n=1$ and $S=1/2$ then the two formulae are clearly related and the answer obtained must be identical.
$\mu_{so}= \sqrt{n(n+2)} \nonumber$
$\mu_{so}= \sqrt{4S(S+1)} \nonumber$
Comparison of calculated spin-only magnetic moments with experimental data for some octahedral complexes
Ion Config μso / B.M. μobs / B.M.
Ti(III) d1 (t2g1) √3 = 1.73 1.6-1.7
V(III) d2 (t2g2) √8 = 2.83 2.7-2.9
Cr(III) d3 (t2g3) √15 = 3.88 3.7-3.9
Cr(II) d4 high spin (t2g3 eg1) √24 = 4.90 4.7-4.9
Cr(II) d4 low spin (t2g4) √8 = 2.83 3.2-3.3
Mn(II)/ Fe(III) d5 high spin (t2g3 eg2) √35 = 5.92 5.6-6.1
Mn(II)/ Fe(III) d5 low spin (t2g5) √3 = 1.73 1.8-2.1
Fe(II) d6 high spin (t2g4 eg2) √24 = 4.90 5.1-5.7
Co(III) d6 low spin (t2g6) 0 0
Co(II) d7 high spin (t2g5 eg2) √15 = 3.88 4.3-5.2
Co(II) d7 low spin (t2g6 eg1) √3 = 1.73 1.8
Ni(II) d8 (t2g6 eg2) √8 = 2.83 2.9-3.3
Cu(II) d9 (t2g6 eg3) √3 = 1.73 1.7-2.2
Comparison of calculated spin-only magnetic moments with experimental data for some tetahedral complexes
Ion Config μso / B.M. μobs / B.M.
Cr(V) d1 (e1) √3 = 1.73 1.7-1.8
Cr(IV) / Mn(V) d2 (e2) √8 = 2.83 2.6 - 2.8
Fe(V) d3 (e2 t21) √15 = 3.88 3.6-3.7
- d4 (e2 t22) √24 = 4.90 -
Mn(II) d5 (e2 t23) √35 = 5.92 5.9-6.2
Fe(II) d6 (e3 t23) √24 = 4.90 5.3-5.5
Co(II) d7 (e4 t23) √15 = 3.88 4.2-4.8
Ni(II) d8 (e4 t24) √8 = 2.83 3.7-4.0
Cu(II) d9 (e4 t25) √3 = 1.73 -
Contributors and Attributions
• The Department of Chemistry, University of the West Indies) | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Crystal_Field_Theory/Magnetic_Moments_of_Transition_Metals.txt |
Movement of an electrical charge (which is the basis of electric currents) generates a magnetic field in a material. Magnetism is therefore a characteristic property of all materials that contain electrically charged particles and for most purposes can be considered to be entirely of electronic origin.
In an atom, the magnetic field is due to the coupled spin and orbital magnetic moments associated with the motion of electrons. The spin magnetic moment is due to the precession of the electrons about their own axes whereas the orbital magnetic moment is due to the motion of electrons around the nucleus. The resultant combination of the spin and orbital magnetic moments of the constituent atoms of a material gives rise to the observed magnetic properties.
Historically, magnetism has been recognized for thousands of years. An account, that is probably apochryphal, tells of a shepherd called Magnes in Crete who around 900 B.C discovered the naturally occurring magnet lodestone (a form of the the spinel magnetite, Fe3O4) in a region later named Magnesia. Supposedly while he was walking over a deposit, the lodestone pulled the nails out of his sandals and the metal tip from his staff.
The Classical Theory of Magnetism
The classical theory of magnetism was well developed before quantum mechanics. Lenz's Law states that when a substance is placed within a magnetic field, $H$, the field within the substance, $B$, differs from $H$ by the induced field, $4πI$, which is proportional to the intensity of magnetization, $I$. That is;
$B = H + 4\pi I \nonumber$
where $B$ is the magnetic field within the substance and $H$ is the applied magnetic field and $I$ is the intensity of magnetization
Lenz's Law (1834)
Lenz's Law can also be written as
$\dfrac{B}{H} = 1 + \dfrac{4π I}{H} \nonumber$
or
$\dfrac{B}{H} = 1 + 4\pi\kappa \nonumber$
where
• $B/H$ is called the magnetic permeability of the material and
• $\kappa$ is the magnetic susceptibility per unit volume, (I/H)
By definition, $\kappa$ in a vacuum is zero, so under those conditions the equation would reduce to $B=H$. It is usually more convenient to measure mass than volume and the mass susceptibility, $χ_g$, is related to the volume susceptibility, $\kappa$, through the density.
$χ_g = \dfrac{\kappa}{ρ} \nonumber$
where $\rho$ is the density.
Finally to get our measured quantity on a basis that can be related to atomic properties, we convert to molar susceptibility
$χ_m =χ_g \times RMM \nonumber$
Since this value includes the underlying diamagnetism of paired electrons, it is necessary to correct for the diamagnetic portion of χm to get a corrected paramagnetic susceptibility.
$\chi'_m = \chi_m + \chi_{dia} \nonumber$
Examples of these corrections are tabulated below.
Table $1$: Table of Diamagnetic Corrections (Pascal's constants, 10-6 c.g.s. units)
Ion DC Ion DC
Na+ 6.8 Co2+ 12.8
K+ 14.9 Co3+ 12.8
NH4+ 13.3 Ni2+ 12.8
Hg2+ 40 VO2+ 12.5
Fe2+ 12.8 Mn3+ 12.5
Fe3+ 12.8 Cr3+ 12.5
Cu2+ 12.8 Cl- 23.4
Br- 34.6 SO42- 40.1
I- 50.6 OH- 12
NO3- 18.9 C2O42- 34
ClO4- 32 OAc- 31.5
IO4- 51.9 pyr 49.2
CN- 13 Me-pyr 60
NCS- 26.2 Acac- 62.5
H2O 13 en 46.3
EDTA4- ~150 urea 33.4
these can be converted to S.I units of m3 mol-1 by multiplying by 4 π x 10-7
There are numerous methods for measuring magnetic susceptibilities, including, the Gouy, Evans and Faraday methods. These all depend on measuring the force exerted upon a sample when it is placed in a magnetic field. The more paramagnetic the sample, the more strongly it will be drawn toward the more intense part of the field.
Determination of Magnetic Susceptibility
• The Gouy Method: The underlying theory of the Gouy method is described here and a form for calculating the magnetic moment from the collected data is available as well.
• The Evans method: The Evans balance measures the change in current required to keep a pair of suspended magnets in place or balanced after the interaction of the magnetic field with the sample. The Evans balance differs from that of the Gouy in that, in the former the permanent magnets are suspended and the position of the sample is kept constant while in the latter the position of the magnet is constant and the sample is suspended between the magnets.
Orbital contribution to magnetic moments
From a quantum mechanics viewpoint, the magnetic moment is dependent on both spin and orbital angular momentum contributions. The spin-only formula used last year was given as:
$\mu_{s.o.} = \sqrt{4S(S+1)} \nonumber$
and this can be modified to include the orbital angular momentum
$\mu_{S+L} = \sqrt{4S(S+1) + L(L+1)} \nonumber$
An orbital angular momentum contribution is expected when the ground term is triply degenerate (i.e. a triplet state). These show temperature dependence as well.
In order for an electron to contribute to the orbital angular momentum the orbital in which it resides must be able to transform into an exactly identical and degenerate orbital by a simple rotation (it is the rotation of the electrons that induces the orbital contribution). For example, in an octahedral complex the degenerate t2g set of orbitals (dxz,dyx,dyz) can be interconverted by a 90o rotation. However the orbitals in the eg subset (dz2,dx2-y2) cannot be interconverted by rotation about any axis as the orbital shapes are different; therefore an electron in the eg set does not contribute to the orbital angular momentum and is said to be quenched. In the free ion case the electrons can be transformed between any of the orbitals as they are all degenerate, but there will still be partial orbital quenching as the orbitals are not identical.
Electrons in the t2g set do not always contribute to the orbital angular moment. For example in the d3, t2g3 case, an electron in the dxz orbital cannot by rotation be placed in the dyz orbital as the orbital already has an electron of the same spin. This process is also called quenching.
Tetrahedral complexes can be treated in a similar way with the exception that we fill the e orbitals first, and the electrons in these do not contribute to the orbital angular momentum. The tables in the links below give a list of all d1 to d9 configurations including high and low spin complexes and a statement of whether or not a direct orbital contribution is expected.
• Octahedral complexes
• Tetrahedral complexes
A and E ground terms
The configurations corresponding to the A1 (free ion S term), E (free ion D term), or A2 (from F term) do not have a direct contribute to the orbital angular momentum. For the A2 and E terms there is always a higher T term of the same multiplicity as the ground term which can affect the magnetic moment (usually by a only small amount).
$μ_{eff} = μ_{s.o.} (1-α λ /Δ) \label{eq10}$
where α is a constant (2 for an E term, 4 for an A2 term) and λ is the spin-orbit coupling constant which is generally only available for the free ion but this does give important information since the sign of the value varies depending on the orbital occupancy.
Table $1$: Some spin-orbit coupling constants for 1st row TM ions
metal ion Ti(III) V(III) Cr(III) Mn(III) Fe(II) Co(II) Ni(II) Cu(II)
d configuration 1 2 3 4 6 7 8 9
λ / cm-1 155 105 90 88 -102 -172 -315 -830
For $d^1$ to $d^4$ the value is positive hence $μ_{eff}$ is less than $μ_{so}$ and for $d^6$ to $d^9$ the value is negative hence $μ_{eff}$ is greater than $μ_{so}$. $Δ$ is the crystal field splitting factor which again is often not available for complexes.
For the tetrahedral Co(II) ion, CoCl42-, the observed experimental magnetic moment, μobs = 4.59 Bohr Magneton (B.M.) The spin-only magnetic moment, μs.o. = 3.88 B.M. which is not in good agreement. How can we improve the analysis?
Since the ground term in the tetrahedral field is split from a 4F to a 4A2 term then we can apply the Equation \ref{eq10}. For an $A$ term the constant α = 4. The spin-orbit coupling constant, λ for the free ion is -172 cm-1 which we can use as an approximation and Δ= 3100 cm-1. Hence
$μ_{eff} = 3.88 \times (1 - (4* -172) / 3100) \nonumber$
which comes out at μeff = 4.73 B.M.
This gives a much better fit than the spin-only formula. In the case of the series;
CoI42-, CoBr42-, CoCl42-, Co(NCS)42-
the magnetic moments have been recorded as 4.77, 4.65, 4.59, 4.40 BM assuming that λ is roughly a constant, then this variation shows the inverse effect of the spectrochemical series on the magnetic moment, since Δ is expected to increase from I- to NCS-.
T ground terms
The configurations corresponding to the T2 term (from D) or a T1 term (from an F term) are those where there is a direct contribution to orbital angular momentum expected. The magnetic moments of complexes with T terms are often found to show considerable temperature dependence. This is as a result of spin-orbit coupling that produces levels whose energy differences are frequently of the order kT, so as a result, temperature will have a direct effect on the population of the levels arising in the magnetic field.
In a Kotani plot μeff is plotted against kT/λ and when this corresponds to a value of 1 then μ equals the "spin-only" value. If this is extrapolated to infinity then the value corresponds to μS+L.
Measuring the magnetic moment at 80 K and 300 K often shows up this variation with temperature.
Example $1$:
Account for the magnetic moments of the complex, (Et4N)2[NiCl4] recorded at 80, 99 and 300 K.
80 K 99 K 300 K
3.25 B.M. 3.43 B.M. 3.89 B.M.
Ni2+ is a d8 metal ion.
The formula suggests a 4 coordinate complex and we can assume that the complex is tetrahedral with a d electron configuration of e4 t24 therefore the spin-only magnetic moment can be calculated as 2.83 BM.
Why did we ignore the possibility of it being square-planar?
The free ion Russell-Saunders ground term is 3F (L=3 and S=1) which will give rise to a lowest energy T term in a tetrahedral field and hence the resultant magnetic moment is expected to be temperature dependent and have a direct orbital contribution. The observed values may be quite different then to the calculated spin only magnetic moment. The value of μS+L can be calculated as:
$mu_{S+L}= \sqrt{4S(S+1)+L(L+1)} \nonumber$
or
$\mu_{S+L}= \sqrt{8+12} \nonumber$
or
$\mu_{S+L} = \sqrt{20} = 4.472\;B.M. \nonumber$
From the observed values it can be seen that the magnetic moment of the d8 Ni2+ complex is intermediate between the μso and μS+L values (probably due to partial quenching of the orbital angular momentum contribution) and is dependent on temperature. Further worked examples and some selected magnetic data are available.
Contributors and Attributions
• The Department of Chemistry, University of the West Indies) | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Crystal_Field_Theory/Magnetism.txt |
Figure 1: The d (left) and dx²−y² (right) orbitals. Images used with permission from Wikipedia.
Let us continue to consider an octahedral complex. The remaining d orbitals, dxy, dxz and dyz see their energy increase to a lesser extent. We refer to the dxy, dxz and dyz orbitals collectively as the t2g d orbitals.
Hybrid orbital theory can be used to describe how metals bond to ligands. When metals bond to ligands, magnetic data shows that some electrons are paired when there is no obvious reason for them to be paired. Molecular orbitals begin to account for this phenomenon by allowing wave functions to interfere in a constructive, low energy (bonding) or destructive, or high energy (antibonding) manner. Thus, the electrons can fill the lowest energy molecular orbitals available to them. However, the electron pairing may be different if the electrons were allowed to fill the lowest energy atomic orbitals available to them.
This diagram shows the field splitting of a metal with ligands in an octahedral configuration. The thick horizontal lines represent atomic orbitals of the metal (left) and ligands (right). The colors correspond to s (black), p (green) and d (red) orbitals. The middle column of horizontal lines represents molecular orbitals made of bonding (lower energy) and antibonding (higher energy) components.
If the ligands are oriented on the cartesian coordinate axes, the metal will still own 3 d orbitals, xy, xz, and xz, which do not intersect (and therefore do not interact) with the ligands. These are considered "nonbonding" orbitals, and are represented by dotted lines in the diagram.
Non-octahedral Complexes
Learning Objectives
• Understand the d-orbital degeneracies of square planar and tetrahedral metal complexes.
Tetragonal and Square Planar Complexes
If two trans- ligands in an octahedral complex are either chemically different from the other four (as in trans-[Co(NH3)4Cl2]+), or at a different distance from the metal than the other four, the result is a tetragonally distorted octahedral complex. The electronic structures of such complexes are best viewed as the result of distorting an octahedral complex. Consider, for example, an octahedral complex such as [Co(NH3)6]3+: two trans- NH3 molecules are slowly removed from the metal along the ±z axes, as shown in the top half of Figure $1$. As the two axial Co–N distances increase simultaneously, the d-orbitals that interact most strongly with the two axial ligands decrease in energy due to a decrease in electrostatic repulsions between the electrons in these orbitals and the negative ends of the ligand dipoles. The affected d orbitals are those with a component along the ±z axes—dz2, dxz, and dyz. These orbitals are not affected equally, however: because the dz2 orbital points directly at the two ligands being removed, its energy will decrease much more rapidly than the degenerate energies of the dxz and dyz, as shown in the bottom half of Figure $1$. In addition, the effective positive charge on the metal increases somewhat as the axial ligands are removed, increasing the attraction between the four remaining ligands and the metal. This increases the extent of their interactions with the other two d orbitals and increases their energies. Again, the two d orbitals are not affected equally: because the dx2−y2 orbital points directly at the four in-plane ligands, its energy increases to a greater extent than the energy of the dxy orbital, which points between the in-plane ligands. If the two axial ligands are moved infinitely far away from the metal, a square planar complex is formed. The energy of the dxy orbital actually surpasses that of the dz2 orbital in the process. The largest orbital splitting in a square planar complex, between the dx2−y2 and dxy energy levels, is identical in magnitude to Δo.
Moving the two axial ligands away from the metal ion along the z axis initially generates an elongated octahedral complex (the center compound of Figure $1$) and eventually produces a square planar complex (right). As shown below the structures, an axial elongation causes the dz2 dxz and dyz orbitals to decrease in energy and the $d_{x^2-y^2}$ and $d_{xy}$ orbitals to increase in energy. The change in energy is not the same for all five d orbitals. The dz2 orbital has a smaller xy component than does the dxy orbital, so it reaches a lower energy level; thus, the order of these orbitals is reversed.
Tetrahedral Complexes
In a tetrahedral arrangement of four ligands around a metal ion, none of the ligands lies on any of the three coordinate axes (illustrated in part (a) in Figure $2$); consequently, none of the five d orbitals points directly at the ligands. Nonetheless, the dxy, dxz, and dyz orbitals interact more strongly with the ligands than do dx2−y2 and dz2; this again results in a splitting of the five d orbitals into two groups. The splitting of the energies of the orbitals in a tetrahedral complex (Δt) is much smaller than that for an octahedral complex (Δo), however, for two reasons: first, the d orbitals interact less strongly with the ligands in a tetrahedral arrangement; second, there are only four negatively-charged regions rather than six, which decreases the electrostatic interactions by one-third if all other factors are equal. It can be shown that for complexes of the same metal ion with the same charge, the same ligands, and the same M–L distance, $\Delta_\textrm t=\frac{4}{9}\Delta_\textrm o$. The relationship between the splitting of the five d orbitals in octahedral and tetrahedral crystal fields imposed by the same ligands is shown schematically in part (b) in Figure $2$.
Δt < Δo because of weaker d-orbital–ligand interactions and decreased electrostatic interactions.
Example $1$: Predicting Structure
For each complex, predict its structure, whether it is high spin or low spin, and the number of unpaired electrons present.
1. $[Cu(NH_3)_4]^{2+}$
2. $[Ni(CN)_4]^{2−}$
Solution
Because Δo is so large for the second- and third-row transition metals, all four-coordinate complexes of these metals are square planar due to the much higher crystal field stabilization energy (CFSE) for square planar versus tetrahedral structures. The only exception is for d10 metal ions such as Cd2+, which have zero CFSE and are therefore tetrahedral as predicted by the VSEPR model. Four-coordinate complexes of the first-row transition metals can be either square planar or tetrahedral; the former is favored by strong-field ligands, whereas the latter is favored by weak-field ligands. For example, the [Ni(CN)4]2− ion is square planar, while the [NiCl4]2− ion is tetrahedral.
1.
The copper in this complex is a d9 ion and it has a coordination number of 4. So it is probably either square planar or tetrahedral. To estimate which, we need to fill in the CFT splitting diagrams for each with 9 electrons and ask which has a lower energy. Comparing the square planer (Figure $1$) splitting diagram with tetrahedral (Figure $2$), suggests that 9 electrons will have a net lower total energy for square planar (since the $d_{x^2-y^2}$ orbital is high in energy, the others are lower).
For the square planar structure, neither high nor low spin states are possible (only one state) with a single unpaired electron.
2.
The nickle in this complex is a d8 ion and it has a coordination number of 4. So it is probably either square planar or tetrahedral. To estimate which, we need to fill in the CFT splitting diagrams for each with 8 electrons and ask which has a lower energy. Comparing the square planer (Figure $1$) splitting diagram with tetrahedral (Figure $2$), suggests that 8 electrons will have a net lower total energy for square planar (since the $d_{x^2-y^2}$ orbital is high in energy, the others are lower).
For the square planar structure, it is a low spin complex since a high spin requires a lot of energy to promote to the $d_{x^2-y^2}$ orbital. Hence, there are no unpaired electrons
Exercise $1$
What are the geometries of the following two complexes?
1. $[\ce{AlCl_4}]^-$
2. $[\ce{Ag(NH_3)_2}]^+$
Answer 1
tetrahedral
Answer 2
linear
Summary
Distorting an octahedral complex by moving opposite ligands away from the metal produces a tetragonal or square planar arrangement, in which interactions with equatorial ligands become stronger. Because none of the d orbitals points directly at the ligands in a tetrahedral complex, these complexes have smaller values of the crystal field splitting energy Δt. In tetrahedral molecular geometry, a central atom is located at the center of four substituents, which form the corners of a tetrahedron. Tetrahedral geometry is common for complexes where the metal has $d^0$ or $d^{10}$ electron configuration. The CFT diagram for tetrahedral complexes has $d_{x^2-y^2}$ and $d_{z^2}$ orbitals equally low in energy because they are between the ligand axis and experience little repulsion. In square planar molecular geometry, a central atom is surrounded by constituent atoms, which form the corners of a square on the same plane. The square planar geometry is prevalent for transition metal complexes with d8 configuration. The CFT diagram for square planar complexes can be derived from octahedral complexes yet the $d_{x^2-y^2}$ level is the most destabilized and is left unfilled. | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Crystal_Field_Theory/Metals_Tetrahedral_and_Octahedral.txt |
How do we tell whether a particular complex is octahedral, tetrahedral, or square planar? Obviously if we know the formula, we can make an educated guess: something of the type ML6 will almost always be octahedral (there is an alternative geometry for 6-coordinate complexes, called trigonal prismatic, but it's pretty rare), whereas something of formula ML4 will usually be tetrahedral unless the metal atom has the d8 electron configuration, in which case it will probably be square planar. But what if we take a particular metal ion and a particular ligand? Can we predict whether it will form an octahedral or a tetrahedral complex, for example? To an extent, the answer is yes... we can certainly say what factors will encourage the formation of tetrahedral complexes instead of the more usual octahedral.
