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In the previous sections, we saw how to predict the approximate geometry around an atom using VSEPR theory, and we learned that lone pairs of electrons slightly distort bond angles from the "parent" geometry. This page discusses the effect of multiple (double and triple) bonds between bonded atoms.
Double and triple bonds are more repulsive than single bonds
VSEPR theory predicts that double and triple bonds have stronger repulsive forces than single bonds. Like lone pairs of electrons, multiple bonds occupy more space around the central atom than a single bond. The result is that bond angles are slightly distorted compared to the parent geometry. Since a multiple bond has a higher electron density than a single bond, its electrons occupy more space than those of a single bond. Double and triple bonds distort bond angles in a way similar to what lone pairs do. Due to the stronger repulsion, double and triple bonds occupy positions similar to those of lone pairs in groups with 5 and 6 electron groups.
CH2O
In a molecule such as CH2O (AX3), whose structure is shown below, the double bond repels the single bonds more strongly than the single bonds repel each other. This causes a deviation from ideal geometry (an H–C–H bond angle of 116.5° rather than 120°).
2-methybutene
In the molecule, CH2C(CH3)2, the methyl—C—methyl bond angle is 115.6°, which is less than the 120° bond angle that would be expected of the parent geometry. On the other hand, the acetyl—C—methyl bond angle is greater than 120°, with an actual bond angle of 122.2°.
Atoms with both lone pairs and multiple bonds
In general, we expect that a lone pair has slightly greater repulsive force than a multiple bond, and that a multiple bond has slightly greater repulsive force than a single bond. The order of expected repulsive force is:
$\text{lone pair} > \text{double or triple bond} > \text{single bond} \nonumber$
IO2F2¯
With five electron groups, the IO2F2 molecule has an approximately trigonal bipyramidal electronic (parent) geometry. The central iodine atom has a single bond to each of the fluorine atoms, a double bond to each oxygen, and a lone pair. The lone pair and the double bonds will occupy more space around iodine than each of the single bonds. Thus, we expect the lone pair and double bonds to occupy equatorial positions around the central iodine. The O=I=O bond angle is 102° (much smaller than the 120° angle expected from the parent geometry). This dramatically smaller angle is a result of the increased repulsion between the lone pair and the double bond.
SeOCl2
With four electron groups, the SeOCl2 molecule has four electron groups and approximately tetrahedral electronic (parent) geometry. It has both a lone pair and a double bond on the central selenium atom. The two chlorine atoms are singly bonded to selenium, while the oxygen is double bonded. The lone pair is most repulsive, followed by the double bonded oxygen, and then the chlorine bonds. This gives a Cl—Se—Cl bond angle of 97° and a Cl—Se—O bond angle of 106°; both angles are less than the 109.5° angles expected for the ideal tetrahedral geometry.
Practice
Exercise $1$
Use VSEPR theory to predict the geometries and draw the structures of the following.
1. XeOF4
2. NO2¯
3. SOCl2
4. IOF3¯
Answer a
This molecule has six electron groups around the central Xe atom (steric number 6), and thus has an approximately octahdral electronic (parent) geometry. There is a double bond to O and a lone pair, both of which are more strongly repulsive than the single bonds to F. The double bond and lone pair will be directly opposite to each other, designated as axial positions. The result is a square pyramidal molecular geometry. VSEPR theory predicts F—Xe—F bond angles of 90°. As a general rule, lone pairs are slightly more repulsive than multiple bonds, and so we might expect the O—Xe—F bond angles to be <90°; however in this case the actual O—Xe—F bond angle is observed to be 91° (you could not have predicted the latter).
Answer b
This molecule has three electron groups around the central atom: one lone pair and two double bonds to oxygen atoms. This results in an approximately trigonal planar electronic (parent geometry). We expect the lone pair to be slightly more repulsive than double bonds, and so we expect the O—N—O bond angle to be slightly less than 120°. The actual O—N—O angle is 115°.
Answer c
This molecule has four electron groups (steric number 4) with an approximately tetrahedral electronic (parent) geometry. Each double bonded oxygen will take up more space around the central S than the single bonded Cl atoms. We should expect the double bonds to repel each other more strongly than they repel each single bond, with the least repulsive interaction being between the two single bonds. The bond angle for Cl—S—Cl is expected to be <109.5° according to VSEPR theory (however, it is actually 111°). We would expect the bond angle for O—S—O to be larger than the angle for Cl—S—Cl and for Cl—S—O. The O—S—O bond angle is predicted to be >109.5° (and the actual bond angle is 120°). The molecular geometry is a distorted tetrahedron.
Answer d
This molecule has five electron groups (steric number 5) with an approximately trigonal bipyramidal electronic (parent) geometry. There is one lone pair, a double bond to O, and three single bonds to F atoms around the central I atom. The lone pair and double bond are most repulsive, and should occupy the less crowded equatorial positions rather than the more crowded axial positions. This results in a seesaw molecular geometry. With the more repulsive lone pair and the strongest equatorial repulsive force being between the double bond and lone pair, we should expect the Fequatorial—I—O bond angle to be less than the 120° angle expected for the parent geometry (it is actually much less, at 98°). Due to repulsion between the axial F atoms and both the lone pair and double bond, we should expect the F—S—F bond angles to be compressed. The Faxial—I—Faxial is actually 168°. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/03%3A_Simple_Bonding_Theory/3.02%3A_Valence_Shell_Electron-Pair_Repulsion/3.2.02%3A_Multiple_Bonds.txt |
Introduction
We saw in previous sections how lone pairs of electrons and multiple bonds distort bond angles between non-central atoms (ligands) around a central atom. This section describes how ligand electronegativity and size also influence bond angles and molecular geometry. Electronegativity is generally correlated with atomic size going down any group of the periodic table. There are some cases where bond angles can be predicted by these correlations. However, size and electronegativity can also work as competing factors in determining bond angles.
Definitions of Electronegativity
Linus Pauling introduced the first electronegativity scale in 1932 in order to explain the extra stability of molecules with polar bonds.[1] The electronegativity of an atom, represented by the Greek letter $χ$ (chi), can be defined as the tendency of an atom to draw electrons to itself in a chemical bond. On the Pauling scale, the electronegativity difference between two atoms A and B was defined in terms of the dissociation energies Ed of the A-A, B-B, and A-B bonds:
$\chi_{A} - \chi_{B} = \sqrt{E_{d}(AB) - [E_{d}(AA) + E_{d}(BB)]/2} \nonumber$
where the energies are expressed in electron volts. Pauling's scale of electronegativity ranges from Fluorine (most electronegative = 4.0) to Francium (least electronegative = 0.7). [2,3] The polarity of bonds and assignment of formal charges is predicted by electronegativity differences.
While directly relevant to the strength of chemical bonds, the Pauling definition of electronegativity has a significant limitation: calculating each value requires data from several compounds in which specific atoms are bonded. This means that the scale cannot be successfully applied to all situations. To overcome this limitation, alternative electronegativity scales were developed based on different thermochemical measurements or calculations. These other scales are listed in the table below. More thorough descriptions of how values in each scale are calculated are described on Wikipedia's Electronegativity page (click) and a concise history of electronegativity scales is summarized in ref 4.
Scale Developer Year first described Characteristics
Linus Pauling[1] 1932 Based on bond energies
Robert S. Mulliken[5,6] 1934 Based on valence electron properties of atoms (electron affinity and ionization energy)
Louis Allred & Eugene G. Rochow[7] 1958 Based on electrostatic force (effective nuclear charge)
Robert T. Sanderson[8] 1983 Based on atomic electron density
Ralph G. Pearson[9] 1985 Related to Hard-Soft Acid-Based theory
Leland C. Allen, Joseph Mann, Terry L. Meek[10,11,12] 1989, 2000
Average ionization energies of valence shell electrons, configuration energies (CE)$\mathrm{CE}=\frac{n \varepsilon_{s}+m \varepsilon_{p}}{n+m} \nonumber$ $n = \text{number of s electrons}$
$m = \text{number of p electrons}$
$\varepsilon_{s}, \varepsilon_{p} = \text{experimental 1-electron s and p energies}$
The primary advantage to this scale developed by Mann, Meek, and Allen[12] is that it is based on configuration energy (CE), the average ionization energies of valence electrons in ground state free atoms. A scale based on ionization energies can be calculated more directly for any element. However, a critique of all electronegativity scales, including this one, is that they are all based on assumptions that fail in some cases.
There is a relationship between electronegativity and atomic size because both are related to ionization energy. Like ionization energy, there is a general trend across the periodic table for both electronegativity and atomic size; in general, smaller atoms toward the top right-hand side of the periodic table have greater electronegativity values and smaller atomic size. We will see below that both electronegativity and atomic size influence bond angles and absolute molecular geometry around a central atom. In some cases, the size and electronegativity affects are aligned; in some cases these effects compete.
Electronegativity and Size Influence Bond Angles
Let's begin by examining the bond angles of several trigonal pyramidal molecules. The molecules shown in Figure $2$ each have three identical "pendant groups" on the central atom (Figure $2$). Pendant atoms or pendant groups are the atoms, or groups of atoms, that are bonded directly to the central atom. The molecules shown here are arranged according to the size of the central and pendant atoms. Central atoms increase in size going down this figure, and pendant atoms increase in size going across from left to right. The bond angles and bond lengths are labeled in each case.
What trends can you identify?
Trends in Size
First, let's examine how the size of the central and pendant groups might influence bond angle. In VSEPR theory, the size of atoms (or groups of atoms) will affect bond angles due to changes in steric interactions between pendant groups.
Size of the central atom
Examine the relationship between the size of the central atom and the bond angles in Figure $2$ (go down any column in the figure). For example, compare NH3, PH3, AsH3, and SbH3, with bond angles 106.6°, 93.2°, 92.1°, and 91.6°, respectively. In this series, the size of the central atom increases from N to Sb while the size of the pendant atom (hydrogen) remains constant. As the size of the central atom increases, the bond angles decrease; thus, we observe a negative relationship between size of the central atom and the bond angle in these molecules. This relationship is explained by sterics. As the central atom increases in size, the bond lengths also increase and the pendant atoms are farther from each other in space. In VSEPR theory, this will reduce steric interactions between the pendant groups. Since the lone pair on these molecules is more repulsive than bonded groups, the decrease in steric interactions between bonded groups results in a decrease in bond angles.
Size of the pendant atoms (or groups)
Examine the relationships between size of the halogen pendant atoms and bond angle in Figure $2$ (go across any row within the shaded region). For example, compare PF3, PCl3, and PBr3, with bond angles 97.8°, 100.3°, and 101.0°, respectively. In this series, the size of the pendant atom increases from F to Br while the central atom remains constant (phosphorous). As the size of pendant atoms increases, the bond angle increases; thus, we observe a direct relationship between size of the pendant group and the bond angle in these molecules. Again, we can explain this using sterics. As the size of the pendant atoms increases, sterics between the pendant groups will increase (despite small changes in bond length). Increased steric interactions between pendant groups will prefer larger bond angles between the groups.
This trend fails, however, if we consider the molecules with hydrogen as the pendant atoms. Notice, for example, that H is the smallest pendant atom. If we consider the trend described above, we should assume that XH3 (where X is a variable atom) would have the smallest bond angle in the series of XH3, XF3, XCl3, and XBr3. However this is not the case of NH3, NF3, and NCl3, with bond angles 106.6°, 102.2°, and 106.8°, respectively. To explain the variation in these bond angles, we need to consider electronegativity.
Trends in Electronegativity
Now let's examine how electronegativity influences bond angles around a central atom. In VSEPR theory, electronegativity of atoms/groups will affect bond angles due to changes in the distribution of electron pairs around the central atom (and thus changes in the severity of electron pair repulsion). This really comes down to bond polarity caused by the difference in electronegativity between the central atom and pendant groups (bond polarity in the context of valence bond theory). In a polar bond, the more electronegative atom will pull electron density towards itself. When a pendant atom is more electronegative, it will pull the bonded electron pair towards itself and away from the central atom; this will reduce the electron pair repulsion between bonded electron pairs on the central atom. A decrease in electron pair repulsion on the central atom should decrease bond angles between the groups (Figure $3$).
Electronegativity is an alternative explanation to the trends we already examined above in Figure $2$ . For example, when we compare the halogen pendant atoms (shaded region in Figure $2$) , the electronegativity of pendant groups decreases, bond polarity of the bonds decreases, and bond angles increase going from left to right and from F to Br. As more electron density remains on the central atom, electron repulsion between the bonded pairs increases and bond angles increase.
The electronegativity argument can also be used to explain the fact that NH3 (106.6°) has a larger bond angle than NF3 (102.2°). This particular case illustrates how electronegativity and size can be competing factors. While electronegativity differences seem to dominate in the case of NH3 (106.6°) compared to NF3 (102.2°), size differences still dominate in the cases of other XH3 and XF3 examples in Figure $2$).
Exercise $1$
Predict the geometry and approximate bond angles. Then put the molecules in each series in order of smallest to largest bond angle. Defend your answer.
1. The X-S-X bond angle in OSF2, OSCl2, OSBr2
2. H2O, H2S, H2Se, H2Te
3. H2O, OF2, OCl2
Answer (a)
Geometry and predicted bond angles: These are molecules with steric number 4. They can be written as two different resonance structures, and the resonance hybrid would have double bond character between S and O.
We would expect a trigonal pyramidal geometry with all bond angles <109.5° because the lone pair is more repulsive than bonds. Because the double bond is more repulsive than single bonds, we should also expect the O-S-X bond angles to be greater than the X-S-X bond angles.
Trend: This is a series of molecules that varies in the identity of the pendant atoms; the variation is within the halogens. The size of the halogens increases, and the electronegativity decreases in the order F, Cl, Br. Both size and electronegativity would lead us to conclude that the X-S-X bond angles would increase in the order OSF2 < OSCl2 < OSBr2
Explanation: (1) Increasing electonegativity of the pendant atom (F > Cl > Br) increases the polarity of the bond and reduces the electron density of the bonded pair on the central atom. This reduces the e-e repulsions of adjacent bonded electron pairs on S, allowing the halogens to become closer. The more electronegative pendant atoms can have smaller bond angles. (2) Increasing size of pendant atoms (F < Cl < Br) increases steric repulsions and increases bond angle. Both explanations lead to the same predicted trend.
The actual measured X-S-X bond angles are OSF2 (92.3°) / OSCl2 (96.2°) / OSBr2 (98.2°). The trend in these bond angles is consistent with the prediction.
Answer (b)
Geometry and predicted bond angles: These are molecules with steric number 4, bent molecular geometry, with predicted bond angles <109.5° because the two lone pairs are each more repulsive than the bonds. There are two lone pairs and two single bonds to H around each central atom.
Trend: This is a series of molecules that varies in the identity of the central atom. The central atom increases in the order O < S < Se < Te, where Te is the largest element and O is the smallest. Arguments based on size would lead us to predict that the pendant groups of H2Te would be less sterically crowded and thus have a smaller bond angle than the pendant groups of H2O. The electronegativity decreases in the order O > S > Se > Te, where O is the most electronegative element and Te is the least. Thus, we expect the bonding electro pairs to be closer to the central atom on O than they would be on Te; we should expect H2Te to have the least electron pair repulsions and thus smallest bond angle in this series, while H2O would have the strongest electron pair repulsions and largest bond angles.
Both arguments lead to the same conclusion, that the order of increasing bond angle is H2Te < H2Se < H2S < H2O.
The actual measured bond angles are H2Te (90.2°) / H2Se (90.6°) / H2S (92.1°) / H2O (104.5). The trend in these bond angles is consistent with prediction.
Answer (c)
Geometry and predicted bond angles: These are molecules with steric number 4, bent molecular geometry, with predicted bond angles <109.5° because the two lone pairs are each more repulsive than the bonds. There are two lone pairs and two single bonds to H around each central atom. (This is similar to the case in (b)).
Trend: This is a series of molecules that varies in the identity of the pendant atoms; two of the molecules have halogens, and the other has hydrogen pendant atoms. This is a case where size and electronegativity will be conflicting factors because trends in electronegativity do not mirror the trend in size.
• Size: The size of pendant atoms increases in the order H < F < Cl where H is smallest and Cl is largest. Prediction of bond angles based on size alone would lead to the predicted order of increasing bond angle H2O < OF2 < OCl2.
• Electronegativity: The electronegativity decreases in the order F > Cl > H where F has the greatest electronegativity and H has the least. Since we expect the most electronegative pendant atoms to have the smallest bond angles, prediction based on electronegativity alone would lead to the predicted order of increasing bond angle OF2 < OCl2 < H2O.
The points above illustrate how the two different arguments would lead to different predictions about the trend in bond angle. This makes it difficult to predict the actual order of increasing bond angle. However, we saw above in the example of NH3 vs NF3 that electronegativity is more important than size; yet in the case of NH3 vs NCl3, the much larger size of the pendant atom is more important. If we apply this lesson to the current problem, we might predict the order OF2 < H2O< OCl2 , and in fact this more nuanced prediction, based on a similar case, matches the actual order for measured bond angles: OF2 (103.3°) < H2O (104.5°) < OCl2 (110.9°).
Group Electronegativities
You probably heard the terms "electron donating group" and "electron withdrawing group" from your coursework in Organic Chemistry. For example, the acid trifluoroacetic acid (TFA) is more acidic than acetic acid due to the electron withdrawing effects of the CF3 group compared to CH3. CF3 is an electron withdrawing group, while CH3 is an electron donating group. In other words, CF3 is more electronegative than CH3.
The electron withdrawing ability (electronegativity) of groups can be estimated and compared, just as they are with atoms. Although there is no one scale that is used for group electronegativities, and published values even for the same groups vary widely, there are reliable trends within similar groups. The same size and electronegativity factors discussed above that affect bond angles for pendant atoms, also can be used to rationalize distorted bond angles for pendant groups around a central atom.
Exercise $2$
Consider the relative electronegativities and sizes of the pendant atoms/groups in the following examples. Is the trend in bond angles what you would expect from the relative group electronegativities and relative sizes? What, if any, is the more dominant factor in determining the trend?
1. N(CH3)3 has an C-N-C bond angle of 110.9°, while N(CF3)3 has a bond angle of 117.9°.
2. The X-S-X bond angles in molecules of the form SO2(X)2 are: SO2(OH)2 101.3°; SO2(CF3)2 102.0°; SO2(CH3)2 102.6°.
Answer (a)
Add texts here. Do not delete this text first.
Answer (b)
Add texts here. Do not delete this text first.
Special case of electronegativity and size in steric number 5 (trigonal bypyramid)
Molecules with steric number 5 are interesting because they posses two different types of positions (equatorial and axial). These positions have unique bond angles and bond lengths. Pendant atoms/groups have different preferences for the axial and equatorial positions that depend somewhat on their electronegativity.
In a previous section, we discussed the preference of lone pairs and multiple bonds for the equatorial positions in trigonal bipyramidal molecules. VSEPR theory rationalizes this by assuming that lone pairs and multiple bonds are more repulsive than the electron pairs in single bonds. Equatorial positions are less crowded (with only two closest neighbors at 90°, and two farther neighbors at 120°) compared to axial positions (with three closest neighbors at 90°), thus the more repulsive groups prefer the less crowded equatorial positions.
Pendant groups that are more electronegative result in weaker electron pair repulsion around the central atom, while groups that are less electronegative result in stronger electron-pair repulsion around the central atom (as described above). The result for steric number 5: bonding pairs to less electronegative elements are more repulsive, and generally prefer equatorial positions. Still, lone pairs and multiple bonds are more repulsive than single bonds and would show a stronger preference for equatorial positions.
Examples and Nuances
Exercise $3$
Some examples of molecule that demonstrate the equatorial preference of less electronegative groups are below. Predict (draw) their structures.
1. PF4Cl, PF3Cl2, and PF2Cl3
2. PF4(CH3) PF3(CH3)2, and PF2(CH3)3
Answer (a)
Cl is less electronegative than F; thus we expect Cl to have stronger preference for the equatorial positions.
Answer (b)
CH3, an electron donating group, is less electronegative than F. We expect CH3 to have a stronger preference for the equatorial positions.
In the exercise above, the structures can be explained by the electronegativity of the pendant groups; less electronegative groups prefer the equatorial positions. There are other similar molecules for which explanations (or predictions) are difficult. For example, the series of molecules below seems to have completely random placement of F and CF3 groups. These structures could not be predicted based on VSEPR theory. It seems that in this case, the more symmetrical arrangements of pendant groups is preferred.
3.2.04: Ligand Close Packing
Introduction
Ligand Close Packing (LCP) theory is complimentary to VSEPR, except that LCP focuses on repulsions between pendant atoms ("outer" atoms that are not directly bonded to one another), rather than focusing on the chemical environment around the central atom in a molecule. Both LCP and the VSEPR models were developed by Robert Gillespie.
The LCP model assumes that ligands (or pendant atoms) "pack" as closely as possible around a central atom. A given ligand will have a specific atomic radius when bound to a central atom. The distance between two ligands around the same central atom is simply the sum of the atomic radii of the pendant atoms. Therefore for a specific set of ligands around a specific central atom, the ligand-ligand distances are constant despite the coordination number or bond angles. In other words, for a series of similar molecules with the same central atom, while bond angles and bond distances of pendant atoms may change, the distances between two pendant atoms remain the same.
Ligand Close Packing Theory
According to LCP, distances between two pendant atoms are similar...
• even when steric number changes
• even when bond angles change
• Reason: pendant atoms "pack" around the central atom.
Ligand Close Packing and Bond Distances
If the distance between non-bonded atoms remains constant even while bond angles change, then the bond length between the pendant and central atom must change to accommodate. For example, $\ce{PF4+}$ and $\ce{PF3O}$ have almost identical $\ce{F - F}$ distances of 238 and 237 pm, respectively. As expected from VSEPR, the $\ce{F - P - F}$ bond angle in $\ce{PF4+}$ is 109.5°. The $\ce{F - P - F}$ bond angle in $\ce{PF3O}$ (101.1°) is smaller due to the increased repulsion of the oxygen double bond. Therefore the P-F bond lengths must be different. In fact, the P-F bond distances are 145.7 pm in $\ce{PF4+}$ and 154 pm in $\ce{PF3O}$.
The relationship between bond angles, bond lengths, and close-packing distance is described by right angle trigonometry (recall SOA-CAH-TOA). In a right triangle, the hypotenuse is the longest side, opposite to the right angle. For either of the other two angles in the triangle, the angle ($\theta$) and lengths of the sides (opposite and adjacent to $\theta$) are related by the trigometric functions shown in Figure $2$ .
In the case of a molecule, we can apply right angle trigonometry by imagining that the bond angle is divided into two right angles, as illustrated in Figure $2$. The hypotenuse of the triangles is the bond length, the opposite side is one half of the close-packing distance, and $\theta$ is one half the bond angle.
Exercise $1$
The distance between F atoms is 212 pm in both $\ce{NF4+}$ and $\ce{NF3}$. The bond angles are 109.5° and 102.3°, respectively. What are the expected $\ce{N - F}$ bond lengths?
Answer
The actual measured bond lengths are 130 pm for $\ce{NF4+}$ and 136.5 pm for $\ce{NF3}$. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/03%3A_Simple_Bonding_Theory/3.02%3A_Valence_Shell_Electron-Pair_Repulsion/3.2.03%3A_Electronegativity_and_Atomic_Size_Effects.txt |
Dipole moments occur when there is a separation of charge. They can occur between two ions in an ionic bond or between atoms in a covalent bond; dipole moments arise from differences in electronegativity. The larger the difference in electronegativity, the larger the dipole moment. The distance between the charge separation is also a deciding factor in the size of the dipole moment. The dipole moment is a measure of the polarity of the molecule.
Introduction
When atoms in a molecule share electrons unequally, they create what is called a dipole moment. This occurs when one atom is more electronegative than another, resulting in that atom pulling more tightly on the shared pair of electrons, or when one atom has a lone pair of electrons and the difference of electronegativity vector points in the same way. One of the most common examples is the water molecule, made up of one oxygen atom and two hydrogen atoms. The differences in electronegativity and lone electrons give oxygen a partial negative charge and each hydrogen a partial positive charge.
Dipole Moment
When two electrical charges, of opposite sign and equal magnitude, are separated by a distance, an electric dipole is established. The size of a dipole is measured by its dipole moment ($\mu$). Dipole moment is measured in Debye units, which is equal to the distance between the charges multiplied by the charge (1 Debye equals $3.34 \times 10^{-30}\; C\, m$). The dipole moment of a molecule can be calculated by Equation $\ref{1}$:
$\vec{\mu} = \sum_i q_i \, \vec{r}_i \label{1}$
where
• $\vec{\mu}$ is the dipole moment vector
• $q_i$ is the magnitude of the $i^{th}$ charge, and
• $\vec{r}_i$ is the vector representing the position of $i^{th}$ charge.
The dipole moment acts in the direction of the vector quantity. An example of a polar molecule is $\ce{H_2O}$. Because of the lone pair on oxygen, the structure of $\ce{H_2O}$ is bent (via VSEPR theory), which means that the vectors representing the dipole moment of each bond do not cancel each other out. Hence, water is polar.
The vector points from positive to negative, on both the molecular (net) dipole moment and the individual bond dipoles. Table A2 shows the electronegativity of some of the common elements. The larger the difference in electronegativity between the two atoms, the more electronegative that bond is. To be considered a polar bond, the difference in electronegativity must be large. The dipole moment points in the direction of the vector quantity of each of the bond electronegativities added together.
It is relatively easy to measure dipole moments: just place a substance between charged plates (Figure $2$); polar molecules increase the charge stored on the plates, and the dipole moment can be obtained (i.e., via the capacitance of the system). Nonpolar $\ce{CCl_4}$ is not deflected; moderately polar acetone deflects slightly; highly polar water deflects strongly. In general, polar molecules will align themselves: (1) in an electric field, (2) with respect to one another, or (3) with respect to ions (Figure $2$).
Equation $\ref{1}$ can be simplified for a simple separated two-charge system like diatomic molecules or when considering a bond dipole within a molecule
$\mu_{diatomic} = Q \times r \label{1a}$
This bond dipole is interpreted as the dipole from a charge separation over a distance $r$ between the partial charges $Q^+$ and $Q^-$ (or the more commonly used terms $δ^+$ - $δ^-$); the orientation of the dipole is along the axis of the bond. Consider a simple system of a single electron and proton separated by a fixed distance. When the proton and electron are close together, the dipole moment (degree of polarity) decreases. However, as the proton and electron get farther apart, the dipole moment increases. In this case, the dipole moment is calculated as (via Equation $\ref{1a}$):
\begin{align*} \mu &= Qr \nonumber \[4pt] &= (1.60 \times 10^{-19}\, C)(1.00 \times 10^{-10} \,m) \nonumber \[4pt] &= 1.60 \times 10^{-29} \,C \cdot m \label{2} \end{align*}
The Debye characterizes the size of the dipole moment. When a proton and electron are 100 pm apart, the dipole moment is $4.80\; D$:
\begin{align*} \mu &= (1.60 \times 10^{-29}\, C \cdot m) \left(\dfrac{1 \;D}{3.336 \times 10^{-30} \, C \cdot m} \right) \nonumber \[4pt] &= 4.80\; D \label{3} \end{align*}
$4.80\; D$ is a key reference value and represents a pure charge of +1 and -1 separated by 100 pm. If the charge separation is increased then the dipole moment increases (linearly):
• If the proton and electron are separated by 120 pm: $\mu = \dfrac{120}{100}(4.80\;D) = 5.76\, D \label{4a}$
• If the proton and electron are separated by 150 pm: $\mu = \dfrac{150}{100}(4.80 \; D) = 7.20\, D \label{4b}$
• If the proton and electron are separated by 200 pm: $\mu = \dfrac{200}{100}(4.80 \; D) = 9.60 \,D \label{4c}$
Example $1$: Water
The water molecule in Figure $1$ can be used to determine the direction and magnitude of the dipole moment. From the electronegativities of oxygen and hydrogen, the difference in electronegativity is 1.2e for each of the hydrogen-oxygen bonds. Next, because the oxygen is the more electronegative atom, it exerts a greater pull on the shared electrons; it also has two lone pairs of electrons. From this, it can be concluded that the dipole moment points from between the two hydrogen atoms toward the oxygen atom. Using the equation above, the dipole moment is calculated to be 1.85 D by multiplying the distance between the oxygen and hydrogen atoms by the charge difference between them and then finding the components of each that point in the direction of the net dipole moment (the angle of the molecule is 104.5˚).