The Crystal Field Stabilization Energy (CFSE) is the additional stabilization gained by the splitting of the orbitals according to the crystal field theory, against the energy of the original five degenerate d orbitals. So, for example, in a d1situation such as [Ti(OH2)6]3+, putting the electron into one of the orbitals of the t2g level gains -0.4 Δo of CFSE. Generally speaking, octahedral complexes will be favored over tetrahedral ones because:
• It is more (energetically) favorable to form six bonds rather than four
• The CFSE is usually greater for octahedral than tetrahedral complexes. Remember that Δo is bigger than Δtet (in fact, Δtet is approximately 4/9 Δo).
If we make the assumption that Δtet = 4/9 Δo, we can calculate the difference in stabilization energy between octahedral and tetrahedral geometries by referencing everything in terms of Δo.
Example $1$: $d^3$ Stabilized Structures
Which is the preferred configuration for a d3 metal: tetrahedral or octahedral?
Solution
To answer this, the Crystal Field Stabilization Energy has to be calculated for a $(d^3$ metal in both configurations. The geometry with the greater stabilization will be the preferred geometry.
• For a d3 octahedral configuration, the Crystal Field Stabilization Energy is
$3 \times -0.4 \Delta_o = -1.2 \Delta_o \nonumber$
• For a d3 tetrahedral configuration (assuming high spin), the Crystal Field Stabilization Energy is
$-0.8 \Delta_{tet} \nonumber$
Remember that because Δtet is less than half the size of Δo, tetrahedral complexes are often high spin. We can now put this in terms of Δo (we can make this comparison because we're considering the same metal ion and the same ligand: all that's changing is the geometry)
So for tetrahedral d3, the Crystal Field Stabilization Energy is:
CFSE = -0.8 x 4/9 Δo = -0.355 Δo.
And the difference in Crystal Field Stabilization Energy between the two geometries will be:
1.2 - 0.355 = 0.845 Δo.
If we do a similar calculation for the other configurations, we can construct a Table of Δo, Δtet and the difference between them (we'll ignore their signs since we're looking for the difference between them).
Table $1$: Crystal Field Stabilization Energies (not splitting parameters). This table compares the values of the CFSE for octahedral and tetrahedral geometries, assuming high spin configurations. The units are Δo, and we're assuming that Δtet = 4/9 Δo.
Octahedral Tetrahedral Difference
d0, d5, d10 0 0 0
d1, d6 0.4 0.27 0.13
d2, d7 0.8 0.53 0.27
d3, d8 1.2 0.36 0.84
d4, d9 0.6 0.18 0.42
These values can be plotted:.
Notice that the Crystal Field Stabilization Energy almost always favors octahedral over tetrahedral in most cases, but the degree of favorability varies with the electronic configuration. In other words, for d1 there's only a small gap between the oct and tet lines, whereas at d3 and d8 there's a big gap. However, for d0, d5 high spin and d10, there is no CFSE difference between octahedral and tetrahedral. The ordering of favorability of octahedral over tetrahedral is:
d3, d8 > d4, d9> d2, d7 > d1, d6 > d0, d5, d10
The units of the graph are Δo. So if we have strong field ligands present, Δo will be bigger anyway (according to the spectrochemical series), and any energy difference between the oct and tet lines will be all the greater for it. A bigger Δo might also push the complexes over to low spin. Similarly, as we saw previously, high oxidation states and metals from the 2nd and 3rd rows of the transition series will also push up Δo.
On the other hand, if large or highly charged ligands are present, they may suffer large interligand repulsions and thus prefer a lower coordination number (4 instead of 6). Consequently if you set out to make something that would have a tetrahedral geometry, you would use large, negatively charged, weak field ligands, and use a metal atom with a d0, d5 or d10 configuration from the first row of the transition series (though of course having weak field ligands doesn't matter in these three configurations because the difference between oct and tet is 0 Δo). As Table 2 shows, you can find tetrahedral complexes for most configurations, but there are very few for d3 and d8.
Table $2$: Tetrahedral complexes of different d electron counts
d0 MnO4- d5 MnCl42-
d1 TiCl4- d6 FeCl42-
d2 Cr(OR)4 d7 CoCl42-
Contributors and Attributions
• Dr Mike Morris, March 2001 | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Crystal_Field_Theory/Octahedral_vs._Tetrahedral_Geometries.txt |
Orgel diagrams are useful for showing the energy levels of both high spin octahedral and tetrahedral transition metal ions. They ONLY show the spin-allowed transitions. For complexes with D ground terms only one electronic transition is expected and the transition energy corresponds directly to D. Hence, the following high spin configurations are dealt with: d1, d4, d6 and d9.
D Orgel diagram
On the left hand side d1, d6 tetrahedral and d4, d9 octahedral complexes are covered and on the right hand side d4, d9 tetrahedral and d1, d6 octahedral. For simplicity, the g subscripts required for the octahedral complexes are not shown.
For complexes with F ground terms, three electronic transitions are expected and D may not correspond directly to a transition energy. The following configurations are dealt with: d2, d3, high spin d7 and d8.
F Orgel diagram
On the left hand side, d2, d7 tetrahedral and d3, d8 octahedral complexes are covered and on the right hand side d3, d8 tetrahedral and d2 and high spin d7 octahedral. Again for simplicity, the g subscripts required for the octahedral complexes are not shown.
On the left hand side, the first transition corresponds to D, the equation to calculate the second contains expressions with both D and C.I. (the configuration interaction from repulsion of like terms) and the third has expressions which contain D, C.I. and the Racah parameter B.
• 4T2g <--- 4A2g transition energy = D
• 4T1g(F) <--- 4A2g transition energy = 9/5 *D - C.I.
• 4T1g(P) <--- 4A2g transition energy = 6/5 *D + 15B' + C.I.
On the right hand side, the first transition can be unambiguously assigned as:
3T2g <--- 3T1g transition energy = 4/5 *D + C.I.
But, depending on the size of the ligand field (D) the second transition may be due to:
3A2g <--- 3T1g transition energy = 9/5 *D + C.I.
for a weak field or
3T1g(P) <--- 3T1g transition energy = 3/5 *D + 15B' + 2 * C.I.
for a strong field.
Tanabe-Sugano Diagrams
An alternative method is to use Tanabe Sugano diagrams, which are able to predict the transition energies for both spin-allowed and spin-forbidden transitions, as well as for both strong field (low spin), and weak field (high spin) complexes. In this method the energy of the electronic states are given on the vertical axis and the ligand field strength increases on the horizontal axis from left to right. Linear lines are found when there are no other terms of the same type and curved lines are found when 2 or more terms are repeated. This is as a result of the "non-crossing rule". The baseline in the Tanabe-Sugano diagram represents the lowest energy or ground term state.
case (not many examples documented)
The electronic spectrum of the V3+ ion, where V(III) is doped into alumina (Al2O3), shows three major peaks with frequencies of: $\nu_1$=17400 cm-1, $\nu_2$=25400 cm-1 and $\nu_3$=34500 cm-1.
These have been assigned to the following spin-allowed transitions.
• 3T2g<---3T1g
• 3T1g(P)<---3T1g
• 3A2g<---3T1g
The ratio between the first two transitions is calculated as $\frac{\nu_1}{\nu_2}$ which is equal to 25400 / 17400 = 1.448.
To calculate the Racah parameter, B, the position on the horizontal axis where the ratio between the lines representing $\nu_2$ and $\nu_1$ is equal to 1.448, has to be determined. On the diagram below, this occurs at D/B=30.9. Having found this value, a vertical line is drawn at this position.
Figure $1$: Tanabe-Sugano diagram for d2 octahedral complexes
On moving up the line from the ground term to where lines from the other terms cross it, we are able to identify both the spin-forbidden and spin-allowed transition and hence the total number of transitions that are possible in the electronic spectrum.
Next, find the values on the vertical axis that correspond to the spin-allowed transitions so as to determine the values of n1/B, n2/B and n3/B. From the diagram above these are 28.78, 41.67 and 59.68 respectively.
Knowing the values of n1, n2 and n3, we can now calculate the value of B. Since n1/B=28.78 and n1 is equal to 17,400 cm-1, then
$B=\dfrac{n_1}{28.78} = \dfrac{17400}{28.78} \nonumber$
or
$B=604.5\;cm^{-1} \nonumber$
Then it is possible to calculate the value of D. Since D/B=30.9, then: D=B*30.9 and hence: D = 604.5 * 30.9 = 18680 cm-1
case
Calculate the value of B and D for the Cr3+ ion in [Cr(H2O)6)]3+ if n1=17000 cm-1, n2=24000 cm-1 and n3=37000 cm-1.
Solution
These values have been assigned to the following spin-allowed transitions.
$^4T_{2g} \leftarrow ^ 4A_{2g} \nonumber$
$^4T_{1g} \leftarrow ^ 4A_{2g} \nonumber$
$^4T_{1g}(P) \leftarrow ^4A_{2g} \nonumber$
From the information given, the ratio $n_2 / n_1 = 24000 / 17000 = 1.412$. Using a Tanabe-Sugano diagram for a d3 system this ratio is found at D/B=24.00
Figure $2$: Tanabe-Sugano diagram for d3 octahedral complexes
Interpolation of the graph to find the Y-axis values for the spin-allowed transitions gives:
• $\dfrac{n_1}{B}=24.00$
• $\dfrac{n_2}{B}=33.90$
• $\dfrac{n_3}{B}=53.11$
Recall that $n_1=17000 \,cm^{-1}$. Therefore for the first spin-allowed transition,
17000 /B =24.00 from which B can be obtained, B=17000 / 24.00 or B=708.3 cm-1.
This information is then used to calculate D.
Since D / B=24.00 then D = B*24.00 = 708.3 * 24.00 = 17000 cm-1.
It is observed that the value of Racah parameter $B$ in the complex is 708.3 cm-1, while the value of B in the free Cr3+ ion is 1030 cm-1. This shows a 31% reduction in the Racah parameter indicating a strong Nephelauxetic effect.
The Nephelauxetic Series is as follows:
F->H2O>urea>NH3 >en~C2O42- >NCS- >Cl-~CN->Br- >S2- ~I-.
Ionic ligands such as F-give small reduction in B, while covalently bonded ligands such as I- give a large reduction in B.
The original paper by Tanabe and Sugano[10] had the d5 and d6 diagrams each missing a T term from excited I states. These diagrams were reproduced in the often quoted text by Figgis[12(a)] and so the errors have been perpetuated. An exception is the text by Purcell and Kotz[15] where the missing T terms have been included, however in their case they have ignored lower lying terms from excited D, F, G and H states which for d5 are the main transitions seen in the spin forbidden spectra of Mn(II) complexes.
A set of qualitative diagrams have been drawn for each configuration (which include the missing T terms) and along with the newest release of "Ligand Field Theory and its applications" by Figgis and Hitchman [12(b)] represent the only examples of Tanabe-Sugano diagrams that provide a comprehensive set of terms for spectral interpretation
Contributors and Attributions
• The Department of Chemistry, University of the West Indies)
Orgel Diagrams
A set of qualitative diagrams have been drawn for each configuration (which include the missing T terms) and along with the newest release of "Ligand Field Theory and its applications" by Figgis and Hitchman [12(b)] represent the only examples of Tanabe-Sugano diagrams that provide a comprehensive set of terms for spectral interpretation. | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Crystal_Field_Theory/Orgel_Diagrams/Qualitative_Orgel_Diagrams.txt |
Tanabe-Sugano diagrams are used in coordination chemistry to predict electromagnetic absorptions of metal coordination compounds of tetrahedral and octahedral complexes. The analysis derived from the diagrams can also be compared to experimental spectroscopic data. Armed with spectroscopic data, an approximation to the crystal field splitting energy (10Dq), generated by ligands attached to a metal center, can be found.
Crystal Field Splitting Energy
Within Crystal Field Theory, the interaction of the metal and ligand arise from the positive charge of the metal and negative charge on the ligands. The theory is developed by looking at the five degenerate d-orbitals and how the energies are changed on being surrounded by the negative point charges of the ligands. As the ligands are moved closer to the metal the repulsion between the electrons of the metal and ligands break the degeneracy of the d-orbitals. In the case of an octahedral complex 6 ligands surround a metal center with a single pair on each axis. This raises the energies of dx^2-y^2, dz^2 relative to those of dxy dxz dyz. This energy split is called Δoct. The tetrahedral energy split is about 4/9Δoct.1
Racah Parameters
Racah parameters were generated as a means to describe the effects of electron-electron repulsion within the metal complexes. The Racah parameters are A, B and C. In the case of Tanabe-Sugano diagrams each electron configuration split has an energy that can be related by the B value. A is ignored because it is roughly the same for any metal center and C generally approximated as being 1/4B. What B represents is an approximation of the bond strength between the ligand and metal.1 Comparisons between tabulated free ion B and B of a coordination complex is called the nephelauxetic ratio (the effect of reducing electron-electron repulsion via ligands). This effect is what gives rise to the spectrochemical series of ligands described later.
$\beta = \dfrac{\beta_{complex}}{\beta_{freeion}} \tag{1}$
Parameters
The x-axis in a Tanabe-Sugano diagram is in terms of the crystal field splitting parameter, 10Dq, or Δoct, scaled by the B Racah Parameter. The y-axis is in terms of energy of a electronic transition, E, scaled by B. Each line represents the energy of an electronic state while varying the strength of octahedral ligand field. And while only a few electronic states are spin allowed the spin forbidden electron transitions are included since spin forbidden transitions can appear in spectrum. Each term symbol is created from the splitting of term symbols from spherical to octahedral symmetry. With the relative energy ordering of the states are determined via Hund's rules.
Diagrams for d4, d5, d6, and d7 metal ions have a discontinuity in energies as the ligand field is varied. The discontinuity, shown with the vertical line, represents complexes changing from high-spin to low-spin complexes. At the line, the spin pairing energy is equal to the crystal field splitting energy. To the left of the line metal complexes are high-spin as the spin pairing energy is greater than that of the ligand field splitting. To the right of the line metal complexes are low-spin as the spin pairing energy is less than that of the ligand field splitting energy.
Diagrams
d2 Tanabe-Sugano diagram d3 Tanabe-Sugano diagram d4 Tanabe-Sugano diagram d5 Tanabe-Sugano diagram
Note: Each of the images
is a thumbnail clicking on
one will expand the image.
d6 Tanabe-Sugano diagram d7 Tanabe-Sugano diagram d8 Tanabe-Sugano diagram
How to use the Diagrams
1. Before looking at the diagrams the d-configuration of the metal ion must be determined.
2. Choose the appropriate Tanabe-Sugano diagram matching the d-configuration (http://chemistry.bd.psu.edu/jircitano/TSdiagram.pdf has full page diagrams necessary for measurements).
3. Take a spectrum of the complex and identify $\lambda_{max}$ for spin-allowed (strong intensity) and spin forbidden (weak intensity) transitions.
4. Convert $\lambda_{max}$ to wavenumbers and generate energy ratios relative to the lowest allowed transition. (i.e. E2/E1and E3/E1)
5. Using a ruler, slide it across the printed diagram until the E/B ratios between lines is equivalent to the ratios found in step 4.
6. Solve for B using the E/B values (y-axis, step 4) and Δoct/B (x-axis, step 5) to yield the ligand field splitting energy 10Dq.
Example $1$: Chromium Splitting
A Cr3+ metal complex has strong transitions and $\lambda_{max}$ at
• 431.03 nm,
• 781.25 nm, and
• 1,250 nm.
Determine the $Δ_{oct}$ for this complex.
Solution
1. Cr has 6 electrons. Cr3+ has three electrons so its has a d-configuration of d3
2. Locate the d3 Tanabe-Sugano diagram
3. Convert to wavenumbers:
$\dfrac{10^7(nm/cm)}{1250\; nm}= 8,000\; cm^{-1} \nonumber$
$\dfrac{10^7(nm/cm)}{781.25\; nm}= 13,600\; cm^{-1} \nonumber$
$\dfrac{10^7(nm/cm)}{431.03\; nm}= 23,200\; cm^{-1} \nonumber$
1. Allowed transitions are $\ce{^4T_{1g}} \leftarrow \ce{ ^4_{\,}A_{2g}}$, $\ce{^4T_{1g} \leftarrow ^4_{\,}A_{2g}}$ and $\ce{^4T_{2g}\leftarrow ^4_{\,}A_{2g}}$.
Transition Energy cm-1 Ratios to lowest
$\ce{^4T_{1g}} \leftarrow \ce{ ^4_{\,}A_{2g}}$ 23,200 2.9
$\ce{^4T_{1g} \leftarrow ^4_{\,}A_{2g}}$ 13,600 1.7
$\ce{^4T_{2g}\leftarrow ^4_{\,}A_{2g}}$ 8,000 1
1. Sliding the ruler perpendicular to the x-axis of the d3 diagram yields the following values:
Δoct/B 10 20 30 40
Height E(ν3)/B 29 45 64 84
Height E(ν2)/B 17 30 40 51
Height E(ν1)/B 10 20 30 40
Ratio E(ν3)/E(ν1) 2.9 2.25 2.13 2.1
Ratio E(ν2)/E(ν1) 1.7 1.5 1.33 1.275
1. Based on the two tables above it should be assessed that the Δoct/B value is 10. B is found by finding the dividing E by the height.
Energy cm-1 Height B
23,200 29 800
13,600 17 800
8,000 10 800
1. Next multiply Δoct/B by B to yield the Δoct energy. $10 \times 800 = 8000\; cm^{-1}=Δ_{oct} \nonumber$
Each problem is of varying complexity as several steps may be needed to find the correct Δoct/B values that yield the proper energy ratios.
Nephelauxetic Effect
Imagine you had an abundance of V(H2O)63+ (d2) which has two absorptions. If you had no other available metal centers, but an abundance of ligands, the complex's absorption spectrum (therefore its color) could be changed via application of the spectrochemical series:
I < Br < S2− < SCN < Cl < NO3 < N3 < F < OH < C2O42− ≈ H2O < NCS < CH3CN < py (pyridine) < NH3 < en (ethylenediamine) < bipy (2,2'-bipyridine) < phen (1,10-phenanthroline) < NO2 < PPh3 < CN ≈ CO
If you wanted higher energy absorptions (shift toward purple colors) you use ligands to the right of H2O and if you wanted redder colors attach ligands to the left of H2O. Here you could measure the nephelauxetic ratio of the ligands to compare how each ligand modifies the B Racah parameter. The series of complexes would also serve as a very good demonstration for a classroom. The reason behind modifying metal ligands has implications for commercial products. Unique colors in ceramics and glass products can be traced back to many metal compounds.
Problems
1. For a d7 metal ion determine the energy ratios for allowed transitions at Δoct/B of 20.
2. For a d6metal ion of Δoct/B = 30 and B=530 cm-1 what would the energies of the 5 allowed transitions be? How many are in the UV-Vis range? How many are in the IR range?
3. Write out the allowed transitions for a d5 metal ion in a E/B> 28 ligand field.
4. A d4 complex exhibits absorptions at 5500 cm-1 (strong) and 31350 cm-1 (weak). What are the transitions that are being exhibited in the complex? What is the corresponding Δoct for the complex?
5. A spectrum of d7 metal complex seemingly exhibits only two intense transitions. What is the Δoct/B that this situation occurs? Please use reference to specific transitions and energy splitting.
Answers
1. Δoct/B of 20 yields E/B values of 38, 32, 18. Ratios then are 2.11 and 1.78
2. Δoct/B = 30 yields E/B heights of 27, 40, 57, 65, 85. Energies are then 14310, 21200, 30210, 34450 and 45050 cm-1. All are in the UV-Vis range. *note you need to infer the E/B value for the last transition as the diagram does not extend that far up.
3. 2A2g<-2T2g, 2T1g<-2T2g, 2Eg<-2T2g, and 2A1g<-2T2g.
4. 31,350/5,500 gives a ratio of 5.7/1. The only Δoct/B value that matches is at 10. B value is then 550 cm-1. Δoctequals 5500 cm-1.
5. Three transitions are generated at low Δoct/B. However, at about a value of Δoct/B = 13 the transitions 4A2g<-4T1g, and 4T1g<-4T1g have the same energies which results in the appearance of only two absorptions.
Contributors and Attributions
• Evan Sarina, UC Davis | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Crystal_Field_Theory/Tanabe-Sugano_Diagrams.txt |
Learning Objectives
• Discuss the d-orbital degeneracy of square planar and tetrahedral metal complexes.
Tetrahedral Geometry
Tetrahedral geometry is a bit harder to visualize than square planar geometry. Tetrahedral geometry is analogous to a pyramid, where each of corners of the pyramid corresponds to a ligand, and the central molecule is in the middle of the pyramid. This geometry also has a coordination number of 4 because it has 4 ligands bound to it. Finally, the bond angle between the ligands is 109.5o. An example of the tetrahedral molecule \(\ce{CH4}\), or methane.
In a tetrahedral complex, \(Δ_t\) is relatively small even with strong-field ligands as there are fewer ligands to bond with. It is rare for the \(Δ_t\) of tetrahedral complexes to exceed the pairing energy. Usually, electrons will move up to the higher energy orbitals rather than pair. Because of this, most tetrahedral complexes are high spin.
Square Planar Complexes
In square planar molecular geometry, a central atom is surrounded by constituent atoms, which form the corners of a square on the same plane. The geometry is prevalent for transition metal complexes with d8 configuration. This includes Rh(I), Ir(I), Pd(II), Pt(II), and Au(III). Notable examples include the anticancer drugs cisplatin (\(\ce{PtCl2(NH3)2}\)).
A square planar complex also has a coordination number of 4. The structure of the complex differs from tetrahedral because the ligands form a simple square on the x and y axes. Because of this, the crystal field splitting is also different (Figure \(1\)). Since there are no ligands along the z-axis in a square planar complex, the repulsion of electrons in the \(d_{xz}\), \(d_{yz}\), and the \(d_{z^2}\) orbitals are considerably lower than that of the octahedral complex (the \(d_{z^2}\) orbital is slightly higher in energy to the "doughnut" that lies on the x,y axis). The \(d_{x^2-y^2}\) orbital has the most energy, followed by the \(d_{xy}\) orbital, which is followed by the remaining orbtails (although \(d_{z^2}\) has slightly more energy than the \(d_{xz}\) and \(d_{yz}\) orbital). This pattern of orbital splitting remains constant throughout all geometries. Whichever orbitals come in direct contact with the ligand fields will have higher energies than orbitals that slide past the ligand field and have more of indirect contact with the ligand fields. So when confused about which geometry leads to which splitting, think about the way the ligand fields interact with the electron orbitals of the central atom.
In square planar complexes \(Δ\) will almost always be large (Figure \(1\)), even with a weak-field ligand. Electrons tend to be paired rather than unpaired because paring energy is usually much less than \(Δ\). Therefore, square planar complexes are usually low spin.
Advanced: \(\ce{[PdCl4]^{2-}}\) is square planar and \(\ce{[NiCl4]^{2-}}\) is tetrahedral
The molecule \(\ce{[PdCl4]^{2−}}\) is diamagnetic, which indicates a square planar geometry as all eight d-electrons are paired in the lower-energy orbitals. However, \(\ce{[NiCl4]^{2−}}\) is also d8 but has two unpaired electrons, indicating a tetrahedral geometry. Why is \(\ce{[PdCl4]^{2−}}\) square planar if \(\ce{Cl^{-}}\) is not a strong-field ligand?
Solution
The geometry of the complex changes going from \(\ce{[NiCl4]^{2−}}\) to \(\ce{[PdCl4]^{2−}}\). Clearly this cannot be due to any change in the ligand since it is the same in both cases. It is the other factor, the metal, that leads to the difference.
Consider the splitting of the d-orbitals in a generic d8 complex. If it were to adopt a square planar geometry, the electrons will be stabilized (with respect to a tetrahedral complex) as they are placed in orbitals of lower energy. However, this comes at a cost: two of the electrons, which were originally unpaired in the tetrahedral structure, are now paired in the square-planer structure:
We can label these two factors as \(ΔE\) (stabilization derived from occupation of lower-energy orbitals) and \(P\) (spin pairing energy) respectively. One can see that:
• If \(ΔE>P\), then the complex will be square planar
• If \(ΔE<P\), then the complex will be tetrahedral.
This is analogous to deciding whether an octahedral complex adopts a high- or low-spin configuration; where the crystal field splitting parameter \(Δ_o\) \(ΔE\) does above. Unfortunately, unlike \(Δ_o\) in octahedral complexes, there is no simple graphical way to represent \(ΔE\) on the diagram above since multiple orbitals are changed in energy between the two geometries.
Interpreting the origin of metal-dependent stabilization energies can be tricky. However, we know experimentally that \(\ce{Pd^{2+}}\) has a larger splitting of the d-orbitals and hence a larger \(\Delta E\) than \(\ce{Ni^{2+}}\) (moreover \(P\) is also smaller).
Practically all 4d and 5d d8 \(\ce{ML4}\) complexes adopt a square planar geometry, irrespective if the ligands are strong-field ligand or not. Other examples of such square planar complexes are \(\ce{[PtCl4]^{2−}}\) and \(\ce{[AuCl4]^{-}}\).
Summary
• In tetrahedral molecular geometry, a central atom is located at the center of four substituents, which form the corners of a tetrahedron.
• Tetrahedral geometry is common for complexes where the metal has d0 or d10electron configuration.
• The CFT diagram for tetrahedral complexes has dx2−y2 and dz2 orbitals equally low in energy because they are between the ligand axis and experience little repulsion.
• In square planar molecular geometry, a central atom is surrounded by constituent atoms, which form the corners of a square on the same plane.
• The square planar geometry is prevalent for transition metal complexes with d8 configuration.
• The CFT diagram for square planar complexes can be derived from octahedral complexes yet the dx2-y2 level is the most destabilized and is left unfilled. | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Crystal_Field_Theory/Tetrahedral_vs._Square_Planar_Complexes.txt |
The energy level splitting of the d-orbitals due to their interaction with the ligands in a complex has important structural and thermodynamic effects on the chemistry of transition-metal complexes. Although these two types of effects are interrelated, they are considered separately here.
Structural Effects
There are two major kinds of structural effects: effects on the ionic radii of metal ions with regular octahedral or tetrahedral geometries, and structural distortions observed for specific electron configurations.
Ionic Radii
Figure $1$ is a plot of the ionic radii of the divalent fourth-period metal ions versus atomic number. The dashed line represents the behavior predicted based on the effects of screening and variation in effective nuclear charge (Zeff), assuming a spherical distribution of the 3d electrons. Because these radii are based on the structures of octahedral complexes and Cr2+ and Cu2+ do not form truly octahedral complexes, the points for these ions are shown as open circles.
A similar effect is observed for the V2+ ion, which has a d3 configuration. Because the three electrons in the t2g orbitals provide essentially no shielding of the ligands from the metal, the ligands experience the full increase of +1 in nuclear charge that occurs from Ti2+ to V2+. Consequently, the observed ionic radius of the V2+ ion is significantly smaller than that of the Ti2+ ion.
Skipping the Cr2+ ion for the moment, consider the d5 Mn2+ ion. Because the nuclear charge increases by +2 from V2+ to Mn2+, Mn2+ might be expected to be smaller than V2+. The two electrons added from V2+ to Mn2+ occupy the eg orbitals, however, which are oriented directly toward the six ligands. Because these electrons are localized directly between the metal ion and the ligands, they are effective at screening the ligands from the increased nuclear charge. As a result, the ionic radius actually increases significantly from V2+ to Mn2+, despite the higher nuclear charge of the latter.
The same effects are observed in the second half of the first-row transition metals. In the Fe2+, Co2+, and Ni2+ ions, the extra electrons are added successively to the t2g orbitals, resulting in increasingly poor shielding of the ligands from the nuclei and in abnormally small ionic radii. Skipping over Cu2+, adding the last two electrons causes a significant increase in the ionic radius of Zn2+, despite its greater nuclear charge.
The Jahn–Teller Effect
Because simple octahedral complexes are not observed for the Cr2+ and Cu2+ ions, only estimated values for their radii are shown in Figure $1$. Since both Cr2+ and Cu2+ ions have electron configurations with an odd number of electrons in the eg orbitals. Because the single electron (in the case of Cr2+) or the third electron (in the case of Cu2+) can occupy either one of two degenerate eg orbitals, both systems have degenerate ground states. The Jahn–Teller theorem states that such non-linear systems are not stable; they undergo a distortion that makes the complex less symmetrical and splits the degenerate states, which decreases the energy of the system. The distortion and resulting decrease in energy are collectively referred to as the Jahn–Teller effect. Neither the nature of the distortion nor its magnitude is specified, and in fact, they are difficult to predict. In principle, Jahn–Teller distortions are possible for many transition-metal ions; in practice, however, they are observed only for systems with an odd number of electrons in the eg orbitals, such as the Cr2+ and Cu2+ ions.
Consider an octahedral Cu2+ complex, [Cu(H2O)6]2+, which is elongated along the z axis. As indicated in Figure $2$, this kind of distortion splits both the eg and t2g sets of orbitals. Because the axial ligands interact most strongly with the dz2 orbital, the splitting of the eg set (δ1) is significantly larger than the splitting of the t2g set (δ2), but both δ1 and δ2 are much, much smaller than the Δo. This splitting does not change the centerpoint of the energy within each set, so a Jahn–Teller distortion results in no net change in energy for a filled or half-filled set of orbitals. If, however, the eg set contains one electron (as in the d4 ions, Cr2+ and Mn3+) or three electrons (as in the d9 ion, Cu2+), the distortion decreases the energy of the system. For Cu2+, for example, the change in energy after distortion is 2(−δ1/2) + 1(δ1/2) = −δ1/2. For Cu2+ complexes, the observed distortion is always an elongation along the z axis by as much as 50 pm; in fact, many Cu2+ complexes are distorted to the extent that they are effectively square planar. In contrast, the distortion observed for most Cr2+ complexes is a compression along the z axis. In both cases, however, the net effect is the same: the distorted system is more stable than the undistorted system.
Jahn–Teller distortions are most important for d9 and high-spin d4 complexes; the distorted system is more stable than the undistorted one.
Increasing the axial metal–ligand distances in an octahedral d9 complex is an example of a Jahn–Teller distortion, which causes the degenerate pair of eg orbitals to split in energy by an amount δ1; δ1 and δ2 are much smaller than Δo. As a result, the distorted system is more stable (lower in energy) than the undistorted complex by δ1/2.
Thermodynamic Effects
As previously noted, crystal field splitting energies (CFSEs) can be as large as several hundred kilojoules per mole, which is the same magnitude as the strength of many chemical bonds or the energy change in most chemical reactions. Consequently, CFSEs are important factors in determining the magnitude of hydration energies, lattice energies, and other thermodynamic properties of the transition metals.
Hydration Energies
The hydration energy of a metal ion is defined as the change in enthalpy for the following reaction:
$M^{2+}_{(g)} + H_2O_{(l)} \rightarrow M^{2+}_{(aq)} \label{1.1.1}$
Although hydration energies cannot be measured directly, they can be calculated from experimentally measured quantities using thermochemical cycles. In Figure $\PageIndex{3a}$, a plot of the hydration energies of the fourth-period metal dications versus atomic number forms a curve with two valleys. Note the relationship between the plot in Figure $\PageIndex{3a}$ and the plot of ionic radii in Figure $1$ the overall shapes are essentially identical, and only the three cations with spherically symmetrical distributions of d electrons (Ca2+, Mn2+, and Zn2+) lie on the dashed lines. In Figure $\PageIndex{3a}$, the dashed line corresponds to hydration energies calculated based solely on electrostatic interactions. Subtracting the CFSE values for the [M(H2O)6]2+ ions from the experimentally determined hydration energies gives the points shown as open circles, which lie very near the calculated curve. Therefore, CFSEs are primarily responsible for the differences between the measured and calculated values of hydration energies.
Lattice Energies
Values of the lattice energies for the fourth-period metal dichlorides are plotted against atomic number in part (b) of Figure $3$. Recall that the lattice energy is defined as the negative of the enthalpy change for the reaction below. Like hydration energies, lattice energies are determined indirectly from a thermochemical cycle.
$M^{2+} (g) + 2Cl^− (g) \rightarrow MCl_2 (s) \label{1.1.2}$
The shape of the lattice-energy curve is essentially the mirror image of the hydration-energy curve in part (a) of Figure $3$, with only Ca2+, Mn2+, and Zn2+ lying on the smooth curve. It is not surprising that the explanation for the deviations from the curve is exactly the same as for the hydration energy data: all the transition-metal dichlorides, except MnCl2 and ZnCl2, are more stable than expected due to CFSE.
Summary
Distorting an octahedral complex by moving opposite ligands away from the metal produces a tetragonal or square planar arrangement, in which interactions with equatorial ligands become stronger. Because none of the d orbitals points directly at the ligands in a tetrahedral complex, these complexes have smaller values of the crystal field splitting energy Δt. The crystal field stabilization energy (CFSE) is the additional stabilization of a complex due to placing electrons in the lower-energy set of d orbitals. CFSE explains the unusual curves seen in plots of ionic radii, hydration energies, and lattice energies versus atomic number. The Jahn–Teller theorem states that a non-linear molecule with a spatially degenerate electronic ground state undergoes a geometrical distortion to remove the degeneracy and lower the overall energy of the system. | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Crystal_Field_Theory/Thermodynamics_and_Structural_Consequences_of_d-Orbital_Splitting.txt |
Covalent Network Solids are giant covalent substances like diamond, graphite and silicon dioxide (silicon(IV) oxide). This page relates the structures of covalent network solids to the physical properties of the substances.
Diamond
Carbon has an electronic arrangement of 2,4. In diamond, each carbon shares electrons with four other carbon atoms - forming four single bonds.
In the diagram some carbon atoms only seem to be forming two bonds (or even one bond), but that's not really the case. We are only showing a small bit of the whole structure. This is a giant covalent structure - it continues on and on in three dimensions. It is not a molecule, because the number of atoms joined up in a real diamond is completely variable - depending on the size of the crystal.
How to draw the structure of diamond
Don't try to be too clever by trying to draw too much of the structure! Learn to draw the diagram given above. Do it in the following stages:
Practice until you can do a reasonable free-hand sketch in about 30 seconds.
Physical Properties of Diamond
• has a very high melting point (almost 4000°C). Very strong carbon-carbon covalent bonds have to be broken throughout the structure before melting occurs.
• is very hard. This is again due to the need to break very strong covalent bonds operating in 3-dimensions.
• doesn't conduct electricity. All the electrons are held tightly between the atoms, and aren't free to move.
• is insoluble in water and organic solvents. There are no possible attractions which could occur between solvent molecules and carbon atoms which could outweigh the attractions between the covalently bound carbon atoms.
Graphite
Graphite has a layer structure which is quite difficult to draw convincingly in three dimensions. The diagram below shows the arrangement of the atoms in each layer, and the way the layers are spaced.
Notice that you cannot really draw the side view of the layers to the same scale as the atoms in the layer without one or other part of the diagram being either very spread out or very squashed. In that case, it is important to give some idea of the distances involved. The distance between the layers is about 2.5 times the distance between the atoms within each layer. The layers, of course, extend over huge numbers of atoms - not just the few shown above.
You might argue that carbon has to form 4 bonds because of its 4 unpaired electrons, whereas in this diagram it only seems to be forming 3 bonds to the neighboring carbons. This diagram is something of a simplification, and shows the arrangement of atoms rather than the bonding.
The Bonding in Graphite
Each carbon atom uses three of its electrons to form simple bonds to its three close neighbors. That leaves a fourth electron in the bonding level. These "spare" electrons in each carbon atom become delocalized over the whole of the sheet of atoms in one layer. They are no longer associated directly with any particular atom or pair of atoms, but are free to wander throughout the whole sheet. The important thing is that the delocalized electrons are free to move anywhere within the sheet - each electron is no longer fixed to a particular carbon atom. There is, however, no direct contact between the delocalized electrons in one sheet and those in the neighboring sheets. The atoms within a sheet are held together by strong covalent bonds - stronger, in fact, than in diamond because of the additional bonding caused by the delocalized electrons.
So what holds the sheets together? In graphite you have the ultimate example of van der Waals dispersion forces. As the delocalized electrons move around in the sheet, very large temporary dipoles can be set up which will induce opposite dipoles in the sheets above and below - and so on throughout the whole graphite crystal.
Graphite has a high melting point, similar to that of diamond. In order to melt graphite, it isn't enough to loosen one sheet from another. You have to break the covalent bonding throughout the whole structure. It has a soft, slippery feel, and is used in pencils and as a dry lubricant for things like locks. You can think of graphite rather like a pack of cards - each card is strong, but the cards will slide over each other, or even fall off the pack altogether. When you use a pencil, sheets are rubbed off and stick to the paper. Graphite has a lower density than diamond. This is because of the relatively large amount of space that is "wasted" between the sheets.
Graphite is insoluble in water and organic solvents - for the same reason that diamond is insoluble. Attractions between solvent molecules and carbon atoms will never be strong enough to overcome the strong covalent bonds in graphite. conducts electricity. The delocalized electrons are free to move throughout the sheets. If a piece of graphite is connected into a circuit, electrons can fall off one end of the sheet and be replaced with new ones at the other end.
Silicon dioxide: SiO2
Silicon dioxide is also known as silica or silicon(IV) oxide has three different crystal forms. The easiest one to remember and draw is based on the diamond structure. Crystalline silicon has the same structure as diamond. To turn it into silicon dioxide, all you need to do is to modify the silicon structure by including some oxygen atoms.
Notice that each silicon atom is bridged to its neighbors by an oxygen atom. Don't forget that this is just a tiny part of a giant structure extending on all 3 dimensions.
Silicon Dioxide has a high melting point - varying depending on what the particular structure is (remember that the structure given is only one of three possible structures), but around 1700°C. Very strong silicon-oxygen covalent bonds have to be broken throughout the structure before melting occurs. Morevoer, it hard due to the need to break the very strong covalent bonds.Silicon Dioxide does not conduct electricity since there aren't any delocalized electrons with all the electrons are held tightly between the atoms, and are not free to move.Silicon Dioxide is insoluble in water and organic solvents. There are no possible attractions which could occur between solvent molecules and the silicon or oxygen atoms which could overcome the covalent bonds in the giant structure. | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Crystal_Lattices/Lattice_Basics/Covalent_Network_Solids.txt |
This page explains how you can decide what sort of structure a substance has by looking at its physical properties.
Melting and boiling points
The best place to start is usually the physical state. Melting point isn't always a good guide to the size of the attractions between particles, because the attractive forces have only been loosened on melting - not broken entirely. Boiling point is a much better guide, because enough heat has now been supplied to break the attractive forces completely. The stronger the attractions, the higher the boiling point.
That being said, melting points are often used to judge the size of attractive forces between particles in solids, but you will find the occasional oddity. Those oddities usually disappear if you consider boiling points instead.
Example 1
You would expect stronger metallic bonding in aluminum than in magnesium, because aluminum has 3 electrons to delocalize into the "sea of electrons" rather than magnesium's 2. The boiling points reflect this: Al 2470 °C vs. Mg 1110 °C. However, aluminum's melting point is only 10 °C higher than magnesium's: Al 660 °C vs. Mg 650 °C.
So, if it is a gas, liquid or low melting point solid, it will consist of covalently bound molecules (except the noble gases which have molecules consisting of single atoms). The size of the melting point or boiling point gives a guide to the strength of the intermolecular forces. That is then the end of the problem. If it is a gas, liquid or low melting point solid then you are talking about a simple molecular substance. Full stop!
If it is a high melting point solid, it will be a giant structure - either ionic, metallic or giant covalent. You now have to sort out which of these it is.
Effect of water
Solubility of a solid in water (without reaction) suggests it is ionic. There are exceptions to this, of course. Sugar (sucrose) is soluble in water despite being a covalent molecule. It is capable of extensive hydrogen bonding with water molecules. And there are a lot of ionic compounds which are insoluble in water, of course. Solubility of a low melting point solid or a liquid in water (without it reacting) suggests a small molecule capable of hydrogen bonding - or, at least, a small very polar molecule.
Conduction of electricity
Conduction of electricity in the solid state suggests delocalized electrons, and therefore either a metal or graphite. The clue as to which you had would usually come from other data - appearance, malleability, etc (see below).
Note
Semiconductors like silicon - a giant covalent structure with the same arrangement of atoms as diamond - also conduct electricity.