The bond moment of the O-H bond =1.5 D, so the net dipole moment is
$\mu=2(1.5) \cos \left(\dfrac{104.5˚}{2}\right)=1.84\; D \nonumber$
Polarity and Structure of Molecules
The shape of a molecule and the polarity of its bonds determine the OVERALL POLARITY of that molecule. A molecule that contains polar bonds might not have any overall polarity, depending upon its shape. The simple definition of whether a complex molecule is polar or not depends upon whether its overall centers of positive and negative charges overlap. If these centers lie at the same point in space, then the molecule has no overall polarity (and is non polar). If a molecule is completely symmetric, then the dipole moment vectors on each molecule will cancel each other out, making the molecule nonpolar. A molecule can only be polar if the structure of that molecule is not symmetric.
A good example of a nonpolar molecule that contains polar bonds is carbon dioxide (Figure $\PageIndex{3a}$). This is a linear molecule and each C=O bond is, in fact, polar. The central carbon will have a net positive charge, and the two outer oxygen atoms a net negative charge. However, since the molecule is linear, these two bond dipoles cancel each other out (i.e. the vector addition of the dipoles equals zero) and the overall molecule has a zero dipole moment ($\mu=0$).
Although a polar bond is a prerequisite for a molecule to have a dipole, not all molecules with polar bonds exhibit dipoles
For $AB_n$ molecules, where $A$ is the central atom and $B$ are all the same types of atoms, there are certain molecular geometries which are symmetric. Therefore, they will have no dipole even if the bonds are polar. These geometries include linear, trigonal planar, tetrahedral, octahedral and trigonal bipyramid.
Example $3$: $\ce{C_2Cl_4}$
Although the C–Cl bonds are rather polar, the individual bond dipoles cancel one another in this symmetrical structure, and $\ce{Cl_2C=CCl_2}$ does not have a net dipole moment.
Example $3$: $\ce{CH_3Cl}$
C-Cl, the key polar bond, is 178 pm. Measurement reveals 1.87 D. From this data, % ionic character can be computed. If this bond were 100% ionic (based on proton & electron),
\begin{align*} \mu &= \dfrac{178}{100}(4.80\; D) \nonumber \[4pt] &= 8.54\; D \nonumber \end{align*}
Although the bond length is increasing, the dipole is decreasing as you move down the halogen group. The electronegativity decreases as we move down the group. Thus, the greater influence is the electronegativity of the two atoms (which influences the charge at the ends of the dipole).
Table $1$: Relationship between Bond length, Electronegativity and Dipole moments in simple Diatomics
Compound Bond Length (Å) Electronegativity Difference Dipole Moment (D)
HF 0.92 1.9 1.82
HCl 1.27 0.9 1.08
HBr 1.41 0.7 0.82
HI 1.61 0.4 0.44 | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/03%3A_Simple_Bonding_Theory/3.03%3A_Molecular_Polarity.txt |
A hydrogen bond is an intermolecular force (IMF) that forms a special type of dipole-dipole attraction when a hydrogen atom bonded to a strongly electronegative atom exists in the vicinity of another electronegative atom with a lone pair of electrons. Intermolecular forces (IMFs) occur between molecules. Other examples include ordinary dipole-dipole interactions and dispersion forces. Hydrogen bonds are are generally stronger than ordinary dipole-dipole and dispersion forces, but weaker than true covalent and ionic bonds.
The evidence for hydrogen bonding
Many elements form compounds with hydrogen. If you plot the boiling points of the compounds of the group 14 elements with hydrogen, you find that the boiling points increase as you go down the group.
The increase in boiling point happens because the molecules are getting larger with more electrons, and so van der Waals dispersion forces become greater. If you repeat this exercise with the compounds of the elements in groups 15, 16, and 17 with hydrogen, something odd happens.
Although the same reasoning applies for group 4 of the periodic table, the boiling point of the compound of hydrogen with the first element in each group is abnormally high. In the cases of $NH_3$, $H_2O$ and $HF$ there must be some additional intermolecular forces of attraction, requiring significantly more heat energy to break the IMFs. These relatively powerful intermolecular forces are described as hydrogen bonds.
Origin of Hydrogen Bonding
The molecules capable of hydrogen bonding include the following:
Notice that in each of these molecules:
• The hydrogen is attached directly to a highly electronegative atoms, causing the hydrogen to acquire a highly positive charge.
• Each of the highly electronegative atoms attains a high negative charge and has at least one "active" lone pair. Lone pairs at the 2-level have electrons contained in a relatively small volume of space, resulting in a high negative charge density. Lone pairs at higher levels are more diffuse and, resulting in a lower charge density and lower affinity for positive charge.
If you are not familiar with electronegativity, you should follow this link before you go on.
Consider two water molecules coming close together.
The $\delta^+$ hydrogen is so strongly attracted to the lone pair that it is almost as if you were beginning to form a co-ordinate (dative covalent) bond. It doesn't go that far, but the attraction is significantly stronger than an ordinary dipole-dipole interaction. Hydrogen bonds have about a tenth of the strength of an average covalent bond, and are constantly broken and reformed in liquid water. If you liken the covalent bond between the oxygen and hydrogen to a stable marriage, the hydrogen bond has "just good friends" status.
Water is an ideal example of hydrogen bonding. Notice that each water molecule can potentially form four hydrogen bonds with surrounding water molecules: two with the hydrogen atoms and two with the with the oxygen atoms. There are exactly the right numbers of $\delta^+$ hydrogens and lone pairs for every one of them to be involved in hydrogen bonding.
This is why the boiling point of water is higher than that of ammonia or hydrogen fluoride. In the case of ammonia, the amount of hydrogen bonding is limited by the fact that each nitrogen only has one lone pair. In a group of ammonia molecules, there are not enough lone pairs to go around to satisfy all the hydrogens. In hydrogen fluoride, the problem is a shortage of hydrogens. In water, two hydrogen bonds and two lone pairs allow formation of hydrogen bond interactions in a lattice of water molecules. Water is thus considered an ideal hydrogen bonded system.
More complex examples of hydrogen bonding
The hydration of negative ions
When an ionic substance dissolves in water, water molecules cluster around the separated ions. This process is called hydration. Water frequently attaches to positive ions by co-ordinate (dative covalent) bonds. It bonds to negative ions using hydrogen bonds.
If you are interested in the bonding in hydrated positive ions, you could follow this link to co-ordinate (dative covalent) bonding.
The diagram shows the potential hydrogen bonds formed with a chloride ion, Cl-. Although the lone pairs in the chloride ion are at the 3-level and would not normally be active enough to form hydrogen bonds, they are made more attractive by the full negative charge on the chlorine in this case.
However complicated the negative ion, there will always be lone pairs that the hydrogen atoms from the water molecules can hydrogen bond to.
Hydrogen bonding in alcohols
An alcohol is an organic molecule containing an -OH group. Any molecule which has a hydrogen atom attached directly to an oxygen or a nitrogen is capable of hydrogen bonding. Hydrogen bonds also occur when hydrogen is bonded to fluorine, but the HF group does not appear in other molecules. Molecules with hydrogen bonds will always have higher boiling points than similarly sized molecules which don't have an -O-H or an -N-H group. The hydrogen bonding makes the molecules "stickier," such that more heat (energy) is required to separate them. This phenomenon can be used to analyze boiling point of different molecules, defined as the temperature at which a phase change from liquid to gas occurs.
Ethanol, $\ce{CH3CH2-O-H}$, and methoxymethane, $\ce{CH3-O-CH3}$, both have the same molecular formula, $\ce{C2H6O}$.
They have the same number of electrons, and a similar length. The van der Waals attractions (both dispersion forces and dipole-dipole attractions) in each will be similar. However, ethanol has a hydrogen atom attached directly to an oxygen; here the oxygen still has two lone pairs like a water molecule. Hydrogen bonding can occur between ethanol molecules, although not as effectively as in water. The hydrogen bonding is limited by the fact that there is only one hydrogen in each ethanol molecule with sufficient + charge.
In methoxymethane, the lone pairs on the oxygen are still there, but the hydrogens are not sufficiently + for hydrogen bonds to form. Except in some rather unusual cases, the hydrogen atom has to be attached directly to the very electronegative element for hydrogen bonding to occur. The boiling points of ethanol and methoxymethane show the dramatic effect that the hydrogen bonding has on the stickiness of the ethanol molecules:
ethanol (with hydrogen bonding) 78.5°C
methoxymethane (without hydrogen bonding) -24.8°C
The hydrogen bonding in the ethanol has lifted its boiling point about 100°C. It is important to realize that hydrogen bonding exists in addition to van der Waals attractions. For example, all the following molecules contain the same number of electrons, and the first two have similar chain lengths. The higher boiling point of the butan-1-ol is due to the additional hydrogen bonding.
Comparing the two alcohols (containing -OH groups), both boiling points are high because of the additional hydrogen bonding; however, the values are not the same. The boiling point of the 2-methylpropan-1-ol isn't as high as the butan-1-ol because the branching in the molecule makes the van der Waals attractions less effective than in the longer butan-1-ol.
Hydrogen bonding in organic molecules containing nitrogen
Hydrogen bonding also occurs in organic molecules containing N-H groups; recall the hydrogen bonds that occur with ammonia. Examples range from simple molecules like CH3NH2 (methylamine) to large molecules like proteins and DNA. The two strands of the famous double helix in DNA are held together by hydrogen bonds between hydrogen atoms attached to nitrogen on one strand, and lone pairs on another nitrogen or an oxygen on the other one.
Donors and Acceptors
In order for a hydrogen bond to occur there must be both a hydrogen donor and an acceptor present. The donor in a hydrogen bond is usually a strongly electronegative atom such as N, O, or F that is covalently bonded to a hydrogen bond.
The hydrogen acceptor is an electronegative atom of a neighboring molecule or ion that contains a lone pair that participates in the hydrogen bond.
Why does a hydrogen bond occur?
Since the hydrogen donor (N, O, or F) is strongly electronegative, it pulls the covalently bonded electron pair closer to its nucleus, and away from the hydrogen atom. The hydrogen atom is then left with a partial positive charge, creating a dipole-dipole attraction between the hydrogen atom bonded to the donor and the lone electron pair of the acceptor. This results in a hydrogen bond.(see Interactions Between Molecules With Permanent Dipoles)
Types of hydrogen bonds
Although hydrogen bonds are well-known as a type of IMF, these bonds can also occur within a single molecule, between two identical molecules, or between two dissimilar molecules.
Intramolecular hydrogen bonds
Intramolecular hydrogen bonds are those which occur within one single molecule. This occurs when two functional groups of a molecule can form hydrogen bonds with each other. In order for this to happen, both a hydrogen donor a hydrogen acceptor must be present within one molecule, and they must be within close proximity of each other in the molecule. For example, intramolecular hydrogen bonding occurs in ethylene glycol (C2H4(OH)2) between its two hydroxyl groups due to the molecular geometry.
Intermolecular hydrogen bonds
Intermolecular hydrogen bonds occur between separate molecules in a substance. They can occur between any number of like or unlike molecules as long as hydrogen donors and acceptors are present in positions where they can interact with one another. For example, intermolecular hydrogen bonds can occur between NH3 molecules alone, between H2O molecules alone, or between NH3 and H2O molecules.
Properties and effects of hydrogen bonds
On Boiling Point
When we consider the boiling points of molecules, we usually expect molecules with larger molar masses to have higher normal boiling points than molecules with smaller molar masses. This, without taking hydrogen bonds into account, is due to greater dispersion forces (see Interactions Between Nonpolar Molecules). Larger molecules have more space for electron distribution and thus more possibilities for an instantaneous dipole moment. However, when we consider the table below, we see that this is not always the case.
Compound Molar Mass Normal Boiling Point
$H_2O$ 18 g/mol 373 K
$HF$ 20 g/mol 292.5 K
$NH_3$ 17 g/mol 239.8 K
$H_2S$ 34 g/mol 212.9 K
$HCl$ 36.4 g/mol 197.9 K
$PH_3$ 34 g/mol 185.2 K
We see that H2O, HF, and NH3 each have higher boiling points than the same compound formed between hydrogen and the next element moving down its respective group, indicating that the former have greater intermolecular forces. This is because H2O, HF, and NH3 all exhibit hydrogen bonding, whereas the others do not. Furthermore, $H_2O$ has a smaller molar mass than HF but partakes in more hydrogen bonds per molecule, so its boiling point is higher.
On Viscosity
The same effect that is seen on boiling point as a result of hydrogen bonding can also be observed in the viscosity of certain substances. Substances capable of forming hydrogen bonds tend to have a higher viscosity than those that do not form hydrogen bonds. Generally, substances that have the possibility for multiple hydrogen bonds exhibit even higher viscosities.
Factors preventing Hydrogen bonding
Electronegativity
Hydrogen bonding cannot occur without significant electronegativity differences between hydrogen and the atom it is bonded to. Thus, we see molecules such as PH3, which do not participate in hydrogen bonding. PH3 exhibits a trigonal pyramidal molecular geometry like that of ammonia, but unlike NH3 it cannot hydrogen bond. This is due to the similarity in the electronegativities of phosphorous and hydrogen. Both atoms have an electronegativity of 2.1, and thus, there is no dipole moment. This prevents the hydrogen atom from acquiring the partial positive charge needed to hydrogen bond with the lone electron pair in another molecule. (see Polarizability)
Atom Size
The size of donors and acceptors can also affect the ability to hydrogen bond. This can account for the relatively low ability of Cl to form hydrogen bonds. When the radii of two atoms differ greatly or are large, their nuclei cannot achieve close proximity when they interact, resulting in a weak interaction.
Hydrogen Bonding in Nature
Hydrogen bonding plays a crucial role in many biological processes and can account for many natural phenomena such as the Unusual properties of Water. In addition to being present in water, hydrogen bonding is also important in the water transport system of plants, secondary and tertiary protein structure, and DNA base pairing.
Plants
The cohesion-adhesion theory of transport in vascular plants uses hydrogen bonding to explain many key components of water movement through the plant's xylem and other vessels. Within a vessel, water molecules hydrogen bond not only to each other, but also to the cellulose chain that comprises the wall of plant cells. Since the vessel is relatively small, the attraction of the water to the cellulose wall creates a sort of capillary tube that allows for capillary action. This mechanism allows plants to pull water up into their roots. Furthermore, hydrogen bonding can create a long chain of water molecules, which can overcome the force of gravity and travel up to the high altitudes of leaves.
Proteins
Hydrogen bonding is present abundantly in the secondary structure of proteins, and also sparingly in tertiary conformation. The secondary structure of a protein involves interactions (mainly hydrogen bonds) between neighboring polypeptide backbones which contain nitrogen-hydrogen bonded pairs and oxygen atoms. Since both N and O are strongly electronegative, the hydrogen atoms bonded to nitrogen in one polypeptide backbone can hydrogen bond to the oxygen atoms in another chain and vice-versa. Though they are relatively weak, these bonds offer substantial stability to secondary protein structure because they repeat many times and work collectively.
In tertiary protein structure, interactions are primarily between functional R groups of a polypeptide chain; one such interaction is called a hydrophobic interaction. These interactions occur because of hydrogen bonding between water molecules around the hydrophobe that further reinforces protein conformation. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/03%3A_Simple_Bonding_Theory/3.04%3A_Hydrogen_Bonding.txt |
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Introduction
The symmetry of a molecule consists of symmetry operations and symmetry elements. A symmetry operation is an operation that is performed to a molecule which leaves it indistinguishable and superimposable on the original position. Symmetry operations are performed with respect to symmetry elements (points, lines, or planes).
An example of a symmetry operation is a 180° rotation of a water molecule in which the resulting position of the molecule is indistinguishable from the original position (see Figure $1$). In this example, the symmetry operation is the rotation and the symmetry element is the axis of rotation.
There are five types of symmetry operations including identity, reflection, inversion, proper rotation, and improper rotation. The improper rotation is the sum of a rotation followed by a reflection. The symmetry elements that correspond to the five types of symmetry operations are listed in Table $1$.
Table $1$: Table of elements and operations
Element Operation Symbol
Identity identity E
Proper axis rotation by (360/n)o Cn
Symmetry plane reflection in the plane σ
Inversion center inversion of a point at (x,y,z) to (-x,-y,-z) i
Improper axis rotation by (360/n)o, followed by reflection in the plane perpendicular to the rotation axis Sn
Symmetry Operations and Elements
Identity (E)
All molecules have the identity element. The identity operation is doing nothing to the molecule (it doesn't rotate, reflect, or invert...it just is).
Proper Rotation and Proper Axis (Cn)
A "proper" rotation is just a simple rotation operation about an axis. The symbol for any proper rotation or proper axis is C(360/n), where n is the degree of rotation. Thus, a 180° rotation is a C2 rotation around a C2 axis, and a 120° rotation is a C3 rotation about a C3 axis.
PRINCIPLE AXIS: The principle axis of a molecule is the highest order proper rotation axis. For example, if a molecule had C2 and C4 axes, the C4 is the principle axis.
Reflection and Symmetry Planes (σ)
Symmetry planes are mirror planes within the molecule. A reflection operation occurs with respect to a plane of symmetry. There are three classes of symmetry elements:
1. σh (horizontal): horizontal planes are perpendicular to principal axis
2. σv (vertical): vertical planes are parallel to the principal axis
3. σd (dihedral): dihedral planes are parallel to the principle axis and bisecting two C2' axes
Inversion and Inversion Center (i)
The inversion operation requires a point of symmetry (a center of symmetry within a molecule). In other words, a point at the center of the molecule that can transform (x,y,z) into (-x,-y,-z) coordinate. Structures of tetrahedrons, triangles, and pentagons lack an inversion center.
Improper rotation (Sn)
Improper rotation is a combination of a rotation with respect to an axis of rotation (Cn), followed by a reflection through a plane perpendicular to that Cn axis. In short, an Sn operation is equivalent to Cn followed by $\sigma_h$.
4.2.01: Groups of Low and High Symmetry
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Introduction
A Point Group describes all the symmetry operations that can be performed on a molecule that result in a conformation indistinguishable from the original. Point groups are used in Group Theory, the mathematical analysis of groups, to determine properties such as a molecule's molecular orbitals.
Assigning Point Groups
While a point group contains all of the symmetry operations that can be performed on a given molecule, it is not necessary to identify all of these operations to determine the molecule's overall point group. Instead, a molecule's point group can be determined by following a set of steps which analyze the presence (or absence) of particular symmetry elements.
Steps for assigning a molecule's point group:
1. Determine if the molecule is of high or low symmetry.
2. If not, find the highest order rotation axis, Cn.
3. Determine whether the molecule has any C2 axes perpendicular to the principal Cn axis. If so, then there are n such C2 axes, and the molecule is in the D set of point groups. If not, it is in either the C or S set of point groups.
4. Determine whether the molecule has a horizontal mirror plane (σh) perpendicular to the principal Cn axis. If so, the molecule is either in the Cnh or Dnh set of point groups.
5. Determine whether the molecule has a vertical mirror plane (σv) containing the principal Cn axis. If so, the molecule is either in the Cnv or Dnd set of point groups. If not, and if the molecule has n perpendicular C2 axes, then it is part of the Dn set of point groups.
6. Determine whether there is an improper rotation axis, S2n, collinear with the principal Cn axis. If so, the molecule is in the S2n point group. If not, the molecule is in the Cn point group.
Example \(1\)
Find the point group of benzene (C6H6).
Answer
Solution
1. Benzene is neither high nor low symmetry
2. Highest order rotation axis: C6
3. There are 6 C2 axes perpendicular to the principal axis
4. There is a horizontal mirror plane (σh)
Benzene is in the D6h point group.
Symmetry Gallery
4.02: Point Groups
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Low Symmetry Point Groups
Low symmetry point groups include the C1, Cs, and Ci groups
Group Description Example
C1 only the identity operation (E) CHFClBr
Cs only the identity operation (E) and one mirror plane C2H2ClBr
Ci only the identity operation (E) and a center of inversion (i) C2H2Cl2Br
High Symmetry Point Groups
High symmetry point groups include the Td, Oh, Ih, C∞v, and D∞h groups. The table below describes their characteristic symmetry operations. The full set of symmetry operations included in the point group is described in the corresponding character table.
Group Description Example
C∞v linear molecule with an infinite number of rotation axes and vertical mirror planes (σv) HBr
D∞h linear molecule with an infinite number of rotation axes, vertical mirror planes (σv), perpendicular C2 axes, a horizontal mirror plane (σh), and an inversion center (i) CO2
Td typically have tetrahedral geometry, with 4 C4 axes, 3 C2 axes, 3 S4 axes, and 6 dihedral mirror planes (σd) CH4
Oh typically have octahedral geometry, with 3 C4 axes, 4 C3 axes, and an inversion center (i) as characteristic symmetry operations SF6
Ih typically have an icosahedral structure, with 6 C5 axes as characteristic symmetry operations B12H122-
Symmetry Gallery
4.2.02: Other Groups
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D Groups
The D set of point groups is classified as Dnh, Dnd, or Dn, where n refers to the principal axis of rotation. Overall, the D groups are characterized by the presence of n C2 axes perpendicular to the principal Cn axis. Further classification of a molecule in the D groups depends on the presence of horizontal or vertical/dihedral mirror planes.
Group Description Example
Dnh n perpendicular C2 axes, and a horizontal mirror plane (σh) benzene, C6H6 is D6h
Dnd n perpendicular C2 axes, and a vertical mirror plane (σv) propadiene, C3H4 is D2d
Dn n perpendicular C2 axes, no mirror planes [Co(en)3]3+ is D3
C Groups
The C set of point groups is classified as Cnh, Cnv, or Cn, where n refers to the principal axis of rotation. The C set of groups is characterized by the absence of n C2 axes perpendicular to the principal Cn axis. Further classification of a molecule in the C groups depends on the presence of horizontal or vertical/dihedral mirror planes.
Group Description Example
Cnh horizontal mirror plane (σh) perpendicular to the principal Cn axis boric acid, H3BO3, is C3h
Cnv vertical mirror plane (σv) containing the principal Cn axis ammonia, NH3, is C3v
Cn no mirror planes P(C6H5)3 is C3
S Groups
The S set of point groups is classified as S2n, where n refers to the principal axis of rotation. The S set of groups is characterized by the absence of n C2 axes perpendicular to the principal Cn axis, as well as the absence of horizontal and vertical/dihedral mirror planes. However, there is an improper rotation (or a rotation-reflection) axis collinear with the principal Cn axis.
Group Description Example
S2n improper rotation (or a rotation-reflection) axis collinear with the principal Cn axis 12-crown-4 is S4
Symmetry Gallery | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/04%3A_Symmetry_and_Group_Theory/4.01%3A_Symmetry_Elements_and_Operations.txt |
Click here to see a lecture on this topic.
Group Multiplication
Now we will investigate what happens when we apply two symmetry operations in sequence. As an example, consider the $NH_3$ molecule, which belongs to the $C_{3v}$ point group. Consider what happens if we apply a $C_3$rotation (120˚ counter-clockwise) followed by a $\sigma_v$ reflection (reflection over the $\sigma_v$ axis). We write this combined operation $\sigma_v$$C_3$ (when written, symmetry operations operate on the thing directly to their right, just as operators do in quantum mechanics – we therefore have to work backwards from right to left from the notation to get the correct order in which the operators are applied). As we shall soon see, the order in which the operations are applied is important.
The combined operation $\sigma_v$$C_3$ is equivalent to $\sigma_v''$ (note the double prime on $\sigma_v''$!), which is also a symmetry operation of the $C_{3v}$ point group. Now let’s see what happens if we apply the operators in the reverse order, i.e., $C_3$$\sigma_v$ is ($\sigma_v$ followed by $C_3$).
Again, the combined operation $C_3$$\sigma_v$ is equivalent to another operation of the point group, this time $\sigma_v'$ (note the single prime on $\sigma_v'$!).
There are two important points that are illustrated by this example:
1. The order in which two operations are applied is important. For two symmetry operations $A$ and $B$, $AB$ is not necessarily the same as $BA$, i.e. symmetry operations do not in general commute. In some groups the symmetry elements do commute; such groups are said to be Abelian.
2. If two operations from the same point group are applied in sequence, the result will be equivalent to another operation from the point group. Symmetry operations that are related to each other by other symmetry operations of the group are said to belong to the same class. In $NH_3$, the three mirror planes $\sigma_v$, $\sigma_v'$ and $\sigma_v''$ belong to the same class (related to each other through a $C_3$ rotation), as do the rotations $C_3^+$ and $C_3^-$ (anticlockwise and clockwise rotations about the principal axis, related to each other by a vertical mirror plane).
Four Properties of Mathematical Groups
Now that we have explored some of the properties of symmetry operations and elements and their behavior within point groups, we are ready to introduce the formal mathematical definition of a group. The definitions below will be put into the context of molecular symmetry.
A mathematical group is defined as a set of elements ($A_1$, $A_2$, $A_3$...) together with a rule for forming combinations $A_i$,$A_j$... For our purposes, $A_1$, $A_2$, $A_3$, etc. are symmetry elements and $A_i$, $A_j$, etc. are symmetry operations described in a previous section. The elements of the group and the rule for combining them must satisfy the following four criteria.
1. The group must include the identity $E$, which commutes with other members of the group. In other terms, $E A_i= A_i$ for all the elements of the group. Application of the identity operation before or after another operation, $A_i$, results in the same outcome as $A_i$ alone.
2. The elements must satisfy the group property that the combination of any pair of elements is also an element of the group. For example, in the $C_{3v}$ point group, a C3 rotation followed by a $\sigma_v$ gives another operation that is already part of the group: a $\sigma_v"$.
3. Each symmetry operation $A_i$ must have an inverse $A_i^{-1}$, which is also an element of the group, such that $A_i A_i^{-1} = A_i^{-1}A_i = E \nonumber$ The inverse $g_i^{-1}$ effectively 'undoes’ the effect of the symmetry operation $g_i$. For example, in the $C_{3v}$ point group, the inverse of $C_3^+$ is $C_3^-$.
4. The rule of combination must be associative $(A_i A_j )(A_k) = A_i(A_jA_k) \nonumber$ Or $A(BC)=(AB)C$. In other words, the order of operations should not matter.
Group theory is an important area in mathematics, and luckily for chemists the mathematicians have already done most of the work for us. Along with the formal definition of a group comes a comprehensive mathematical framework that allows us to carry out a rigorous treatment of symmetry in molecular systems and learn about its consequences.
Many problems involving operators or operations (such as those found in quantum mechanics or group theory) may be reformulated in terms of matrices. Any of you who have come across transformation matrices before will know that symmetry operations such as rotations and reflections may be represented by matrices. It turns out that the set of matrices representing the symmetry operations in a group obey all the conditions laid out above in the mathematical definition of a group, and using matrix representations of symmetry operations simplifies carrying out calculations in group theory. Before we learn how to use matrices in group theory, it will probably be helpful to review some basic definitions and properties of matrices.
*This page was adapted from here (click).
4.03: Properties and Representations of Groups
The symmetry of molecules is essential for understanding the structures and properties of organic and inorganic compounds. The properties of chemical compounds are often easily explained by consideration of symmetry. For example, the symmetry of a molecule determines whether the molecule has a permanent dipole moment or not. The theories that describe optical activity, infrared and ultraviolet spectroscopy, and crystal structure involve the application of symmetry considerations. Matrix algebra is the most important mathematical tool in the description of symmetry.
The properties of symmetry groups are organized in character tables (discussed later in this chapter). Character tables are constructed based on matrices. This page is a brief description of matrices and matrix multiplication.
What is a matrix?
An $m\times n$ matrix $\mathbf{A}$ is a rectangular array of numbers with $m$ rows and $n$ columns. The numbers $m$ and $n$ are the dimensions of $\mathbf{A}$. The numbers in the matrix are called its entries. The entry in row $i$ and column $j$ is called $a_{ij}$.
Some types of matrices have special names:
• A square matrix:$\begin{pmatrix} 3 &-2 &4 \ 5 &3i &3 \ -i & 1/2 &9 \end{pmatrix} \nonumber$ with $m=n$
• A rectangular matrix:$\begin{pmatrix} 3 &-2 &4 \ 5 &3i &3 \end{pmatrix}\nonumber$ with $m\neq n$
• A column vector:$\begin{pmatrix} 3 \ 5\ -i \end{pmatrix}\nonumber$ with $n=1$
• A row vector:$\begin{pmatrix} 3 &-2 &4 \ \end{pmatrix}\nonumber$ with $m=1$
• The identity matrix:$\begin{pmatrix} 1 &0 &0 \ 0 &1 &0 \ 0&0 &1 \end{pmatrix}\nonumber$ with $a_{ij}=\delta_{i,j}$, where $\delta_{i,j}$ is a function defined as $\delta_{i,j}=1$ if $i=j$ and $\delta_{i,j}=0$ if $i\neq j$.
• A diagonal matrix:$\begin{pmatrix} a &0 &0 \ 0 &b &0 \ 0&0 &c \end{pmatrix}\nonumber$ with $a_{ij}=c_i \delta_{i,j}$.