If a substance doesn't conduct electricity as a solid, but undergoes electrolysis when it is molten, that would confirm that it was ionic. Electrolysis is the splitting up of a compound using electricity. For example, molten sodium chloride conducts electricity and is split into sodium and chlorine in the process
Appearance
Don't forget about obvious things like the shiny appearance of most metals, and their ease of working. Metals are malleable (can be bent or beaten into shape easily) and ductile (can be pulled out into wires). By contrast, giant ionic or giant covalent structures tend to be brittle - shattering rather than bending. | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Crystal_Lattices/Lattice_Basics/How_to_Decide_What_Type_of_Structure.txt |
This page explains the relationship between the arrangement of the ions in a typical ionic solid like sodium chloride and its physical properties - melting point, boiling point, brittleness, solubility and electrical behavior. It also explains why cesium chloride has a different structure from sodium chloride even though sodium and cesium are both in Group 1 of the Periodic Table.
The structure of a typical ionic solid - sodium chloride
Sodium chloride is taken as a typical ionic compound. Compounds like this consist of a giant (endlessly repeating) lattice of ions. So sodium chloride (and any other ionic compound) is described as having a giant ionic structure.
You should be clear that giant in this context does not just mean very large. It means that you can't state exactly how many ions there are. There could be billions of sodium ions and chloride ions packed together, or trillions, or whatever - it simply depends how big the crystal is. That is different from, say, a water molecule which always contains exactly 2 hydrogen atoms and one oxygen atom - never more and never less. A small representative bit of a sodium chloride lattice looks like this:
If you look at the diagram carefully, you will see that the sodium ions and chloride ions alternate with each other in each of the three dimensions. This diagram is easy enough to draw with a computer, but extremely difficult to draw convincingly by hand. We normally draw an "exploded" version which looks like this:
Only those ions joined by lines are actually touching each other. The sodium ion in the center is being touched by 6 chloride ions. By chance we might just as well have centered the diagram around a chloride ion - that, of course, would be touched by 6 sodium ions. Sodium chloride is described as being 6:6-coordinated. This diagram represents only a tiny part of the whole sodium chloride crystal; the pattern repeats in this way over countless ions.
How to draw this structure
Draw a perfect square:
Now draw an identical square behind this one and offset a bit. You might have to practice a bit to get the placement of the two squares right. If you get it wrong, the ions get all tangled up with each other in your final diagram.
Turn this into a perfect cube by joining the squares together:
Now the tricky bit! Subdivide this big cube into 8 small cubes by joining the mid point of each edge to the mid point of the edge opposite it. To complete the process you will also have to join the mid point of each face (easily found once you've joined the edges) to the mid point of the opposite face.
Now all you have to do is put the ions in. Use different colors or different sizes for the two different ions, and don't forget a key. It does not matter whether you end up with a sodium ion or a chloride ion in the center of the cube - all that matters is that they alternate in all three dimensions.
You should be able to draw a perfectly adequate free-hand sketch of this in under two minutes - less than one minute if you're not too fussy!
Why is sodium chloride 6:6-coordinated?
The more attraction there is between the positive and negative ions, the more energy is released. The more energy that is released, the more energetically stable the structure becomes. That means that to gain maximum stability, you need the maximum number of attractions. So why does each ion surround itself with 6 ions of the opposite charge? That represents the maximum number of chloride ions that you can fit around a central sodium ion before the chloride ions start touching each other. If they start touching, you introduce repulsions into the crystal which makes it less stable.
The different structure of cesium chloride
We'll look first at the arrangement of the ions and then talk about why the structures of sodium chloride and cesium chloride are different afterwards. Imagine a layer of chloride ions as shown below. The individual chloride ions aren't touching each other. That's really important - if they were touching, there would be repulsion.
Now let's place a similarly arranged layer of cesium ions on top of these.
Notice that the cesium ions aren't touching each other either, but that each cesium ion is resting on four chloride ions from the layer below.
Now let's put another layer of chloride ions on, exactly the same as the first layer. Again, the chloride ions in this layer are NOT touching those in the bottom layer - otherwise you are introducing repulsion. Since we are looking directly down on the structure, you can't see the bottom layer of chloride ions any more, of course.
If you now think about a cesium ion sandwiched between the two layers of chloride ions, it is touching four chloride ions in the bottom layer, and another four in the top one. Each cesium ion is touched by eight chloride ions. We say that it is 8-coordinated. If we added another layer of cesium ions, you could similarly work out that each chloride ion was touching eight cesium ions. The chloride ions are also 8-coordinated. Overall, then, cesium chloride is 8:8-coordinated.
The final diagram in this sequence takes a slightly tilted view of the structure so that you can see how the layers build up. These diagrams are quite difficult to draw without it looking as if ions of the same charge are touching each other. They aren't!
Diagrams of ionic crystals are usually simplified to show the most basic unit of the repeating pattern. For cesium chloride, you could, for example, draw a simple diagram showing the arrangement of the chloride ions around each cesium ion:
By reversing the colors (green chloride ion in the center, and orange cesium ions surrounding it), you would have an exactly equivalent diagram for the arrangement of cesium ions around each chloride ion.
Why are the cesium chloride and sodium chloride structures different?
When attractions are set up between two ions of opposite charges, energy is released. The more energy that can be released, the more stable the system becomes. That means that the more contact there is between negative and positive ions, the more stable the crystal should become.
If you can surround a positive ion like cesium with eight chloride ions rather than just six (and vice versa for the chloride ions), then you should have a more stable crystal. So why does not sodium chloride do the same thing? Look again at the last diagram:
Now imagine what would happen if you replaced the cesium ion with the smaller sodium ion. Sodium ions are, of course, smaller than cesium ions because they have fewer layers of electrons around them.
You still have to keep the chloride ions in contact with the sodium. The effect of this would be that the whole arrangement would shrink, bringing the chloride ions into contact with each other - and that introduces repulsion.
Any gain in attractions because you have eight chlorides around the sodium rather than six is more than countered by the new repulsions between the chloride ions themselves. When sodium chloride is 6:6-coordinated, there are no such repulsions - and so that is the best way for it to organize itself.
Which structure a simple 1:1 compound like NaCl or CsCl crystallizes in depends on the radius ratio of the positive and the negative ions. If the radius of the positive ion is bigger than 73% of that of the negative ion, then 8:8-coordination is possible. Less than that (down to 41%) then you get 6:6-coordination.
In CsCl, the cesium ion is about 93% of the size of the chloride ion - so is easily within the range where 8:8-coordination is possible. But with NaCl, the sodium ion is only about 52% of the size of the chloride ion. That puts it in the range where you get 6:6-coordination.
The physical properties of sodium chloride
Sodium chloride is taken as typical of ionic compounds, and is chosen rather than, say, cesium chloride, because it is found on every syllabus at this level.
Sodium chloride has a high melting and boiling point
There are strong electrostatic attractions between the positive and negative ions, and it takes a lot of heat energy to overcome them. Ionic substances all have high melting and boiling points. Differences between ionic substances will depend on things like:
• The number of charges on the ions: Magnesium oxide has exactly the same structure as sodium chloride, but a much higher melting and boiling point. The 2+ and 2- ions attract each other more strongly than 1+ attracts 1-.
• The sizes of the ions: If the ions are smaller they get closer together and so the electrostatic attractions are greater. Rubidium iodide, for example, melts and boils at slightly lower temperatures than sodium chloride, because both rubidium and iodide ions are bigger than sodium and chloride ions. The attractions are less between the bigger ions and so less heat energy is needed to separate them.
Sodium chloride crystals are brittle
Brittleness is again typical of ionic substances. Imagine what happens to the crystal if a stress is applied which shifts the ion layers slightly.
Ions of the same charge are brought side-by-side and so the crystal repels itself to pieces!
Sodium chloride is soluble in water
Many ionic solids are soluble in water - although not all. It depends on whether there are big enough attractions between the water molecules and the ions to overcome the attractions between the ions themselves. Positive ions are attracted to the lone pairs on water molecules and co-ordinate (dative covalent) bonds may form. Water molecules form hydrogen bonds with negative ions.
Sodium chloride is insoluble in organic solvents
This is also typical of ionic solids. The attractions between the solvent molecules and the ions are not big enough to overcome the attractions holding the crystal together.
The electrical behavior of sodium chloride
Solid sodium chloride does not conduct electricity, because there are no electrons which are free to move. When it melts, sodium chloride undergoes electrolysis, which involves conduction of electricity because of the movement and discharge of the ions. In the process, sodium and chlorine are produced. This is a chemical change rather than a physical process.
The positive sodium ions move towards the negatively charged electrode (the cathode). When they get there, each sodium ion picks up an electron from the electrode to form a sodium atom. These float to the top of the melt as molten sodium metal. (And assuming you are doing this open to the air, this immediately catches fire and burns with an orange flame.)
$Na^+ + e^- \rightarrow Na \nonumber$
The movement of electrons from the cathode onto the sodium ions leaves spaces on the cathode. The power source (the battery or whatever) moves electrons along the wire in the external circuit to fill those spaces. That flow of electrons would be seen as an electric current (the external circuit is all the rest of the circuit apart from the molten sodium chloride.)
Meanwhile, chloride ions are attracted to the positive electrode (the anode). When they get there, each chloride ion loses an electron to the anode to form an atom. These then pair up to make chlorine molecules. Chlorine gas is produced. Overall, the change is . . .
$2Cl^- \rightarrow Cl_2 + 2e^- \nonumber$
The new electrons deposited on the anode are pumped off around the external circuit by the power source, eventually ending up on the cathode where they will be transferred to sodium ions. Molten sodium chloride conducts electricity because of the movement of the ions in the melt, and the discharge of the ions at the electrodes. Both of these have to happen if you are to get electrons flowing in the external circuit. In solid sodium chloride, of course, that ion movement ca not happen and that stops any possibility of any current flow in the circuit. | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Crystal_Lattices/Lattice_Basics/Ionic_Structures.txt |
This page describes the structure of metals, and relates that structure to the physical properties of the metal.
The structure of metals
Metals are giant structures of atoms held together by metallic bonds. "Giant" implies that large but variable numbers of atoms are involved - depending on the size of the bit of metal.
12-coordination
Most metals are close packed - that is, they fit as many atoms as possible into the available volume. Each atom in the structure has 12 touching neighbors. Such a metal is described as 12-coordinated.
Each atom has 6 other atoms touching it in each layer.
There are also 3 atoms touching any particular atom in the layer above and another 3 in the layer underneath.
This second diagram shows the layer immediately above the first layer. There will be a corresponding layer underneath (there are actually two different ways of placing the third layer in a close packed structure, but that goes beyond the topic here).
8-coordination
Some metals (notably those in Group 1 of the Periodic Table) are packed less efficiently, having only 8 touching neighbors. These are 8-coordinated.
The left hand diagram shows that no atoms are touching each other within a particular layer . They are only touched by the atoms in the layers above and below. The right hand diagram shows the 8 atoms (4 above and 4 below) touching the darker colored one.
Crystal grains
It would be misleading to suppose that all the atoms in a piece of metal are arranged in a regular way. Any piece of metal is made up of a large number of "crystal grains", which are regions of regularity. At the grain boundaries atoms have become misaligned.
The physical properties of metals
Melting points and boiling points
Metals tend to have high melting and boiling points because of the strength of the metallic bond. The strength of the bond varies from metal to metal and depends on the number of electrons which each atom delocalizes into the sea of electrons, and on the packing. Group 1 metals like sodium and potassium have relatively low melting and boiling points mainly because each atom only has one electron to contribute to the bond - but there are other problems as well:
• Group 1 elements are also inefficiently packed (8-coordinated), so that they aren't forming as many bonds as most metals.
• They have relatively large atoms (meaning that the nuclei are some distance from the delocalized electrons) which also weakens the bond.
Electrical conductivity
Metals conduct electricity. The delocalized electrons are free to move throughout the structure in 3-dimensions. They can cross grain boundaries. Even though the pattern may be disrupted at the boundary, as long as atoms are touching each other, the metallic bond is still present. Liquid metals also conduct electricity, showing that although the metal atoms may be free to move, the delocalization remains in force until the metal boils.
Thermal conductivity
Metals are good conductors of heat. Heat energy is picked up by the electrons as additional kinetic energy (it makes them move faster). The energy is transferred throughout the rest of the metal by the moving electrons.
Malleability and ductility
Metals are described as malleable (can be beaten into sheets) and ductile (can be pulled out into wires). This is because of the ability of the atoms to roll over each other into new positions without breaking the metallic bond. If a small stress is put onto the metal, the layers of atoms will start to roll over each other. If the stress is released again, they will fall back to their original positions. Under these circumstances, the metal is said to be elastic.
If a larger stress is put on, the atoms roll over each other into a new position, and the metal is permanently changed.
The hardness of metals
This rolling of layers of atoms over each other is hindered by grain boundaries because the rows of atoms don't line up properly. It follows that the more grain boundaries there are (the smaller the individual crystal grains), the harder the metal becomes.
Offsetting this, because the grain boundaries are areas where the atoms aren't in such good contact with each other, metals tend to fracture at grain boundaries. Increasing the number of grain boundaries not only makes the metal harder, but also makes it more brittle.
Controlling the size of the crystal grains
If you have a pure piece of metal, you can control the size of the grains by heat treatment or by working the metal. Heating a metal tends to shake the atoms into a more regular arrangement - decreasing the number of grain boundaries, and so making the metal softer. Banging the metal around when it is cold tends to produce lots of small grains. Cold working therefore makes a metal harder. To restore its workability, you would need to reheat it.
You can also break up the regular arrangement of the atoms by inserting atoms of a slightly different size into the structure. Alloys such as brass (a mixture of copper and zinc) are harder than the original metals because the irregularity in the structure helps to stop rows of atoms from slipping over each other.
Contributor
Jim Clark (Chemguide.co.uk) | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Crystal_Lattices/Lattice_Basics/Metallic_Structures.txt |
The physical properties of substances having molecular structures varies with the type of intermolecular attractions: hydrogen bonding or van der Waals forces.
Molecules are made of fixed numbers of atoms joined together by covalent bonds, and can range from the very small (even down to single atoms, as in the noble gases) to the very large (as in polymers, proteins or even DNA). The covalent bonds holding the molecules together are very strong, but these are largely irrelevant to the physical properties of the substance. Physical properties are governed by the intermolecular forces - forces attracting one molecule to its neighbors - van der Waals attractions or hydrogen bonds.
Melting and boiling points
Molecular substances tend to be gases, liquids or low melting point solids, because the intermolecular forces of attraction are comparatively weak. You don't have to break any covalent bonds in order to melt or boil a molecular substance. The size of the melting or boiling point will depend on the strength of the intermolecular forces. The presence of hydrogen bonding will lift the melting and boiling points. The larger the molecule the more van der Waals attractions are possible - and those will also need more energy to break.
Solubility in water
Most molecular substances are insoluble (or only very sparingly soluble) in water. Those which do dissolve often react with the water, or else are capable of forming hydrogen bonds with the water.
Why doesn't methane dissolve in water?
The methane, CH4, itself is not the problem. Methane is a gas, and so its molecules are already separate - the water doesn't need to pull them apart from one another. The problem is the hydrogen bonds between the water molecules. If methane were to dissolve, it would have to force its way between water molecules and so break hydrogen bonds. That costs a reasonable amount of energy.
The only attractions possible between methane and water molecules are the much weaker van der Waals forces - and not much energy is released when these are set up. It simply isn't energetically profitable for the methane and water to mix.
Why does ammonia dissolve in water?
Ammonia has the ability to form hydrogen bonds. When the hydrogen bonds between water molecules are broken, they can be replaced by equivalent bonds between water and ammonia molecules. Some of the ammonia also reacts with the water to produce ammonium ions and hydroxide ions.
$NH_3 + H_2O \rightleftharpoons NH_4^+ + OH^- \nonumber$
The reversible arrows show that the reaction doesn't go to completion. At any one time only about 1% of the ammonia has actually reacted to form ammonium ions. The solubility of ammonia is mainly due to the hydrogen bonding and not the reaction. Other common substances which are freely soluble in water because they can hydrogen bond with water molecules include ethanol (alcohol) and sucrose (sugar).
Solubility in organic solvents
Molecular substances are often soluble in organic solvents - which are themselves molecular. Both the solute (the substance which is dissolving) and the solvent are likely to have molecules attracted to each other by van der Waals forces. Although these attractions will be disrupted when they mix, they are replaced by similar ones between the two different sorts of molecules.
Electrical conductivity
Molecular substances will not conduct electricity. Even in cases where electrons may be delocalized within a particular molecule, there isn't sufficient contact between the molecules to allow the electrons to move through the whole solid or liquid. | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Crystal_Lattices/Lattice_Basics/Molecular_Structures.txt |
An ideal crystalline solid exhibit a periodic crystal structure with the positions of atoms or molecules occurring on repeating fixed distances, determined by the unit cell parameters. However, the arrangement of atoms or molecules in most crystalline materials is not perfect and the regular patterns are interrupted by crystallographic defects. Lattice structures (or crystals) are more prone to defects when their temperature is greater than 0 K.
Lattice Defects
The Frenkel defect (also known as the Frenkel pair/disorder) is a defect in the lattice crystal where an atom or ion occupies a normally vacant site other than its own. As a result the atom or ion leaves its own lattice site vacant.
The Frenkel Defect in a Molecule
The Frenkel Defect explains a defect in the molecule where an atom or ion (normally the cation) leaves its own lattice site vacant and instead occupies a normally vacant site (Figure $1$). The cation leaves its own lattice site open and places itself between the area of all the other cations and anions. This defect is only possible if the cations are smaller in size when compared to the anions.
The number of Frenkel Defects can be calculated using the equation:
$\sqrt{NN^*}\, e^{\frac{\Delta H}{2RT}} \nonumber$
where $N$ is the number of normally occupied positions, $N^*$ is the number of available positions for the moving ion, the $\Delta H$ of formation is the enthalpy formation of one Frenkel defect, and $R$ is the gas constant. Frenkel defects are intrinsic defects because the existence causes the Gibbs energy of a crystal to decrease, which means it is favorable to occur.
Solids Found with a Frenkel Defect
The crystal lattices are relatively open and the coordination number is low.
Problems
1. What requirements are needed in order for the Frenkel defect to occur in an atom?
2. What are the differences between the Schottky defect and the Frenkel defect?
Answers
1. A low coordination number as well as having the crystal lattices open for the molecule.
2. The Frenkel defect causes an cation to leave its own lattice and go to another, while Sckhotty defect depicts that an equal number of cations and anions must be absent to maintain charge neutraility.
Contributors and Attributions
• Stanley Hsia, UC Davis
Metal Oxides
Discussion Questions
• How does metal oxides protect a metal from further oxidation?
• How does oxidation rate progress on metal surfaces?
• How is iron erroded?
• What are the major iron oxides and their structures?
Defects in Metal Oxides
Metal oxides are very common commodities, widely applied, and have many different varieties. For example, zinc oxide sintered together with other metal oxide additives have been made into nonlinear resistors, which are called Varistors for surge suppressing function. The suppressing function has been applied for switching and for protection of random voltage protections. Iron oxide and other metal oxides are used in thermite reactions, and this has been applied in many ways, including welding in spaceship repairs. Iron oxides are also the raw material for all magnets and magnetic materials used for computer disks and recording tapes.
How does metal oxides protect a metal from further oxidation?
Metals are protected from further oxidation by forming a hard scale of oxides when it is being oxidized. Not all metal oxides form a scale. In general, when the oxide formed is not very dense, it is not under stress, and the oxide layer forms a scale. Usually, a mole of metal oxide should occupy more volume than a mole of the metal itself. If this is true, the oxide is not under stress, and a protective scale is formed. In general, if the volume of the metal oxide per mole of metal is greater than the molar volume of the metal, the oxide will form a protective scale.
On the other hand, if the oxide formed occupies a smaller volume than the volume occupied by the metal itself, the oxide layer will be under tension and at some point it will crack. Thus, the oxide offers no protection for further oxidation. The molar volume is easily calculated by dividing the molar mass by the density:
$\text{molar volume} =\dfrac{\text{Molar mass}}{density} \nonumber$
Example 1
The densities of $Mg$ and $MgO$ are 1.74 and 3.58 g/mL respectively. Calculate their molar volumes.
Solution
The molar volumes are given below:
Molar volume of Mg = 24.31 (g/mol) / 1.74 (g/mL) = 14.0 mL / mol
Molar volume of MgO = (24.31+16.00) (g/mol) / 3.58 (g/mL) = 11.3 mL / mol
DISCUSSION
Since the molar volume of the oxide is less than that of the metal, the oxide does not form a protective scale.
Example 2
The densities of $Al$ and $Al_2O_3$ are 2.702 and 3.965 g/mL respectively. Calculate their volumes per mole of Al.
Solution
The volumes per mole of La are given below:
Molar volume of Al = 26.98 (g/mol) / 2.702 (g/mL) = 9.985 mL / mol
Volume of Al2O3 = (26.98+24.00) (g/mol) / 3.965 (g/mL =12.86 mL / mol
DISCUSSION
It has been a well known fact that aluminum oxide forms a protective scale. These data confirm the fact, and now you have an explanation for corrosion. However, we should realize that sometimes the metal oxide does not form a protective layer even if the oxide is not under tension.
How does oxidation rate progress on metal surfaces?
A major technological concern of metals is the corrosion due to oxidation. The rate of oxidation is usually expressed in terms of depth of the oxidation layer. Several models have been proposed to express the thickness of the oxide layer y as a function of time t. In the following discussion, k is a rate constant.
The linear rule. When the oxide offers absolutely no protection, the progress of oxidation is a linear relationship with time. This has been called the rectilinear rate law by Swaddle.
$dy = k dt \nonumber$
or
$y = k t \nonumber$
When the oxide layer gives some protection, the parabolic law apply. This law is formulated with the consideration,
$y dy = k dt \nonumber$
or
$y_2 = k' t + c \nonumber$
with $k' = 2 k$ and c is another constant
or
$y = k" t 1/2 + c" \nonumber$
with k" and c" other constants
Oxidation of copper has been shown to follow this rule. The thicker the oxide layer, the more protection the oxide offers in this case. When the oxide layer forms a protective layer, but large flakes crack and leas to faster oxidation as a result. Then, the rate is a combination of the linear rule and the parabolic law.
When the oxide forms a good protective layer, the logarithmic rate law and the inverse logarithmic rate law) have been applied. Some suggested formulas are:
$\dfrac{1}{y} = a \ln (k t +1) \nonumber$
or
$y = b ln (k t + 1) \nonumber$
The a, b, and k are just some parameters to be determined by experimental methods. The properties of metals and its oxide play roles in the rate of corrosion. There is no set rate laws, and every case must be studied carefully. So far, we have hinted a few has models for exploration and analysis of corrosion problems.
How is iron corroded?
The problem with iron as well as many other metals is that the oxide formed by oxidation does not firmly adhere to the surface of the metal and flakes off easily causing "pitting". Extensive pitting eventually causes structural weakness and disintegration of the metal. Most people consider the oxidation of iron results in the formation of a film of iron sesquioxide, which is a term given to red iron(III) oxide, Fe2O3, which is also called ferric oxide. In reality, the oxidation and corrosion of iron is a very complicated process.
The corrosion takes place due to oxidation and galvanic actions. The rust formed is generally represented by
$Fe_2O_3 \cdot xH_2O \nonumber$
where x is an unspecific amount of water. Depending on the amount of water, the oxides appear in various colors. The pH of water and electrolytes present in the water affect the rate of iron corrosion, because the presence of electrolytes increases the conductivity of the solution.
What are the major iron oxides and their structures?
As mentioned above, iron can be at oxidation state II or III in the form of Fe2+ or Fe3+ in its oxides. As a result, iron oxides tends to be somewhat non-stoichiometric.
The common iron oxides are:
ferrous oxide FeO
hematite a-Fe2O3
maghemite g-Fe2O3
magnetite Fe3O4
However, when water is involved, several oxyhydroxides are possible:
goethite a-FeOOH
akaganeite b-FeOOH
lepidocrocite g-FeOOH
There is no need to memorize all these minerals, but you get the idea of varieties of iron oxides and oxyhydroxides.
The basic structures of iron oxides and iron oxy hydroxides can be described as a close packing of oxygen (or hydroxide) with iron ions in the octahedral sites. The structures of goetite and hematite may be describe as an approximate hcp close packing of O2- or HO- ions with some of the octahedral sites occupied by iron ions. Thus, these two types of structures are usually designated as a phases.
The structures of lepidocrocite and maghemite may be described as an approximate ccp (fcc) packing of O2- or HO- ions with some of the octahedral sites occupied by iron ions. Thus, these two types of structures are usually designated as g phases. Some of the iron ions can be replaced by other metal ions, forming solid solutions in a process known as isomorphous substitution in terms of structural chemistry.
Questions
1. What is the condition for a metal oxide to form a protective layer to prevent the metal from further oxidation?
2. Which occupies a larger volume, 1 mole of MgO or 1 mole of Mg metal?
3. Which occupies a larger volume, 0.5 mole of Al2O3 or 1 mole of Al metal?
4. If the rate of errosion is expressed by the equation,
y = b ln (k t + 1)
What is the differential rate law?
Solutions
1. Hint: The volume of the metal oxide per mole of metal is greater than the molar volume of the metal.
Skill -
Explain protection of metal from corrosion.
2. Hint: 1 mol of Mg metal.
3. Hint: 0.5 mol of Al2O3
Skill -
Explain protection of metal from corrosion.
4. Hint: Work out dy/dt | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Crystal_Lattices/Lattice_Defects/Frenkel_Defect.txt |
Lattice structures are not perfect; in fact most of the time they experience defects. Lattice structures (or crystals) are prone to defects especially when their temperature is greater than 0 K [1]. One of these defects is known as the Schottky defect, which occurs when oppositely charged ions vacant their sites [1].
Introduction
Like the human body, lattice structures (most commonly known as crystals) are far from perfection. Our body works hard to keep things proportional but occasionally our right foot is bigger than our left; similarly, crystals may try to arrange it's ions under a strict layout, but occasionally an ion slips to another spot or simply goes missing. Realistically speaking, it should be expected that crystals will depart itself from order (not surprising considering defects occurs at temperature greater than 0 K). There are many ways a crystal can depart itself from order (thus experiences defects); these defects can be grouped in different categories such as Point Defects, Line Defects, Planar Defects, or Volume or Bulk Defects [2]. We will focus on Point Defects, specifically the defect that occurs in ionic crystal structures (i.e. NaCl) called the Schottky Defect.
Point Defects
Lattice structures (or crystals) undergoing point defects experience one of two types:
1. atoms or ions leaving their spot (thus creating vacancies).
2. atoms or ions slipping into the little gaps in between other atoms or ions; those little gaps are known as interstitials--since atoms or ions in the crystals are occupying interstitials, they inherently become (create) interstitials.
By the simplest definition, the Schottky defect is defined by type one, while type two defects are known as the Frenkel defect. The Schottky defect is often visually demonstrated using the following layout of anions and cations:
+ - + - + - + - + - +
- + - + - + - + (vacant) + -
+ - + - + - + - + - +
- + - (vacant) - + - + - + -
+ - + - + - + - + - +
- + - + - + - + - + -
Figure $1$: The positive symbols represents cations (i.e. Na+) and the negative symbol represents anions (i.e. Cl-).
In addition, this layout is applicable only for ionic crystal compounds of the formula MX--layout for ionic crystals with formula MX2 and M2X3 will be discussed later--where M is metal and X is nonmetal. Notice the figure has exactly one cation and one anion vacating their sites; that is what defines a (one) Schottky Defect for a crystal of MX formula--for every cation that vacant its site, the same number of anion will follow suit; essentially the vacant sites come in pairs. This also means the crystal will neither be too positive or too negative because the crystal will always be in equilibrium in respect to the number of anions and cations.
It is possible to approximate the number of Schottky defects (ns) in a MX ionic crystal compound by using the equation:
$N= \exp^{-\dfrac{\Delta H}{2RT}} \label{3}$
where
• $\Delta{H}$ is the enthalpy of defect formation,
• $R$ is the gas constant,
• $T$ is the absolute temperature (in K), and
N can be calculated by:
$N = \dfrac{\text{density of the ionic crystal compound} \times N_A}{\text{molar mass of the ionic crystal compound}} \label{4}$
From Equation $\ref{3}$, it is also possible to calculate the fraction of vacant sites by using the equation:
$\dfrac{n_s}{N} = \exp^{-\dfrac{\Delta H}{2RT}} \label{5}$
Schottky defects for $MX_2$ and $M_2X_3$
As mentioned earlier, a Schottky defect will always result a crystal structure in equilibrium--where no crystal is going to be too positive or too negative; thus in the case of:
• MX2: one Schottky defect equals one cation and two anion vacancy.
• M2X3: one Schottky defect equals two cation and three anion vacancy.
Problems
1. How does an ionic crystal structure maintain electrical neutrality despite undergoing a Schottky defect?
2. How is a Schottky defect defined for a compound with a MX formula? MX2? M2X3?
3. Given that the enthalpy of defect formation for LiCl is 3.39 x 10-19 J and the density of LiCl is 2.068 g/cm-3. Calculate the number of Schottky defect at 873 K.
4. Using the number of Schottky defect solved for question 3, determine the fraction of vacant site for LiCl.
5. If a anion and a cation vacant its site and occupies a space between other anions and cations, is it still a Schottky defect?
Answers
1. For a MX compound: one anion and one cation vacant their sites, so the overall charge will remain balanced. This is the same for MX2 and M2X3 because appropriate numbers of anions and cations vacant their site thus leaving the overall charge neutral.
2. MX compound: one Schottky defect is when one anion and one cation leave their sites. MX2 compound: one Schottky defect is when one anion and two cations leave their sites. M2X3 is when two anions and three cations leave their sites. | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Crystal_Lattices/Lattice_Defects/Schottky_Defects.txt |
Discussion Questions
• What are crystal defects and how are they classified?
• How do impurities affect the structure and properties of a solid?
• What are color centers and how do they affect electric conductivity of solids?
Few, if any, crystals are perfect in that all unit cells consist of the ideal arrangement of atoms or molecules and all cells line up in a three dimensional space with no distortion. Some cells may have one or more atoms less whereas others may have one or more atoms than the ideal unit cell. The imperfection of crystals are called crystal defects. Crystal defects are results of thermodynamic equilibrium contributed also by the increase in entropy TS term of the Gibb's free energy:
$\Delta G = \Delta H - T \Delta S \nonumber$
Only at the unattainable absolute zero K will a crystal be perfect, in other words, no crystals are absolutely perfect. However, the degree of imperfection vary from compound to compound. On the other hand, some solid-like structure called flickering clusters also exist in a liquid. For example, the density of water is the highest at 277 K. The flickering clusters increase as temperature drops below 277 K, and the water density decreases as a result. The missing and lacking of atoms or ions in an ideal or imaginary crystal structure or lattice and the misalignment of unit cells in real crystals are called crystal defects or solid defects. Crystal defects occur as points, along lines, or in the form of a surface, and they are called point, line, or plane defects respectively.
Point Defects
Point defects can be divided into Frenkel defects and Schottky defects, and these often occur in ionic crystals. The former are due to misplacement of ions and vacancies. Charges are balanced in the whole crystal despite the presence of interstitial or extra ions and vacancies. On the other hand, when only vacancies of cation and anions are present with no interstitial or misplaced ions, the defects are called Schottky defects.
Point defects are common in crystals with large anions such as AgBr, AgI, RbAgI4. Due to the defects, the ions have some freedom to move about in crystals, making them relatively good conductors. These are called ionic conductors, unlike metals in which electrons are responsible for electric conductivity. Recently, ionic conductors have attracted a lot of attention because the fuel cell and battery technologies require conducting solids to separate the electrodes.
Line Defects
Line defects are mostly due to misalignment of ions or presence of vacancies along a line. When lines of ions are missing in an otherwise perfect array of ions, an edge dislocation appeared. Edge dislocation is responsible for the ductility and malleability. In fact the hammering and stretching of materials often involve the movement of edge dislocation. Movements of dislocations give rise to their plastic behavior. Line dislocations usually do not end inside the crystal, and they either form loops or end at the surface of a single crystal.
A dislocation is characterized by its Burgers vector: If you imagine going around the dislocation line, and exactly going back as many atoms in each direction as you have gone forward, you will not come back to the same atom where you have started. The Burgers vector points from start atom to the end atom of your journey (This "journey" is called Burgers circuit in dislocation theory).
In this electron microscope image of the surface of a crystal, you see point defects and a Burger journey around an edge dislocation. The dislocation line is in the crystal, and the image shows its ending at the surface. A Burger vector is approximately perpendicular to the dislocation line, and the missing line of atoms is somewhere within the block of the Buerger journey.
If the misalignment shifts a block of ions gradually downwards or upwards causing the formation of a screw like deformation, a screw dislocation is formed. The diagram here shows the idealized screw dislocation.
Line defects weakens the structure along a one-dimensional space, and the defects type and density affects the mechanical properties of the solids. Thus, formation and study of dislocations are particularly important for structural materials such as metals. This link gives some impressive images of dislocations. Chemical etching often reveal pits which are visible under small magnifications.
Example $1$
The Table of X-ray Crystallographic Data of Minerals (The CRC Handbook of Chemistry and Physics) list the following for bunsenite (NiO): Crystal system: cubic, structure type: rock salt, a = 4.177*10-8 cm. In the table of Physical Constant of Inorganic Compounds, the density of bunsenite (NiO) is 6.67g / cm3. From these values, evaluate the cell volume (volume of the unit cell), sum of Ni and O radii (rNi + rO), 2 Molar volumes, (X-ray) density, and the Schottky defect vacancy rate.
Solution
Actually, most of the required values have been listed in the table, but their evaluations illustrate the methods. These values are evaluated below:
Cell volume = a3
= (4.177*10-8)3
= 72.88*10-24 cm3
$r_{Ni}+r_O = \dfrac{a}{ 2} \nonumber$
$= 2.088 \times 10^{-8} \nonumber$
X-ray density = 4*(58.69+16.00) / (6.023*1023*72.88*10-24)
= 6.806 g / cm3,
Compared to the observed density = 6.67 g / cm3. The molar volumes (58.69+16.00) / density are thus
74.69 / 6.806 = 10.97 cm3; and 74.69 / 6.67 = 11.20 cm3
The vacancy rate = 6.806 - 6.67 / 6.806
= 0.02 (or 2%)
DISCUSSION
These methods are hints to assignments.
From the given conditions, we cannot calculate individual radii of Ni and O, but their sum is calculable.
Plane Defects
Plane defects occur along a 2-dimensional surface. The surface of a crystal is an obvious imperfection, because these surface atoms are different from those deep in the crystals. When a solid is used as a catalyst, the catalytic activity depends very much on the surface area per unit mass of the sample. For these powdery material, methods have been developed for the determination of unit areas per unit mass.
Another surface defects are along the grain boundaries. A grain is a single crystal. If many seeds are formed when a sample starts to crystallize, each seed grow until they meet at the boundaries. Properties along these boundaries are different from the grains. A third plane defects are the stacking faults. For example, in the close packing arrangement, the adjancent layers always have the AB relationship. In a ccp (fcc) close packing sequence, ...ABCABC..., one of the layer may suddenly be out of sequence, and become ..ABABCABC.... Similarly, in the hcp sequence, there is a possibility that one of the layer accidentally startes in the C location and resulting in the formation of a grain boundary.
How do impurities affect the structure and properties of a solid?
You already know that to obtain a perfectly pure substance is almost impossible. Purification is a costly process. In general, analytical reagent-grade chemicals are of high purity, and yet few of them are better than 99.9% pure. This means that a foreign atom or molecule is present for every 1000 host atoms or molecules in the crystal. Perhaps the most demanding of purity is in the electronic industry. Silicon crystals of 99.999 (called 5 nines) or better are required for IC chips productions. These crystal are doped with nitrogen group elements of P and As or boron group elements B, Al etc to form n- aand p-type semiconductors. In these crystals, the impurity atom substitute atoms of the host crystals.
Presence minute foreign atoms with one electron more or less than the valence four silicon and germanium host atoms is the key of making n- and p-type semiconductors. Having many semiconductors connected in a single chip makes the integrated circuit a very efficient information processor. The electronic properties change dramatically due to these impurities. This is further described in Inorganic Chemistry by Swaddle.
In other bulk materials, the presence of impurity usually leads to a lowering of melting point. For example, Hall and Heroult tried to electrolyze natural aluminum compounds. They discovered that using a 5% mixture of Al2O3 (melting point 273 K) in cryolite Na3AlF6 (melting point 1273 K) reduced the melting point to 1223 K, and that enabled the production of aluminum in bulk. Recent modifications lowered melting temperatures below 933 K. Some types of glass are made by mixing silica (SiO2), alumina (Al2O3), calcium oxide (CaO), and sodium oxide (Na2O). They are softer, but due to lower melting points, they are cheaper to produce.
Color centers and how do they affect electric conductivity of solids?
Color centers are imperfections in crystals that cause color (defects that cause color by absorption of light). Due to defects, metal oxides may also act as semiconductors, because there are many different types of electron traps. Electrons in defect region only absorb light at certain range of wavelength. The color seen are due to lights not absorbed. For example, a diamond with C vacancies (missing carbon atoms) absorbs light, and these centers give green color as shown here. Replacement of Al3+ for Si4+ in quartz give rise to the color of smoky quartz.
A high temperature phase of ZnOx, (x < 1), has electrons in place of the O2- vacancies. These electrons are color centers, often referred to as F-centers (from the German word farben meaning color). Similarly, heating of ZnS to 773 K causes a loss of sulfur, and these material fluoresces strongly in ultraviolet light. Some non-stoichiometric solids are engineered to be n-type or p-type semiconductors. Nickel oxide NiO gain oxygen on heating in air, resulting in having Ni3+ sites acting as electron trap, a p-type semiconductor. On the other hand, ZnO lose oxygen on heating, and the excess Zn metal atoms in the sample are ready to give electrons. The solid is an n-type semiconductor.
Questions
1. Why does the density of water decreases when the temperature decreases from 277 to 273 K?
2. What type of defects is due to misplaced atoms or ions and vacancies of the same in a crystalline material.
3. What defects reduce the density of a solid?
4. What type of material do the fuel cell and battery technologies need to separate the electrodes?
Solutions
1. Hint: Due to the formation of flickering clusters.
Skill -
Correlate properties of a material to its structure.
2. Hint: Frenkel defects.
Skill -
Explain Frendel and Schottky defects.
3. Hint: Vacancies of Schottky defects reduce the density.
Skill -
Correlate properties of a material to its structure.
4. Hint: Ionic conductors.
Skill -
Specify the desirable properties of a material. | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Crystal_Lattices/Lattice_Defects/Solid_Defects.txt |
Metals are composed of atoms in ordered layers that form a three-dimensional, crystalline structure. That means that the individual units within the solid -- the atoms, in this case -- are organized in a regularly repeating pattern.
Metal Lattices
For comparison, maybe there is another set of atoms, also in a simple square layer. Suppose they are well-separated from each other; maybe they are far enough apart that you could fit an extra atom between each pair if you wanted to. If the free electron is in the same place -- the middle of the nearest hole -- you can see that it is much farther from the nucleus in this case. The force of attraction is much lower in this case, and the overall energy is not as low.
• Most metals pack very efficiently together to form a solid.
• Efficient packing leads to stronger bonding interactions.
That first case, with atoms packed more tightly together, may be preferable, because of the stronger interaction between the metal nucleus and the free electron. For reasons like this, understanding the packing efficiency in a crystal can be very important.
3. Metal Packing: Three Dimensions
Note that this cube has an atom in the center of each face; hence the name, face-centered cubic.
• A cubic close-packed takes the name "cubic" from the presence of a cubic unit cell.
• This cubic unit cell is oriented diagonally to the close-packed layers.
• This cubic unit cell has an atom in the center of each face; it is also called "face-centered cubic".
Body Centered Packing
The Model for Polonium Crystal Structure
Discussion -
O O | If we cut the cube along a face diagonal, the arrangement of
O | atoms look like the diagram on the left.
O O | Thus, bd = 4 R
• If the edge of the cube has a length represented by a, and if the radius of the atoms is R, express a as a function of R.
Hint: $a = \dfrac{4 R}{\sqrt3}$
Discussion -
a = 4 R / √3
Metal Structures
Copper is the most common mentioned metal that has the fcc structure. This element is one of the noble metals. It has been widely used for door knobs and other tools, and has been widely recognized. Yet, most noble metals (\(\ce{Cu}\), \(\ce{Ag}\), \(\ce{Au}\), \(\ce{Ni}\), \(\ce{Pd}\), \(\ce{Pt}\), \(\ce{Rh}\), and \(\ce{Ir}\)) have fcc type structures. Among the group 2 elements, only \(\ce{Ca}\) and \(\ce{Sr}\) have the fcc structure, whereas \(\ce{Be}\) and \(\ce{Mg}\) have hcp structures. So do \(\ce{Zn}\), \(\ce{Cd}\), \(\ce{Sc}\), \(\ce{Y}\), \(\ce{Lu}\), \(\ce{Ti}\), \(\ce{Zr}\), \(\ce{Hf}\), \(\ce{Tc}\), \(\ce{Re}\), \(\ce{Ru}\), \(\ce{Os}\) and most rare earth elements. These are mentioned to bring your attention to these two common types of structures, and you are encouraged to at least be able to give a few examples for each type.
Amazingly, the hcp and fcc structure are very similar in many aspects, but nature knows best. The structures adopted by various metals occur by their design. When crystallization takes place, the atoms arrange themselves according to their structure types.
Exercises 1
1. Copper has a specific gravity of 8.92; evaluate its atomic radius.
2. The atomic radius of silver \(\ce{Ag}\) is listed as 145 pm. Evaluate its density.
Simple Cubic
Learning Objectives
• Recognize a unit cell.