• An upper triangular matrix:$\begin{pmatrix} a &b &c \ 0 &d &e \ 0&0 &f \end{pmatrix}\nonumber$ All the entries below the main diagonal are zero.
• A lower triangular matrix:$\begin{pmatrix} a &0 &0 \ b &c &0 \ d&e &f \end{pmatrix}\nonumber$ All the entries above the main diagonal are zero.
• A triangular matrix is one that is either lower triangular or upper triangular.
The Trace of a Matrix
The trace of an $n\times n$ square matrix $\mathbf{A}$ is the sum of the diagonal elements, and formally defined as $Tr( \mathbf{A})=\sum_{i=1}^{n}a_{ii}$.
For example,
$\mathbf{A}=\begin{pmatrix} 3 &-2 &4 \ 5 &3i &3 \ -i & 1/2 &9 \end{pmatrix}\; ; Tr(\mathbf{A})=12+3i \nonumber$
Singular and Nonsingular Matrices
A square matrix with nonzero determinant is called nonsingular. A matrix whose determinant is zero is called singular. (Note that you cannot calculate the determinant of a non-square matrix).
For a 2x2 matrix,
$\mathbf{B}=\begin{pmatrix} a & b \ c & d \end{pmatrix}\; ; det(\mathbf{B})= ad - cb \nonumber$
For a 3x3 matrix,
$\mathbf{C}=\begin{pmatrix} a & b & c \ d & e & f \ g & h & i \end{pmatrix} \nonumber$
$det(\mathbf{C})=a\begin{pmatrix} e & f \ h & i \end{pmatrix} - b\begin{pmatrix} d & f \ g & i \end{pmatrix} + c\begin{pmatrix} d & e \ g & h \end{pmatrix} \nonumber$
$det(\mathbf{C})= aei - ahf -bdi + bgf + cdh - cge \nonumber$
The Matrix Transpose
The matrix transpose, most commonly written $\mathbf{A}^T$, is the matrix obtained by exchanging $\mathbf{A}$’s rows and columns. It is obtained by replacing all elements $a_{ij}$ with $a_{ji}$. For example:
$\mathbf{A}=\begin{pmatrix} 3 &-2 &4 \ 5 &3i &3 \end{pmatrix}\rightarrow \mathbf{A}^T=\begin{pmatrix} 3 &5\ -2 &3i\ 4&3 \end{pmatrix} \nonumber$
Matrix Multiplication
To multiply two matrices, the number of vertical columns in the first matrix must be the same as the number of rows in the second matrix. If $\mathbf{A}$ has dimensions $m\times n$ and $\mathbf{B}$ has dimensions $n\times p$, then the product $\mathbf{AB}$ is defined, and has dimensions $m\times p$.
$c_{ij} = \sum a_{ij} \times b_{ij} \nonumber$
The entry $a_{ij} \times b_{ij}$ is obtained by multiplying row $i$ of $\mathbf{A}$ by column $j$ of $\mathbf{B}$, which is done by multiplying corresponding entries together and then adding the results:
Example $1$
Calculate the product
$\begin{pmatrix} 1 &-2 &4 \ 5 &0 &3 \ 0 & 1/2 &9 \end{pmatrix}\begin{pmatrix} 1 &0 \ 5 &3 \ -1 &0 \end{pmatrix} \nonumber$
Solution
We need to multiply a $3\times 3$ matrix by a $3\times 2$ matrix, so we expect a $3\times 2$ matrix as a result.
$\begin{pmatrix} 1 &-2 &4 \ 5 &0 &3 \ 0 & 1/2 &9 \end{pmatrix}\begin{pmatrix} 1 &0 \ 5 &3 \ -1 &0 \end{pmatrix}=\begin{pmatrix} a&b \ c&d \ e &f \end{pmatrix} \nonumber$
To calculate $a$, which is entry (1,1), we use row 1 of the matrix on the left and column 1 of the matrix on the right:
$\begin{pmatrix} {\color{red}1} &{\color{red}-2} &{\color{red}4} \ 5 &0 &3 \ 0 & 1/2 &9 \end{pmatrix}\begin{pmatrix} {\color{red}1} &0 \ {\color{red}5} &3 \ {\color{red}-1} &0 \end{pmatrix}=\begin{pmatrix} {\color{red}a}&b \ c&d \ e &f \end{pmatrix}\rightarrow a=1\times 1+(-2)\times 5+4\times (-1)=-13 \nonumber$
To calculate $b$, which is entry (1,2), we use row 1 of the matrix on the left and column 2 of the matrix on the right:
$\begin{pmatrix} {\color{red}1} &{\color{red}-2} &{\color{red}4} \ 5 &0 &3 \ 0 & 1/2 &9 \end{pmatrix}\begin{pmatrix} 1&{\color{red}0} \ 5&{\color{red}3} \ -1&{\color{red}0} \end{pmatrix}=\begin{pmatrix} a&{\color{red}b} \ c&d \ e &f \end{pmatrix}\rightarrow b=1\times 0+(-2)\times 3+4\times 0=-6 \nonumber$
To calculate $c$, which is entry (2,1), we use row 2 of the matrix on the left and column 1 of the matrix on the right:
$\begin{pmatrix} 1&-2&4\ {\color{red}5} &{\color{red}0} &{\color{red}3} \ 0 & 1/2 &9 \end{pmatrix}\begin{pmatrix} {\color{red}1} &0 \ {\color{red}5} &3 \ {\color{red}-1} &0 \end{pmatrix}=\begin{pmatrix} a&b \ {\color{red}c}&d \ e &f \end{pmatrix}\rightarrow c=5\times 1+0\times 5+3\times (-1)=2 \nonumber$
To calculate $d$, which is entry (2,2), we use row 2 of the matrix on the left and column 2 of the matrix on the right:
$\begin{pmatrix} 1&-2&4\ {\color{red}5} &{\color{red}0} &{\color{red}3} \ 0 & 1/2 &9 \end{pmatrix}\begin{pmatrix} 1&{\color{red}0} \ 5&{\color{red}3} \ -1&{\color{red}0} \end{pmatrix}=\begin{pmatrix} a&b \ c&{\color{red}d} \ e &f \end{pmatrix}\rightarrow d=5\times 0+0\times 3+3\times 0=0 \nonumber$
To calculate $e$, which is entry (3,1), we use row 3 of the matrix on the left and column 1 of the matrix on the right:
$\begin{pmatrix} 1&-2&4\ 5&0&3 \ {\color{red}0} &{\color{red}1/2} &{\color{red}9} \end{pmatrix}\begin{pmatrix} {\color{red}1} &0 \ {\color{red}5} &3 \ {\color{red}-1} &0 \end{pmatrix}=\begin{pmatrix} a&b \ c&d \ {\color{red}e} &f \end{pmatrix}\rightarrow e=0\times 1+1/2\times 5+9\times (-1)=-13/2 \nonumber$
To calculate $f$, which is entry (3,2), we use row 3 of the matrix on the left and column 2 of the matrix on the right:
$\begin{pmatrix} 1&-2&4\ 5&0&3 \ {\color{red}0} &{\color{red}1/2} &{\color{red}9} \end{pmatrix}\begin{pmatrix} 1&{\color{red}0} \ 5&{\color{red}3} \ -1&{\color{red}0} \end{pmatrix}=\begin{pmatrix} a&b \ c&d \ e&{\color{red}f} \end{pmatrix}\rightarrow f=0\times 0+1/2\times 3+9\times 0=3/2 \nonumber$
The result is:
$\displaystyle{\color{Maroon}\begin{pmatrix} 1 &-2 &4 \ 5 &0 &3 \ 0 & 1/2 &9 \end{pmatrix}\begin{pmatrix} 1 &0 \ 5 &3 \ -1 &0 \end{pmatrix}=\begin{pmatrix} -13&-6 \ 2&0 \ -13/2 &3/2 \end{pmatrix}} \nonumber$
Example $2$
Calculate
$\begin{pmatrix} 1 &-2 &4 \ 5 &0 &3 \ \end{pmatrix}\begin{pmatrix} 1 \ 5 \ -1 \end{pmatrix}\nonumber$
Solution
We are asked to multiply a $2\times 3$ matrix by a $3\times 1$ matrix (a column vector). The result will be a $2\times 1$ matrix (a vector).
$\begin{pmatrix} 1 &-2 &4 \ 5 &0 &3 \ \end{pmatrix}\begin{pmatrix} 1 \ 5 \ -1 \end{pmatrix}=\begin{pmatrix} a \ b \end{pmatrix}\nonumber$
$a=1\times1+(-2)\times 5+ 4\times (-1)=-13\nonumber$
$b=5\times1+0\times 5+ 3\times (-1)=2\nonumber$
The solution is:
$\displaystyle{\color{Maroon}\begin{pmatrix} 1 &-2 &4 \ 5 &0 &3 \ \end{pmatrix}\begin{pmatrix} 1 \ 5 \ -1 \end{pmatrix}=\begin{pmatrix} -13 \ 2 \end{pmatrix}}\nonumber$
Need help? The link below contains solved examples: Multiplying matrices of different shapes (three examples): http://tinyurl.com/kn8ysqq
External links:
The Commutator
Matrix multiplication is not, in general, commutative. For example, we can perform
$\begin{pmatrix} 1 &-2 &4 \ 5 &0 &3 \ \end{pmatrix}\begin{pmatrix} 1 \ 5 \ -1 \end{pmatrix}=\begin{pmatrix} -13 \ 2 \end{pmatrix} \nonumber$
but cannot perform
$\begin{pmatrix} 1 \ 5 \ -1 \end{pmatrix}\begin{pmatrix} 1 &-2 &4 \ 5 &0 &3 \ \end{pmatrix} \nonumber$
Even with square matrices, that can be multiplied both ways, multiplication is not commutative. In this case, it is useful to define the commutator, defined as:
$[\mathbf{A},\mathbf{B}]=\mathbf{A}\mathbf{B}-\mathbf{B}\mathbf{A} \nonumber$
Example $3$
Given $\mathbf{A}=\begin{pmatrix} 3&1 \ 2&0 \end{pmatrix}$ and $\mathbf{B}=\begin{pmatrix} 1&0 \ -1&2 \end{pmatrix}$
Calculate the commutator $[\mathbf{A},\mathbf{B}]$
Solution
$[\mathbf{A},\mathbf{B}]=\mathbf{A}\mathbf{B}-\mathbf{B}\mathbf{A}\nonumber$
$\mathbf{A}\mathbf{B}=\begin{pmatrix} 3&1 \ 2&0 \end{pmatrix}\begin{pmatrix} 1&0 \ -1&2 \end{pmatrix}=\begin{pmatrix} 3\times 1+1\times (-1)&3\times 0 +1\times 2 \ 2\times 1+0\times (-1)&2\times 0+ 0\times 2 \end{pmatrix}=\begin{pmatrix} 2&2 \ 2&0 \end{pmatrix}\nonumber$
$\mathbf{B}\mathbf{A}=\begin{pmatrix} 1&0 \ -1&2 \end{pmatrix}\begin{pmatrix} 3&1 \ 2&0 \end{pmatrix}=\begin{pmatrix} 1\times 3+0\times 2&1\times 1 +0\times 0 \ -1\times 3+2\times 2&-1\times 1+2\times 0 \end{pmatrix}=\begin{pmatrix} 3&1 \ 1&-1 \end{pmatrix}\nonumber$
$[\mathbf{A},\mathbf{B}]=\mathbf{A}\mathbf{B}-\mathbf{B}\mathbf{A}=\begin{pmatrix} 2&2 \ 2&0 \end{pmatrix}-\begin{pmatrix} 3&1 \ 1&-1 \end{pmatrix}=\begin{pmatrix} -1&1 \ 1&1 \end{pmatrix}\nonumber$
$\displaystyle{\color{Maroon}[\mathbf{A},\mathbf{B}]=\begin{pmatrix} -1&1 \ 1&1 \end{pmatrix}}\nonumber$
Multiplication of a vector by a scalar
The multiplication of a vector $\vec{v_1}$ by a scalar $n$ produces another vector of the same dimensions that lies in the same direction as $\vec{v_1}$;
$n\begin{pmatrix} x \ y \end{pmatrix}=\begin{pmatrix} nx \ ny \end{pmatrix} \nonumber$
The scalar can stretch or compress the length of the vector, but cannot rotate it (figure [fig:vector_by_scalar]).
Multiplication of a square matrix by a vector
The multiplication of a vector $\vec{v_1}$ by a square matrix produces another vector of the same dimensions of $\vec{v_1}$. For example, we can multiply a $2\times 2$ matrix and a 2-dimensional vector:
$\begin{pmatrix} a&b \ c&d \end{pmatrix}\begin{pmatrix} x \ y \end{pmatrix}=\begin{pmatrix} ax+by \ cx+dy \end{pmatrix} \nonumber$
For example, consider the matrix
$\mathbf{A}=\begin{pmatrix} -2 &0 \ 0 &1 \end{pmatrix} \nonumber$
The product
$\begin{pmatrix} -2&0 \ 0&1 \end{pmatrix}\begin{pmatrix} x \ y \end{pmatrix} \nonumber$
is
$\begin{pmatrix} -2x \ y \end{pmatrix} \nonumber$
We see that $2\times 2$ matrices act as operators that transform one 2-dimensional vector into another 2-dimensional vector. This particular matrix keeps the value of $y$ constant and multiplies the value of $x$ by -2 (Figure $3$).
Notice that matrices are useful ways of representing operators that change the orientation and size of a vector. An important class of operators that are of particular interest to chemists are the so-called symmetry operators.
Attribution
This page was adapted from Matrices (click here), contributed by Marcia Levitus, Associate Professor (Biodesign Institute) at Arizona State University.
Curated or created by Kathryn Haas | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/04%3A_Symmetry_and_Group_Theory/4.03%3A_Properties_and_Representations_of_Groups/4.3.01%3A_Matrices.txt |
Symmetry Operations: Matrix Representations
A symmetry operation, such as a rotation around a symmetry axis or a reflection through a plane, is an operation that, when performed on an object, results in a new orientation of the object that is indistinguishable from the original. For example, if we rotate a square in the plane by $\pi/2$ or $\pi$, the new orientation of the square is superimposable on the original one (Figure $1$).
If rotation by an angle $\theta$ of a molecule (or object) about some axis results in an orientation of the molecule (or object) that is superimposable on the original, the axis is called a rotation axis. The molecule (or object) is said to have an $n$-fold rotational axis, where $n$ is $2\pi/\theta$. The axis is denoted as $C_n$. The square of Figure $1$ has a $C_4$ axis perpendicular to the plane because a $90^{\circ}$ rotation leaves the figure indistinguishable from the initial orientation. This axis is also a $C_2$ axis because a $180^{\circ}$ degree rotation leaves the square indistinguishable from the original square. In addition, the figure has several other $C_2$ axis that lie on the same plane as the square:
A symmetry operation moves all the points of the object from one initial position to a final position, and that means that symmetry operators are $3\times 3$ square matrices (or $2\times 2$ in two dimensions). Each symmetry operation can be expressed as a transformation matrix where the vector $(x',y',z')$ represents the new coordinates of the point $(x,y,z)$ after the symmetry operation.
$[\text{New Coordinates} (x',y',z')] = [\text{Transformation Matrix}] \times [\text{Old Coordinates} (x,y,z)] \nonumber$
We will use the example of water, which is in the $C_{2v}$ point group, to illustrate how transformation matrices can be used to represent the symmetry of a group.
Figure $2$ shows the three symmetry elements of the molecule of water (H$_2$O). This molecule has only one rotation axis, which is 2-fold, and therefore we call it a “$C_2$ axis.” It also has two mirror planes, one that contains the two hydrogen atoms ($\sigma_{yz}$), and another one perpendicular to it ($\sigma_{xz}$). Both planes contain the C$_2$ axis.
Transformation Matrix of $C_2$ rotation
A 2-fold rotation around the $z-$axis changes the location of a point $(x,y,z)$ to $(-x,-y,z)$ (see Figure $3$). By convention, rotations are always taken in the counterclockwise direction.
What is the matrix that represents the $C_2$ rotation? The matrix transforms the vector $(x,y,z)$ into $(-x,-y,z)$, so
${C_2}(x,y,z)=(-x,-y,z) \nonumber$
$\begin{pmatrix} a_{11}&a_{12}&a_{13} \ a_{21}&a_{22}&a_{23} \ a_{31}&a_{32}&a_{33} \end{pmatrix}\begin{pmatrix} x \ y \ z \end{pmatrix}=\begin{pmatrix} -x \ -y \ z \end{pmatrix} \nonumber$
We know the matrix is a $3\times 3$ square matrix because it needs to multiply a 3-dimensional vector. In addition, we write the vector as a vertical column to satisfy the requirements of matrix multiplication.
$\begin{pmatrix} a_{11}&a_{12}&a_{13} \ a_{21}&a_{22}&a_{23} \ a_{31}&a_{32}&a_{33} \end{pmatrix}\begin{pmatrix} x \ y \ z \end{pmatrix} \nonumber$
$a_{11}x+a_{12}y+a_{13}z=-x \nonumber$
$a_{21}x+a_{22}y+a_{23}z=-y \nonumber$
$a_{31}x+a_{32}y+a_{33}z=z \nonumber$
and we conclude that $a_{11}=-1$, $a_{12}=a_{13}=0$, $a_{22}=-1$, $a_{21}=a_{23}=0$ and $a_{33}=1$, $a_{31}=a_{32}=0$. The transformation matrix of the $C_2$ operation of the $C_{2v}$ point group is:
$C_2=\begin{pmatrix} -1&0&0 \ 0&-1&0 \ 0&0&1 \end{pmatrix} \nonumber$
Transformation Matrix of $\sigma_{xz}$ reflection
Rotations are not the only symmetry operations we can perform on a molecule. Figure $4$ illustrates the reflection of a point through the $xz$ plane. This operation transforms the vector $(x,y,z)$ into the vector $(x,-y,z)$. Symmetry operators involving reflections through a plane are usually denoted with the letter $\sigma$, so the operator that reflects a point through the $xz$ plane is $\hat{\sigma}_{xz}$:
$\sigma_{xz}(x,y,z)=(x,-y,z) \nonumber$
Following the same logic we used for the rotation matrix, we can write the $\sigma_{xz}$ transformation matrix as:
$\sigma_{x,z}=\begin{pmatrix} 1&0&0 \ 0&-1&0 \ 0&0&1 \end{pmatrix} \nonumber$
This is true because
$\begin{pmatrix} 1&0&0 \ 0&-1&0 \ 0&0&1 \end{pmatrix}\begin{pmatrix} x \ y \ z \end{pmatrix}=\begin{pmatrix} x \ -y \ z \end{pmatrix} \nonumber$
Exercise $1$
Find the transformation matrix of the identity (E) and the $\sigma_{y,z}$ operations under the $C_{2v}$ point group.
Answer
The transformation matrix for $E$ is $\begin{pmatrix} 1&0&0 \ 0&1&0 \ 0&0&1 \end{pmatrix} \nonumber$.
The transformation matrix for $\sigma_{v(yz)"}$ is $\begin{pmatrix} -1&0&0 \ 0&1&0 \ 0&0&1 \end{pmatrix} \nonumber$.
Characters
For a square matrix, the character is the trace of the matrix. For the $C_2$ operation, with the transformation matrix
$\begin{pmatrix} -1&0&0 \ 0&-1&0 \ 0&0&1 \end{pmatrix}, \nonumber$
the trace is $(-1) + (-1) + 1 = -1$.
The set of characters for a point group is called a reducible representation ($\Gamma$). The reducible representation for the $C_{2v}$ point group is:
$\begin{array}{l|llll} C_{2v} & E & C_2 & \sigma_v & \sigma_v' \ \hline \Gamma & 3 & -1 & 1 & 1 \end{array} \nonumber$
Exercise $1$
Prove that the characters in the reducible representation for $C_{2v}$ are correct:
$\begin{array}{l|llll} C_{2v} & E & C_2 & \sigma_v & \sigma_v' \ \hline \Gamma & 3 & -1 & 1 & 1 \end{array} \nonumber$
Answer
For the $E$ operation, with the transformation matrix $\begin{pmatrix} 1&0&0 \ 0&1&0 \ 0&0&1 \end{pmatrix} \nonumber$, the trace is $1 + 1 + 1 =$ $3$.
For the $C^z_2$ operation, with the transformation matrix $\begin{pmatrix} -1&0&0 \ 0&-1&0 \ 0&0&1 \end{pmatrix} \nonumber$, the trace is $(-1) + (-1) + 1 =$ $-1$.
For the $\sigma_{v(xz)'}$ operation, with the transformation matrix $\begin{pmatrix} 1&0&0 \ 0&-1&0 \ 0&0&1 \end{pmatrix} \nonumber$, the trace is $1 + (-1) + 1 =$ $1$.
For the $\sigma_{v(yz)"}$ operation, with the transformation matrix $\begin{pmatrix} -1&0&0 \ 0&1&0 \ 0&0&1 \end{pmatrix} \nonumber$, the trace is $-1 + 1 + 1 =$ $1$.
This gives the reducible representation $\begin{array}{l|llll} C_{2v} & E & C_2 & \sigma_v & \sigma_v' \ \hline \Gamma & 3 & -1 & 1 & 1 \end{array}$
Reducible and Irreducible Representations
Let us now go back and look in more detail at the transformation matrices of the $C_{2v}$ point group that we derived above. If we look at the matrices carefully we see that they all take the same block diagonal form (a square matrix is said to be block diagonal if all the elements are zero except for a set of submatrices lying along the diagonal).
$E=\begin{pmatrix} {\color{red}[1]}&0&0 \ 0&{\color{blue}[1]}&0 \ 0&0&{\color{green}[1]} \end{pmatrix}, \, C_2=\begin{pmatrix} {\color{red}[-1]}&0&0 \ 0&{\color{blue}[-1]}&0 \ 0&0&{\color{green}[1]} \end{pmatrix} ,\, \sigma_{v(xz)}'=\begin{pmatrix} {\color{red}[1]}&0&0 \ 0&{\color{blue}[-1]}&0 \ 0&0&{\color{green}[1]} \end{pmatrix} ,\, \sigma_{v(yz)"}=\begin{pmatrix} \color{red}[-1]&0&0 \ 0&{\color{blue}[1]}&0 \ 0&0&{\color{green}[1]} \end{pmatrix} \nonumber$
All the non-zero elements become 1x1 matrices that each represent individual ${\color{red}x}, {\color{blue}y}, {\color{green}z}$ coordinates. In other words, the element $\color{red}a_{11}$ represents $\color{red}x$, $\color{blue}a_{22}$ represents $\color{blue}y$, and $\color{green}a_{33}$ represents $\color{green}z$. The matrix elements for x from each transformation matrix combine to form an irreducible representation of the $C_{2v}$ point group. Likewise, the matrix elements for y combine to form a second irreducible representation, and the same is true for z elements. These irreducible representations are shown below:
$\begin{array}{l|llll|l} C_{2v} & E & C_2 & \sigma_v & \sigma_v' & \text{Coordinate Used}\ \hline & \color{red}1 & \color{red}-1&\color{red}1&\color{red}-1 & \color{red}x \ & \color{blue}1 & \color{blue}-1 & \color{blue}-1 & \color{blue}1 & \color{blue}y \ & \color{green}1 & \color{green}1 & \color{green}1 & \color{green}1 & \color{green}z\ \hline \Gamma & 3 & -1 & 1 & 1 \end{array} \nonumber$
The irreducible representations add to form the reducible representation, $\Gamma$. This $\Gamma$, which is the set of 3x3 matrices, can be reduced to the set of 1x1 matrices of the irreducible representations. The irreducible representations cannot be reduced further, hence their name. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/04%3A_Symmetry_and_Group_Theory/4.03%3A_Properties_and_Representations_of_Groups/4.3.02%3A_Representations_of_Point_Groups.txt |
Introduction to Character Tables, using $C_{2v}$ as example
A character table is the complete set of irreducible representations of a symmetry group. In the previous section, we derived three of the four irreducible representations for the $C_{2v}$ point group. These three irreducible representations are labeled $A_1$, $B_1$, and $B_2$. The fourth irreducible representation, $A_2$, can be derived using the properties (or "rules") for irreducible representations listed below.
Properties of Characters of Irreducible Representations in Point Groups
1. There is always a totally symmetric representation in which all the characters are 1.
e.g. In $C_{2v}$, $A_1$ is totally symmetric.
2. The order of the group ($h$) is the total number of symmetry operations in the group.
e.g. In $C_{2v}$, $h=4$
3. Similar operations are listed as classes (R) and appear as columns in the table.
e.g. In $C_{2v}$, there are four classes of operations, $E$, $C_2$, $\sigma_{v(xz)}$, and $\sigma_{v(yz)}'$
4. The number of irreducible representations (rows) must equal the number of classes (columns). This results in all character tables being square.
e.g. In $C_{2v}$, there are four classes and four irreducible representations.
5. The sum of squares of all characters under $E$ is equal to the order of the group: $h = \sum [\chi_i]^2$
e.g. In $C_{2v}$, $h = 1^2 + 1^2 +1^2 +1^2 = 4$
6. For any irreducible representation ($i$), the sum of squares of its characters multiplied by the number of operations in the class is the order of the group: $h = \sum [\chi_i(R)]^2$
e.g. For $A_2$ in $C_{2v}$, $h = (1\times 1)^2 + (1\times 1)^2 +(-1\times 1)^2 +(-1\times 1)^2 = 4$
7. Irreducible representations are orthogonal. For any two representations ($i$ and $j$): $\sum [\chi_i*(R)\chi_j(R)] = 0$
e.g. For $\color{red}B_1$ and $\color{blue}B_2$ of $C_{2v}$, $[{\color{red}1} \times {\color{blue}1}] + [{\color{red}-1} \times {\color{blue}-1}] + [{\color{red}1} \times {\color{blue}-1}] + [{\color{red}-1} \times {\color{blue}1}] = 0$
The complete character table for $C_{2v}$ is given below.
$\begin{array}{l|llll|l|l} C_{2v} & E & C_2 & \sigma_v & \sigma_v' & h=4\ \hline \color{green}A_1 & \color{green}1 & \color{green}1 & \color{green}1 & \color{green}1 & \color{green}z & x^2,y^2,z^2\ \color{purple}A_2 & \color{purple}1 & \color{purple}1 & \color{purple}-1 & \color{purple}-1 & R_z & xy \ \color{red}B_1 & \color{red}1 & \color{red}-1&\color{red}1&\color{red}-1 & {\color{red}x},R_y & xz \ \color{blue}B_2 & \color{blue}1 & \color{blue}-1 & \color{blue}-1 & \color{blue}1 & {\color{blue}y} ,R_x & yz \end{array}$
The various sections of the table are as follows:
• The first element in the table gives the name of the point group, usually in Schoenflies ($C_{2v}$) notation.
• Along the first row are the symmetry operations of the group, $E$, $C_2$, $\sigma_v$ and $\sigma_v"$, followed by the order of the group, $h$.
• In the first column are the irreducible representations of the group, represented by Mulliken Labels. In $C_{2v}$ the irreducible representations are $A_1$, $A_2$, $B_1$ and $B_2$. The Mulliken labels indicate the symmetry of each representation (explained further below).
• The characters ($\chi$) of the irreducible representations under each symmetry operation are given in the bulk of the table.
• The final column(s) of the table lists a number of functions that transform as the various irreducible representations of the group. These are the Cartesian axes $\begin{pmatrix} x, y, z \end{pmatrix}$, the Cartesian products $\begin{pmatrix} z^2, x^2 + y^2, xy, xz, yz \end{pmatrix}$, and the rotations $\begin{pmatrix} R_x, R_y, R_z \end{pmatrix}$ (explained further below).