• Use geometry to correlate atomic radius with unit cell edge length.
• Calculate parameters relevant to crystal chemistry.
Solids
Representation of 3-dimensional structures should usually be made using models, and their representation on a flat surface is difficult. For simplicity, we use a 2-dimensional pattern (plane) to illustrate a 2-dimensional (planar) crystal structure.
``` # @ # @ # @ # @ # @ # @ # @ # @
@ # @ # @ # @ # @ # @ # @ # @ #
# @ # @ # @ # @ # @ # @ # @ # @
@ # @ # @ # @ # @ # @ # @ # @ #
# @ # @ # @ # @ # @ # @ # @ # @
@ # @ # @ # @ # @ # @ # @ # @ #
# @ # @ # @ # @ # @ # @ # @ # @
@ # @ # @ # @ # @ # @ # @ # @ #
```
This pattern or crystal structure is generated by using a unit marked by the centers of any four @ or # signs. The choice is up to you in this case. Such a unit is called a primitive unit. The pattern has a square (or rectangular on some screens) appearance, and to preserve the square, we may use a square unit of
``` # @ # @ # @
@ # @ or # @ #
# @ # @ # @
```
as our unit cell. Such choices result in having two @ and # per unit cell, and these are called centered cells.
Thus, if we know the arrangement of a unit cell, we can use our imagination to build a crystal structure, or use symbols or models to represent a crystal structure. Since each pattern has features shared by many structures, often such a pattern is called a lattice. For example, the diamond, zinc blende, wurtzite, and \(\ce{NaCl}\) structures have been called lattices; however, the word lattice has a more formal definition by crystal physics and chemists. The above site gives a gallery of lattices.
Definition: Crystals
Inorganic Chemistry by Swaddle defined crystals as packed regular arrays of atoms, ions, or molecules in a pattern repeated periodically ad infinitum.
Sphere Packing
Learning Objectives
• Use the packing of spheres to model the fcc crystal structure.
• Practice solving problems related to crystal structures.
Tetrahedral and Octahedral Sites
Learning Objectives
• To construct a model for the tetrahedral and octahedral holes of closest packing.
• To calculate geometric properties of the model.
• Apply these properties to interpret crystal chemistry. | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Crystal_Lattices/Metal_Lattices/2._Metal_Packing%3A_Layers.txt |
Lattice enthalpy is a measure of the strength of the forces between the ions in an ionic solid. The greater the lattice enthalpy, the stronger the forces. Those forces are only completely broken when the ions are present as gaseous ions, scattered so far apart that there is negligible attraction between them.
• Lattice Energy
The Lattice energy, U, is the amount of energy requried to separate a mole of the solid (s) into a gas (g) of its ions.
• Lattice Energy: The Born-Haber cycle
Ionic solids tend to be very stable compounds. The enthalpies of formation of the ionic molecules cannot alone account for this stability. These compounds have an additional stability due to the lattice energy of the solid structure. However, lattice energy cannot be directly measured. The Born-Haber cycle allows us to understand and determine the lattice energies of ionic solids.
• Lattice Enthalpies and Born Haber Cycles
Lattice enthalpy is a measure of the strength of the forces between the ions in an ionic solid. The greater the lattice enthalpy, the stronger the forces. Those forces are only completely broken when the ions are present as gaseous ions, scattered so far apart that there is negligible attraction between them. You can show this on a simple enthalpy diagram.
• The Born-Lande' equation
The Born-Landé equation is a concept originally formulated in 1918 by the scientists Born and Lande and is used to calculate the lattice energy (measure of the strength of bonds) of a compound. This expression takes into account both the Born interactions as well as the Coulomb attractions.
Thermodynamics of Lattices
Ionic solids tend to be very stable compounds. The enthalpies of formation of the ionic molecules cannot alone account for this stability. These compounds have an additional stability due to the lattice energy of the solid structure. However, lattice energy cannot be directly measured. The Born-Haber cycle allows us to understand and determine the lattice energies of ionic solids.
Introduction
This module will introduce the idea of lattice energy, as well as one process that allows us to calculate it: the Born-Haber Cycle. In order to use the Born-Haber Cycle, there are several concepts that we must understand first.
Lattice Energy
Lattice Energy is a type of potential energy that may be defined in two ways. In one definition, the lattice energy is the energy required to break apart an ionic solid and convert its component atoms into gaseous ions. This definition causes the value for the lattice energy to always be positive, since this will always be an endothermic reaction. The other definition says that lattice energy is the reverse process, meaning it is the energy released when gaseous ions bind to form an ionic solid. As implied in the definition, this process will always be exothermic, and thus the value for lattice energy will be negative. Its values are usually expressed with the units kJ/mol.
Lattice Energy is used to explain the stability of ionic solids. Some might expect such an ordered structure to be less stable because the entropy of the system would be low. However, the crystalline structure allows each ion to interact with multiple oppositely charge ions, which causes a highly favorable change in the enthalpy of the system. A lot of energy is released as the oppositely charged ions interact. It is this that causes ionic solids to have such high melting and boiling points. Some require such high temperatures that they decompose before they can reach a melting and/or boiling point.
Born-Haber Cycle
There are several important concept to understand before the Born-Haber Cycle can be applied to determine the lattice energy of an ionic solid; ionization energy, electron affinity, dissociation energy, sublimation energy, heat of formation, and Hess's Law.
• Ionization Energy is the energy required to remove an electron from a neutral atom or an ion. This process always requires an input of energy, and thus will always have a positive value. In general, ionization energy increases across the periodic table from left to right, and decreases from top to bottom. There are some excepts, usually due to the stability of half-filled and completely filled orbitals.
• Electron Affinity is the energy released when an electron is added to a neutral atom or an ion. Usually, energy released would have a negative value, but due to the definition of electron affinity, it is written as a positive value in most tables. Therefore, when used in calculating the lattice energy, we must remember to subtract the electron affinity, not add it. In general, electron affinity increases from left to right across the periodic table and decreases from top to bottom.
• Dissociation energy is the energy required to break apart a compound. The dissociation of a compound is always an endothermic process, meaning it will always require an input of energy. Therefore, the change in energy is always positive. The magnitude of the dissociation energy depends on the electronegativity of the atoms involved.
• Sublimation energy is the energy required to cause a change of phase from solid to gas, bypassing the liquid phase. This is an input of energy, and thus has a positive value. It may also be referred to as the energy of atomization.
• The heat of formation is the change in energy when forming a compound from its elements. This may be positive or negative, depending on the atoms involved and how they interact.
• Hess's Law states that the overall change in energy of a process can be determined by breaking the process down into steps, then adding the changes in energy of each step. The Born-Haber Cycle is essentially Hess's Law applied to an ionic solid.
Using the Born-Haber Cycle
The values used in the Born-Haber Cycle are all predetermined changes in enthalpy for the processes described in the section above. Hess' Law allows us to add or subtract these values, which allows us to determine the lattice energy.
Step 1
Determine the energy of the metal and nonmetal in their elemental forms. (Elements in their natural state have an energy level of zero.) Subtract from this the heat of formation of the ionic solid that would be formed from combining these elements in the appropriate ration. This is the energy of the ionic solid, and will be used at the end of the process to determine the lattice energy.
Step 2
The Born-Haber Cycle requires that the elements involved in the reaction are in their gaseous forms. Add the changes in enthalpy to turn one of the elements into its gaseous state, and then do the same for the other element.
Step 3
Metals exist in nature as single atoms and thus no dissociation energy needs to be added for this element. However, many nonmetals will exist as polyatomic species. For example, Cl exists as Cl2 in its elemental state. The energy required to change Cl2 into 2Cl atoms must be added to the value obtained in Step 2.
Step 4
Both the metal and nonmetal now need to be changed into their ionic forms, as they would exist in the ionic solid. To do this, the ionization energy of the metal will be added to the value from Step 3. Next, the electron affinity of the nonmetal will be subtracted from the previous value. It is subtracted because it is a release of energy associated with the addition of an electron.
*This is a common error due to confusion caused by the definition of electron affinity, so be careful when doing this calculation.
Step 5
Now the metal and nonmetal will be combined to form the ionic solid. This will cause a release of energy, which is called the lattice energy. The value for the lattice energy is the difference between the value from Step 1 and the value from Step 4.
--------------------------------------------------------------------------------------------------------------------------------------------
The diagram below is another representation of the Born-Haber Cycle.
Equation
The Born-Haber Cycle can be reduced to a single equation:
Heat of formation= Heat of atomization+ Dissociation energy+ (sum of Ionization energies)+ (sum of Electron affinities)+ Lattice energy
*Note: In this general equation, the electron affinity is added. However, when plugging in a value, determine whether energy is released (exothermic reaction) or absorbed (endothermic reaction) for each electron affinity. If energy is released, put a negative sign in front of the value; if energy is absorbed, the value should be positive.
Rearrangement to solve for lattice energy gives the equation:
Lattice energy= Heat of formation- Heat of atomization- Dissociation energy- (sum of Ionization energies)- (sum of Electron Affinities)
Problems
1. Define lattice energy, ionization energy, and electron affinity.
2. What is Hess' Law?
3. Find the lattice energy of KF(s).
Note: Values can be found in standard tables.
4. Find the lattice energy of MgCl2(s).
5. Which one of the following has the greatest lattice energy?
1. A) MgO
2. B) NaC
3. C) LiCl
4. D) MgCl2
6. Which one of the following has the greatest Lattice Energy?
1. NaCl
2. CaCl2
3. AlCl3
4. KCl
Solutions
1. Lattice energy: The difference in energy between the expected experimental value for the energy of the ionic solid and the actual value observed. More specifically, this is the energy gap between the energy of the separate gaseous ions and the energy of the ionic solid.
Ionization energy: The energy change associated with the removal of an electron from a neutral atom or ion.
Electron affinity: The release of energy associated with the addition of an electron to a neutral atom or ion.
2. Hess' Law states that the overall energy of a reaction may be determined by breaking down the process into several steps, then adding together the changes in energy of each step.
3. Lattice Energy= [-436.68-89-(0.5*158)-418.8-(-328)] kJ/mol= -695.48 kJ/mol
4. Lattice Energy= [-641.8-146-243-(737.7+1450.6)-(2*-349)] kJ/mol= -2521.1 kJ/mol
5. MgO. It has ions with the largest charge.
6. AlCl3. According to the periodic trends, as the radius of the ion increases, lattice energy decreases. | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Crystal_Lattices/Thermodynamics_of_Lattices/Lattice_Energy%3A_The_Born-Haber_cycle.txt |
Discussion Questions
• How is lattice energy estimated using Born-Haber cycle?
• How is lattice energy related to crystal structure?
The Lattice energy, $U$, is the amount of energy required to separate a mole of the solid (s) into a gas (g) of its ions.
$\ce{M_{a} L_{b} (s) \rightarrow a M^{b+} (g) + b X^{a-} (g) } \label{eq1}$
This quantity cannot be experimentally determined directly, but it can be estimated using a Hess Law approach in the form of Born-Haber cycle. It can also be calculated from the electrostatic consideration of its crystal structure. As defined in Equation \ref{eq1}, the lattice energy is positive, because energy is always required to separate the ions. For the reverse process of Equation \ref{eq1}:
$\ce{ a M^{b+} (g) + b X^{a-} (g) \rightarrow M_{a}L_{b}(s) } \nonumber$
the energy released is called energy of crystallization ($E_{cryst}$). Therefore,
$U_{lattice} = - E_{cryst} \nonumber$
Values of lattice energies for various solids have been given in literature, especially for some common solids. Some are given here.
Table $1$: Comparison of Lattice Energies (U in kJ/mol) of Some Salts
Solid U Solid U Solid U Solid U
LiF 1036 LiCl 853 LiBr 807 LiI 757
NaF 923 NaCl 786 NaBr 747 NaI 704
KF 821 KCl 715 KBr 682 KI 649
MgF2 2957 MgCl2 2526 MgBr2 2440 MgI2 2327
The following trends are obvious at a glance of the data in Table $1$:
• As the ionic radii of either the cation or anion increase, the lattice energies decrease.
• The solids consists of divalent ions have much larger lattice energies than solids with monovalent ions.
How is lattice energy estimated using Born-Haber cycle?
Estimating lattice energy using the Born-Haber cycle has been discussed in Ionic Solids. For a quick review, the following is an example that illustrate the estimate of the energy of crystallization of NaCl.
Hsub of Na = 108 kJ/mol (Heat of sublimation)
D of Cl2 = 244 (Bond dissociation energy)
IP of Na(g) = 496 (Ionization potential or energy)
EA of Cl(g) = -349 (Electron affinity of Cl)
Hf of NaCl = -411 (Enthalpy of formation)
The Born-Haber cycle to evaluate Elattice is shown below:
-----------Na+ + Cl(g)--------
|
| |-349
|496+244/2 ¯
| Na+(g) + Cl-(g)
| |
Na(g) + 0.5Cl2(g) |
|
|108 |
| |Ecryst= -788
Na(s) + 0.5Cl2(l) |
| |
|-411 |
¯ ¯
-------------- NaCl(s) --------------
Ecryst = -411-(108+496+244/2)-(-349) kJ/mol
= -788 kJ/mol.
Discussion
The value calculated for U depends on the data used. Data from various sources differ slightly, and so is the result. The lattice energies for NaCl most often quoted in other texts is about 765 kJ/mol.
Compare with the method shown below
Na(s) + 0.5 Cl2(l) ® NaCl(s) - 411 Hf
Na(g) ® Na(s) - 108 -Hsub
Na+(g) + e ® Na(g) - 496 -IP
Cl(g) ® 0.5 Cl2(g) - 0.5 * 244 -0.5*D
Cl-(g) ® Cl(g) + 2 e 349 -EA
Add all the above equations leading to
Na+(g) + Cl-(g) ® NaCl(s) -788 kJ/mol = Ecryst
Lattice Energy is Related to Crystal Structure
There are many other factors to be considered such as covalent character and electron-electron interactions in ionic solids. But for simplicity, let us consider the ionic solids as a collection of positive and negative ions. In this simple view, appropriate number of cations and anions come together to form a solid. The positive ions experience both attraction and repulson from ions of opposite charge and ions of the same charge.
As an example, let us consider the the NaCl crystal. In the following discussion, assume r be the distance between Na+ and Cl- ions. The nearest neighbors of Na+ are 6 Cl- ions at a distance 1r, 12 Na+ ions at a distance 2r, 8 Cl- at 3r, 6 Na+ at 4r, 24 Na+ at 5r, and so on. Thus, the energy due to one ion is
$E = \dfrac{Z^2e^2}{4\pi\epsilon_or} M \label{6.13.1}$
The Madelung constant, $M$, is a poorly converging series of interaction energies:
$M= \dfrac{6}{1} - \dfrac{12}{2} + \dfrac{8}{3} - \dfrac{6}{4} + \dfrac{24}{5} ... \label{6.13.2}$
with
• $Z$ is the number of charges of the ions, (e.g., 1 for NaCl),
• $e$ is the charge of an electron ($1.6022 \times 10^{-19}\; C$),
• $4\pi \epsilon_o$ is 1.11265x10-10 C2/(J m).
The above discussion is valid only for the sodium chloride (also called rock salt) structure type. This is a geometrical factor, depending on the arrangement of ions in the solid. The Madelung constant depends on the structure type, and its values for several structural types are given in Table 6.13.1.
A is the number of anions coordinated to cation and C is the numbers of cations coordinated to anion.
Table $2$: Madelung Constants
Compound
Crystal Lattice
M
A : C Type
NaCl NaCl 1.74756 6 : 6 Rock salt
CsCl CsCl 1.76267 6 : 6 CsCl type
CaF2 Cubic 2.51939 8 : 4 Fluorite
CdCl2 Hexagonal 2.244
MgF2 Tetragonal 2.381
ZnS (wurtzite) Hexagonal 1.64132
TiO2 (rutile) Tetragonal 2.408 6 : 3 Rutile
bSiO2 Hexagonal 2.2197
Al2O3 Rhombohedral 4.1719 6 : 4 Corundum
A is the number of anions coordinated to cation and C is the numbers of cations coordinated to anion.
Madelung constants for a few more types of crystal structures are available from the Handbook Menu. There are other factors to consider for the evaluation of energy of crystallization, and the treatment by M. Born led to the formula for the evaluation of crystallization energy $E_{cryst}$, for a mole of crystalline solid.
$E_{cryst} = \dfrac{N Z^2e^2}{4\pi \epsilon_o r} \left( 1 - \dfrac{1}{n} \right)\label{6.13.3a}$
where N is the Avogadro's number (6.022x10-23), and n is a number related to the electronic configurations of the ions involved. The n values and the electronic configurations (e.c.) of the corresponding inert gases are given below:
n = 5 7 9 10 12
e.c. He Ne Ar Kr Xe
The following values of n have been suggested for some common solids:
n = 5.9 8.0 8.7 9.1 9.5
e.c. LiF LiCl LiBr NaCl NaBr
Example $1$
Estimate the energy of crystallization for $\ce{NaCl}$.
Solution
Using the values giving in the discussion above, the estimation is given by Equation \ref{6.13.3a}:
\begin{align*} E_cryst &= \dfrac{(6.022 \times 10^{23} /mol (1.6022 \times 10 ^{-19})^2 (1.747558)}{ 4\pi \, (8.854 \times 10^{-12} C^2/m ) (282 \times 10^{-12}\; m} \left( 1 - \dfrac{1}{9.1} \right) \[4pt] &= - 766 kJ/mol \end{align*}
Discussion
Much more should be considered in order to evaluate the lattice energy accurately, but the above calculation leads you to a good start. When methods to evaluate the energy of crystallization or lattice energy lead to reliable values, these values can be used in the Born-Haber cycle to evaluate other chemical properties, for example the electron affinity, which is really difficult to determine directly by experiment.
Exercise $1$
Which one of the following has the largest lattice energy? LiF, NaF, CaF2, AlF3
Answer
Skill: Explain the trend of lattice energy.
Exercise $2$
Which one of the following has the largest lattice energy? LiCl, NaCl, CaCl2, Al2O3
Answer
Corrundum Al2O3 has some covalent character in the solid as well as the higher charge of the ions.
Exercise $3$
Lime, CaO, is know to have the same structure as NaCl and the edge length of the unit cell for CaO is 481 pm. Thus, Ca-O distance is 241 pm. Evaluate the energy of crystallization, Ecryst for CaO.
Answer
Energy of crystallization is -3527 kJ/mol
Skill: Evaluate the lattice energy and know what values are needed.
Exercise $4$
Assume the interionic distance for NaCl2 to be the same as those of NaCl (r = 282 pm), and assume the structure to be of the fluorite type (M = 2.512). Evaluate the energy of crystallization, Ecryst .
Answer
-515 kJ/mol
Discussion: This number has not been checked. If you get a different value, please let me know. | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Crystal_Lattices/Thermodynamics_of_Lattices/Lattice_Energy.txt |
Lattice enthalpy is a measure of the strength of the forces between the ions in an ionic solid. The greater the lattice enthalpy, the stronger the forces. This page introduces lattice enthalpies (lattice energies) and Born-Haber cycles.
Defining Lattice Enthalpy
There are two different ways of defining lattice enthalpy which directly contradict each other, and you will find both in common use. In fact, there is a simple way of sorting this out, but many sources do not use it. Lattice enthalpy is a measure of the strength of the forces between the ions in an ionic solid. The greater the lattice enthalpy, the stronger the forces. Those forces are only completely broken when the ions are present as gaseous ions, scattered so far apart that there is negligible attraction between them. You can show this on a simple enthalpy diagram.
For sodium chloride, the solid is more stable than the gaseous ions by 787 kJ mol-1, and that is a measure of the strength of the attractions between the ions in the solid. Remember that energy (in this case heat energy) is released when bonds are made, and is required to break bonds.
So lattice enthalpy could be described in either of two ways.
• It could be described as the enthalpy change when 1 mole of sodium chloride (or whatever) was formed from its scattered gaseous ions. In other words, you are looking at a downward arrow on the diagram.
• Or, it could be described as the enthalpy change when 1 mole of sodium chloride (or whatever) is broken up to form its scattered gaseous ions. In other words, you are looking at an upward arrow on the diagram.
Both refer to the same enthalpy diagram, but one looks at it from the point of view of making the lattice, and the other from the point of view of breaking it up. Unfortunately, both of these are often described as "lattice enthalpy".
Definitions
• The lattice dissociation enthalpy is the enthalpy change needed to convert 1 mole of solid crystal into its scattered gaseous ions. Lattice dissociation enthalpies are always positive.
• The lattice formation enthalpy is the enthalpy change when 1 mole of solid crystal is formed from its separated gaseous ions. Lattice formation enthalpies are always negative.
This is an absurdly confusing situation which is easily resolved by never using the term "lattice enthalpy" without qualifying it.
• You should talk about "lattice dissociation enthalpy" if you want to talk about the amount of energy needed to split up a lattice into its scattered gaseous ions. For NaCl, the lattice dissociation enthalpy is +787 kJ mol-1.
• You should talk about "lattice formation enthalpy" if you want to talk about the amount of energy released when a lattice is formed from its scattered gaseous ions. For NaCl, the lattice formation enthalpy is -787 kJ mol-1.
That immediately removes any possibility of confusion.
Factors affecting Lattice Enthalpy
The two main factors affecting lattice enthalpy are
• The charges on the ions and
• The ionic radii (which affects the distance between the ions).
The charges on the ions
Sodium chloride and magnesium oxide have exactly the same arrangements of ions in the crystal lattice, but the lattice enthalpies are very different.
You can see that the lattice enthalpy of magnesium oxide is much greater than that of sodium chloride. That's because in magnesium oxide, 2+ ions are attracting 2- ions; in sodium chloride, the attraction is only between 1+ and 1- ions.
The Radius of the Ions
The lattice enthalpy of magnesium oxide is also increased relative to sodium chloride because magnesium ions are smaller than sodium ions, and oxide ions are smaller than chloride ions. That means that the ions are closer together in the lattice, and that increases the strength of the attractions.
This effect of ion size on lattice enthalpy is clearly observed as you go down a Group in the Periodic Table. For example, as you go down Group 7 of the Periodic Table from fluorine to iodine, you would expect the lattice enthalpies of their sodium salts to fall as the negative ions get bigger - and that is the case:
Attractions are governed by the distances between the centers of the oppositely charged ions, and that distance is obviously greater as the negative ion gets bigger. And you can see exactly the same effect if as you go down Group 1. The next bar chart shows the lattice enthalpies of the Group 1 chlorides.
Calculating Lattice Enthalpy
It is impossible to measure the enthalpy change starting from a solid crystal and converting it into its scattered gaseous ions. It is even more difficult to imagine how you could do the reverse - start with scattered gaseous ions and measure the enthalpy change when these convert to a solid crystal. Instead, lattice enthalpies always have to be calculated, and there are two entirely different ways in which this can be done.
1. You can can use a Hess's Law cycle (in this case called a Born-Haber cycle) involving enthalpy changes which can be measured. Lattice enthalpies calculated in this way are described as experimental values.
2. Or you can do physics-style calculations working out how much energy would be released, for example, when ions considered as point charges come together to make a lattice. These are described as theoretical values. In fact, in this case, what you are actually calculating are properly described as lattice energies.
Born-Haber Cycles
Standard Atomization Enthalpies
Before we start talking about Born-Haber cycles, we need to define the atomization enthalpy, $\Delta H^o_a$. The standard atomization enthalpy is the enthalpy change when 1 mole of gaseous atoms is formed from the element in its standard state. Enthalpy change of atomization is always positive. You are always going to have to supply energy to break an element into its separate gaseous atoms. All of the following equations represent changes involving atomization enthalpy:
$\dfrac{1}{2} Cl_2 (g) \rightarrow Cl(g) \;\;\;\; \Delta H^o_a=+122\, kJ\,mol^{-1} \nonumber$
$\dfrac{1}{2} Br_2 (l) \rightarrow Br(g) \;\;\;\; \Delta H^o_a=+122\, kJ\,mol^{-1} \nonumber$
$Na (s) \rightarrow Na(g) \;\;\;\; \Delta H^o_a=+107\, kJ\,mol^{-1} \nonumber$
Notice particularly that the "mol-1" is per mole of atoms formed - NOT per mole of element that you start with. You will quite commonly have to write fractions into the left-hand side of the equation. Getting this wrong is a common mistake.
Example $1$: Born-Haber Cycle for $\ce{NaCl}$
Consider a Born-Haber cycle for sodium chloride, and then talk it through carefully afterwards. You will see that I have arbitrarily decided to draw this for lattice formation enthalpy. If you wanted to draw it for lattice dissociation enthalpy, the red arrow would be reversed - pointing upwards.
Focus to start with on the higher of the two thicker horizontal lines. We are starting here with the elements sodium and chlorine in their standard states. Notice that we only need half a mole of chlorine gas in order to end up with 1 mole of NaCl. The arrow pointing down from this to the lower thick line represents the enthalpy change of formation of sodium chloride.
The Born-Haber cycle now imagines this formation of sodium chloride as happening in a whole set of small changes, most of which we know the enthalpy changes for - except, of course, for the lattice enthalpy that we want to calculate.
• The +107 is the atomization enthalpy of sodium. We have to produce gaseous atoms so that we can use the next stage in the cycle.
• The +496 is the first ionization energy of sodium. Remember that first ionization energies go from gaseous atoms to gaseous singly charged positive ions.
• The +122 is the atomization enthalpy of chlorine. Again, we have to produce gaseous atoms so that we can use the next stage in the cycle.
• The -349 is the first electron affinity of chlorine. Remember that first electron affinities go from gaseous atoms to gaseous singly charged negative ions.
• And finally, we have the positive and negative gaseous ions that we can convert into the solid sodium chloride using the lattice formation enthalpy.
Now we can use Hess' Law and find two different routes around the diagram which we can equate. As drawn, the two routes are obvious. The diagram is set up to provide two different routes between the thick lines. So, from the cycle we get the calculations directly underneath it . . .
-411 = +107 + 496 + 122 - 349 + LE
LE = -411 - 107 - 496 - 122 + 349
LE = -787 kJ mol-1
How would this be different if you had drawn a lattice dissociation enthalpy in your diagram? Your diagram would now look like this:
The only difference in the diagram is the direction the lattice enthalpy arrow is pointing. It does, of course, mean that you have to find two new routes. You cannot use the original one, because that would go against the flow of the lattice enthalpy arrow. This time both routes would start from the elements in their standard states, and finish at the gaseous ions.
-411 + LE = +107 + 496 + 122 - 349
LE = +107 + 496 + 122 - 349 + 411
LE = +787 kJ mol-1
Once again, the cycle sorts out the sign of the lattice enthalpy.
Theoretical Estimates of Lattice Energies
Let's assume that a compound is fully ionic. Let's also assume that the ions are point charges - in other words that the charge is concentrated at the center of the ion. By doing physics-style calculations, it is possible to calculate a theoretical value for what you would expect the lattice energy to be. Calculations of this sort end up with values of lattice energy, and not lattice enthalpy. If you know how to do it, you can then fairly easily convert between the two.
There are several different equations, of various degrees of complication, for calculating lattice energy in this way. There are two possibilities:
• There is reasonable agreement between the experimental value (calculated from a Born-Haber cycle) and the theoretical value. Sodium chloride is a case like this - the theoretical and experimental values agree to within a few percent. That means that for sodium chloride, the assumptions about the solid being ionic are fairly good.
• The experimental and theoretical values do not agree. A commonly quoted example of this is silver chloride, AgCl. Depending on where you get your data from, the theoretical value for lattice enthalpy for AgCl is anywhere from about 50 to 150 kJ mol-1 less than the value that comes from a Born-Haber cycle. In other words, treating the AgCl as 100% ionic underestimates its lattice enthalpy by quite a lot.
The explanation is that silver chloride actually has a significant amount of covalent bonding between the silver and the chlorine, because there is not enough electronegativity difference between the two to allow for complete transfer of an electron from the silver to the chlorine. Comparing experimental (Born-Haber cycle) and theoretical values for lattice enthalpy is a good way of judging how purely ionic a crystal is.
Example $2$: Born-Haber Cycle for $\ce{MgCl2}$
The question arises as to why, from an energetics point of view, magnesium chloride is MgCl2 rather than MgCl or MgCl3 (or any other formula you might like to choose). It turns out that MgCl2 is the formula of the compound which has the most negative enthalpy change of formation - in other words, it is the most stable one relative to the elements magnesium and chlorine.
Let's look at this in terms of Born-Haber cycles of and contrast the enthalpy change of formation for the imaginary compounds MgCl and MgCl3. That means that we will have to use theoretical values of their lattice enthalpies. We ca not use experimental ones, because these compounds obviously do not exist! I'm taking theoretical values for lattice enthalpies for these compounds that I found on the web. I can't confirm these, but all the other values used by that source were accurate. The exact values do not matter too much anyway, because the results are so dramatically clear-cut.
1. The Born-Haber cycle for MgCl
We will start with the compound MgCl, because that cycle is just like the NaCl one we have already looked at.
$\ce{Mg(s) + 1/2 Cl_2(g) \rightarrow MgCl (s)} \nonumber$
Find two routes around this without going against the flow of any arrows. That's easy:
ΔHf = +148 + 738 + 122 - 349 - 753
ΔHf = -94 kJ mol-1
So the compound MgCl is definitely energetically more stable than its elements. I have drawn this cycle very roughly to scale, but that is going to become more and more difficult as we look at the other two possible formulae. So I am going to rewrite it as a table. You can see from the diagram that the enthalpy change of formation can be found just by adding up all the other numbers in the cycle, and we can do this just as well in a table.
kJ
atomization enthalpy of Mg +148
1st IE of Mg +738
atomization enthalpy of Cl +122
electron affinity of Cl -349
lattice enthalpy -753
calculated ΔHf -94
2. The Born-Haber cycle for MgCl2
The equation for the enthalpy change of formation this time is
$\ce{Mg (s) + Cl2 (g) \rightarrow MgCl2 (s)} \nonumber$
So how does that change the numbers in the Born-Haber cycle?
• You need to add in the second ionization energy of magnesium, because you are making a 2+ ion.
• You need to multiply the atomization enthalpy of chlorine by 2, because you need 2 moles of gaseous chlorine atoms.
• You need to multiply the electron affinity of chlorine by 2, because you are making 2 moles of chloride ions.
• You obviously need a different value for lattice enthalpy.
kJ
atomization enthalpy of Mg +148
1st IE of Mg +738
2nd IE of Mg +1451
atomization enthalpy of Cl (x 2) +244
electron affinity of Cl (x 2) -698
lattice enthalpy -2526
calculated ΔHf -643
You can see that much more energy is released when you make MgCl2 than when you make MgCl. Why is that? You need to put in more energy to ionize the magnesium to give a 2+ ion, but a lot more energy is released as lattice enthalpy. That is because there are stronger ionic attractions between 1- ions and 2+ ions than between the 1- and 1+ ions in MgCl. So what about MgCl3? The lattice energy here would be even greater.
3. The Born-Haber cycle for MgCl3
The equation for the enthalpy change of formation this time is
$\ce{Mg(s) + 3/2 Cl_2(g) \rightarrow MgCl_3 (s)} \nonumber$
So how does that change the numbers in the Born-Haber cycle this time?
• You need to add in the third ionization energy of magnesium, because you are making a 3+ ion.
• You need to multiply the atomization enthalpy of chlorine by 3, because you need 3 moles of gaseous chlorine atoms.
• You need to multiply the electron affinity of chlorine by 3, because you are making 3 moles of chloride ions.
• You again need a different value for lattice enthalpy.
kJ
atomization enthalpy of Mg +148
1st IE of Mg +738
2nd IE of Mg +1451
3rd IE of Mg +7733
atomization enthalpy of Cl (x 3) +366
electron affinity of Cl (x 3) -1047
lattice enthalpy -5440
calculated ΔHf +3949
This time, the compound is hugely energetically unstable, both with respect to its elements, and also to other compounds that could be formed. You would need to supply nearly 4000 kJ to get 1 mole of MgCl3 to form!
Look carefully at the reason for this. The lattice enthalpy is the highest for all these possible compounds, but it is not high enough to make up for the very large third ionization energy of magnesium.
Why is the third ionization energy so big? The first two electrons to be removed from magnesium come from the 3s level. The third one comes from the 2p. That is closer to the nucleus, and lacks a layer of screening as well - and so much more energy is needed to remove it. The 3s electrons are screened from the nucleus by the 1 level and 2 level electrons. The 2p electrons are only screened by the 1 level (plus a bit of help from the 2s electrons).
Conclusion
Magnesium chloride is MgCl2 because this is the combination of magnesium and chlorine which produces the most energetically stable compound - the one with the most negative enthalpy change of formation.
The Born-Lande' equation
The Born-Landé equation is a concept originally formulated in 1918 by the scientists Born and Landé and is used to calculate the lattice energy (measure of the strength of bonds) of a compound. This expression takes into account both the Born interactions as well as the Coulomb attractions.
Introduction
Due to its high simplicity and ease, the Born-Landé equation is commonly used by chemists when solving for lattice energy. This equation proposed by Max Born and Alfred Landé states that lattice energy can be derived from ionic lattice based on electrostatic potential and the potential energy due to repulsion. To solve for the Born-Landé equation, you must have a basic understanding of lattice energy:
• Lattice energy decreases as you go down a group (as atomic radii goes up, lattice energy goes down).
• Going across the periodic table, atomic radii decreases, therefore lattice energy increases.
The Born-Landé equation was derived from these two following equations. the first is the electrostatic potential energy:
$\Delta U = - \dfrac{N_A M\left | Z^+ \right | \left | Z^- \right |e^2}{4\pi\epsilon_o r} \label{1}$
with
• $M_A$ is Avogadro's constant ($6.022 \times 10^{23}$)
• $M$ is the Madelung Constant (a constant that varies for different structures)
• $e$ is the charge of an electron ($1.6022 \times 10^{-19}$ C)
• $Z^+$ is the cation charge
• $Z^-$ is the anion charge
• $\epsilon_o$ is the permittivity of free space
The second equation is the repulsive interaction:
$\Delta U = \dfrac{N_A B}{r^n} \label{2}$
with
• $B$ is the repulsion coefficient and
• $n$ is the Born Exponent (typically ranges between 5-12) that is used to measure how much a solid compresses
These equations combine to form:
$\Delta U (0K) = \dfrac{N_A M\left | Z^+ \right | \left | Z^- \right |e^2}{4\pi\epsilon_or_o} \left ( 1- \dfrac{1}{n} \right) \label{3}$
with
• $r_0$ is the closest ion distance
Calculate Lattice Energy
Lattice energy, based on the equation from above, is dependent on multiple factors. We see that the charge of ions is proportional to the increase in lattice energy. In addition, as ions come into closer contact, lattice energy also increases.
Example $1$
Which compound has the greatest lattice energy?
• AlF3
• NACl
• LiF
• CaCl2
Solution
This question requires basic knowledge of lattice energy. Since F3 gives the compound a +3 positive charge and the Al gives the compound a -1 negative charge, the compound has large electrostatic attraction. The bigger the electrostatic attraction, the greater the lattice energy.
Example $2$
What is the lattice energy of NaCl? (Hint: you must look up the values for the constants for this compound)
Solution
-756 kJ/mol (again, this value is found in a table of constants)
Example $3$
Calculate the lattice energy of NaCl. | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Crystal_Lattices/Thermodynamics_of_Lattices/Lattice_Enthalpies_and_Born_Haber_Cycles.txt |
Contributors
Biological Crystallography
An application of Patterson methods for solution of crystal structures, typically proteins with heavy-atom derivatives, where the Patterson function is calculated using structure-factor coefficients based on the difference between the heavy-atom derivative and the native molecule.
Maximum likelihood
An approach to structure refinement in which the parameters of a structural model are modified to optimize the statistical probability of generating a set of observed intensities. The technique is often used in the refinement of structures of biological macromolecules, where the unfavorable parameter-to-observation ratio often leads to overfitted data and consequent systematic errors in least-squares minimization procedures.
Molecular Replacement
An approach to solving the phase problem by concentrating on phase relationships that arise through X-ray diffraction from similar molecular components. The components can be molecular fragments related through noncrystallographic symmetry (e.g. icosahedral subunits of a virus) or a similar molecule such as a homologous protein with high sequence identity.
Discussion
The technique is most commonly used in biological crystallography. Typically, an attempt is made to fit the structure under investigation (the 'target' structure) to a previously solved ('probe') structure. The probe may be a different crystal form of the same protein; or it may be a different protein with a high level of sequence identity, which correlates well with structural resemblance. As a rule of thumb, molecular replacement is often straightforward if the probe is well characterized and shares at least 40% sequence identity with the target. Rotation and transformation matrices must be applied. Patterson methods are suitable for determining the rotation function R, since intramolecular vectors are all shifted to the origin of a Patterson map. The orientational fit (i.e. the quality of the calculated rotation function) has usually been assessed by monitoring R factors or real-space correlation coefficients. Increasingly, modern programs use maximum likelihood based algorithms.
The translation function T is then determined to shift the now correctly orientated probe model to the correct coordinates within the asymmetric unit. Space-group symmetries can be used to help reduce the computationally intense calculations required.
History
Early applications of the technique concentrated on large virus structures with icosahedral symmetry. In the 1960s the early development of the molecular replacement technique was aimed primarily at ab initio phase determination. It was only in the 1970s, when more structures became available, that it was possible to use the technique to solve homologous structures with suitable search models .
See also
1. Noncrystallographic symmetry. D. M. Blow. International Tables for Crystallography (2006). Vol. F, ch. 13.1, pp. 263-268
2. Rossmann, M. G. & Blow, D. M. (1962). Acta Cryst. 15, 24–31. The detection of sub-units within the crystallographic asymmetric unit.
Multiwavelength anomalous diffraction (MAD)
An approach to solving the phase problem in protein structure determination by comparing structure factors collected at different wavelengths, including the absorption edge of a heavy-atom scatterer. Also known as multiple-wavelength anomalous diffraction or multiwavelength anomalous dispersion.
Discussion
The 'normal' atomic scattering factor $f^0$ describes the strength of X-rays scattered from the electrons in an atom assuming that they are free oscillators. Because the scattering electrons are in fact bound in atomic orbitals, they act instead as a set of damped oscillators with resonant frequencies matched to the absorption frequencies of the electron shells. The total atomic scattering factor f is then a complex number, and is represented by the sum of the normal factor and real and imaginary 'anomalous' components:
$f = f^0 + f' + if''. \nonumber$
A consequence of the wavelength dependence of anomalous dispersion is that the structure factors will be significantly perturbed, both in amplitude and in phase, by resonant scattering off an absorption edge. Hence, if diffraction is carried out at a wavelength matching the absorption edge of a scattering atom, and again at a wavelength away from the absorption edge, comparison of the resulting diffraction patterns will allow information to be extracted about the phase differences. For suitable species, the effect is of comparing a native molecule with a strictly isomorphous derivative (and in such cases phase determination and improvement are similar to isomorphous replacement methods).
The technique, often using tunable synchrotron radiation, is particularly well suited to proteins where methionine residues can be readily replaced by selenomethionine derivatives; selenium has a sufficiently strong anomalous scattering effect that it allows phasing of a macromolecule.
History
This technique was introduced by W. Hendrickson (Hendrickson, W. A., 1991, Determination of macromolecular structures from anomalous diffraction of synchrotron radiation. Science, 254, 51–58.)
See also
MAD and MIR. J. L. Smith, W. A. Hendrickson, T. C. Terwilliger and J. Berendzen. International Tables for Crystallography (2006). Vol. F, ch. 14.2, pp. 299-309
Noncrystallographic symmetry
A symmetry operation that is not compatible with the periodicity of a crystal pattern (in two or three dimensions) is called a noncrystallographic symmetry. Rotations other than 1, 2, 3, 4, and 6 (in E2 and E3) belong to this type of symmetry. Rotations 5, 8, 10 and 12 are compatible with a translation in higher-dimensional spaces, but they are commonly considered noncrystallographic. For example、in quasicrystals fivefold or tenfold rotational axes are incapable of tiling space through the application of three-dimensional lattice translations, but they act as normal symmetry axes in a higher-dimensional space. Continuous rotations, which give rise to the Curie groups contained in the cylindrical system and in the spherical system, are noncrystallographic in any dimension. | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Crystallography/Biological_Crystallography/Difference_Patterson_map.txt |
By analogy with optical observations, e.g. in astronomy, the term optical resolution in structure determination,$d_{opt}$, is used to describe the expected minimum distance between two resolved peaks in an electron-density map.
$d_{opt}=[2(\sigma^2_{Patt}+\sigma^2_{sph})]^{\frac{1}{2}} \nonumber$
where σPatt is the standard deviation of the Gaussian function fitted to the Patterson origin peak, σsph is the standard deviation of the Gaussian function fitted to the origin peak of the spherical interference function, representing the Fourier transform of a sphere with radius 1 / dmin, and dmin is the nominal resolution.
Real-space correlation coefficient
The real-space correlation coefficient, RSCC, is a measure of the similarity between an electron-density map calculated directly from a structural model and one calculated from experimental data. An advantage of techniques for evaluating goodness of fit in real space is that they can be performed for arbitrary sets of atoms. They are therefore used most often in the refinement of biological macromolecular structures to improve the model fit on a per-residue basis.