Another example: $C_{3v}$
Th $C_{3v}$ point group has three classes of operations: $E$, $C_{3}$, and $\sigma_{v(xz)}$. The derivation of transformation matrices for E and $\sigma_{v(xz)}$ is similar to the case for $C_{2v}$. However, the $C_{3v}$ operation does not give simple 1 or -1 characters. If we carry out a rotation about $z$ by an angle $\theta$, our $x$ and $y$ axes are transformed onto new axes $x'$ and $y'$. The new axes can each be written as a linear combination of our original $x$ and $y$ axes. The derivation of the rotation matrices will not be covered in this text, but is described elsewhere:
$\begin{array}{ccc}x' & = & x\cos\theta + y\sin\theta \ y' & = & -x\sin\theta + y\cos\theta \end{array} \nonumber$
For a $C_3$ rotation counterclockwise through 120° (or $\frac{2\pi}{3}$):
$\begin{array}{ccccc}x' & = & x\cos(2\pi/3) + y\sin(2\pi/3) & = & {\color{orange}-\frac{1}{2}x-\frac{\sqrt{3}}{2}y} \ y' & = & -x\sin(2\pi/3) + y\cos(2\pi/3) & = & {\color{violet}\frac{\sqrt{3}}{2}x-\frac{1}{2}y} \end{array} \nonumber$
The transformation matrices for symmetry operations of $C_{3v}$ are as follows:
$E=\begin{pmatrix} 1&0&0 \ 0&1&0 \ 0&0&1 \end{pmatrix} \nonumber$ $C_3=\begin{pmatrix} {\color{orange}[-\frac{1}{2}}&{\color{orange}-\frac{\sqrt{3}}{2}]}&0 \ {\color{violet}[\frac{\sqrt{3}}{2}}&{\color{violet}-\frac{1}{2}]}&0 \ 0&0&1 \end{pmatrix} \nonumber$ $\sigma_{v(xz)}'=\begin{pmatrix} 1&0&0 \ 0&-1&0 \ 0&0&1 \end{pmatrix} \nonumber$
The $C_3$ transformation matrix contains off-diagonal entries, and therefore it cannot be block diagonalized as 1x1 matrices. However, the first two lines can be diagonalized as a 2x2 and the last line as a 1x1 matrix (Figure $1$):
The character from a 2x2 matrix is the sum of the trace of that matrix. So, for the $C_3$ operation, the 2x2 matrix gives the character -1 (from ${\color{orange}-\frac{1}{2}} + {\color{violet}-\frac{1}{2}}$).
The character table for $C_{3v}$ is shown below.
$\begin{array}{l|lll|l} \hline C_{3v} & E & 2C_3 & 3\sigma_v & h=6 \ \hline A_1 & 1 & 1 & 1 & z, z^2, x^2+y^2 \ A_2 & 1 & 1 & -1 & R_z \ E & 2 & -1 & 0 & \begin{pmatrix} x, y \end{pmatrix}, \begin{pmatrix} xy, x^2+y^2 \end{pmatrix}, \begin{pmatrix} xz, yz \end{pmatrix}, \begin{pmatrix} R_x, R_y \end{pmatrix} \ \hline \end{array}$
Additional features of character tables
Additional Features of Character Tables
1. Symmetry operations of the same class are grouped into the same column (class) in the character table and not listed separately.
e.g. In the $C_{3v}$ point group, there are four operations: $E$, $C_3$, $C_3^2$, and $\sigma_{v}$. The $C_3$ and $C_3^2$ operations are listed together in the character table as $2C_3$.
2. IF there are multiple $C_2$ axes (in a $D$ group), the $C_2$ axes that are perpendicular to the principle axis are labeled with primes (e.g. $C_2'$ and $C_2''$); when there are multiple types of perpendicular $C_2$ axes, one prime ($C_2'$) means that it passes through more atoms, while a double prime ($C_2''$) means it goes between atoms.
3. Mirror planes that are perpendicular to the principle axis are "horizontal" mirror planes and are designated with an $h$ subscript ($\sigma_h$). Mirror planes that are in-plane with the principle axis are "vertical" mirror planes, $\sigma_v$. When there are two types of vertical mirror planes, those that run through more atoms are $\sigma_v$ while those that run between atoms are "dihedral", $\sigma_d$.
4. Matching the symmetry operations listed in the character table to the symmetry operations of a molecule can confirm its point group.
5. Irreducible representations are each assigned a Mulliken label, listed in the left-hand column, that indicates the symmetry of that representation as follows:
$\begin{array}{l|l} \hline \textbf{Mulliken Labels} & \textbf{meaning}\ \hline A & \text{singly degenerate (1x1), symmetric to principle axis} \ B & \text{singly degenerate (1x1), antisymmetric to principle axis} \ E & \text{doubly degenerate (2x2)} \ T & \text{triply degenerate (3x3)} \ \hline \textbf{Subscripts and superscripts} & \textbf{meaning} \ \hline 1 & \text{symmetric to } \sigma_v \text{or perpendicular to } C_2 \ 2 & \text{anti-symmetric to } \sigma_v \text{or perpendicular to } C_2 \ g & \text{symmetric to inversion center} \ u & \text{anti-symmetric to inversion center} \ ' & \text{symmetric to } \sigma_h \ " & \text{anti-symmetric to } \sigma_h \ \hline \end{array}$
6. The right-hand columns of the character table list a number of functions that transform as the various irreducible representations of the group. These are the Cartesian axes $\begin{pmatrix} x, y, z \end{pmatrix}$, the Cartesian products $\begin{pmatrix} z^2, x^2 + y^2, xy, xz, yz \end{pmatrix}$, and the rotations $\begin{pmatrix} R_x, R_y, R_z \end{pmatrix}$. These expressions indicate the properties of orbitals within the symmetry group. The $s$-orbital, which is totally symmetric, corresponds to the irreducible representation that possesses symmetry of $x^2$, $y^2$ and $z^2$ combined. The $p$-orbitals each possess the symmetry of the corresponding axis (e.g. $p_x$ corresponds to the $x$ axis). Each of the $d$-orbitals possess the symmetry of the corresponding binary product (e.g. $d_{xy}$ corresponds to the binary product, $xy$, in the character table).
The functions listed in the final column of the table are important in many chemical applications of group theory, particularly in spectroscopy. For example, by looking at the transformation properties of $x$, $y$ and $z$ (sometimes given in character tables as $T_x$, $T_y$, $T_z$) we can discover the symmetry of translations along the $x$, $y$, and $z$ axes. Similarly, $R_x$, $R_y$ and $R_z$ represent rotations about the three Cartesian axes. The transformation properties of $x$, $y$, and $z$ can be used to determine whether or not a molecule is IR-active or whether or not it can absorb a photon of $x$-, $y$-, or $z$-polarized light and undergo a spectroscopic transition. The Cartesian products play a similar role in determining selection rules for Raman transitions, which involve two photons.
A visual summary of the sections and their significance is given in Figure $2$. Character tables for common point groups are given in the References section of LibreTexts Bookshelves. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/04%3A_Symmetry_and_Group_Theory/4.03%3A_Properties_and_Representations_of_Groups/4.3.03%3A_Character_Tables.txt |
The purpose of learning group theory is that molecular symmetry can be applied to understand the properties of molecules. This chapter introduces two applications of symmetry: chirality and molecular vibrations (in the context of analyzing IR and Raman spectra). An additional application of symmetry, to understand molecular bonding, deserves its own chapter and will be explored in Chapter 5.
• 4.4.1: Chirality
• 4.4.2: Molecular Vibrations
Symmetry and group theory can be applied to understand molecular vibrations. This is particularly useful in the contexts of predicting the number of peaks expected in the infrared (IR) and Raman spectra of a given compound. We will use water as a case study to illustrate how group theory is used to predict the number of peaks in IR and Raman spectra.
4.04: Examples and Applications of Symmetry
Introduction
Around the year 1847, the French scientist Louis Pasteur provided an explanation for the optical activity of tartaric acid salts. When he carried out a particular reaction, Pasteur observed that two types of crystals precipitated. Patiently and carefully using tweezers, Pasteur was able to separate the two types of crystals. Pasteur noticed that the types rotated plane-polarized light by the same amount but in different directions. These two compounds are called enantiomers.
What is chirality?
A molecule is chiral (or dissymetric) if it is non-superimposable on its mirror image. The two mirror images of a chiral molecule are called enantiomers. Enantiomers have the same physical properties (e.g., melting point, etc.). They differ in their ability to rotate plane polarized light and in their reactivity with other chiral molecules. Due to their ability to rotate plane polarized light, they are referred to as being optically active.
Using Symmetry to Determine Chirality
There are some general rules of thumb that help determine whether a molecule is chiral or achiral. The point group of the molecule, and the symmetry operations within that point group, can give clues as to whether the molecule is chiral.
Symmetry operations of chiral molecules
A chiral molecule cannot possess a plane of symmetry ($\sigma$), a center of inversion ($i$), or an improper rotation ($S_n$). Due to the fact that all groups that lack both $\sigma$ and $i$ also lack $S_n$, a molecule that belongs to any group that lacks $S_n$ is chiral.
An example of an inorganic coordination complex is tris(ethylenediamine)cobalt(III) (Figure $1$). Figure $1$ shows the two enantiomers of tris(ethylenediamine)cobalt(III): the $\Delta$ and $\Lambda$ isomers.
You can visualize the $\Lambda$ and $\Delta$ isomers by imagining that the ligands around the metal centers are blades of a fan. To push the air toward you, you would need to rotate the $\Delta$ isomer clockwise, and the $\Lambda$ isomer counter-clockwise.
Exercise $1$
Which point groups are possible for chiral molecules?
Answer
As a result of the previous discussion, there are a few classes of point groups that lack an improper axis. Those classes are $C_{1}$, $C_{n}$, and $D_{n}$.
4.4.02: Molecular Vibrations
Symmetry and group theory can be applied to understand molecular vibrations. This is particularly useful in the context of predicting the number of peaks expected in the infrared (IR) and Raman spectra of a given compound.
We will use water as a case study to illustrate how group theory is used to predict the number of peaks in IR and Raman spectra.
How many IR and Raman peaks would we expect for $H_2O$?
To answer this question with group theory, a pre-requisite is that you assign the molecule's point group and assign an axis system to the entire molecule. By convention, the $z$ axis is collinear with the principle axis, the $x$ axis is in-plane with the molecule or the most number of atoms. It is a good idea to stick with this convention (see Figure $1$).
What is the point group for $H_2O$? (click to see answer)
$H_2O$ has the following operations: $E$, $C_2$, $\sigma_v$, $\sigma_v'$. The point group is $C_{2v}$.
Now that we know the molecule's point group, we can use group theory to determine the symmetry of all motions in the molecule, or the symmetry of each of its degrees of freedom. Then we will subtract rotational and translational degrees of freedom to find the vibrational degrees of freedom. The number of degrees of freedom depends on the number of atoms ($N$) in a molecule. Each atom in the molecule can move in three dimensions ($x,y,z$), and so the number of degrees of freedom is three dimensions times $N$ number of atoms, or $3N$. The total degrees of freedom include a number of vibrations, three translations (in $x$, $y$, and $z$), and either two or three rotations. Linear molecules have two rotational degrees of freedom, while non-linear molecules have three. The vibrational modes are represented by the following expressions:
$\begin{array}{ccc} \text{Linear Molecule Degrees of Freedom} & = & 3N - 5 \ \text{Non-Linear Molecule Degrees of Freedom} & = & 3N-6 \end{array} \nonumber$
Our goal is to find the symmetry of all degrees of freedom, and then determine which are vibrations that are IR- and Raman-active.
STEP 1: Find the reducible representation for all normal modes $\Gamma_{modes}$.
The first major step is to find a reducible representation ($\Gamma$) for the movement of all atoms in the molecule (including rotational, translational, and vibrational degrees of freedom). We'll refer to this as $\Gamma_{modes}$. To find normal modes using group theory, assign an axis system to each individual atom to represent the three dimensions in which each atom can move. Each axis on each atom should be consistent with the conventional axis system you previously assigned to the entire molecule (see Figure $1$).
$\Gamma_{modes}$ is the sum of the characters (trace) of the transformation matrix for the entire molecule (in the case of water, there are 9 degrees of freedom and this is now a 9x9 matrix). Let's walk through this step-by-step. The transformation matrix of $E$ and $C_2$ are shown below:
$E=\begin{pmatrix} \color{red}1&0&0&0&0&0&0&0&0 \ 0&\color{red}1&0&0&0&0&0&0&0 \ 0&0&\color{red}1&0&0&0&0&0&0 \0&0&0&\color{red}1&0&0&0&0&0 \ 0&0&0&0&\color{red}1&0&0&0&0 \ 0&0&0&0&0&\color{red}1&0&0&0 \ 0&0&0&0&0&0&\color{red}1&0&0 \ 0&0&0&0&0&0&0&\color{red}1&0 \ 0&0&0&0&0&0&0&0&\color{red}1 \ \end{pmatrix} \begin{pmatrix} x_{oxygen} \ y_{oxygen} \ z_{oxygen} \ x_{hydrogen-a} \ y_{hydrogen-a} \ z_{hydrogen-a} \ x_{hydrogen-b} \ y_{hydrogen-b} \ z_{hydrogen-b} \end{pmatrix} = \begin{pmatrix} x'_{oxygen} \ y'_{oxygen} \ z'_{oxygen} \ x'_{hydrogen-a} \ y'_{hydrogen-a} \ z'_{hydrogen-a} \ x'_{hydrogen-b} \ y'_{hydrogen-b} \ z'_{hydrogen-b} \end{pmatrix}, \chi=9 \nonumber$
$C_2=\begin{pmatrix} \color{red}-1&0&0&0&0&0&0&0&0 \ 0&\color{red}-1&0&0&0&0&0&0&0 \ 0&0&\color{red}1&0&0&0&0&0&0 \0&0&0&\color{red}0&0&0&-1&0&0 \ 0&0&0&0&\color{red}0&0&0&-1&0 \ 0&0&0&0&0&\color{red}0&0&0&1 \ 0&0&0&-1&0&0&\color{red}0&0&0 \ 0&0&0&0&-1&0&0&\color{red}0&0 \ 0&0&0&0&0&1&0&0&\color{red}0 \ \end{pmatrix} \begin{pmatrix} x_{oxygen} \ y_{oxygen} \ z_{oxygen} \ x_{hydrogen-a} \ y_{hydrogen-a} \ z_{hydrogen-a} \ x_{hydrogen-b} \ y_{hydrogen-b} \ z_{hydrogen-b} \end{pmatrix} = \begin{pmatrix} x'_{oxygen} \ y'_{oxygen} \ z'_{oxygen} \ x'_{hydrogen-a} \ y'_{hydrogen-a} \ z'_{hydrogen-a} \ x'_{hydrogen-b} \ y'_{hydrogen-b} \ z'_{hydrogen-b} \end{pmatrix}, \chi=1 \nonumber$
It is unnecessary to find the transformation matrix for each operation since it is only the TRACE that gives us the character, and any off-diagonal entries do not contribute to $\Gamma_{modes}$. The values that contribute to the trace can be found simply by performing each operation in the point group and assigning a value to each individual atom to represent how it is changed by that operation. If the atom moves away from itself, that atom gets a character of zero (this is because any non-zero characters of the transformation matrix are off of the diagonal). If the atom remains in place, each of its three dimensions is assigned a value of $\cos \theta$. For the example of $H_2O$ under the $C_{2v}$ point group, the axes that remain unchanged ($\theta = 0^{\circ}$) are assigned a value of $\cos(0^{\circ})=1$, while those that are moved into the negative of themselves (rotated or reflected to $\theta = 180^{\circ}$) are assigned $\cos(180^{\circ}) = -1$. The character for $\Gamma$ is the sum of the values for each transformation.
Let's walk through the steps to assign characters of $\Gamma_{modes}$ for $H_2O$ to illustrate how this works:
For the operation $E$, performed on $H_2O$, all three atoms remain in place. The three axes $x,y,z$ on each atom remain unchanged. Thus, each of the three axes on each of three atom (nine axes) is assigned the value $\cos(0^{\circ})=1$, resulting in a sum of $\chi=9$ for the $\Gamma_{modes}$.
For the operation $C_2$, the two hydrogen atoms are moved away from their original position, and so the hydrogens are assigned a value of zero. The oxygen remains in place; the $z$-axis on oxygen is unchanged ($\cos(0^{\circ})=1$), while the $x$ and $y$ axes are inverted ($\cos(180^{\circ})$). The sum of these characters gives $\chi=-1$ in the $\Gamma_{modes}$.
Now you try! Find the characters of $\sigma_{v(xz)}$ and $\sigma_{v(yz)}$ under the $C_{2v}$ point group. Compare what you find to the $\Gamma_{modes}$ for all normal modes given below.
$\begin{array}{l|llll} C_{2v} & E & C_2 & \sigma_v & \sigma_v' \ \hline \Gamma_{modes} & 9 & -1 & 3 & 1 \end{array} \label{gammamodes}$
STEP 2: Break $\Gamma_{modes}$ into its component irreducible representations.
Now that we've found the $\Gamma_{modes}$ ($\ref{gammamodes}$), we need to break it down into the individual irreducible representations ($i,j,k...$) for the point group. We can do this systematically using the following formula:
$\text{# of } i = \frac{1}{h}\sum(\text{# of operations in class)}\times(\chi_{\Gamma}) \times (\chi_i) \label{irs}$
In other words, the number of irreducible representations of type $i$ is equal to the sum of the number of operations in the class $\times$ the character of the $\Gamma_{modes}$ $\times$ the character of $i$, and that sum is divided by the order of the group ($h$).
Using equation $\ref{irs}$, we find that for all normal modes of $H_2O$:
$\Gamma_{modes}=3A_1+1A_2+3B_1+2B_2 \label{water}$.
Notice there are 9 irreducible representations in Equation \ref{water}. These irreducible representations represent the symmetries of all 9 motions of the molecule: vibrations, rotations, and translations.
Exercise $1$: Derive the irreducible representation in equation $\ref{water}$.
Derive the nine irreducible representations of $\Gamma_{modes}$ for $H_2O$, expression $\ref{water}$.
Hint
To find the number of each kind of irreducible representation that combine to form the $\Gamma_{modes}$, we need the characters of $\Gamma_{modes}$ that we found above ($\ref{gammamodes}$), the $C_{2v}$ character table (below), and equation $\ref{irs}$.
$\begin{array}{l|llll|l|l} C_{2v} & {\color{red}1}E & {\color{red}1}C_2 & {\color{red}1}\sigma_v & {\color{red}1}\sigma_v' & \color{orange}h=4\ \hline \color{green}A_1 & \color{green}1 & \color{green}1 & \color{green}1 & \color{green}1 & \color{green}z & \color{green}x^2,y^2,z^2\ \color{green}A_2 & \color{green}1 & \color{green}1 & \color{green}-1 & \color{green}-1 & \color{green}R_z & \color{green}xy \ \color{green}B_1 & \color{green}1 & \color{green}-1&\color{green}1&\color{green}-1 & \color{green}x,R_y & \color{green}xz \ \color{green}B_2 & \color{green}1 & \color{green}-1 & \color{green}-1 & \color{green}1 & {\color{green}y} ,\color{green}R_x & \color{green}yz \end{array}$
In the $C_{2v}$ point group, each class has only one operation, so the number of operations in each class (from equation $\ref{irs}$) is ${\color{red}1}$ for each class. This has been explicitly added to the character table above for emphasis.
Answer
The number of $A_1$ = $\frac{1}{\color{orange}4} \left[ ({\color{green}1} \times 9 \times {\color{red}1}) + ({\color{green}1} \times (-1) \times {\color{red}1}) + ({\color{green}1} \times 3 \times {\color{red}1}) + ({\color{green}1} \times 1 \times {\color{red}1})\right] = 3A_1$
The number of $A_2$ = $\frac{1}{\color{orange}4} \left[ ({\color{green}1} \times 9 \times {\color{red}1}) + ({\color{green}1} \times (-1) \times {\color{red}1}) + ({\color{green}(-1)} \times 3 \times {\color{red}1}) + ({\color{green}(-1)} \times 1 \times {\color{red}1})\right] = 1A_2$
The number of $B_1$ = $\frac{1}{\color{orange}4} \left[ ({\color{green}1} \times 9 \times {\color{red}1}) + ({\color{green}(-1)} \times (-1) \times {\color{red}1}) + ({\color{green}1} \times 3 \times {\color{red}1}) + ({\color{green}(-1)} \times 1 \times {\color{red}1})\right] = 3B_1$
The number of $B_2$ = $\frac{1}{\color{orange}4} \left[ ({\color{green}1} \times 9 \times {\color{red}1}) + ({\color{green}(-1)} \times (-1) \times {\color{red}1}) + ({\color{green}(-1)} \times 3 \times {\color{red}1}) + ({\color{green}1} \times 1 \times {\color{red}1})\right] = 2B_2$
STEP 3: Subtract rotations and translations to find vibrational modes.
Because we are interested in molecular vibrations, we need to subtract the rotations and translations from the total degrees of freedom.
$\text{Vibrations } = \Gamma_{modes}-\text{ Rotations } - \text{ Translations } \nonumber$
In the example of $H_2O$, the total degrees of freedom are given above in equation $\ref{water}$, and therefore the vibrational degrees of freedom can be found by:
$H_2O\text{ vibrations} = \left(3A_1 + 1A_2 + 3B_1 + 2B_2\right) - \text{ Rotations } - \text{ Translations } \label{watervib}$
But which of the irreducible representations are ones that represent rotations and translations? The symmetry of rotational and translational degree modes can be found by inspecting the right-hand columns of any character table. Rotational modes correspond to irreducible representations that include $R_x$, $R_y$, and $R_z$ in the table, while each of the three translational modes has the same symmetry as the $x$, $y$ and $z$ axes. For a non-linear molecule, subtract three rotational irreducible representations and three translational irreducible representations from the total $\Gamma_{modes}$.
In the specific case of water, we refer to the $C_{2v}$ character table:
$\begin{array}{l|llll|l|l} C_{2v} & E & C_2 & \sigma_v & \sigma_v' & h=4\ \hline A_1 &1 & 1 & 1 & 1 & \color{red}z & x^2,y^2,z^2\ A_2 & 1 & 1 & -1 & -1 & \color{red}R_z & xy \ B_1 &1 & -1&1&-1 & \color{red}x,R_y &xz \ B_2 & 1 & -1 &-1 & 1 & \color{red}y ,R_x & yz \end{array} \nonumber$
In $C_{2v}$, translations correspond to $B_1$, $B_2$, and $A_1$ (respectively for $x,yz$), and rotations correspond to $B_2$, $B_1$, and $A_1$ (respectively for $R_x,R_y,R_z$). Subtracting these six irreducible representations from $\Gamma_{modes}$ will leave us with the irreducible representations for vibrations.
$\begin{array}{lll} H_2O\text{ vibrations} &=& \Gamma_{modes} - \text{ Rotations } - \text{ Translations }\ &=& \left(3A_1 + 1A_2 + 3B_1 + 2B_2\right) - (A_1 + B_1 + B_2) -(A_2 + B_1 + B_2) \ &=& 2A_1 + 1B_1 \end{array} \nonumber$
The three vibrational modes for $H_2O$ are $2A_1 + 1B_1$. Note that we have the correct number of vibrational modes based on the expectation of $3N-6$ vibrations for a non-linear molecule.
STEP 4: Determine which of the vibrational modes are IR-active and Raman-active.
The next step is to determine which of the vibrational modes is IR-active and Raman-active. To do this, we apply the IR and Raman Selection Rules below:
Infrared selection rules:
If a vibration results in the change in the molecular dipole moment, it is IR-active. In the character table, we can recognize the vibrational modes that are IR-active by those with symmetry of the $x,y$, and $z$ axes.
In $C_{2v}$, any vibrations with $A_1$, $B_1$ or $B_2$ symmetry would be IR-active.
Raman selection rules:
If a vibration results in a change in the molecular polarizability, it will be Raman-active. In the character table, we can recognize the vibrational modes that are Raman-active by those with symmetry of any of the binary products ($xy$, $xz$, $yz$, $x^2$, $y^2$, and $z^2$) or a linear combination of binary products (e.g. $x^2-y^2$).
In $C_{2v}$, any vibrations with $A_1$, $A_2$, $B_1$ or $B_2$ symmetry would be Raman-active.
In our $H_2O$ example, we found that of the three vibrational modes, two have $A_1$ and one has $B_1$ symmetry. Both $A_1$ and $B_1$ are IR-active, and both are also Raman-active. There are two possible IR peaks and three possible Raman peaks expected for water.*
*It is important to note that this prediction tells only what is possible, but not what we might actually see in the IR and Raman spectra. For example, if the two IR peaks overlap, we might actually notice only one peak in the spectrum. Or, if one or more peaks is off-scale, we wouldn't see it in actual data. Group theory tells us what is possible and allows us to make predictions or interpretations of spectra.
Summary of Analysis for Water
Each molecular motion for water, or any molecule, can be assigned a symmetry under the molecule's point group. For water, we found that there are a total of 9 molecular motions; $3A_1 + A_2 +3B_1 + 2B_2$. Six of these motions are not the translations and rotations. The remaining motions are vibrations; two with $A_1$ symmetry and one with $B_1$ symmetry.
We can tell what these vibrations would look like based on their symmetries. The two $A_1$ vibrations must be completely symmetric, while the $B_1$ vibration is antisymmetric with respect to the principal $C_2$ axis.
Table $1$: Summary of the Symmetry of Molecular Motions for Water
All Motions (step 2 above) Translations (x,y,z) Rotations ($R_x,R_y,R_z$) Remaining Vibrations Description of Vibration
$3A_1$ $1A_1$ $2A_1$ One is a symmetric stretch. The other is a symmetric bend. Both are IR-active and Raman-active
$A_2$ $1A_2$
$3B_1$ $1B_1$ $1B_1$ $1B_1$ Antisymmetric stretch that is IR-active and Raman-active.
$2B_2$ $1B_2$ $1B_2$
Exercise $2$
Find the symmetries of all motions of the square planar complex, tetrachloroplatinate (II). Determine which are rotations, translations, and vibrations. Determine which vibrations are IR and Raman active.
Answer
The point group of $\ce{[PtCl4]^2-}$ is $D_{4h}$ (refer to its character table). There are five atoms and 15 vectors ($x,y,z$ for each atom $\times$ 5 atoms).
STEP 1: The first major step is to find a reducible representation ($\Gamma$) for the movement of all atoms in the molecule.
$\begin{array}{l|rrrrrrrrrr} C_{2v} & E & 2C_4 & C_2 & 2C_2' & 2C_2" & i & 2S_4 & \sigma_h & 2\sigma_v & 2\sigma_d \ \hline \Gamma_{modes} & 15 & 1 & -1 & -3 & -1 & -3 & -1 & 5 & 3 & 1 \end{array} \label{gammamodes2}$.
STEP 2: Break $\Gamma_{modes}$ into its component irreducible representations.
Following the process described earlier, we come to $A_{1g} + A_{2g} + B_{1g} + B_{2g} + E_g + 2A_{2u} + B_{2u} + 3E_u$. This accounts for all modes of movement, including rotations and translations.
STEP 3: Subtract rotations and translations to find vibrational modes.
The translations are $A_{2u}+E_u$ and the rotations are $A_{2g}+E_g$.
The remaining normal modes are: $A_{1g} + B_{1g} + B_{2g} + A_{2u} + B_{2u} + 2 E_u$
STEP 4: Determine which of the vibrational modes are IR-active and Raman-active:
$A_{2u} + E_u$ are IR-active. Since $A_{2u}$ is singly degenerate and $E_u$ is doubly degenerate, we expect three possible IR bands.
$A_{1g} + B_{1g} + B_{2g}$ are Raman-Active. Each of these is singly degenerate, so we expect three possible Raman bands.
Selected Vibrational Modes
The interpretation of CO stretching vibrations in an IR spectrum is particularly useful. Symmetry and group theory can be applied to predict the number of CO stretching bands that appear in a vibrational spectrum for a given metal coordination complex. A classic example of this application is in distinguishing isomers of metal-carbonyl complexes. For example, the cis- and trans- isomers of square planar metal dicarbonyl complexes (ML2(CO)2) have a different number of IR stretches that can be predicted and interpreted using symmetry and group theory. Another example is the case of mer- and fac- isomers of octahedral metal tricarbonyl complexes (ML3(CO)3). Structures of the two types of metal carbonyl configurations and their isomers are shown in Figure $1$. The isomers in each case can be distinguished using vibrational spectroscopy.
EXAMPLE 1: Distinguishing cis- and trans- isomers of square planar metal dicarbonyl complexes
General structures of the cis- and trans- isomers of square planar metal dicarbonyl complexes (ML2(CO)2) are shown in the left box in Figure $1$. We can use symmetry and group theory to predict how many carbonyl stretches we should expect for each isomer following the steps below.
Step 1: Assign the point group and Cartesian coordinates for each isomer.
The cis-isomer has $C_{2v}$ symmetry and the trans-isomer has $D_{2h}$ symmetry. We assign the Cartesian coordinates so that $z$ is colinear with the principle axis in each case. For the $D_2{h}$ isomer, there are several orientations of the $z$ axis possible. The axes shown in Figure $2$ will be used here.
Step 2: Produce a reducible representation ($\Gamma$) for CO stretches in each isomer
First, assign a vector along each C—O bond in the molecule to represent the direction of C—O stretching motions, as shown in Figure $2$ (red arrows ). These vectors are used to produce a reducible representation ($\Gamma$) for the C—O stretching motions in each molecule. Using the symmetry operations under the appropriate character table, assign a value of 1 to each vector that remains in place during the operation, and a value of 0 if the vector moves out of place. There will be no occasion where a vector remains in place but is inverted, so a value of -1 will not occur.
cis- ML2(CO)2:
For cis- ML2(CO)2, the point group is $C_{2v}$ and so we use the operations under the $C_{2v}$ character table to create the $\Gamma_{cis-CO}$. $\begin{array}{|c|cccc|} \hline \bf{C_{2v}} & E & C_2 &\sigma_v (xz) & \sigma_v' (yz) \ \hline \bf{\Gamma_{cis-CO}} & 2 & 0 & 2 & 0 \ \hline \end{array} \nonumber$
trans- ML2(CO)2:
For trans- ML2(CO)2, the point group is $D_{2h}$ and so we use the operations under the $D_{2h}$ character table to create the $\Gamma_{trans-CO}$. $\begin{array}{|c|cccccccc|} \hline \bf{D_{2h}} & E & C_2(z) & C_2(y) &C_2(x) & i &\sigma(xy) & \sigma(xz) & \sigma(yz) \ \hline \bf{\Gamma_{trans-CO}} & 2 & 0 & 0 & 2 & 0 & 2 & 2 & 0\ \hline \end{array} \nonumber$
Step 3: Break each $\Gamma$ into its component irreducible representations
Each $\Gamma$ can be reduced using inspection or by the systematic method described previously.
In the case of the cis- ML2(CO)2, the CO stretching vibrations are represented by $A_1$ and $B_1$ irreducible representations: $\begin{array}{|c|cccc|cc|} \hline \bf{C_{2v}} & E & C_2 &\sigma_v (xz) & \sigma_v' (yz) \ \hline \bf{\Gamma_{cis-CO}} & 2 & 0 & 2 & 0 & & \ \hline A_1 & 1 & 1 & 1 & 1 & z & x^2, y^2, z^2 \ B_1 & 1 & -1 & 1 & -1 & x, R_y & xz \ \hline \end{array} \label{c2v}$
In the case of trans- ML2(CO)2, the CO stretching vibrations are represented by $A_1$ and $B_{3u}$ irreducible representations:
$\begin{array}{|c|cccccccc|cc|} \hline \bf{D_{2h}} & E & C_2(z) & C_2(y) &C_2(x) & i &\sigma(xy) & \sigma(xz) & \sigma(yz) \ \hline \bf{\Gamma_{trans-CO}} & 2 & 0 & 0 & 2 & 0 & 2 & 2 & 0 & & \ \hline A_{g} & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & & x^2, \; y^2, \; z^2\ B_{3u} & 1 & -1 & -1 & 1 & -1 & 1 & 1 & -1 & x & \ \hline \end{array} \nonumber$
These irreducible representations correspond to the symmetries of only the selected C—O vibrations. Since these motions are isolated to the C—O group, they do not include any rotations or translations of the entire molecule, and so we do not need to find and subtract rotationals or translations (unlike the previous cases where all motions were considered).
Step 4: Determine which vibrational modes are IR-active and/or Raman-active
Apply the infrared selection rules described previously to determine which of the CO vibrational motions are IR-active and Raman-active. The two isomers of ML2(CO)2 are described below.
In the case of the cis- ML2(CO)2, the CO stretching vibrations are represented by $A_1$ and $B_1$ irreducible representations. The characters of both representations and their functions are shown above, in \ref{c2v} (and can be found in the $C_{2v}$ character table). Under $C_{2v}$, both the $A_1$ and $B_1$ CO vibrational modes are IR-active and Raman-active. Therefore, two bands in the IR spectrum and two bands in the Raman spectrum are possible.
In the case of the trans- ML2(CO)2, the CO stretching vibrations are represented by $A_g$ and $B_{3u}$ irreducible representations. The characters of both representations and their functions are shown above, in \ref{c2v} (and can be found in the $D_{2h}$ character table). Under $D_{2h}$, the $A_g$ vibrational mode is is Raman-active only, while the $B_{3u}$ vibrational mode is IR-active only. Therefore, only one IR band and one Raman band are possible for this isomer.
Summary
It is possible to distinguish between the two isomers of square planar ML2(CO)2 using either IR or Raman vibrational spectroscopy. The cis- ML2(CO)2 can produce two CO stretches in an IR or Raman spectrum, while the trans- ML2(CO)2 isomer can produce only one band in either type of vibrational spectrum. If a sample of ML2(CO)2 produced two CO stretching bands, we could rule out the possibility of a pure sample of trans-ML2(CO)2.
Exercise $3$
Repeat the steps outlined above to determine how many CO vibrations are possible for mer-ML3(CO)3 and fac-ML3(CO)3 isomers (see Figure $1$) in both IR and Raman spectra. Could either of these vibrational spectroscopies be used to distinguish the two isomers?
Answer
Step 1: Assign the point group and Cartesian coordinates for each isomer.
The fac-isomer is $C_{3v}$. The mer-isomer is $C_{2v}$.
Step 2: Produce a reducible representation for CO stretches in each isomer.
Step 3: Break each into its component irreducible representations.
Step 4: Determine which vibrational modes are IR-active and/or Raman-active.
For fac-ML3(CO)3, the point group is $C_{3v}$ and so we use the operations under the $C_{3v}$ character table to create the $\Gamma_{fac-CO}$. Then break it into its irreducible representations and determine which are IR and Raman active: $\begin{array}{|c|ccc|} \hline \bf{C_{3v}} & E & 2C_2 &3\sigma_v \ \hline \bf{\Gamma_{fac-CO}} & 3 & 0 & 1 \ \hline \end{array} \nonumber$
This reduces to $A_1 + E$. Both of these are IR active, and since one is singly degenerate while the other is doubly degenerate, we expect three possible IR bands from this isomer. Both vibrational modes are also Raman active, and again we would expect three possible bands in the Raman spectrum.
For mer-ML3(CO)3, the point group is $C_{2v}$ and so we use the operations under the $C_{2v}$ character table to create the $\Gamma_{mer-CO}$. Then break it into its irreducible representations and determine which are IR and Raman active:
$\begin{array}{|c|cccc|} \hline \bf{C_{2v}} & E & C_2 &\sigma_v (xz) & \sigma_v' (yz) \ \hline \bf{\Gamma_{mer-CO}} & 3 & 1 & 3 & 1 \ \hline \end{array} \nonumber$
This reduces to $2A_1+B_1$; both $A_1$ and $B_1$ are IR and Raman active. So this isomer would have three possible IR bands and three possible Raman Bands.
These two isomers have the same number of possible bands in both IR and Raman spectroscopy. It would not be straightforward to distinguish them from each other based on the number of possible bands in the vibrational spectrum. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/04%3A_Symmetry_and_Group_Theory/4.04%3A_Examples_and_Applications_of_Symmetry/4.4.01%3A_Chirality.txt |
Problems
1. Chlorophyll a is a green pigment that is found in plants. Its molecular formula is C55H77O5N4Mg. How many degrees of freedom does this molecule possess? How many vibrational degrees of freedom does it have?
2. CCl4 was commonly used as an organic solvent until its severe carcinogenic properties were discovered. How many vibrational modes does CCl4 have? Are they IR and/or Raman active?
3. The same vibrational modes in H2O are IR and Raman active. WF6- has IR active modes that are not Raman active and vice versa. Explain why this is the case.
4. How many IR peaks do you expect from SO3? Estimate where these peaks are positioned in an IR spectrum.
5. Calculate the symmetries of the normal coordinates of planar BF3.
Answers to Problems
1. Chlorophyll a has 426 degrees of freedom and 420 vibrational modes.
2. The point group is Td, Tvib = a1 + e + 2t2; a1 and e are Raman active, t2 is both IR and Raman active.
3. For molecules that possess a center of inversion i, modes cannot be simultaneously IR and Raman active.
4. Point group is D3h; one would expect three IR active peaks. Asymmetric stretch highest (1391 cm-1), two bending modes (both around 500 cm-1). The symmetric stretch is IR inactive.
5. T3N = A1' + A2' + 3E' + 2 A2" + E" and Tvib= A1' + 2E' + A2"
Contributors and Attributions
• Kristin Kowolik, University of California, Davis
Problems
1. The water molecule H2O was used as an example; it was mentioned that when water was rotated 180 degrees around an axis bisecting the oxygen, the molecule was superimposable on the original water molecule. How about CO2? Will it be like the water molecule since CO2 also has 2 atoms of oxygen?
Of course not, because every molecule has a different molecular shape. To recognize the symmetry of any molecule, the structure and the molecular shape of that molecule should be defined. The water molecule is bent but CO2 is not, and if CO2 is rotated 360 degrees around the axis bisecting the C atom, then it can be superimposed on the original molecule. We then see the symmetry for the CO2.
2. Why should all of the five symmetry elements be done on a molecule in order to find the point group the molecule belongs to? Why is performing only one or two of the symmetry elements not enough for recognizing the point group?
One or two of the symmetry elements will not be able to tell us everything about the molecule's symmetry, since those one or two properties do not tell us everything about the molecule. Also, while different molecules may have one or two symmetrical properties in common, the five properties will not be the same for all molecules.
3. What does the symbol Cn stand for and what does n represent? Why is it important to identify n?
C is the axis of rotation and n is the order of the axis.
4. How are the character tables helpful?
The character table tells us about all the operational elements performed on the molecule and indicates whether we have forgotten to perform any of the symmetry elements. The tables serve as a checklist because all the operational elements should be done on the molecule in order to find the point group of the molecule.
5. Why is important to find symmetry in molecules?
Symmetry tells us about bonding for that molecule. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/04%3A_Symmetry_and_Group_Theory/4.P%3A_Problems_%28under_construction%29.txt |
Molecular Orbital Theory
Molecular Orbital (MO) Theory is a sophisticated bonding model. It is generally considered to be more powerful than Lewis and Valence Bond Theories for predicting molecular properties; however, this power comes at the price of complexity. In its full development, MO Theory requires complex mathematics, though the ideas behind it are simple. Atomic orbitals (AOs) that are localized on individual atoms combine to make molecular orbitals (MOs) that are distributed over the molecule. The simplest example is the molecule dihydrogen (H2), in which two independent hydrogen 1s orbitals combine to form the $\sigma$ bonding MO and the $\sigma$ antibonding MO of the dihydrogen molecule (see figure). The MO’s are also called Linear Combinations of Atomic Orbitals (LCAO).
• 5.1: Formation of Molecular Orbitals from Atomic Orbitals
Molecular orbital theory extends from quantum theory and the atomic orbital wavefunctions ( ψ ) described by the Schrödinger equation. While the Schrödinger equation defines a Ψ for electrons in individual atoms, we can approximate a the molecular wavefunction ( Ψ would look like if we combined the ψ of individual atoms. The addition or subtraction of wavefunctions is termed linear combination of atomic orbitals (LCAO). Molecular orbital theory applied LCAO to describe bonding.
• 5.2: Homonuclear Diatomic Molecules
In this section you will be introduced to the molecular orbital diagrams of several homonuclear diatomic molecules. Homonuclear diatomic molecules are molecules made of exactly two identical atoms, and they are relatively simple.
• 5.3: Heteronuclear Diatomic Molecules
• 5.4: Larger (Polyatomic) Molecules
We can extend the method we used for diatomic molecules to draw the molecular orbitals of more complicated, polyatomic molecules (molecules with more than two atoms). To combine several different atoms in a molecular orbital diagram, we will group orbitals from different atoms into sets that match the symmetry of a central atom. These group orbitals are also referred to as symmetry adapted linear combinations (SALCs).
• 5.P: Problems
05: Molecular Orbitals
Molecular orbital theory extends from quantum theory and the atomic orbital wavefunctions ($\psi$) described by the Schrödinger equation. While the Schrödinger equation defines a $\Psi$ for electrons in individual atoms, we can approximate the molecular wavefunction (what $\Psi$ would look like if we combined the $\psi$ of individual atoms). The addition or subtraction of wavefunctions is termed linear combination of atomic orbitals (LCAO). Molecular orbital theory is applied to LCAO to describe bonding.
The LCAO for the wavefunction of two atoms ($\psi_a$ and $\psi_b$) is represented by the general expression below. The coefficients $c_a$ and $c_b$ quantify the contribution of each atomic $\psi$ to the molecular $\Psi$.
$\Psi=c_{a} \psi_{a}+c_{b} \psi_{b} \nonumber$
For two atomic $\psi$s to form a bond, three conditions must be satisfied:
• First, the distance between the atoms must be small enough to provide good overlap. This is because the two orbitals must be able to overlap using regions of $\psi^2$ where the probability of finding electrons is significant. If atoms are not close enough, $\psi_a$ and $\psi_b$ cannot interact productively. At the same time, the atoms must be far enough apart that their nuclei do not repel each other.
• The symmetry of the orbitals must be compatible such that regions of $\psi_a$ and $\psi_b$ with the same sign constructively interfere more than regions with opposite sign destructively interfere. In other words, for a productive interaction, there must be a relatively high probability of finding an electron between the two nuclei, and this depends on the symmetry (and sign of the $\psi$) of the atomic orbitals.
• Third, the energies of the atomic orbitals must be similar. Orbitals with similar energy combine to make the most stable bonding molecular orbitals. Atomic orbitals with very different energies form less stable bonding interactions, and the larger the difference in the atomic orbitals, the weaker the bonding interaction will be.
Atomic orbitals combine to form molecular orbitals when these three conditions are met. The result is a set of bonding molecular orbitals that are lower in energy than the original atomic orbital energies.
5.01: Formation of Molecular Orbitals from Atomic Orbitals
In the case of the hydrogen molecule, we took two atomic orbitals and combined them to form two molecular orbitals. These new molecular orbitals had different wavelengths than the two atomic orbitals: one had a longer wavelength and was a little lower in energy, while the other had a shorter wavelength and was a little higher in energy. If we take into account the energy of the two original atomic wavefunctions, and compare them to the total energy of the two new molecular wavefunctions, there is no change overall.
We started with two atomic orbitals, and by combining them we produced two molecular orbitals. Both of these ideas are useful in considering the formation of more complex molecules from individual atoms.
• The average energy of the orbitals has remained almost constant.
• Also, the number of wavefunctions has remained constant.
Of course, from the point of view of the two real electrons, some remarkable changes have occurred. Both of these electrons have adopted a longer wavelength and a lower energy and that has made all the difference. There is an occupied molecular orbital and an unoccupied molecular orbital; only the occupied orbital makes a real energetic contribution to the overall stability of the molecule. The unoccupied orbital is completely imaginary.
A bonding picture of He2 would look exactly the same, because it would also involve the overlap of 1s electrons on one atom with 1s electrons on the other atom. There would be a different electronic energy, however. That difference would affect the prospects of helium-helium bond formation.
• The electrons have lower kinetic energy in the bond than they had before bonding.
• Electronic energy has decreased. A stable bond has formed.
Exercise \(1\)
Construct molecular orbital diagrams for the following diatomic species and discuss the likelihood of bond formation in each case.
1. He2.
2. Li2.
3. Be2.
Answer | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/05%3A_Molecular_Orbitals/5.01%3A_Formation_of_Molecular_Orbitals_from_Atomic_Orbitals/5.1.01%3A_Molecular_Orbitals_from_s_Orbitals.txt |
Bonding with P-orbitals
Sigma bonding with p-orbitals
Other diatomic molecules in the upper right corner of the periodic table can be constructed in a similar way. Look at dinitrogen, N2.
We can think about how dinitrogen would form if two nitrogen atoms were placed close enough together to share electrons. Nitrogen has more electrons than hydrogen, so this interaction is more complicated.
In our qualitative examination of bonding in main group diatomics, we will take the approach used in Lewis structures and just look at the valence electrons. A quantitative molecular orbital calculation with a computer would not take this shortcut, but would include all of the electrons in the atoms that are bonding together.
Nitrogen has five valence electrons, and these electrons are found in the 2s and 2p levels. There are three possible atomic orbitals in the 2p level where some of these electrons could be found: px, py and pz. We need to look at the interaction between the s and px, py and pz orbitals on one nitrogen atom with the s and px, py and pz orbitals on the other nitrogen. That process could be extremely complicated, but:
• Orbital interactions are governed by symmetry.
Orbitals interact most easily with other orbitals that have the same element of symmetry. For now, we can simplify and say that orbitals on one atom only interact with the same type of orbitals on the other atom.
• s orbitals interact with s orbitals. We can already see how that will work out in dinitrogen, because that is what happened in dihydrogen.
• px orbitals interact with px orbitals.
• py orbitals interact with py orbitals.
• pz orbitals interact with pz orbitals.
Another complication here is that the s and p orbitals do not start out at the same energy level. When the orbitals mix, one combination goes up in energy and one goes down. Does the s antibonding combination go higher in energy than the combinations from p orbitals? Do the p bonding combinations go lower in energy than the combinations from s orbitals? We will simplify and assume that the s and p levels remain completely separate from each other. This is not always true, but the situation varies depending on what atoms we are dealing with.
• The combination of one s orbital with another is just like in hydrogen. Two original orbitals will combine and rearrange to produce two new orbitals.
• There is a bonding combination in which the orbitals are in phase. The new orbital produced has a longer wavelength than the original orbital. It is lower in energy.
• There is an antibonding combination in which the orbitals are out of phase. The new orbital produced has a shorter wavelength than the original orbital. It is higher in energy.
In considering the interaction of two p orbitals, we have to keep in mind that p orbitals are directional. A p orbital lies along a particular axis: x, y or z. The three p orbitals on nitrogen are all mutually perpendicular (or orthogonal) to each other. That situation is in contrast to s orbitals, which are spherical and thus look the same from any direction.
We first need to define one axis as lying along the N-N bond. It does not really matter which one. We arbitrarily say the N-N bond lies along the z axis. The pz orbitals have a different spatial relationship to each other compared to the py and px. The pz orbitals lie along the bond axis, whereas the py and px are orthogonal to it.
As the nitrogen atoms are brought together, one lobe on one pz orbital overlaps strongly with one lobe on the other pz orbital. The other lobes point away from each other and do not interact in any obvious way.
As with the s orbital, the pz orbitals can be in-phase or out-of-phase. The in-phase combination results in constructive interference. (Here, "in-phase" means the lobes that overlap are in-phase; for that to happen the two p orbitals are actually completely out-of-phase with each other mathematically, so that one orbital is the mirror image of the other.) This combination is at a longer wavelength than the original orbital. It is a lower energy combination.
The out-of-phase combination (meaning in this case that the overlapping lobes are out-of-phase) results in destructive interference. This combination is at shorter wavelength than the original orbital. It is a higher energy combination.
As a result, we have two different combinations stemming from two different p orbitals coming together in two different ways. We get a low-energy, in-phase, bonding combination and a high-energy, out-of-phase, antibonding combination.
What about those other p orbitals, the ones that do not lie along the bond axis? We'll take a look at that problem on the next page.
Exercise \(2\)
Draw an MO cartoon of a sigma bonding orbital formed by the overlap of two p orbitals between two oxygen atoms. Label the positions of the oxygen nuclei with the symbol "O". Label the O-O bond axis.
Answer
Exercise \(3\)
Chemical reactions can be described by MO diagrams too! Consider the following reaction in which a new sigma bond is formed.
Draw an MO diagram for the reaction above. In other words, start from the one frontier MO on each reactant to build the MO's of the new sigma bond in the product.
o Draw the orbital from the base (hydroxide) that is likely to donate its electrons.
o Draw the orbital from the acid (aluminum chloride) that is likely to accept electrons.
o Complete the MO mixing diagram of these two orbitals:
• Label the electron donating orbital
• Label the electron accepting orbital
• Populate the MO mixing diagram with electrons
o Draw a cartoon showing each reactant and product MO that contributes to the bonding interaction.
Answer
Pi-bonding with p-orbitals
Earlier, we saw that p orbitals that lie along the same axis can interact to form bonds.
Parallel, but not collinear, p orbitals can also interact with each other. They would approach each other side by side, above and below the bond axis between the two atoms. They can be close enough to each other to overlap, although they do not overlap as strongly as orbitals lying along the bond axis. They can make an in-phase combination, as shown below.
They could also make an out-of-phase combination, as shown below.
• parallel p orbitals can overlap to produce bonding and antibonding combinations.
• the resulting orbitals contain nodes along the bond axis.
• the electron density is found above and below the bond axis.
• this is called a p (pi) bond.
The illustration above is for one set of p orbitals that are orthogonal to the bond axis. The second picture shows the result of the constructive (or destructive) interference. A similar picture could be shown for the other set of p orbitals.
In a main group diatomic species like dinitrogen, one p orbital lying along the bond axis can engage in s bonding. The two p orbitals orthogonal to the bond axis can engage in p bonding. There will be both bonding and antibonding combinations.
Just as the sigma-bonding orbitals display progressively shorter wavelengths along the bonding axis as they go to higher energy, so do the pi bonding orbitals. In other words, there are more nodes in the higher-energy orbitals than in the lower-energy ones.
An important consequence of the spatial distribution or "shape" of a p orbital is that it is not symmetric with respect to the bond axis. An s orbital is not affected when the atom at one end of the bond is rotated with respect to the other. A p orbital is affected by such a rotation. If one atom turns with respect to the other, the p orbital would have to stretch to maintain the connection. The orbitals would not be able to overlap, so the connection between the atoms would be lost.
Exercise \(4\)
The combinations of ______________ atomic orbitals leads to σ orbitals.
Draw pictures.
Answer
The combinations of s + s OR s + p OR p + p OR s + d OR p + d atomic orbitals can lead to σ orbitals.
Exercise \(5\)
The combinations of ______________ atomic orbitals leads to π orbitals.
Draw pictures.
Answer
The combinations of side by side p + p or p + d atomic orbitals leads to π orbitals.
Exercise \(6\)
Which molecular orbital is typically the highest in energy?
a. p
b. σ
c. π*
d. π
e. σ*
Answer
e) σ*
Exercise \(7\)
Why would a core 1s orbital not interact with a valence 2s orbital?
Hint: Why is a Li2O bond stronger than a K2O bond?
Answer
Li+ and O2- are more similar in size than K+ and O2-, so the bond between Li+ and O2- is stronger.
The energy difference between any core orbitals and valence orbitals is too large, so they cannot interact. In order for orbitals to interact, the orbitals need to have the same symmetry, be in the same plane, and be similar in energy.
Exercise \(8\)
Add a few words to explain the ideas conveyed in these drawings.
Answer
When two parallel p orbitals combine out-of-phase, destructive interference occurs.
There is a node between the atoms.
The energy of the electrons increases.
When two parallel p orbitals combine in-phase, constructive interference occurs.
There is no node between the atoms; the electrons are found above and below the axis connecting the atoms.
The energy of the electrons decreases.
Attribution
Curated or created by Kathryn Haas | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/05%3A_Molecular_Orbitals/5.01%3A_Formation_of_Molecular_Orbitals_from_Atomic_Orbitals/5.1.02%3A_Molecular_Orbitals_from_p_Orbitals.txt |
In transition metals and other heavier elements, the $d$ orbitals may combine with other orbitals of compatible symmetry (and energy) to form molecular orbitals. Generally, there are three types of bonding and antibonding interactions that may occur with $d$ orbitals: sigma ($\sigma$), pi ($\pi$), and delta ($\delta$) bonds.
Sigma ($\sigma$) bonding with $d$ orbitals
$\sigma$ bonds are symmetric with respect to the inter-nuclear axis (in a diatomic molecule, this is the $z$ axis). An example of a $\sigma$ bond formed by $d$ orbitals is that of two $d_{z^2}$ orbitals (see Figure $2$). If a bonded atom is in a position other than on the $z$ axis (in an octahedral geometry, for example), $\sigma$ bonds can also form. For example, two $d_{x^2-y^2}$ orbitals on atoms bonded along the $x$ or $y$ axes could also form a $\sigma$ bond.
$d$ orbitals can also form $\sigma$ bonds with other types of orbitals with the appropriate symmetry. Examples of orbitals with appropriate symmetry are the $s$ orbital and certain $p$ orbitals on another atom, as shown below in Figure $2$.
Pi ($\pi$) bonding with $d$ orbitals
$\pi$ bonds are those with one node that is in-plane with the internuclear axis. A $\pi$ bond can form between two $d$ orbitals or between $d$ orbitals and other types of orbitals with comparable symmetry. An example of a $\pi$ bond between two $d$ orbitals is that formed by two $d_{xz}$ orbitals along the $z$ axis (shown in Figure $3$). $d$ orbitals can also form $\pi$ bonds using $p$ orbitals with compatible symmetry, as shown in Figure $3$.
Delta ($\delta$) bonding with $d$ orbitals
$\delta$ bonds are those with two nodes that are in-plane with the internuclear axis. $\delta$ bonds can form between two $d$ orbitals with appropriate symmetry. For example, when two atoms bond along the $z$ axis, the $d_{xy}$ orbitals and the two $d_{x^2-y^2}$ orbitals can form $\delta$ bonds (Figure $4$).
Incompatible orbitals
In the descriptions above, we focused on how bonds (and antibonds) can be formed with $d$ orbitals. All bonding and non-bonding interactions require that orbitals have compatible symmetry to form productive interactions. It is worth mentioning that orbitals with symmetry that is incompatible with the $d$ orbitals will not have bonding or antibonding interactions with $d$ orbitals. The figure below shows several sets of orbitals that are incompatible for bonding.
Curated or created by Kathryn Haas
5.1.04: Nonbonding Orbitals and Other Factors
The simplest case is when there is an even number of atomic orbitals that all combine to form strong bonding and antibonding orbitals. But what if there is an uneven number of atomic orbitals? Or what if there are some orbitals that don't meet the criteria for bonding? Or what if the bonding interactions are weak? In these cases, there will be molecular orbitals on the molecule that have non-bonding character.
It is important to note that the bonding, non-bonding, and antibonding nature of orbitals exist on a spectrum. Some bonding and anti-bonding orbitals may have some non-bonding character depending on where their energies lie with respect to the original atomic orbital energies. When molecular orbitals have energies similar to their original atomic orbitals, they will have some non-bonding character. The closer the energies of atomic and molecular orbitals, the more non-bonding the molecular orbitals.
Two major factors to consider
Uneven number of atomic orbitals: In the case that there is an uneven number of atomic orbitals with compatible symmetry, orbitals with non-bonding character will form. For example, in the case where three atomic orbitals combine, the most common result is formation of a low-energy bonding orbital, a high energy antibonding orbital, and a non-bonding orbital of intermediate energy (Figure \(1\)).
Differences in energy: Combination of orbitals with different energies may lead to orbitals with non-bonding character. Atomic orbitals that have similar energies will have the strongest interactions, and result in bonding molecular orbitals with much lower energies than the component atomic orbitals. On the other hand, atomic orbitals with very unequal energies have a weaker interaction because the molecular orbitals are closer in energy to the atomic orbital energies, thus there is less energy benefit to putting electrons in the bonding molecular orbitals (Figure \(2\)). When bonding or antibonding orbitals are close to the energies of the contributing atomic orbitals, those molecular orbitals may have some non-bonding character. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/05%3A_Molecular_Orbitals/5.01%3A_Formation_of_Molecular_Orbitals_from_Atomic_Orbitals/5.1.03%3A_Molecular_orbitals_from_d_orbitals.txt |
In this section you will be introduced to the molecular orbital diagrams of several homonuclear diatomic molecules. Homonuclear diatomic molecules are molecules made of exactly two identical atoms, and they are relatively simple.
• 5.2.1: Molecular Orbitals
There are several cases where our more elementary models of bonding (like Lewis Theory and Valence Bond Theory) fail to predict the actual molecular properties and reactivity. A classic example is the case of O₂ and its magnetic properties. At very cold temperatures, O₂ is attracted to a magnetic field, and thus it must be paramagnetic (unpaired electrons give rise to magnetism, see video). However, both its Lewis structure and Valance Bond Theory predict that O₂ is diamagnetic.
• 5.2.2: Orbital Mixing
Orbitals of compatible symmetry can combine, or mix, even when they have different energies. When sets of orbitals mix, it has the effect of decreasing the energy of the lower-energy set and increasing the energy of the higher-energy set.
• 5.2.3: Diatomic Molecules of the First and Second Periods
• 5.2.4: Photoelectron Spectroscopy
5.02: Homonuclear Diatomic Molecules
There are several cases where our more elementary models of bonding (like Lewis Theory and Valence Bond Theory) fail to predict the actual molecular properties and reactivity. A classic example is the case of $\ce{O_2}$ and its magnetic properties. At very cold temperatures, $\ce{O_2}$ is attracted to a magnetic field, and thus it must be paramagnetic (unpaired electrons give rise to magnetism). Watch the video below!
The magnetic properties of $\ce{O_2}$ are easily rationalized by its molecular orbital diagram. A molecular orbital diagram is a diagram that shows the relative energies and identities of each molecular orbital in a molecule. Figure $1$ shows a simplified and generic molecular orbital diagram for a second-row homonuclear diatomic molecule. The diagram is simplified in that it assumes that interactions are limited to degenerate orbitals from two atoms (see next section).