$\text{RSCC} =\dfrac{\sum | \rho_{obs} - \langle \rho_{obs} \rangle| \sum | \rho_{calc} - \langle \rho_{calc} \rangle|}{\sqrt{\sum | \rho_{obs} - \langle \rho_{obs} \rangle|^2 \sum | \rho_{calc} - \langle \rho_{calc} \rangle|^2}} \nonumber$
(ρ's are electron density values at grid points that cover the residue in question, obs and calc refer to experimental and model electron density, respectively.)
This metric is similar to the real-space residual RSR, but does not require that the two densities be scaled against each other.
Real-space residual
The real-space residual, RSR, is a measure of the similarity between an electron-density map calculated directly from a structural model and one calculated from experimental data. An advantage of techniques for evaluating goodness of fit in real space is that they can be performed for arbitrary sets of atoms. They are therefore used most often in the refinement of biological macromolecular structures to improve the model fit on a per-residue basis.
$\text{RSR}=\dfrac{\sum | \rho_{obs} - \rho_{calc}| }{\sum | \rho_{obs} +\rho_{calc}| } \nonumber$
(ρ's are electron density values at grid points that cover the residue in question, obs and calc refer to experimental and model electron density, respectively.)
The measure of similarity is often provided in the form of a graph of RSR values against residue number, showing clearly which residues give best and worst agreement with the experimental electron-density map. For nucleic acid structures, RSR may also be calculated separately for base, sugar and phosphate moieties of the nucleic acid monomer. RSR is generally considered an excellent model-validation tool.
Sohncke groups
Sohncke groups are called the 65 three-dimensional space groups containing only operations of the first kind (rotations, rototranslations, translations). It is very generally accepted that enantiomerically-pure compounds (e.g. proteins) crystallize in these groups. The term comes from Leonhard Sohncke (Halle, 22 February 1842 – München, 1 November 1897), German mathematician.
Diadochy
• Diadochy
Diadochy is the capability of an atom to replace another in a particular crystal structure. The degree of substitution depends on the physical conditions (temperature and pressure) as well as the atomic size, charge, and electronic structure of the diadochic partners.
• Isomorphous Crystals
• Isostructural Crystals
Crystal Chemistry
Diadochy is the capability of an atom to replace another in a particular crystal structure. The degree of substitution depends on the physical conditions (temperature and pressure) as well as the atomic size, charge, and electronic structure of the diadochic partners.
Isomorphous Crystals
Two crystals are said to be isomorphous if (a) both have the same space group and unit-cell dimensions and (b) the types and the positions of atoms in both are the same except for a replacement of one or more atoms in one structure with different types of atoms in the other (isomorphous replacement), such as heavy atoms, or the presence of one or more additional atoms in one of them (isomorphous addition). Isomorphous crystals can form solid solutions.
Isostructural Crystals
Two crystals are said to be isostructural if they have the same structure, but not necessarily the same cell dimensions nor the same chemical composition, and with a 'comparable' variability in the atomic coordinates to that of the cell dimensions and chemical composition. For instance, calcite CaCO3, sodium nitrate NaNO3 and iron borate FeBO3 are isostructural. One also speaks of isostructural series, or of isostructural polymorphs or isostructural phase transitions.
The term isotypic is synonymous with isostructural.
Abelian group
An abelian group, also called a commutative group, is a group (G, * ) such that
$g_1 * g_2 = g_2 * g_1 \nonumber$
for all $g_1$ and $g_2$ in $G$, where $*$ is a binary operation in $G$. This means that the order in which the binary operation is performed does not matter, and any two elements of the group commute.
Groups that are not commutative are called non-abelian (rather than non-commutative).
Abelian groups are named after Niels Henrik Abel.
Absolute structure
The spatial arrangement of the atoms of a physically identified non-centrosymmetric crystal and its description by way of unit-cell dimensions, space group, and representative coordinates of all atoms.
Affine Isomorphism
Each symmetry operation of crystallographic group in E3 may be represented by a 3×3 matrix W (the linear part) and a vector w. Two crystallographic groups G1 = {(W1i,w1i)} and G2 = {(W2i,w2i)} are called affine isomorphic is there exists a non-singular 3×3 matrix A and a vector a such that:
G2 = {(A,a)(W1i,w1i)(A,a)-1}
Two crystallographic groups are affine isomorphic if and only if their arrangement of symmetry elements may be mapped onto each other by an affine mapping of E3. Two affine isomorphic groups are always isomorphic. | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Crystallography/Biological_Crystallography/Optical_resolution.txt |
An affine mapping is any mapping that preserves collinearity and ratios of distances: if three points belong to the same straight line, their images under an affine transformation also belong to the same line. Moreover, the middle point is also conserved under the affine mapping. By contrast, angles and lengths in general are not kept constant by an affine mapping.
Under an affine mapping:
• parallel lines remain parallel;
• concurrent lines remain concurrent (images of intersecting lines intersect);
• the ratio of length of segments of a given line remains constant;
• the ratio of areas of two triangles remains constant;
• ellipses, parabolas and hyperbolas remain ellipses, parabolas and hyperbolas respectively;
• barycenters of polygons map into the corresponding barycenters.
Geometric contraction, expansion, dilation, reflection, rotation, shear, similarity transformations, spiral similarities, and translation are all affine transformations, as are their combinations. Affine mappings that keep also distances and angles are called Euclidean mappings.
Aperiodic crystal
A periodic crystal is a structure with, ideally, sharp diffraction peaks on the positions of a reciprocal lattice. The structure then is invariant under the translations of the direct lattice. Periodicity here means lattice periodicity. Any structure without this property is aperiodic. For example, an amorphous system is aperiodic. An aperiodic crystal is a structure with sharp diffraction peaks, but without lattice periodicity. Therefore, amorphous systems are not aperiodic crystals. The positions of the sharp diffraction peaks of an aperiodic crystal belong to a vector module of finite rank. This means that the diffraction wave vectors are of the form
$\mathbf{k}=\sum_{i-1}^{n}h_i\mathbf{a}_i^*, (integer\,h_i) \nonumber$
The basis vectors $a_i^*$ are supposed to be independent over the rational numbers, i.e. when a linear combination of them with rational coefficients is zero, all coefficients are zero. The minimum number of basis vectors is the rank of the vector module. If the rank n is larger than the space dimension, the structure is not periodic, but aperiodic.
Aristotype
An aristotype is a high-symmetry structure type that can be viewed as an idealized version of a lower-symmetry structure. It was introduced by Helen Megaw in relation to perovskites, where it is still mostly used. The cubic perovskite structure (which is adopted at most by half a dozen compounds) is regarded as the aristotype for the vast array of other lower-symmetry perovskites. The lower-symmetry structure is called hettotype.
After Buerger, aristotypes are also known as basic structures and hettotypes as derivative structures.
Originally, an aristotype is a printing-out process using paper coated with silver chloride in gelatin; now, any such process using silver salts in either collodion or gelatin; also, a print so made.
Arithmetic crystal class
The arithmetic crystal classes are obtained in an elementary fashion by combining the geometric crystal classes and the corresponding types of Bravais lattices. For instance, in the monoclinic system, there are three geometric crystal classes, 2, m and 2/m, and two types of Bravais lattices, P and C. There are therefore six monoclinic arithmetic crystal classes. Their symbols are obtained by juxtaposing the symbol of the geometric class and that of the Bravais lattice, in that order: 2P, 2C, mP, mC, 2/mP, 2/mC (note that in the space group symbol the order is inversed: P2, C2, etc...). In some cases, the centering vectors of the Bravais lattice and some symmetry elements of the crystal class may or may not be parallel; for instance, in the geometric crystal class mm with the Bravais lattice C, the centering vector and the two-fold axis may be perpendicular or coplanar, giving rise to two different arithmetic crystal classes, mm2C and 2mmC (or mm2A, since it is usual to orient the two-fold axis parallel to c), respectively. There are 13 two-dimensional arithmetic crystal classes and 73 three-dimensional arithmetic crystal classes that are listed in the attached table. Space groups belonging to the same geometric crystal class and with the same type of Bravais lattice belong to the same arithmetic crystal class; these are therefore in one to one correspondence with the symmorphic space groups.
The group-theoretical definition of the arithmetic crystal classes is given in Section 8.2.3 of International Tables of Crystallography, Volume A. | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Crystallography/Fundamental_Crystallography/Affine_mapping.txt |
An asymmetric unit of a space group is a simply connected smallest closed part of space from which, by application of all symmetry operations of the space group, the whole space is filled. This implies that:
• mirror planes must form boundary planes of the asymmetric unit;
• rotation axes must form boundary edges of the asymmetric unit;
• inversion centers must either form vertices of the asymmetric unit or be located at the midpoints of boundary planes or boundary edges.
These restrictions do not hold for screw axes and glide planes.
The term "asymmetric unit" does not mean that this region has an asymmetric shape. In mathematics it is called fundamental region or fundamental domain.
Automorphism
An isomorphism from a group (G,*) to itself is called an automorphism of this group. It is a bijection f : GG such that
f (g) * f (h) = f (g * h)
An automorphism preserves the structural properties of a group, e.g.:
• The identity element of G is mapped to itself.
• Subgroups are mapped to subgroups, normal subgroups to normal subgroups.
• Conjugacy classes are mapped to conjugacy classes (the same or another).
• The image f(g) of an element g has the same order as g.
The composition of two automorphisms is again an automorphism, and with composition as binary operation the set of all automorphisms of a group G, denoted by Aut(G), forms itself a group, the automorphism group of G.
Binary Operation
A binary operation on a set S is a mapping f from the Cartesian product S × S to S. A mapping from K x S to S, where K need not be S, is called an external binary operation.
Many binary operations are commutative (i.e. f(a,b) = f(b,a) holds for all a, b in S) or associative (i.e. f(f(a,b), c) = f(a, f(b,c)) holds for all a,b,c in S). Many also have identity elements and inverse elements. Typical examples of binary operations are the addition (+) and multiplication (*) of numbers and matrices as well as composition of functions or symmetry operations.
Examples of binary operations that are not commutative are subtraction (-), division (/), exponentiation(^), super-exponentiation(@), and composition.
Binary operations are often written using infix notation such as a * b, a + b, or a · b rather than by functional notation of the form f(a,b). Sometimes they are even written just by concatenation: ab.
Bragg's Law
Bragg's law provides the condition for a plane wave to be diffracted by a family of lattice planes:
$2 d \sin θ = n λ. \nonumber$
where d is the lattice spacing, θ the angle between the wavevector of the incident plane wave, ko, and the lattice planes, λ its wave length and n is an integer, the order of the reflection. It is equivalent to the diffraction condition in reciprocal space and to the Laue equations.
Direct Derivation of Bragg's Law
• Reflection from the first plane: The scattered waves will be in phase whatever the distribution of the point scatterers in the first plane if the angle of the reflected wave vector, kh, is also equal to θ. This is Snell-Descartes' law of reflection.
• Reflection from the second plane: Since the phase of the reflected waves is independent of the position of the point scatterer in the plane, the phase difference between the waves reflected by two successive lattice planes is obtained by choosing arbitrarily a scattering point, A, on the first plane and a scattering point, b on the second plane such that AB is normal to the planes. If C and d are the projections of A on the incident and reflected wave vectors passing through B, it is clear from Figure 1 that the path difference between the waves reflected at A and B, respectively, is: $CB + BD = 2 d \sin θ \nonumber$ and that the two waves will be in phase if this path difference is equal to n λ where n is an integer.
• Reflection from the third, etc., planes: If Bragg's relation is satisfied for the first two planes, the waves reflected with wave vector kh will be in phase fo all the planes of the family.
Order of the Reflection
Bragg's law may also be written:
$2 (d/n) sin θ = λ. \nonumber$
One may then say that a Bragg reflection of order n on a family of lattice planes or order n is equivalent to reflection of order 1 on a family of fictitious, or imaginary, planes of lattice spacing:
$d_{hkl} = \dfrac{d}{n} \nonumber$
This fictitious family is associated to the reciprocal lattice vector OH where OH = n/d = 1/dhkl. The indices of the reflection are:hkl. For instance, the dashed blue lines in Figure 1 correspond to the fistitious planes associated to the second order, n = 2.
Extinctions or Systematic Absences
If there is a glide plane or a screw axis normal to the lattice planes, the spacing of the actual reflecting planes is d/2 for a glide plane and (d p/q) for a qp screw axis. Bragg's law should then be written:
$2 \left(\dfrac{d}{2}\right) \sin θ = n λ \nonumber$
$2 d \sin θ = 2n λ \nonumber$
for a glide plane and
$2 (d p/q) \sin θ = n λ ⇒ 2 d \sin θ = (q/p)n λ \nonumber$
for a screw axis qp.
The reflections of odd order for a glide plane and of order different from (q/p)n for a screw axis are then absent. One speaks of extinctions or systematic absences related to the presence of glide or screw components.
As an example, Figure 2 shows the case of a 21 screw axis: the reflecting planes are the blue planes and the green planes deduced from the latter by the action of the screw axis. Reflections of odd order will be sytematically absent.
Influence of deformation
A deformation that leaves a family of lattice planes (hkl) undistorted and its lattice spacing d unchanged will not affect the Bragg angle of kkl reflections, e.g. lattice planes parallel to a screw dislocations.
History
Bragg (1890-1971) presented his derivation of the reflection condition at a meeting of the Cambridge Philosophical Society on 11 November 1912. His paper was published in 1913 (Bragg W.L., 1913, The Diffraction of Short Electromagnetic Waves by a Crystal, Proc. Cambridge Phil. Soc., 17, 43-57. For details, see P. P. Ewald, 1962, IUCr, 50 Years of X-ray Diffraction, Section 5, page 64. | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Crystallography/Fundamental_Crystallography/Asymmetric_Unit.txt |
The Bravais-Miller indices are used in the case of hexagonal lattices. In that case, one uses four axes, a1, a2, a3, c and four Miller indices, (hkil), where h, k, i, l are prime integers inversely proportional to the intercepts OP, OQ, OS, OR of a plane of the family with the four axes. The indices h, k, i are cyclically permutable and related by h + k + i = 0
Bravais class
An arithmetic crystal class with matrix group of lattices is called a Bravais arithmetic crystal class, or Bravais class for short.
Each lattice is associated with a Bravais class, and each matrix group of a Bravais class represents the point group of a lattice referred to an appropriate primitive basis.
There exist 5 Bravais classes in two dimensions:
• 2p
• 2mmp
• 2mmc
• 4mmp
• 6mmh
There exist 14 Bravais classes in three dimensions:
• ${\bar 1}$P
• 2/mP
• 2/mS
• mmmP
• mmmS
• mmmI
• mmmF
• 4/mmmP
• 4/mmmI
• ${\bar 3}$mR
• 6/mmmP
• m${\bar 3}$mP
• m${\bar 3}$mI
• m${\bar 3}$mF
Bravais flock
Space groups that are assigned to the same Bravais class belong to the same Bravais flock of space groups.
The introduction of the concept of Bravais flock is necessary in order to classify space groups on the basis of their Bravais type of lattice, independently from any accidental metric of the lattice.
An orthorhombic crystal may have accidentally a = b have a tetragonal lattice not because of symmetry restrictions but just by accident. If the Bravais types of lattices were used directly to classify space group, such a crystal would be go in another category with respect to an orthorhombic crystal without specialized metric. The concept of Bravais flock solves this ambiguity.
A space group G is assigned to a Bravais class on the basis of the corresponding point group P and the arithmetic crystal class associated with it:
• If the arithmetic crystal class of G is a Bravais class, then G is assigned to that Bravais class;
• if the arithmetic crystal class of G is not a Bravais class, then the Bravais class to which G is assigned is obtained as follows:
• those Bravais classes are retained whose point group B is such that P is a subgroup of B;
• G is assigned to that Bravais class, among those selected above, for which the ratio of the order of the point groups of B and of P is minimal.
For example, a space group of type I41 belongs to the arithmetic crystal class 4I, to which two Bravais class can be associated, 4/mmmI and $m{\bar 3}mI$. The second condition uniquely assigns G to the Bravais flock of 4/mmmI, despite the fact that the Bravais class of the lattice may be $m{\bar 3}mI$ as a result of accidental symmetry.
Bravais lattice
The current nomenclature adopted by the IUCr prefers to use the expression Bravais types of lattices to emphasize that Bravais lattices are not individual lattices but types or classes of all lattices with certain common properties .
All vector lattices whose matrix groups belong to the same Bravais class correspond to the same Bravais type of lattice.
Brillouin zones
A Brillouin Zone is a particular choice of the unit cell of the reciprocal lattice. It is defined as the Wigner-Seitz cell (also called Dirichlet or Voronoi Domain) of the reciprocal lattice. It is constructed as the set of points enclosed by the Bragg planes, the planes perpendicular to a connection line from the origin to each lattice point and passing through the midpoint. Alternatively, it is defined as the set of points closer to the origin than to any other reciprocal lattice point. The whole reciprocal space may be covered without overlap with copies of such a Brillouin Zone.
For high-symmetry lattices one introduces sometimes the notion of n-th Brillouin Zone. This is the set of points one reaches with a straight line from the origin and passing through n-1 Bragg Planes. In this terminology, the Brillouin Zone defined above is the first Brillouin Zone. The n-th Brillouin Zone is a shell around lower Brillouin Zones and its shape becomes for higher values of n rapidly rather complicated.
Vectors in the Brillouin Zone or on its boundary characterize states in a system with lattice periodicity, e.g. phonon or electron states. States are non-equivalent if they belong to different vectors in a unit cell of the reciprocal lattice. This is not necessarily the Brillouin Zone. Especially, for low-symmetry systems Brillouin Zones are sometimes difficult to visualize, and another choice of unit cell may be useful, e.g. the parallelepiped spanned by the basis vectors.
The Brillouin Zone boundary consists of pieces of Bragg planes. Since two points on the boundary may differ by a reciprocal lattice vector, only a part of the boundary may be used for characterization of states. There is no simple rule for that choice. Dispersion curves at the zone boundary have either a maximum or minimum there or they merge with other curves giving rise to a change of degeneracy.
See also
Section 2.2 of International Tables of Crystallography, Volume D | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Crystallography/Fundamental_Crystallography/Bravais-Miller_indices.txt |
The Cartesian product is a direct product of sets.
The Cartesian product of two sets $X$ and $Y$, denoted $X \times Y$, is the set of all possible ordered pairs whose first component is a member of $X$ and whose second component is a member of $Y$:
$X\times Y = \{(x,y) | x\in X\;\mathrm{and}\;y\in Y\}.$
A Cartesian product of two finite sets can be represented by a table, with one set as the rows and the other as the columns, and forming the ordered pairs, the cells of the table, by choosing the element of the set from the row and the column.
Center
The center (or centre) of a group G is the set Z(G) = { a in G : a*g = g*a for all g in G } of elements commuting with all elements of G. The center is an Abelian group.
The center of a group G is always a normal subgroup of G, namely the kernel of the homomorphism mapping an element a of G to the inner automorphism fa: gaga-1.
Centered lattices
When the unit cell does not reflect the symmetry of the lattice, it is usual in crystallography to refer to a 'conventional', non-primitive, crystallographic basis, ac, bc, cc instead of a primitive basis, a, b, c. This is done by adding lattice nodes at the center of the unit cell or at one or three faces. The vectors joining the origin of the unit cell to these additional nodes are called 'centering vectors'. In such a lattice ac, bc and cc with all their integral linear combinations are lattice vectors again, but there exist other lattice vectors tL, t = t1 ac + t2 bc + t3 cc; with at least two of the coefficients t1, t2, t3 being fractional. The table below gives the various types of centering vectors and the corresponding types of centering. Each one is described by a letter, called the Bravais letter, which is to be found in the Hermann-Mauguin symbol of a space group.
The 'multiplicity', m, of the centered cell is the number of lattice nodes per unit cell (see table).
The volume of the unit cell, Vc = (ac, bc, cc) is given in terms of the volume of the primitive cell, V = (a, b, c), by:
Vc = m V
Centralizer
The centralizer CG(g) of an element g of a group G is the set of elements of G which commute with g:
CG(g) = {x ∈ G : xg = gx}.
If H is a subgroup of G, then CH(g) = CG(g) ∩ H.
More generally, if S is any subset of G (not necessarily a subgroup), the centralizer of S in G is defined as
CG(S) = {x ∈ G : ∀ s ∈ S, xs = sx}.
If S = {g}, then C(S) = C(g).
C(S) is a subgroup of G; in fact, if x, y are in C(S), then xy −1s = xsy−1 = sxy−1.
Complex
A complex is a subset obtained from a group by choosing part of its elements in such a way that the closure property of groups is not respected. Therefore, a complex is not a group itself.
A typical example of complexes is that of cosets. In fact, a coset does not contain the identity and therefore it is not a group.
A subgroup is a particular case of complex that obeys the closure property and is a group itself. | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Crystallography/Fundamental_Crystallography/Cartesian_product.txt |
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