There are some things you should note as you inspect Figure $1$ (and keep these in mind as you draw your own diagrams!). First, notice that there are the same number of molecular orbitals as there are atomic orbitals. Second, notice that each orbital in the diagram is rigorously labeled using labels ($\sigma$ and $\pi$) that include the subscripts $u$ and $g$. These labels and subscripts indicate the symmetry of the orbitals. The $\sigma$ symbol indicates that the orbital is symmetric with respect to the internuclear axis, while the $\pi$ label indicates that there is one node along that axis. The $g$ and $u$ stand for gerade and ungerade, the German words for even and uneven, respectively. The subscript $g$ is given to orbitals that are even, or symmetric, with respect to an inversion center. The subscript $u$ is given to orbitals that are uneven, or antisymetric, with respect to an inversion center. The pictures of calculated molecular orbitals are shown in Figure $1$ to illustrate the symmetry of each orbital.
Another important thing to notice is that the diagram in Figure $1$ lacks electrons (because it is generic for any second-row diatomic molecule). If this were a complete molecular orbital diagram it would include the electrons for each atom and for the molecule. Electrons in molecular orbitals are filled in the same way an atomic orbital diagram would be filled, where electrons occupy lower energy orbitals before higher energy orbitals, and electrons occupy empty degenerate orbitals before pairing. A complete molecular orbital diagram would show whether the molecule is diamagnetic or paramagnetic. It can also be used to calculate the bond order of the molecule (the number of bonds between atoms) using the formula below:
$\text{Bond order } =\frac{1}{2}\left[\left(\begin{array}{c}\text { number of electrons } \ \text { in bonding orbitals }\end{array}\right)-\left(\begin{array}{c}\text { number of electrons } \ \text { in antibonding orbitals }\end{array}\right)\right] \nonumber$
In general, non-valence (core) electrons can be ignored because they contribute nothing to the bond order. In fact, many molecular orbital diagrams will ignore the core orbitals, as they are insignificant for bonding interactions and reactivity.
Now, to see how this molecular orbital diagram can explain the magnetic behaviour of $\ce{O2}$, complete the example below.
Example $1$
Let's change Figure $1$ to make it specific for $\ce{O2}$. Re-draw the MO diagram (no need to draw the shapes of orbitals). Fill in the correct number of electrons for each oxygen atom on either side of the diagram. Then, fill in the total molecular electrons in the center. Calculate the bond order and determine whether it is diamagnetic or paramagnetic.
Solution
The diagram in Figure $1$ includes core orbitals (the 1s) and valence electrons (2s, 2p). Therefore, we will consider all the electrons in an oxygen atom and a dioxygen molecule. An oxygen atom has eight total electrons. So we fill eight electrons into the atomic orbitals for the oxygen atom on the right, and eight electrons into the atomic orbitals for oxygen on the left. The total number of electrons for the molecule is sixteen, so fill in 16 electrons into the molecular orbitals, being sure to apply Hund's rule and the Aufbau principle. The result is a diagram that looks like the one drawn below in Figure $2$.
The bond order is calculated using the molecular orbitals (we can ignore atomic orbitals). There are 10 electrons in binding orbitals and 6 electrons in antibonding orbitals). This gives a bond order of $\frac{1}{2}(10-6)=2$. This bond order is consistent with valence bond theory!
This diagram indicates that dioxygen is paramagnetic; it has two unpaired electrons in the $\pi^*$ orbitals. This paramagnetic electron configuration explains why dioxygen is attacted to magnetic fields!
Exercise $1$
Draw the molecular orbital diagram for $\ce{F2}$; be sure to label your orbitals with the appropriate symmetry and count your orbitals to make sure that the total number of atomic orbitals and molecular orbitals is the same. As a shortcut, include only the valence orbitals and electrons. What is the bond order? Is the molecule diamagnetic or paramagnetic?
Answer
The valence orbitals on an F atom are 2s and 2p. And, there are seven valence electrons in F. This gives fourteen total valence electrons. The MO diagram of the valence molecular orbitals can be constructed by combining the valence 2s and valence 2p orbitals from each F atom. The bond order is 1 and the molecule is diamagnetic.
5.2.02: Orbital Mixing
In the previous section, we introduced a simplified molecular orbital (MO) diagram, assuming that interactions were limited to degenerate orbitals of compatible symmetry. A version of that simplified diagram for second-row homonuclear diatomics is shown on the left side of Figure $1$ (only valence orbitals shown).
In reality, orbitals of compatible symmetry can combine, or mix, even when they have different energies. When sets of orbitals mix, it has the effect of decreasing the energy of the lower-energy set and increasing the energy of the higher-energy set. There are two ways to explain mixing, described in the points below. The diagram on the right of side of Figure $1$ shows how energy levels are affected by orbital mixing.
• Starting from the simplified MO diagram and mixing molecular orbitals of like symmetry: If we start with the "no mixing" simplified diagram shown on the left, we can pick out the orbitals with like symmetry and consider what will happen if these molecular orbitals mix. For example, starting from the MO diagram on the left side of Figure $1$, we see that there are two $\sigma_g$ orbitals that have identical symmetry (hence they have identical symmetry labels). Upon mixing, the lower-energy $\sigma_g(2s)$ will decrease in energy while the higher-energy $\sigma_g(2p)$ will increase in energy. Likewise, there are two $\sigma_u^{*}$ orbitals that will mix, decreasing the energy of $\sigma_u^{*}(2s)$ while increasing the energy of $\sigma_u^{*}(2p)$.
• Starting from and mixing atomic orbitals of compatible symmetry: Starting from the atomic orbitals, we can select the orbitals that have compatible symmetry to make productive interactions, and combine them as a set to make molecular orbitals. From the two sets of atomic orbitals on the right panel in Figure $1$, there are two sets of four orbitals with compatable symmetry; one is the set of two $2s$ and two $2p_z$ orbitals. These four orbitals can combine to form molecular orbitals with $\sigma$ symmetry. They combine to form one lowest-energy bonding $\sigma_g$, a highest-energy $\sigma_u^{*}$, and two orbitals with intermediate energy ($\sigma_g$, and $\sigma_u^{*}$). This treatment can be expressed as the linear combination of four atomic orbitals: $\Psi=c_{1} \psi\left(2 s_{a}\right) \pm c_{2} \psi\left(2 s_{b}\right) \pm c_{3} \psi\left(2 p_{a}\right) \pm c_{4} \psi\left(2 p_{b}\right) \nonumber$ where the coefficients $c_1=c_2$ and $c_3=c_4$ for homonuclear diatomic molecules.
In the case of homonuclear diatomic molecules of the second row, orbital mixing has important consequences for the energetic order of the $\sigma_g(2p)$ and $\pi_u(2p)$ orbitals. This will be discussed in the next section. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/05%3A_Molecular_Orbitals/5.02%3A_Homonuclear_Diatomic_Molecules/5.2.01%3A_Molecular_Orbitals.txt |
First Period Homonuclear Diatomic Molecules
In the first row of the periodic table, the valence atomic orbitals are $1s$. There are two possible homonuclear diatomic molecules of the first period:
Dihydrogen, H2 $[\sigma_g^2(1s)]$: This is the simplest diatomic molecule. It has only two molecular orbitals ($\sigma_g$ and $\sigma_u^{*}$), two electrons, a bond order of 1, and is diamagnetic. Its bond length is 74 pm. MO theory would lead us to expect bond order to decrease and bond length to increase if we either add or subtract one electron. The calculated bond length for the $\ce{H_2^{+}}$ ion is approximately 105 pm.1
Dihelium, He2 $[\sigma_g^2\sigma_u^{*2}(1s)]$: This molecule has a bond order of zero due to equal number of electrons in bonding and antibonding orbitals. Like other nobel gases, He exists in the atomic form and does not form bonds at ordinary temperatures and pressures.
Exercise $1$
Draw the complete molecular orbital diagrams for $\ce{H2}$ and for $\ce{He2}$. Include sketches of the atomic and molecular orbitals.
Answer
A complete molecular orbital diagram includes all atomic orbitals and molecular orbitals, their symmetry labels, and electron filling.
Second Period Homonuclear Diatomic Molecules
The second period elements span from Li to Ne. The valence orbitals are 2s and 2p. In their molecular orbital diagrams, non-valence orbitals (1s in this case) are often disregarded in molecular orbital diagrams.
Orbital mixing has significant consequences for the magnetic and spectroscopic properties of second period homonuclear diatomic molecules because it affects the order of filling of the $\sigma_g(2p)$ and $\pi_u(2p)$ orbitals. Early in period 2 (up to and including nitrogen), the $\pi_u(2p)$ orbitals are lower in energy than the $\sigma_g(2p)$ (see Figure $1)$. However, later in period 2, the $\sigma_g(2p)$ orbitals are pulled to a lower energy. This lowering in energy of $\sigma_g(2p)$ is not unique; all of the $\sigma$ orbitals in the molecule are pulled to lower energy due to the increasing positive charge of the nucleus. The $\pi$ orbitals in the molecule are also affected, but to a much lesser extent than $\sigma$ orbitals. The reason has to do with the high penetration of $s$ atomic orbitals compared to $p$ atomic orbitals (recall our previous discussion on penetration and shielding, and its effect on periodic trends). The $\sigma$ molecular orbitals have more $s$ character and thus their energy is more influenced by increasing nuclear charge. As nuclear charge increases, the energy of the $\sigma_g(2p)$ orbital is lowered significantly more than the energy of the $\pi_u(2p)$ orbitals (Figure $1)$.
Dilithium, Li2 $[\sigma_g^2(2s)]$: This molecule has a bond order of one and is observed experimentally in the gas phase to have one Li-Li bond.
Diberylium, Be2 $[\sigma_g^2\sigma_u^{*2}(2s)]$: This molecule has a bond order of zero due to the equal number of electrons in bonding and antibonding orbitals. Although Be2 does not exist under ordinary conditions, it can be produced in a laboratory and its bond length measured (Figure $2$). Although the bond is very weak, its bond length is surprisingly ordinary for a covalent bond of the second period elements.2
Diboron, B2 $[\sigma_g^2\sigma_u^{*2}(2s)\pi_u^1\pi_u^1(2p)]$: The case of diboron is one that is much better described by molecular orbital theory than by Lewis structures or valence bond theory. This molecule has a bond order of one. The molecular orbital description of diboron also predicts, accurately, that diboron is paramagnetic. The paramagnetism is a consequence of orbital mixing, resulting in the $\sigma_g$ orbital's being at a higher energy than the two degenerate $\pi_u^*$ orbitals.
Dicarbon, C2 $[\sigma_g^2\sigma_u^{*2}(2s)\pi_u^2\pi_u^2(2p)]$: This molecule has a bond order of two. Molecular orbital theory predicts two bonds with $\pi$ symmetry, and no $\sigma$ bonding. C2 is rare in nature because its allotrope, diamond, is much more stable.
Dinitrogen, N2 $[\sigma_g^2\sigma_u^{*2}(2s)\pi_u^2\pi_u^2\sigma_g^2(2p)]$: This molecule is predicted to have a triple bond. This prediction is consistent with its short bond length and bond dissociation energy. The energies of the $\sigma_g(2p)$ and $\pi_u(2p)$ orbitals are very close, and their relative energy levels have been a subject of some debate (see next section for discussion).
Dioxygen, O2 $[\sigma_g^2\sigma_u^{*2}(2s)\sigma_g^2\pi_u^2\pi_u^2\pi_g^{*1}\pi_g^{*1}(2p)]$: This is another case where valence bond theory fails to predict actual properties. Molecular orbital theory correctly predicts that dioxygen is paramagnetic, with a bond order of two. Here, the molecular orbital diagram returns to its "normal" order of orbitals where orbital mixing could be somewhat ignored, and where $\sigma_g(2p)$ is lower in energy than $\pi_u(2p)$.
Difluorine, F2 $[\sigma_g^2\sigma_u^{*2}(2s)\sigma_g^2\pi_u^2\pi_u^2\pi_g^{*2}\pi_g^{*2}(2p)]$: This molecule has a bond order of one and like oxygen, the $\sigma_g(2p)$ is lower in energy than $\pi_u(2p)$.
Dineon, Ne2 $[\sigma_g^2\sigma_u^{*2}(2s)\sigma_g^2\pi_u^2\pi_u^2\pi_g^{*2}\pi_g^{*2}\sigma_u^{*2}(2p)]$: Like other noble gases, Ne exists in the atomic form and does not form bonds at ordinary temperatures and pressures. Like Be2, Ne2 is an unstable species that has been created in extreme laboratory conditions and its bond length has been measured (Figure $2$)
Exercise $2$
Draw the complete molecular orbital diagram for O2. Show calculation of its bond order and tell whether it is diamagnetic or paramagnetic.
Answer
O2 is paramagnetic with a bond order of 2. Its $\sigma_g(2p)$ molecular orbital is lower in energy that the set of $\pi_{u}(2p)$ orbitals.
Bond order $=\frac{1}{2}\left[\left(\begin{array}{c}\text { 8 electrons in} \ \text { valence bonding orbitals }\end{array}\right)-\left(\begin{array}{c}\text {4 electrons in} \ \text { valence antibonding orbitals }\end{array}\right)\right]$
Exercise $3$: The Peroxide Ion
Use a qualitative molecular orbital energy-level diagram to predict the electron configuration, the bond order, and the number of unpaired electrons in the peroxide ion (O22).
Answer
This diagram looks similar to that of $\ce{O2}$, except that there are two additional electrons.
$\left ( \sigma _{g}(2s) \right )^{2}\left ( \sigma_u ^{\star }(2s) \right )^{2}\left ( \sigma _g(2p) \right )^{2}\left ( \pi _{u}(2p) \right )^{4}\left ( \pi _g(2p) \right )^{4}$; bond order of 1; no unpaired electrons.
Bond Lengths in Homonuclear Diatomic Molecules
The trends in experimental bond lengths are predicted by molecular orbital theory, specifically by the calculated bond order. The values of bond order and experimental bond lengths for the second period diatomic molecules are given in Figure $1$, and shown in graphical format on the plot in Figure $2$. From the plot, we can see that bond length correlates well with bond order, with a minimum bond length occurring where the bond order is greatest ($\ce{N2}$). The shortest bond distance is at $\ce{N2}$ due to its high bond order of 3. From $\ce{N2}$ to $\ce{F2}$ the bond distance increases despite the fact that atomic radius decreases.
5.2.04: Photoelectron Spectroscopy
A photoelectron spectrum can show the relative energies of occupied molecular orbitals by ionization. The ionization energy is a direct measure of the energy required to just remove the electron concerned from its initial level to the vacuum level (free electron). Photoelectron spectroscopy measures the relative energies of the ground and excited positive ion states that are obtained by removal of single electrons from the neutral molecule.
$A + \text{photon} \rightarrow A^+ + e^- \nonumber$
The information obtained from photoelectron spectroscopy is typically discussed in terms of the electronic structure and bonding in the ground states of neutral molecules, with ionization of electrons occurring from bonding molecular orbitals, lone pairs, antibonding molecular orbitals, or atomic cores. These descriptions reflect the relationship of ionization energies to the molecular orbital model of electronic structure.
Ionization energies are directly related to the energies of molecular orbitals (by Koopmans' theorem).
Example: Photoelectron spectrum of dihydrogen
The molecular orbital description of dihydrogen involves two $1s$ atomic orbitals generating two molecular orbitals: a bonding $\sigma_g$ and an antibonding $\sigma_u^*$. The two electrons occupy the $\sigma_g$ bonding orbital, leaving the molecule with a bond order of one (Figure $1$). The PES spectrum of dihydrogen (Figure $1$) has a single band that corresponds to the ionization of one electron from the $\sigma_g$. The multiple peaks are due to electrons ejecting from a range of stimulated vibrational energy levels. When extensive vibrational structure is resolved in a PES molecular orbital, then the removal of an electron from that molecular orbital induces a significant change in the bonding (in this case an increase in the bond length due to decrease in bond order).
Example: Photoelectron spectrum of dinitrogen
Diatomic nitrogen is more complex than hydrogen since multiple molecular orbitals are occupied. Five molecular orbitals are occupied; two of them are degenerate. Three bands in the photoelectron spectrum correspond to ionization of an electron in $\sigma_g(2p)$, $\pi_u(2p)$ and $\sigma_u^*(2s)$ molecular orbitals. Ionization of the fourth type of orbital, $\sigma_g(2s)$, does not appear in Figure $2$ because it is either off scale or because the incident light $h\nu$ used did not have sufficient energy to ionize electrons in that deeply stabilized molecular orbital. Note that extensive vibrational structure for the $\pi_u(2p)$ band indicates that the removal of an electron from this molecular orbital causes a significant change in the bonding.
From the photoelectron spectrum of dinitrogen, we can see that the electrons in the $\sigma_g(2p)$ orbital can be ionized using less energy than required to ionize electrons in the $\pi_u(2p)$ orbital. This is evidence for $\sigma_g(2p)$ existing at a higher energy than the $\pi_u(2p)$ orbitals. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/05%3A_Molecular_Orbitals/5.02%3A_Homonuclear_Diatomic_Molecules/5.2.03%3A_Diatomic_Molecules_of_the_First_and_Second_Periods.txt |
Diatomic molecules with two non-identical atoms are called heteronuclear diatomic molecules. When atoms are not identical, the molecule forms by combining atomic orbitals of unequal energies. The result is a polar bond in which atomic orbitals contribute unevenly to each molecular orbital.
The application of molecular orbital theory to heteronuclear diatomic molecules is similar to the case of homonuclear diatomics, except that the atomic orbitals from each atom have different energies and contribute unequally to molecular orbitals. Recall that atomic orbitals must have compatible symmetry and similar energy to combine into molecular orbitals. In the case where atomic orbitals of like symmetry have different energies, they combine less favorably than orbitals that are closer to one another in energy. As a general rule, orbitals that have energy differences of greater than 10-14 eV do not combine favorably. In the molecular orbital diagram, the closer a molecular orbital is to an atomic orbital, the more that atomic orbital contributes to the molecular orbital. This last point is helpful for back-of-the napkin estimations of what the molecular orbitals "look" like.
In this section, you should learn how to generate molecular orbital diagrams of heteronuclear diatomic molecules. To approach such a problem, we must start with a knowledge of the relative energies of electrons in different atomic orbitals. In other words, we need knowledge of the orbital potential energies (or orbital ionization energies).
Sources:
• Gray, Harry. Electrons and Chemical Bonding, Benjamin, 1964.
• Miessler, Gary L, and Donald A. Tarr. Inorganic Chemistry. Upper Saddle River, N.J: Pearson Education, 2014. Print.
5.03: Heteronuclear Diatomic Molecules
To generate molecular orbital diagrams of heteronuclear diatomic molecules, we must start with a knowledge of the relative energies of electrons in different atomic orbitals. In other words, we need knowledge of the orbital potential energies (or orbital ionization energies).
Orbital Ionization Energies
There are two approaches you can use to "know" or estimate the atomic orbital energy levels.
1. Use a table of atomic orbital ionization energies, like those found in Table $1$.
2. When you do not have access to a table of values like the one below, use periodic trends in electronegativity and/or ionization energies as your guide to approximate relative values for different atoms.
Table $1$: These are one-electron ionization energies of the valence orbitals calculated by average energies of both the ground-state and ionized-state configurations. (From Harry Gray, “Electrons and Chemical Bonding,” Benjamin, 1964, Appendix)
$\begin{array} {|c|ccccccc|c|ccc|} \hline Atom & 1s & 2s & 2p & 3s & 3p & 4s & 4p & Atom & 3d & 4s & 4p \ \hline H & -13.64 & & & & & & &Sc & -4.71 & -5.70 & -3.22 \ He & -24.55 & & & & & & & Ti & -5.58 & -6.08 & -3.35 \ Li & & -5.46 & & & & & & V & -6.32 & -6.32 & -3.47\ Be & & -9.30 & & & & & & Cr & -7.19 & -6.57 & -3.47 \ B & & -14.01 & -8.31 & & & & & Mn & -7.93 & -6.82 & -3.60 \ C & & -19.47 & -10.66 & & & & & Fe & -8.68 & -7.07 & -3.72\ N & & -25.54 & -13.14 & & & & & Ni & -10.04 & -7.56 & -3.84\ O & & -32.36 & -15.87 & & & & & Cu & -10.66 & -7.69 & -3.97 \ F & & -46.37 & -18.72 & & & & \ Ne & & -48.48 & -21.57 & & & & \ Na & & & & -5.21 & & & \ Mg & & & & -7.69 & & & \ Al & & & & -11.28 & -5.95 & & \ Si & & & & -15.00 & -7.81 & & \ P & & & & -18.72 & -10.17 & & \ S & & & & -20.71 & -11.65 & & \ Cl & & & & -25.29 & -13.76 & & \ Ar & & & & -29.26 & -15.87 & & \ K & & & & & & -4.34 & \ Ca & & & & & & -6.08 & \ Zn & & & & & & -9.42 & \ Ga & & & & & & -12.65 & -5.95 \ Ge & & & & & & -15.62 & -7.56 \ As & & & & & & -17.61 & -9.05 \ Se & & & & & & -20.83 & -10.79 \ Br & & & & & & -24.05 & -12.52 \ Kr & & & & & & -27.52 & -14.26 \ \hline \end{array} \nonumber$ | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/05%3A_Molecular_Orbitals/5.03%3A_Heteronuclear_Diatomic_Molecules/5.3.01%3A_Orbital_ionization_energies.txt |
Molecular orbital diagrams for heteronuclear diatomic molecules
The molecular orbital diagram of a heteronuclear diatomic molecule is approached in a way similar to that of a homonuclear diatomic molecule. The orbital diagrams may also look similar. A major difference is that the more electronegative atom will have orbitals at a lower energy level. Two examples of heteronuclear diatomic molecules will be explored below as illustrative examples.
Carbon monoxide MO diagram
Carbon monoxide is an example of a heteronuclear diatomic molecule where both atoms are second-row elements. The valence molecular orbitals in both atoms are the $2s$ and $2p$ orbitals. The molecular orbital diagram for carbon monoxide (Figure $1$) is constructed in a way similar to how you would construct dicarbon or dioxygen, except that the oxygen orbitals have a lower potential energy than analogous carbon orbitals. The labeling of molecular orbitals in this diagram follows a convention by which orbitals are given serial labels according to type of orbital ($\sigma$, $\pi$, etc.). The lowest energy orbitals of any type are assigned a value of 1 and higher energy orbitals of the same type are assigned by increasing intervals (..2, 3, 4...). The orbital labeling system described previously is inappropriate for heteronuclear diatomic molecules that cannot be assigned $g$ and $u$ subscripts.
A consequence of unequal atomic orbital energy levels is that orbital mixing is significant. Notice the order of the molecular orbitals labeled $1\pi$ and $3\sigma$ in Figure $1$. This is a similar order of $\pi$ and $\sigma$ orbitals to the one we saw in the case of the $\sigma_g$ and $\pi_u$ orbitals of $N_2$ and lighter diatomics of the second period. Because the oxygen $2p_z$ orbital is close in energy to both the carbon $2p_z$ and carbon $2s$, these three orbitals will have significant interaction (mixing). The result is an increase in the energy of the $3\sigma$ orbital and a decrease in energy of the $2\sigma^*$ orbital, resulting in the diagram shown in Figure $1$.
In the case of carbon monoxide (Figure $1$), atomic orbitals contribute unequally to each molecular orbital. For example, because the $2s$ orbital of oxygen is very close in energy to the $2\sigma$ moelcular orbital, it contributes to that molecular orbital more than the $2s$ orbital from carbon. Notice the shape of this $2\sigma$ orbital and how it is unevenly distributed over the two atoms; it is more heavily distributed on the oxygen because it is most like the oxygen $2s$. This is in line with the assumption that electron density is distributed more on oxygen because it is more electronegative than carbon. Likewise, the $1\pi$ orbitals are unevenly distributed, with more distribution close to the oxygen.
Exercise $1$
Examine the shape of the $3\sigma$ orbital of carbon monoxide in Figure $1$. Describe what ways this shape is different from the shape of the $\sigma_g$ orbitals from second period homonuclear diatomic molecules (see Fig. 5.2.1.1). Rationalize these differences. Both orbitals are re-created below for convenience.
Answer
The 3$\sigma$ orbital is like the $\sigma_g$ in that it has three lobes and two nodes distributed along the internuclear bond. They are different in their distribution. The two external lobes of $\sigma_g$ are evenly distributed because they are an equal combination of two $p_z$ orbitals (one from each atom). The $3\sigma$ orbital is more heavily distributed toward the carbon atom, the less electronegative atom, than toward the oxygen. The unequal distribution of $3\sigma$ is apparent in the unequal sizes of its exterior lobes and the uneven shape of the interior lobe. The heavier distribution within the exterior lobe on carbon is caused by the mixing of the carbon $2s$ orbital with carbon and oxygen $2p_z$ orbitals. The uneven shape of the interior lobe, where it leans toward oxygen, is best explained by the fact that the $3\sigma$ orbital is closer in energy to the oxygen $2p_z$ than the carbon $2p_z$.
Hydrogen fluoride MO diagram
Hydrogen fluoride is an example of a heteronuclear diatomic molecule in which the two atoms are from different periods. In this case, the valence orbital of H is $1s$ while those of F are $2s$ and $2p$. The molecular orbital diagram for HF is shown in Figure $2$.
Three of these orbitals have compatible symmetry for mixing; these are the hydrogen $1s$, fluorine $2s$, and fluorine $2p$. However, the extent to which they will interact depends on their relative energies. Fluorine is more electronegative than H, and the fluorine atom has a higher first ionization energy than does hydrogen. From these trends, we can expect that the fluorine valence orbitals are lower in energy than that of hydrogen. From Table 5.3.1, we find that the 1s orbital of H (-13.6 eV) is higher in energy than both fluorine orbitals (-18.7 and -40.2 eV, respectively for $2p$ and $2s$). The energies of hydrogen $1s$ and fluorine $2p$ are a good match for combination; however, the fluorine $2s$ orbital is much too different to create a productive interaction. Therefore, we expect that the fluorine $2s$ will create a non-bonding molecular orbital, while the $1s$ and $2p_z$ orbitals combine to make $\sigma$ bonding and $\sigma^*$ antibonding molecular orbitals. The remaining $2p_x$ and $2p_y$ orbitals do not have compatible symmetry for bonding with hydrogen, and they will form non-bonding $\pi$ molecular orbitals. The non-bonding orbitals will have similar energy and character as their component atomic orbitals.
Chemical reactions take place at the HOMO and LUMO orbitals
Knowledge of molecular orbital diagrams, and the shapes of molecular orbitals, can be used to accurately explain and predict chemical reactivity. Chemical reactions take place using the highest occupied molecular orbitals (HOMO) of a nucleophile or Lewis base, and the lowest unoccupied molecular orbital (LUMO) of an electrophile or Lewis acid.
Lewis bases react using electrons in the HOMO, while Lewis acids react using the empty LUMO.
Example: Reactivity of CO with metal ions
CO is an excellent ligand for many metal ions. In fact, the strong affinity between CO and the heme iron (Fe) ions in hemoglobin can explain the mechanism of carbon monoxide poisoning. When CO binds in place of $O_2$ to hemoglobin, that hemoglobin can no longer carry $O_2$ to tissue cells. CO binding to hemoglobin is strong and practically irreversible. When CO binds to metal ions, it does so through the carbon atom. This is contrary to expectations based on the Lewis structure and the known bond polarity, where electron density is polarized toward oxygen. The distribution of the electron density of the HOMO of CO can explain this observation!
Exercise $2$
Refer to the MO diagram for CO. Identify the HOMO and explain why CO bonds to metal ions through the carbon atom rather than through the oxygen atom.
Answer
In the interaction between CO and a metal ion, CO would act as a Lewis base; thus it will react using electrons in its HOMO. The MO diagram for CO is shown in Figure $1$: the HOMO is the $3\sigma$ orbital (also discussed in Exercise $1$). The electron density of that MO is centered around the carbon atom, thus the carbon atom will be a better Lewis base than the O atom.
5.3.03: Ionic Compounds and Molecular Orbitals
Ionic interactions lie at one extreme on a spectrum of bonding. On the opposite end of the spectrum are the non-polar covalent bonds (e.g., homonuclear diatomics). In these molecules, molecular orbitals are formed by equal-energy atomic orbitals, resulting in electron density evenly distributed over the molecule. In the middle of the spectrum are the cases of polar covalent bonds (e.g., heteronuclear diatomics), in which atomic orbitals of unequal energies contribute unequally to molecular orbitals, resulting in uneven distribution of electron density across the molecule. In the case of polar bonds, the electron density is shifted toward the more electronegative atom since that atom contributes more to the lowest energy bonding molecular orbitals. Molecular orbital diagrams can be drawn for ionic compounds as if they are extremely polar bonds in which electrons are not only shifted toward, but are transferred completely to the more electronegative atom.
Example: NaCl
In $\ce{NaCl}$, the sodium $3s$ orbital (-5.2 eV) is significantly higher in energy than the chlorine valence orbitals. The chlorine $3s$ and $3p_z$ orbitals have compatible symmetry, yet only the $3p_z$ orbital (-13.8 eV) is close enough in energy to interact with the Na $3s$; still, the energy difference is large enough to make bonding weak. The Na $3s$ orbital combines with Cl $3p_z$ to form the molecular orbitals labeled $4\sigma$ and $4\sigma^*$ in Figure $1$. The $4\sigma$ orbital is weakly bonding, but is very close in energy to the Cl $3p_z$ orbital, and is mostly Cl-like in character. Notice that all $\sigma$ orbitals look very much like either $s$ or $p$ orbitals centered on the $\ce{Cl}$ atom, while the $4\sigma^*$ orbital is centered almost entirely on Na. The lack of molecular orbitals that are distributed over both atoms at once is consistent with a lack of significant covalent bond character in $\ce{NaCl}$. The bonding here is characterized by transfer of one electron from $\ce{Na}$ to $\ce{Cl}$ and is almost entirely electrostatic. Bonding that is mostly electrostatic in character is non-directional, unlike true covalent bonding.
Exercise $1$
Draw the molecular orbital diagram for $\ce{LiF}$. Make sure to label all molecular orbitals appropriately, and specify whether they are mostly bonding, non-bonding, or antibonding. Identify the HOMO and LUMO. Sketch the approximate shapes of all orbitals.
Answer
We expect $\ce{LiF}$ to be an ionic compound because the energy difference in valence orbitals is at least 10-14 eV. (See Table 5.3.1) There are a variety of ways used to label molecular orbitals. In the figure below, we are using the convention of labeling each type of orbital with numbers starting from the lowest-energy orbitals. The $1\sigma$ orbital would be mostly F $1s$ in character and is not shown. The $2\sigma$ orbital is mostly non-bonding in nature, although it has a very small contribution from $2s$ of Li due to compatible symmetry. The $3\sigma$ orbital is slightly bonding but it is mostly F$2p$ in character. The two $1\pi$ orbitals are completely non-bonding. The $3\sigma^*$ orbital is antibonding.
• HOMO is $1\pi$
• LUMO is $3\sigma^*$ | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/05%3A_Molecular_Orbitals/5.03%3A_Heteronuclear_Diatomic_Molecules/5.3.02%3A_Polar_bonds.txt |
We can extend the method we used for diatomic molecules to draw the molecular orbitals of more complicated, polyatomic molecules (molecules with more than two atoms). To combine several different atoms in a molecular orbital diagram, we will group orbitals from different atoms into sets that match the symmetry of a central atom. These group orbitals are also referred to as symmetry adapted linear combinations (SALCs). We'll use a stepwise approach to do this as summarized below.
Symmetry adapted linear combinations (SALCs)
We need SALCs (aka group orbitals) to draw molecular orbital (MO) diagrams of polyatomic molecules. SALCs are groups of orbitals on pendant atoms. These groups of orbitals much match the symmetry of valence orbitals on the central atom in order to create a productive interaction. When we combine SALCs with the atomic orbitals on the central atom, we can generate an MO diagram that gives us information about the molecule’s bonding and electronic states.
Below is a set of steps you can follow to find SALCs and draw an MO Diagram. Each step will be illustrated in detail through the examples in the subsections that follow this page.
1. Find the point group of the molecule and assign Cartesian coordinates so that $z$ is the principal axis.
Sometimes we can simplify things by looking only at the point group of the relevant orbitals.* Only simplify when instructed to do so.
2. Identify and count the pendant atoms' valence orbitals.
Is there more than 1 type for each atom (ex. just s, or s and p?) We expect 1 SALC for each ligand orbital.
3. Generate a reducible representation ($\Gamma$) for the group of pendant atom orbitals using the appropriate character table.
You have to do this for each set of orbital types. If you only have s orbitals for each ligand, you only need to generate 1 reducible representation. If you have $p$ orbitals, you would generate additional $\Gamma$'s for $p_x$, $p_y$, and $p_z$.
4. Break the $\Gamma$ into its component irreducible representations from the character table. Note the symmetry of each irreducible representation, its associated orbitals, and their degeneracy.
5. If you are asked to sketch the shapes of SALCs, determine what they “look like” using one of the following strategies.
(a) Shortcut: if the reducible representations are listed with s, p, or d orbitals in the character table, just draw them (… and skip steps b-c)
(b) Systematic (Projection Operator) method: Draw an expanded character table and draw your molecule with each ligand identified by letters or numbers (a,b,c... or i,j,k... or 1,2,3...) Determine where each pendant atom/orbital ends up under each of the operations of your expanded table. Project the values for each irreducible representation onto the chart you created and then add up the values. Positive and negative values are opposite signed orbitals.
6. Draw the MO diagram by combining SALCs with AO’s of like symmetry. When drawing SALC energy levels, remember that the more nodal planes in your SALC orbital drawing, the higher the energy for that SALC orbital.
* When we are focused on orbitals, as in the case for finding group orbitals and drawing molecular orbital diagrams, the symmetry of the orbitals is what we are interested in. In the case of high-symmetry point groups, like $D_{\infty h}$ and $C_{\infty v}$, even though the molecule may have a $C_\infty$ axis, the orbitals do not necessarily retain this symmetry element. For example, the $p_x$ orbital would not have a $C_\infty$ axis, but rather a $C_2$ axis. It is sufficient and useful to substitute $D_{2h}$ for $D_{\infty h}$ and $C_{2v}$ for $C_{\infty v}$ to simplify the problem.
• 5.4.1: Bifluoride anion
• 5.4.2: Carbon Dioxide
Carbon dioxide is another linear molecule. This example is slightly more complex than the previous example of the bifluoride anion. While bifluoride had only one valence orbital to consider in its central H atom (the 1s orbital), carbon dioxide has a larger central atom, and thus more valence orbitals that will interact with SALCs.
• 5.4.3: H₂O
Water is a bent molecule, and so it is important to remember that interactions of pendant ligands are dependent on their positions in space. You should consider the positions of the three atoms in water to be essentially fixed in relation to each other. The process for constructing the molecular orbital diagram for a non-linear molecule, like water, is similar to the process for linear molecules. We will walk through the steps below to construct the molecular orbital diagram of water.
• 5.4.4: NH₃
• 5.4.5: CO₂ (Revisted with Projection Operators)
• 5.4.6: BF₃
BF₃ is more complex than previous examples because it is the first case in which there are multiple types of valence orbitals on the pendant atoms. BF₃ possesses s and p orbitals on both the central atom and all of the pendant atoms. We can follow the same steps that we have previously to derive other molecular orbital diagrams; however, there is one important difference: we will treat each type of pendant orbital as an individual set of SALCs.
5.04: Larger (Polyatomic) Molecules
Finding SALCs and drawing the MO diagram for [F-H-F]$^-$
The linear anion [F-H-F]$^-$ is a good place to start as an example to illustrate the process of generating pendant atom SALCs and then constructing a molecular orbital diagram for a polyatomic molecule. We will proceed using the steps outlined on the previous page for generating SALCs and a molecular orbital diagram.
Step 1. Find the point group of the molecule and assign Cartesian coordinates so that z is the principal axis.
We begin by assigning the appropriate point group for this molecule: $D_{\infty h}$ (Figure $1$). As mentioned in the previous page, it is useful to substitute $D_{2h}$ for $D_{\infty h}$ when generating SALCs and molecular orbital diagrams. The $z$ axis is assigned to be colinear with the principal axis, and in this case is the same as the $C_\infty$ axis (Figure $1$).
Step 2. Identify and count the pendant atoms' valence orbitals.
The next step is to identify the valence orbitals on the pendant F atoms that will form SALCs. In most cases, you should consider all of the valence orbitals. In this case, each of the fluorine atoms has four valence orbitals ($2s$, $2p_x$, $2p_y$, and $2p_z$). From these eight fluorine valence orbitals, we should expect eight group orbitals (SALCs).
3. - 5. Generate SALCS (shortcut)
To draw these SALCs for this molecule is rather simple, and we don't need to follow all the steps of finding the $\Gamma$'s and reducing them. Rather, you can proceed as if you are creating bonding and antibonding molecular orbitals between the two F atoms, except that the F orbitals are separated by the H atom. Approximate sketches of the eight SALCs from F valence orbitals are shown in Figure $2$.
6. Draw the MO diagram by combining SALCs with AO’s of like symmetry.
SALCs can productively interact with the central atom only when symmetry is compatible. Just some of the groups of orbitals in Figure $2$ possess appropriate symmetry to combine with the hydrogen atom valence orbital (the H $1s$). In this simple case, you can decide whether orbitals have compatible symmetry by visually inspecting the shapes of the group orbitals. The F $2p_y$ and $2p_x$ orbitals do not have appropriate symmetry to bond to the H $1s$ orbital because the nodes of these orbitals run through the center of the H $1s$ orbital, thus we can eliminate all SALCs composed from F $2p_x$ and $2p_y$ (these are SALCs numbered 3-6 in Figure $2$). On the other hand, the F $2s$ and $2p_z$ orbitals individually do have appropriate shape and direction in space for productive interaction with an H $1s$ orbital. However only SALCs where the entire group has appropriate symmetry will combine with H $1s$ to produce bonding or antibonding molecular orbitals. Only the SALCs labeled with numbers 1 and 7 can combine with an $s$ orbital in the center of the group. The ways in which these SALCs are able to combine with H $1s$ are illustrated in Figure $3$.
Before we assume that both SALC-1 and SALC-7 will combine with the H $1s$ orbital, we must consider the energies of all atomic orbitals. The F $2p_z$ orbital has a potential energy of $-18.7$ eV (see Table 5.3.1). This is a good match for the H $1s$ orbital (-13.6 eV). However, the F $2s$ orbital has a much lower energy of $-46.37$ eV and would have weak interaction with the H $1s$.
The molecular orbital diagram for [F-H-F]$^-$ is shown in Figure $4$. Notice that all atomic orbitals, group orbitals (SALCs), and molecular orbitals in Figure $4$ are assigned a symmetry label that corresponds to each element's symmetry under the $D_{2h}$ point group. The molecular orbital labels correspond to lower-case Mulliken Labels of individual reducible representations from the $D_{2h}$ character table. Upper-case symbols are used to indicate symmetry and irreducible representations, while lower-case symbols are used to indicate the identity of an orbital with that symmetry. The labeling methods described for simple diatomic linear molecules ($\sigma, \pi$) are not sufficient to indicate the more complex symmetries of the molecular orbitals in polyatomic molecules. The use of lower-case Mulliken symbols is the most rigorous way to label the orbitals. Refer to your instructor for how you should label orbitals for any graded work.
Constructing the MO diagram
After identifying the atomic orbitals and constructing SALCs, place the atomic orbitals of H on one side of the diagram and all SALCs from F on the other side. Molecular orbitals are in the center.
The lowest-energy fluorine SALCs will be those composed of the lowest-energy atomic orbitals of fluorine; these would be SALC-1 ($a_{1g}$) and SALC-2 ($a_{1u}$) that are constructed from fluorine $s$ atomic orbitals. Of these two orbitals, the one with zero nodes would be slightly lower in energy than the one with one node. When these two orbitals form molecular orbitals, the completely symmetric SALC-1 will be mostly non-bonding with slight bonding character from minor combination with the hydrogen $1s$ orbital. SALC-2, however is symmetrically incompatible with hydrogen $1s$ and will be a truly non-bonding orbital distributed over both F atoms.
The six SALCs constructed of fluorine $p$ orbitals will have higher energy since the fluorine $p$ atomic orbitals are higher in energy than the $s$ orbitals (SALC-3 through SALC-8). Again, we expect SALCs with more nodes (SALC-4, -6, -8) to have slightly higher energy than those with fewer nodes (SALC-3, -5, -7). SALC-7 will form a bonding and antibonding interaction with the hydrogen $1s$ orbital. All other SALCS are truly non-bonding, but are non-degenerate non-bonding orbitals.
In bifluoride, there is an important distinction between the Lewis structure and molecular orbital description of lone pairs. In Lewis theory, the lone pairs are localized to individual fluorine atoms, while in the molecular orbital description each lone pair is distributed over both fluorine atoms at once (see surface depiction of each molecular orbital in Figure $4$). | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/05%3A_Molecular_Orbitals/5.04%3A_Larger_%28Polyatomic%29_Molecules/5.4.01%3A_Bifluoride_anion.txt |
Construct SALCs and the molecular orbital diagram for CO$_2$.
Carbon dioxide is another linear molecule. This example is slightly more complex than the previous example of the bifluoride anion. While bifluoride had only one valence orbital to consider in its central H atom (the $1s$ orbital), carbon dioxide has a larger central atom, and thus more valence orbitals that will interact with SALCs.
Preliminary Steps
Step 1. Find the point group of the molecule and assign Cartesian coordinates so that z is the principal axis.
The CO$_2$ molecule is linear and its point group is $D_{\infty h}$. The $z$ axis is collinear with the $C_\infty$ axis. We will use the $D_{2h}$ point group as a substitute since the orbital symmetries are retained in the $D_{2h}$ point group.
Step 2. Identify and count the pendant atoms' valence orbitals.
Each of the two pendant oxygen atoms has four valence orbitals; $2s$, $2p_x$, $2p_y$, and $2p_z$. Thus, we can expect a total of eight SALCs.
Generate SALCs
The SALCs for CO$_2$ are identical in shape and symmetry to those described in the previous example for the bifluoride anion. But instead of cutting to the shortcut, we will systematically derive the SALCs here to demonstrate the process.
Step 3. Generate the $\Gamma$'s
Use the $D_{2h}$ character table to generate four reducible representations ($\Gamma's$); one for each of the four types of pendant atom orbitals ($s, \;p_x, \;p_y, \;p_z$). For each $s$ orbital, assign a value of 1 if it remains in place during the operation or zero if it moves out of its original place. For each $p$ orbital, assign 1 if there is no change, -1 if it remains in place but is inverted, and 0 if it moves out of its original position. The four $\Gamma$'s are given below:
$\begin{array}{|c|cccccccc|} \hline \bf{D_{2h}} & E & C_2(z) & C_2(y) &C_2(x) & i &\sigma(xy) & \sigma(xz) & \sigma(yz)\ \hline \bf{\Gamma_{2s}} & 2 & 2 & 0 & 0 & 0 & 0 & 2 & 2 \ \bf{\Gamma_{2p_x}} & 2 & -2 & 0 & 0 & 0 & 0 & 2 & -2 \ \bf{\Gamma_{2p_y}} & 2 & -2 & 0 & 0 & 0 & 0 & -2 & 2 \ \bf{\Gamma_{2p_z}} & 2 & 2 & 0 & 0 & 0 & 0 & 2 & 2 \ \hline \end{array} \nonumber$
Step 4. Break $\Gamma$'s into irreducible representations for individual SALCs
Reduce each $\Gamma$ into its component irreducible representations. There are two strategies that can be used to do this. The quick and easy way is to do it "by inspection", but this only works well for simple cases. The other is by using the systematic approach for breaking a $\Gamma$ into reducible representations described previously in section 4.4.2 using the following formula:
$\text{# of } i = \frac{1}{h}\sum(\text{# of operations in class)}\times(\chi_{\Gamma}) \times (\chi_i) \label{irs}$
In other words, the number of irreducible representations of type $i$ is equal to the sum of the number of operations in the class $\times$ the character of the $\Gamma_{modes}$ $\times$ the character of $i$, and that sum is divided by the order of the group ($h$).
Using either approach results in the following eight irreducible representations ($2A_{g} + 2B_{1u} + B_{2g} + B_{3u} + B_{3g} + B_{2u}$):
$\begin{array}{|c|c|cccccccc|} \hline \bf{D_{2h}} & E & C_2(z) & C_2(y) &C_2(x) & i &\sigma(xy) & \sigma(xz) & \sigma(yz)\ \hline \bf{\Gamma_{2s} = A_g + B_{1u}} &\bf{\Gamma_{2s}} & \bf{2} & \bf{2} & \bf{0} & \bf{0} & \bf{0} & \bf{0} & \bf{2} & \bf{2}\ & A_{g} & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \ & B_{1u} & 1 & 1 & -1 & -1 & -1 & -1 & 1 & 1 \ \hline \ \hline \bf{\Gamma_{2p_x} = B_{2g} + B_{3u}} &\bf{\Gamma_{2p_x}} & \bf{2} & \bf{-2} & \bf{0} & \bf{0} & \bf{0} & \bf{0} & \bf{2} & \bf{-2} \ & B_{2g} & 1 & -1 & 1 & -1 & 1 & -1 & 1 & -1 \ & B_{3u} & 1 & -1 & -1 & 1 & -1 & 1 & 1 & -1 \ \hline \ \hline \bf{\Gamma_{2p_y}=B_{3g}+B_{2u}} & \bf{\Gamma_{2p_y}} & \bf2 & \bf-2 & \bf0 & \bf0 & \bf0 & \bf0 & \bf-2 & \bf2 \ & B_{3g} & 1 & -1 & -1 & 1 & 1 & -1 & -1 & 1 \ & B_{2u} & 1 & -1 & 1 & -1 & -1 & 1 & -1 & 1 \ \hline \ \hline \bf{\Gamma_{2p_z} = A_{g}+B_{1u}} & \bf{\Gamma_{2p_z}} & \bf{2} & \bf{2} & \bf{0} & \bf{0} & \bf{0} & \bf{0} & \bf{2} & \bf{2}\ & A_{g} & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \ & B_{1u} & 1 & 1 & -1 & -1 & -1 & -1 & 1 & 1 \ \hline \end{array} \nonumber$
Step 5. Sketch the SALCs
From the systematic process above, you have found the symmetries (the irreducible representations) of all eight SALCs under the $D_{2h}$ point group. To sketch the SALC that corresponds to each irreducible representation, again we use the $D_{2h}$ character table, and specifically the functions listed on the right side columns of the table.
Two $A_g$ SALCs (one from $s$ and one from $p_z$): The $A_g$ SALCs are each singly degenerate and symmetric with respect to both the principal axis ($z$) and the inversion center ($i$) (based on its Mulliken Label). We can look at the functions in the $D_{2h}$ character table that correspond to $A_g$ and see that it is completely symmetric under the group (because the combination of $x^2,y^2,z^2$ shows that it is totally symmetric). This would be the same symmetry as an $s$ orbital on the central atom. From this information, we know that these SALCs must have symmetry compatible with an $s$ orbital on the central atom, and we can draw the two $A_g$ SALCs shown in Figure $2$.
Two $B_{1u}$ SALCs (one from $s$ and one from $p_z$): The Mulliken Label tells us that the $B_{1u}$ SALCs are each antisymmetric with respect to both the principal axis and the inversion center. The function, $z$, appearing with $B_{1u}$ in the character table tells us that these SALCs have the same symmetry as the $z$ axis, or a $p_z$ orbital on the central atom. From this information, we know that these two SALCs should be compatible with a $p_z$ orbital on the central atom, and we can draw the two $B_{1u}$ SALCs shown in Figure $2$.
All other SALCs shown below are derived using a strategy similar to the one in the two cases described above.
Draw the MO diagram for $CO_2$
Step 6. Combine SALCs with AO’s of like symmetry.
First we identify the valence orbitals on carbon: there are four including $2s$, $2p_x$, $2p_y$, and $2p_z$. Now we identify the symmetry of each using the $D_{2h}$ character table. The symmetry of a central $s$ orbital corresponds to the combination of functions $x^2$, $y^2$, and $z^2$ in the character table; this is $A_g$. The $p_z$ orbital corresponds to the symmetry of the linear function $z$ in the character table; this is $B_{1u}$. And so on... The symmetries of the C valence orbitals are listed below. $2s =A_g \ 2p_x = B_{3u} \ 2p_y = B_{2u} \ 2p_z = B_{1u} \nonumber$
Now that we have identified the symmetries of the eight oxygen SALCs and the four valence orbitals on carbon, we know which atomic orbitals and SALCs may combine based on compatible symmetries. We also need to know the relative orbital energy levels so that we can predict the relative strength of orbital interactions. The orbital ionization energies are listed in Section 5.3.
With knowledge of both orbital symmetries and energies, we can construct the molecular orbital diagram. The carbon atom goes on one side of the diagram while the oxygen SALCs are drawn on the opposite side. Molecular orbitals are drawn in the center column of the diagram:
The SALCs on the right side of the diagram above are constructed from groups of either the $2s$ atomic orbitals of oxygen, or the $2p$ atomic orbitals of oxygen. Each SALC will combine with the atomic orbitals of carbon that have compatible symmetry; but the strength of the interaction depends on their relative energies. The individual valence atomic orbitals of oxygen have energies of $-32.36 eV$ ($2s$) and $-15.87 eV$ ($2p$), while the valence orbitals of carbon have energies of $-19.47 eV$ ($2s$) and $-10.66 eV$ ($2p$). The $2s$ orbitals of the oxygen are far lower in energy ($-32.36 eV$) than all other valence orbitals, and so we should expect that molecular orbitals that are constructed from these oxygen $2s$ will be mostly non-bonding. These mostly non-bonding orbitals are the $a_{1g}$ and $b_{1g}$ orbitals shown at the bottom of Figure $3$. The $a_g$ orbital is lower in energy than the $b_{1u}$ due to slight mixing with other orbitals of $a_g$ symmetry.
The bonding orbitals of $CO_2$ include two $sigma$ molecular orbitals of $a_g$ and $b_{1u}$ symmetry and two $\pi$ molecular orbitals of $b_{2u}$ and $b_{3u}$ symmetry. Each of these bonding molecular orbitals possesses a higher-energy antibonding partner. The $b_{2g}$ and $b_{3g}$ SALCs that are formed from oxygen $2p_x$ and $2p_y$ orbitals have no compatible match with carbon valence orbitals; thus they form two $b_{2g}$ and $b_{3g}$ molecular orbitals, which are truly non-bonding and mostly oxygen in character. Still, notice that each orbital is spread across both oxygen atoms at once, and again we see that each non-bonding electron pair in the HOMO is very different in molecular orbital theory compared to Lewis theory. Each non-bonding pair is distributed over both oxygen atoms at once in molecular orbital theory, while in Lewis theory each lone pair is isolated to one atom or to localized bonds attached to that atom. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/05%3A_Molecular_Orbitals/5.04%3A_Larger_%28Polyatomic%29_Molecules/5.4.02%3A_Carbon_Dioxide.txt |
Construct SALCs and the molecular orbital diagram for H$_2$O.
This is the first example so far that is not a linear molecule. Water is a bent molecule, and so it is important to remember that interactions of pendant ligands are dependent on their positions in space. You should consider the positions of the three atoms in water to be essentially fixed in relation to each other. The process for constructing the molecular orbital diagram for a non-linear molecule, like water, is similar to the process for linear molecules. We will walk through the steps below to construct the molecular orbital diagram of water.
Preliminary Steps
Step 1. Find the point group of the molecule and assign Cartesian coordinates so that z is the principal axis.
The H$_2$O molecule is bent and its point group is $C_{2v}$. The $z$ axis is collinear with the principal axis, the $C_2$ axis. There is no need to simplify this problem, as we had done for previous examples. The $C_{2v}$ point group is simple enough.
Step 2. Identify and count the pendant atoms' valence orbitals.
Each of the two pendant hydrogen atoms has one valence orbital, the $1s$. Thus, we can expect a total of two SALCs from these two atoms.
Generate SALCs
The SALCs for H$_2$O are quite simple, yet we will systematically derive the SALCs here to demonstrate the process.
Step 3. Generate the $\Gamma$'s
Use the $C_{2v}$ character table to generate one reducible representation ($\Gamma$); in this case we need only one $\Gamma$ because there is only one type of valence orbital (the $1s$). For each $s$ orbital, assign a value of 1 if it remains in place during the operation or zero if it moves out of its original place. The $\Gamma$ is given below:
$\begin{array}{|c|cccc|} \hline \bf{C_{2v}} & E & C_2 &\sigma_v (xz) & \sigma_v' (yz) \ \hline \bf{\Gamma_{1s}} & 2 & 0 & 2 & 0 \ \hline \end{array} \nonumber$
Step 4. Break $\Gamma$'s into irreducible representations for individual SALCs
Reduce each $\Gamma$ into its component irreducible representations. Using either of the processes described previously, we find that the $\Gamma$ reduces to the two irreducible representations $A_1$ and $B_1$ under the $C_{2v}$ point group.
$\begin{array}{|c|cccc|} \hline \bf{C_{2v}} & E & C_2 &\sigma_v (xz) & \sigma_v' (yz) \ \hline \bf{\Gamma_{1s}} & \bf 2 & \bf 0 & \bf 2 & \bf 0 \ A_{1} & 1 & 1 & 1 & 1 \ B_{1} & 1 & -1 & 1 & -1 \ \hline \end{array} \nonumber$
Step 5. Sketch the SALCs
From the systematic process above, you have found the symmetries (the irreducible representations) of both SALCs under the $C_{2v}$ point group. To sketch the SALC that corresponds to each irreducible representation, again we use the $C_{2v}$ character table, and specifically the functions listed on the right side columns of the table.
One $A_1$ SALC: The $A_1$ SALC is singly degenerate and symmetric with respect to both the principle axis ($z$) and the inversion center ($i$) (based on its Mulliken Label). We can look at the functions in the $C_{2v}$ character table that correspond to $A_1$ and see that it is completely symmetric under the group (because the combination of $x^2,y^2,z^2$ shows that it is totally symmetric). This would be the same symmetry as an $s$ orbital on the central oxygen atom. From this information, we know that this SALC must have symmetry compatible with an $s$ orbital on the central atom. We can also see from the character table that the $z$ axis, and thus a $p_z$ orbital on oxygen, also possesses $A_1$ symmetry. This tells us that in addition to being compatible with the oxygen $s$ orbital, it should also be compatible with the oxygen $p_z$ orbital. From this information we can draw the $A_1$ SALC shown in Figure $2$. If it is not obvious how the $A_1$ sketch in Fig. $2$ is compatible with the $s$ and $p_z$ orbitals of oxygen, inspect the drawings corresponding to molecular orbitals $2a_1$ and $3a_1$ in Fig. $3$.
One $B_{1}$ SALC: The Mulliken Label tells us that the $B_{1}$ SALC is singly degenerate and antisymmetric with respect to both the principal axis and the inversion center. The function, $x$, appearing with $B_{1}$ in the character table tells us that this SALC has the same symmetry as the $x$ axis, or a $p_x$ orbital on the central oxygen atom. From this information, we know that this SALC should be compatible with a $p_x$ orbital on the central atom, and we can draw the $B_{1}$ SALCs shown in Figure $2$. If it is not obvious how the $B_1$ sketch in Fig. $2$ is compatible with the $p_x$ orbital of oxygen, inspect the drawing corresponding to molecular orbital $1b_1$ in Fig. $3$.
Not convinced about the sketches of these SALCs? You can convince yourself by putting them to the test. Try the exercise below.
Exercise $1$
Perform all operations of the $C_{2v}$ point group on the two sketches of H SALCs shown in Figure $2$, and convince yourself that each sketch does possess the $A_1$ and $B_1$ symmetries assigned to them, respectively, under the $C_{2v}$ point group.
Answer
Add texts here. Do not delete this text first.
Draw the MO diagram for $H_2O$
Step 6. Combine SALCs with AO’s of like symmetry.
First we must identify the valence orbitals on the central oxygen: there are four including $2s$, $2p_x$, $2p_y$, and $2p_z$. Now we identify the symmetry of each using the $C_{2v}$ character table. The symmetry of a central $2s$ orbital corresponds to the combination of functions $x^2$, $y^2$, and $z^2$ in the character table; this is $A_1$. The $p_z$ orbital also corresponds to $A_1$. And so on... The symmetries of oxygen valence orbitals are listed below. $2s =A_1 \ 2p_x = B_1 \ 2p_y = B_2 \ 2p_z = A_1 \nonumber$
Now that we have identified the symmetries of the two hydrogen SALCs and the four valence orbitals on oxygen, we know which atomic orbitals and SALCs may combine based on compatible symmetries. We also need to know the relative orbital energy levels so that we can predict the relative strength of orbital interactions. The orbital ionization energies are listed in Section 5.3.1.
With knowledge of both orbital symmetries and energies, we can construct the molecular orbital diagram. The valence orbitals of oxygen go on one side of the diagram while the hydrogen group orbitals are drawn on the opposite side. Molecular orbitals are drawn in the center column of the diagram, as shown in Figure $3$:
Example $1$
In the previous examples shown for the molecular orbital diagrams of the bifluoride anion and carbon dioxide, we discussed differences in the understanding of those molecules from molecular orbital theory compared to Lewis structures. Water contains two lone pairs in its Lewis structure. Compare the predictions about water's lone pairs of electrons and its reactivity based on (1) the combination of elementary models (Lewis, Valence Bond, and Hybridized Orbital theories) and (2) Molecular Orbital theory. Specifically address and explain how the elementary models differ from molecular orbital theory in the following respects:
1. Where are the lone pairs in water?
2. Are the two lone pairs equivalent or are they different?
3. Where are sites on the molecule that will undergo reaction with electrophiles and nucleophiles?
Solution
1) Elementary models: The Lewis structure predicts that two lone pairs are (a) localized on the oxygen atom of water and that (b) both lone pairs are equivalent. The Lewis structure, combined with Valence Bond Theory, would predict that lone pairs occupy two equivalent hybridized $sp^3$ atomic orbitals on oxygen. (c) Lewis theory would predict that the oxygen lone pairs are nucleophiles; thus an electrophile would react with the oxygen atom lone pairs. The polarized O-H bond leaves the H atoms as the most electrophilic locations on the water molecule; thus nucleophiles would react at the H atoms of water.
2) Molecular orbital theory: The molecular orbitals predict that (a) the two lone pairs of water are not equivalent, and that (b) each is distributed over the entire molecule. One lone pair is in the truly non-bonding $1b_2$ orbital (Figure $3$), which is also the HOMO. There is not another truly non-bonding orbital in the molecule, but the lowest-energy $2a_1$ orbital could be considered mostly non-bonding due to the large energy difference between it and other valence orbitals. (c) Although the $2a_1$ orbital can be considered mostly non-bonding, we cannot expect the electrons in $2a_1$ to react readily, as they are in the lowest-energy molecular orbital. Molecular orbital theory predicts that reactions occur at the HOMO and LUMO orbitals. The HOMO reacts with electrophiles, and in this case, the HOMO is distributed over the top and bottom faces of the molecule, and is centered on the oxygen atom. The LUMO would react with nucleophiles. The LUMO is the orbital labeled $3a_1^*$ in Figure $3$, and has a major lobe that is distributed more heavily over the H atoms, and from this we should predict H atoms to be the preferred site of reaction for nucleophiles.
Expressing molecular orbitals in terms of $\Psi$
The general expression for a molecular orbital, or the linear combination of atomic orbitals (LCAO), was given previously as $\Psi=c_{a} \psi_{a}+c_{b} \psi_{b}$. In this expression, the wavefunction of two atoms ($\psi_a$ and $\psi_b$) is combined to form the wavefunction of the molecular orbital. The coefficients $c_a$ and $c_b$ quantify the contribution of each atomic $\psi$ to the molecular $\Psi$.
$\Psi=c_{a} \psi_{a}+c_{b} \psi_{b} \nonumber$
In the case of polyatomic orbitals, each LCAO is constructed of atomic orbitals from a central atom and group orbitals (SALCs) from the pendent atoms. In the case of water, the expression above could be modified to give the expression below.
$\Psi = c_{oxygen}\left(\psi_{oxygen}\right) + c_{SALCs}\left(N (\psi_{H_a} \pm \psi_{H_b})\right) \nonumber$
N represents the normalizing requirement that was mentioned previously in reference to the requirements for electron wavefunctions and probability functions. The normalizing requirement simply stated is the requirement that the probability of finding the electron in any orbital, including group orbitals, is 1. In general, the value of N for a group orbital is... $N=\frac{1}{\sum{c_i^2} } \nonumber$ ...where $c_i$ is the coefficient of each unique atomic orbital that contributes to the group. For the hydrogen SALCs of water, there are two atomic orbitals that contribute to equally to each SALC, and so the coefficient on each of the two orbitals is 1. This gives $N=\left(\frac{1}{\sqrt{1^2 + 1^2}}\right) = \frac{1}{\sqrt{2}}$ for the H SALCs of water.
There are only two hydrogen SALCs: the one in which the hydrogen wavefunctions are added ($N(\psi_{H_a} + \psi_{H_b})$) has $A_1$ symmetry, and the one in which the hydrogen wavefunctions are subtracted ($N(\psi_{H_a} - \psi_{H_b})$) has $B_1$ symmetry.
The molecular orbitals from Figure $3$ are expressed below in terms of their LCAO of individual wavefunctions. Yet, note that these expressions are simplifications that ignore orbital mixing. For example, as expressed below, they ignore contributions of oxygen $2s$ to the higher energy moelcular orbitals with $A_1$ symmetry (the $3a_1$ and $3a_1^*$ in Figure $3$).
$\begin {array}{|rcccc|l|} \hline MO & & Oxygen AO & & Hydrogen SALC & Description \ \hline \Psi_{2a_1} & = & c_{(ox1)}\psi_{(2s)} & + & c_{(hy1)} [N(\psi_{H_a} + \psi_{H_b})] & c_{(hy1)} \text{ is positive; bonding, slightly nonbonding} \ \Psi_{1b_1} & = & c_{(ox2)}\psi_{(2p_x)} & + & c_{(hy2)} [N(\psi_{H_a} - \psi_{H_b})] & c_{(hy2)} \text{ is positive; bonding} \ \Psi_{3a_1} & = & c_{(ox3)}\psi_{(2p_z)} & + & c_{(hy3)} [N(\psi_{H_a} + \psi_{H_b})] & c_{(hy3)} \text{ is positive; bonding} \ \Psi_{1b_2} & = & \psi_{(2p_y)} & & & \text{nonbonding} \ \Psi_{3a_1}^* & = & c_{(ox4)}\psi_{(2p_z)} & + & c_{(hy4)} [N(\psi_{H_a} + \psi_{H_b})] & c_{(hy4)} \text{ is negative; antibonding} \ \Psi_{1b_1}^* & = & c_{(ox5)}\psi_{(2p_x)} & + & c_{(hy5)} [N(\psi_{H_a} - \psi_{H_b})] & c_{(hy5)} \text{ is negative; antibonding} \ \hline \end {array} \nonumber$ | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/05%3A_Molecular_Orbitals/5.04%3A_Larger_%28Polyatomic%29_Molecules/5.4.03%3A_HO.txt |
Construct SALCs and the molecular orbital diagram for $\ce{NH3}$
This is the first example so far that has more than two pendant atoms and the first example in which the molecule has atoms that lie in three dimensions (i.e., it is not flat). Ammonia is a trigonal pyramidal molecule, with three pendant hydrogen atoms. The three-dimensional shape and the odd number of pendant atoms makes this example more complicated than the previous cases of water, carbon dioxide, and bifluoride. In this case, sketching the shapes (step 5) of pendant atom SALCs is less straightforward; rather, an alternative method, the projection operator method, is preferred for generating pictorial representations of the SALCs.
As in previous examples, it is important to remember that interactions of pendant ligands are dependent on their positions in three-dimensional space. You should consider the positions of the four atoms in ammonia to be essentially fixed in relation to each other. We will walk through the steps used to construct the molecular orbital diagram of ammonia. The first few steps are the same as you've seen before:
Step 1. Find the point group of the molecule and assign Cartesian coordinates so that z is the principal axis.
The NH$_3$ molecule is trigonal pyramidal and its point group is $C_{3v}$. The $z$ axis is collinear with the principal axis, the $C_3$ axis.
Step 2. Identify and count the pendant atoms' valence orbitals.
Each of the three pendant hydrogen atoms has one valence orbital; the $1s$. Thus, we can expect a total of three SALCs from these three atoms.
Step 3. Generate the $\Gamma$'s
Use the $C_{3v}$ character table to generate one reducible representation ($\Gamma$); in this case we need only one $\Gamma$ because there is only one type of valence orbital (the $1s$). For each $s$ orbital, assign a value of 1 if it remains in place during the operation or zero if it moves out of its original place. The $\Gamma$ is given below:
$\begin{array}{|c|ccc|} \hline \bf{C_{3v}} & E & 2C_3 & 3\sigma_v \ \hline \bf{\Gamma_{1s}} & 3 & 0 & 1 \ \hline \end{array} \nonumber$
Step 4. Break $\Gamma$'s into irreducible representations for individual SALCs
Reduce each $\Gamma$ into its component irreducible representations. Using either of the processes described previously, we find that the $\Gamma$ reduces to the two irreducible representations $A_1$ and $E$ under the $C_{3v}$ point group.
$\begin{array}{|c|ccc|} \hline \bf{C_{3v}} & E & 2C_3 & 3\sigma_v \ \hline \bf{\Gamma_{1s}} & \bf 3 & \bf 0 & \bf 2 \ A_{1} & 1 & 1 & 1 \ E & 2 & -1 & 0 \ \hline \end{array} \nonumber$
Notice that we only found TWO irreducible representations. But, in fact we have THREE different SALCs. The $E$ irreducible representation is doubly degenerate, which in this context means that it corresponds to two degenerate SALCs. Thus, we have already found the symmetries of the three SALCs for ammonia: Two of the SALCs are degenerate with $E$ symmetry under the $C_{3v}$ point group, while the third SALC has $A_1$ symmetry.
Step 5. Sketch the SALCs using the PROJECTION OPERATOR METHOD
From the first four steps (described above), you have found the symmetries (the irreducible representations) of all three SALCs under the $C_{3v}$ point group. To sketch the SALC that corresponds to each irreducible representation, again we use the $C_{3v}$ character table. But now we will introduce the projection operator method to derive the surface representation of each SALC.
Step 5.1: Label the pendent atoms
In the projection operator method, first label each of the pendant atoms so that we can distinguish identical atoms from one another; for example, identify the three hydrogen atoms on ammonia as $H_a$, $H_b$, and $H_c$ (Figure $2$).
Step 5.2: Create an expanded character table with one pendant atom's projected position after each operation
We will perform each operation of the $C_{3v}$ point group on this labeled molecule and follow where one of the atoms is projected after the operation is complete. We will arbitrarily choose H$_a$. For example, upon performing the identity operation, E, the H$_a$ atom is projected onto itself. On the other hand, H$_a$ is projected onto $H_b$ upon a clockwise $C_3$ rotation (as drawn in Figure $2$). We take into account the result of each operation using an expanded character table (refer to the table below, \ref{expanded1}). In the expanded character table, each operation within each class is written separately (i.e., $2C_3$ is accounted for separately as $C_3$ and $C_3^{-1}$).
$\text{Table }\ref{expanded1} \text{: The expanded character table, and the projection of $H_a$ by each operation is shown below.} \nonumber$\begin{array}{|r|cccccc|} \hline \bf{C_{3v}} & E & C_3 & C_3^{-1} & \sigma_v(a) & \sigma_v(b) & \sigma_v(c) \ \hline \bf{\text{Projection of }H_a} & H_a & H_b & H_c & H_a & H_c & H_b \ \hline \end{array} \label{expanded1}$
Step 5.3: Find the contribution of each pendant atom to each SALC
Next, create a linear combination of the projections for each of the SALCs (the irreducible representations found in step 4). For each of the irreducible representations, multiply the projection by the respective character of the operation.
$\text{Contribution of each atom to the SALC } = \sum(\text{Projection of }H_a \times \chi) \nonumber$
The linear combination for all irreducible representations of $C_3v$ is shown below.
$\text{Table }\ref{expanded2} \text{: The symmetry adapted linear combination (SALC) for each irreducible representation of $C_3v$ is shown.} \nonumber$\begin{array}{|r|cccccc|l|} \hline \bf{C_{3v}} & E & C_3 & C_3^{-1} & \sigma_v(a) & \sigma_v(b) & \sigma_v(c) & \text{Linear Combination} \ \hline \bf{\text{Projection of }H_a} & \bf H_a & \bf H_b & \bf H_c & \bf H_a & \bf H_c & \bf H_b & \ A_1 & 1 & 1 & 1 & 1 & 1 & 1 & = \bf 2H_a + 2H_b + 2H_c \ A_2 & 1 & 1 & 1 & -1 & -1 & -1 & = 0 \ E & 2 & -1 & -1 & 0 & 0 & 0 & = \bf 2H_a - H_b - H_c \ \hline \end{array} \label{expanded2}$
Notice that there are only two irreducible representations that produce SALCs, and these are the same that were found in Step 4, above. This illustrates the fact that only the irreducible representations found through the reduction of the $\Gamma$ will produce SALCs. We can ignore any irreducible representations that were not found in Step 4. Or, you can check your work in Step 4 by applying the projection operator to any irreducible representations not found in Step 4, and finding that they produce a sum of zero.
Step 5.4: Sketch the SALCs
The meaning of the linear combinations found in Table \ref{expanded2} is as follows:
• Sketch the SALC with $A_1$ symmetry: The linear combination $2H_a + 2H_b + 2H_c$ indicates all contributions to this SALC are of the same sign (of the wavefunction). Qualitatively, this means there is no node in this SALC. We can take this SALC almost literally to assume that all three H 1s wavefunctions contribute equally to the SALC. We can also use the fact that the $A_1$ representation possesses the full symmetry of the $C_3v$ point group, is compatible with an s orbital on N, and thus is a totally symmetric SALC.
Quantitatively, we can apply the normalizing factor, N, for this SALC. The linear combination for the $A_1$ SALC in Table \ref{expanded2} shows us that the coefficient for each orbital is 1. Thus, the normalizing factor for the $A_1$ SALC is $N=\left(\frac{1}{\sqrt{1^2 + 1^2 + 1^2}}\right) = \frac{1}{\sqrt{3}}$. This tells us that $H_a$, $H_b$, and $H_c$ each contribute $\frac{1}{\sqrt{3}}$ to the normalized $A_1$ group orbital: $A_1 \text{ group orbital } = \frac{1}{\sqrt{3}}\left[\psi_{H_{a}}+\psi_{H_{b}}+\psi_{H_{c}}\right] \nonumber$ This is shown visually in Figure $3$.
• Sketch two SALCs with E symmetry: The linear combination $2H_a - H_b - H_c$ indicates that there are contributions with both positive and negative sign to the wavefunction for each of the $E$ group orbitals. Qualitatively, this tells us that there is a node within each of these two SALCs, and that the total contribution of the positive portion of the wavefunction is equal to the contribution of the negative portion.
• Finding the first E SALC:
For one of the SALCs, we can take the linear combination from Table \ref{expanded2} almost literally: The contribution from $H_a$ is equal and opposite to the sum of the contributions from $H_b + H_c$. This would result in a SALC as shown in Figure $3$ (E, left), that would have symmetry of the $x$ axis and that would be compatible with the $p_x$ orbital on N. The node exists in the $yz$ plane and there is an equal contribution of positive and negative parts of the total wavefunction.
We can also apply the normalizing factor, N, to find this SALC. The linear combination for the $E$ SALC in Table \ref{expanded2} shows us that the coefficients for orbital contributions are 2, -1, and -1. Thus, the normalizing factor for the first $E$ SALC is $N=\left(\frac{1}{\sqrt{2^2 + (-1)^2 + (-1)^2}}\right) = \frac{1}{\sqrt{6}}$. This normalized $E$ group orbital is: $\text{First } E \text{ group orbital } = \frac{1}{\sqrt{6}}\left[2\psi_{H_{a}}-\psi_{H_{b}}-\psi_{H_{c}}\right] \nonumber$ This is shown visually in Figure $3$ (E, left).
• Finding the second E SALC:
The second SALC is less obvious at first glance. We must use clues from the character table to help us determine what it should "look like". Since the $E$ representation transforms as the group, ($x,y$), we know that one of the SALCs will have the symmetry of the $x$ axis and $p_x$ orbital on the central nitrogen atom, while the other should possess the symmetry of the $y$ axis and be compatible with the $p_y$ orbital on nitrogen. This alone could lead you to the sketch of the second SALC with E symmetry shown in Figure $3$ (E, right). This second E SALC must have a node in the $xz$ plane, and because $H_a$ lies in this node, $H_a$ cannot contribute to this group orbital. Retaining the fact that the positive and negative contributions of the linear combination must be equal, we can arrive at a SALC that has equal but opposite contributions of $H_b$ and $H_c$, and no contribution by $H_c$ (Figure $3$, right).
We can also apply the normalizing factor, N, to find this SALC. But we must apply the fact that $H_a$ cannot contribute to the SALC with "$y$" symmetry. We remove the contribution of $H_a$, leaving us with coefficients of 0 for $H_a$ and $1$ for $H_b$ and $H_c$. Thus, the normalizing factor for the second $E$ SALC is $N=\left(\frac{1}{\sqrt{0^2 + 1^2 + 1^2}}\right) = \frac{1}{\sqrt{2}}$. To this, we must add positive and negative coefficients so that the entire wavefunction of this SALC is normalized. The result for this normalized $E$ group orbital is: $\text{Second } E \text{ group orbital } = \frac{1}{\sqrt{2}}\left[\psi_{H_{b}}-\psi_{H_{c}}\right] \nonumber$ This is shown visually in Figure $3$.
Step 6. Combine SALCs with AO’s of like symmetry to draw the MO diagram for $\ce{NH_3}$
First we must identify the valence orbitals on the central nitrogen: there are four including $2s$, $2p_x$, $2p_y$, and $2p_z$. Now we identify the symmetry of each using the $C_{3v}$ character table. The symmetry of a central $2s$ orbital corresponds to the combination of functions $x^2$, $y^2$, and $z^2$ in the character table; this is $A_1$. The $p_z$ orbital also corresponds to $A_1$. And so on... The symmetries of nitrogen valence orbitals are listed below. $2s =A_1 \ (2p_x, \; 2p_y) = E \ 2p_z = A_1 \nonumber$
The next step is to get a sense of the relative energies of valence atomic orbitals for nitrogen and hydrogen, and then construct the molecular orbital diagram. The nitrogen $2s$ orbital is about 12 eV lower in energy than a H $1s$. This difference is an acceptable energy-match for the H orbitals to interact, but would not usually result in a strong interaction. On the other hand, the nitrogen 2p orbitals have an energy very close to the hydrogen 1s, different by only 0.5 eV. Knowledge of the relative energies of the atomic orbitals tells us that, when constructing a diagram, we can place the H SALCs at an energy similar to that of the N $2p$ orbitals, while the N $2s$ is much lower (Figure $4$).
The $2s$ and $2p_z$ orbitals have $A_1$ symmetry and can combine with the $A_1$ SALC. These orbitals will give three molecular orbitals. The $2s$ orbital will not have strong interactions with the other orbitals of $A_1$ symmetry due to the large energy difference. The $2p_z$ orbital will not have good overlap with the wavefunctions for the three hydrogen orbitals due to their positions in space: The $2p_z$ orbital has half of its angular distribution pointed away from the hydrogen orbitals, and half pointed toward the center of a triangle formed by the three hydrogen atoms. The nitrogen $2s$ and $2p_z$ will combine with the hydrogen SALC of $A_1$ symmetry to give three molecular orbitals; one low-energy bonding orbital, one mid-energy non-bonding orbital, and one high-energy antibonding orbital. The mid-level non-bonding $a_1$ molecular orbital is close in energy to, and resembles, the nitrogen $2p_z$ orbital. The lowest energy $a_1$ molecular orbital will be lower in energy, but be similar to the nitrogen $2s$.
The remaining two $2p$ orbitals are degenerate and possess $E$ symmetry under the $C_{3v}$ point group. Although these orbitals are a good energy match, again, their orientation and positions in space do not allow for good orbital overlap, and their bonding and antibonding interactions are relatively weak compared to what we might otherwise expect from such a good energy match. The $e$ atomic orbitals of nitrogen will combine with the $e$ SALCs to give a set of two degenerate bonding molecular orbitals and a set of two degenerate antibonding orbitals (four total molecular orbitals of $e$ symmetry). The MO diagram for $\ce{NH3}$ is shown in Figure $4$, with calculated electron density surfaces of each MO shown. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/05%3A_Molecular_Orbitals/5.04%3A_Larger_%28Polyatomic%29_Molecules/5.4.04%3A_NH.txt |
The process for finding SALCs and constructing the molecular orbital diagram for carbon dioxide was described in a previous section (Section 5.4.2). However, we also just presented an alternative strategy, the projection operator method, to finding the shapes of SALCs in Section 5.4.4. Let's revisit carbon dioxide and demonstrate how you can use the projection operator method to find SALCs for carbon dioxide.
Just as before, we can simplify the problem by approximating the $D_{\infty h}$point group using $D_{2h}$. The first four steps for constructing the MO diagram would be the same as described in Section 5.4.2. and we will pick up from that point (with Step 5) to find what the SALCs look like using the projection operator method.
The projection operator method applied to $\ce{CO2}$
Step 5.1: Label the pendent atoms.
We can label the pendant atoms as $\ce{O}_a$ and $\ce{O}_b$, as in Figure $1$.
Step 5.2: Create an expanded character table with one pendant atom's projected position for each operation
The $D_{2h}$ character table needs no expansion because each operation is in its own class. We will arbitrarily choose to determine the new position of $\ce{O}_a$ after each operation.
$\begin{array}{|c|cccccccc|} \hline \bf{D_{2h}} & E & C_2(z) & C_2(y) &C_2(x) & i &\sigma(xy) & \sigma(xz) & \sigma(yz)\ \hline \bf{\text{Projection of }\ce{O}_a} & \ce{O}_a & \ce{O}_a & \ce{O}_b & \ce{O}_b & \ce{O}_b & \ce{O}_b & \ce{O}_a & \ce{O}_a \ \hline \end{array} \nonumber$
Step 5.3: Find the contribution of each pendant atom to each SALC
Create a linear combination of the projections for each or the SALCs. The irreducible representations were found in step 4 in Section 5.4.2. Eight irreducible representations were found to be: $2A_{g} + 2B_{1u} + B_{2g} + B_{3u} + B_{3g} + B_{2u}$. For each of the irreducible representations, multiply the projection by the respective character of the operation.
$\text{Contribution of each atom to the SALC } = \sum(\text{Projection of }H_a \times \chi) \nonumber$
The linear combination for all irreducible representations of $D_{2h}$ are shown below.
$\text{Table }\ref{expanded2} \text{: The symmetry adapted linear combination (SALC) for each irreducible representations of $C_3v$ are shown.} \nonumber$\begin{array}{|c|cccccccc|l|} \hline \bf{D_{2h}} & E & C_2(z) & C_2(y) &C_2(x) & i &\sigma(xy) & \sigma(xz) & \sigma(yz) & \text{Linear Combination} \ \hline \bf{\text{Projection of }\ce{O}_a} &\bf \ce{O}_a &\bf \ce{O}_a &\bf \ce{O}_b &\bf \ce{O}_b &\bf \ce{O}_b &\bf \ce{O}_b &\bf \ce{O}_a &\bf \ce{O}_a \ A_{g} & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & = \bf 4\ce{O}_a + 4\ce{O}_b \ B_{1u} & 1 & 1 & -1 & -1 & -1 & -1 & 1 & 1 & = \bf 4\ce{O}_a - 4\ce{O}_b \ B_{2g} & 1 & -1 & 1 & -1 & 1 & -1 & 1 & -1 & = \bf 0 \ B_{3u} & 1 & -1 & -1 & 1 & -1 & 1 & 1 & -1 & = \bf 0\ B_{3g} & 1 & -1 & -1 & 1 & 1 & -1 & -1 & 1 & = \bf 0\ B_{2u} & 1 & -1 & 1 & -1 & -1 & 1 & -1 & 1 & = \bf 0\ \hline \end{array} \label{expanded2}$
Notice that there are only two irreducible representations that produce SALCs, the $A_g$ and the $B_{1u}$. Since we found two of each ($2A_{g} + 2B_{1u}$) when we found the irreducible representations, we know that there will be two SALCs of $A_g$ symmetry and two of $B_{1u}$ symmetry; that is four total SALCs from the pendant oxygens. This is the same as what we found in Section 5.4.2 using a different method.
Step 5.4: Sketch the SALCs
The SALCs with $A_g$ symmetry: Quantitatively, we can apply the normalizing factor, N, for these SALCs. Each atom contributes equally to the SALCs, thus the normalizing factor for the $A_g$ SALC is $N=\left(\frac{1}{\sqrt{1^2 + 1^2}}\right) = \frac{1}{\sqrt{2}}$. This tells us that each oxygen contributes $\frac{1}{\sqrt{2}}$ to each of the normalized $A_g$ group orbitals: $A_g \text{ group orbital } = \frac{1}{\sqrt{2}}\left[\psi_{O_{a}}+\psi_{O_{b}}\right] \nonumber$
The linear combination $4\ce{O}_a + 4\ce{O}_b$ indicates that wavefunctions from each oxygen atom contribute equally to each of the two SALCs, with the same sign of the wavefunction for each. Because we know that $A_g$ is totally symmetric (all 1's in the characters), we can assume that the SALC's will look like a symmetric arrangement of oxygen orbitals. But which orbitals contribute? We could peek at the character table to get a hint. But we'll treat this problem systematically.
The SALCs with $B_{1u}$ symmetry: The linear combination $4\ce{O}_a - 4\ce{O}_b$ indicates that each oxygen atom contributes equally, but that they have opposite wavefunctions. Application of the normalizing factor would give us: $B_{1u} \text{ group orbital } = \frac{1}{\sqrt{2}}\left[\psi_{O_{a}}+\psi_{O_{b}}\right] \nonumber$ But how do we know what the SALCs look like? Again, the character table can give us some hints, but we'll use a systematic process, below.
What do the SALCs look like? One way to systematically derive the SALC shapes is to perform the projection (Step 5.3) on specific groups of orbitals on the oxygen atom. For example, if we replicate Table \ref{expanded2} using just the group of oxygen 2s orbitals and the $A_g$ and $B_{1u}$ representations, we get the following:
$\begin{array}{|c|cccccccc|l|} \hline \bf{D_{2h}} & E & C_2(z) & C_2(y) &C_2(x) & i &\sigma(xy) & \sigma(xz) & \sigma(yz) & \text{Linear Combination} \ \hline \bf{\text{Projection of 2s orbital of }\ce{O}_a} &\bf \ce{O}1s_a &\bf \ce{O}1s_a &\bf \ce{O}1s_b &\bf \ce{O}1s_b &\bf \ce{O}1s_b &\bf \ce{O}1s_b &\bf \ce{O}1s_a &\bf \ce{O}1s_a \ A_{g} & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & = \bf 4\ce{O}1s_a + 4\ce{O}1s_b \ B_{1u} & 1 & 1 & -1 & -1 & -1 & -1 & 1 & 1 & = \bf 4\ce{O}1s_a - 4\ce{O}1s_b \ \hline \end{array} \label{expanded3}$
This tells us that one of the $A_g$ SALCs looks like two oxygen 1s orbitals with equal signs and contributions. It also tells us that one of the $B_{1u}$ SALCs looks like two $s$ orbitals with opposite signs and equal magnitudes. These two SALCs are shown in Figure $2$.
If we do the same for the $2p_z$ orbitals, we need to pay attention to the orientation of the $p$ orbital lobes. If the orbital is moved into the opposite orientation, it would get a negative sign in the projection (see below).
$\begin{array}{|c|cccccccc|l|} \hline \bf{D_{2h}} & E & C_2(z) & C_2(y) &C_2(x) & i &\sigma(xy) & \sigma(xz) & \sigma(yz) & \text{Linear Combination} \ \hline \bf{\text{Projection of 2px orbital of }\ce{O}_a} &\bf +\ce{O}2pz_a &\bf +\ce{O}2pz_a &\bf -\ce{O}2pz_b &\bf +\ce{O}2pz_b &\bf -\ce{O}2pz_b &\bf -\ce{O}2pz_b &\bf +\ce{O}2pz_a &\bf +\ce{O}2pz_a \ A_{g} & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & = \bf 4\ce{O}2pz_a - 4\ce{O}2pz_b \ B_{1u} & 1 & 1 & -1 & -1 & -1 & -1 & 1 & 1 & = \bf 4\ce{O}2pz_a + 4\ce{O}2pz_b \ \hline \end{array} \label{expanded4}$
This tells us that one of the $A_g$ SALCs is derived from two $2p_z$ orbitals of opposite sign but equal magnitude. In other words, the two $p$ orbitals are pointing in opposite directions, but the lobes that are facing one another are of the same sign (see Figure $3$, left). This also tells us that one of the $B_{1u}$ orbitals is derived from two $p_z$ orbitals of equal magnitude and equal sign (pointing in the same directions, with lobes facing one another of opposite sign!) (see Figure $3$, right).
Now we have reached the same SALCs that we found previously in Section 5.4.2, but by an alternate method. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/05%3A_Molecular_Orbitals/5.04%3A_Larger_%28Polyatomic%29_Molecules/5.4.05%3A_CO_%28Revisted_with_Projection_Operators%29.txt |
The case of boron trifluoride ($\ce{BF_3}$) is an example of a molecule with one more layer of complexity than the other examples we have seen in previous sections in this chapter. ($\ce{BF_3}$) is more complex than previous examples because it is the first case in which there are multiple types of valence orbitals on the pendant atoms. ($\ce{BF_3}$) possesses $s$ and $p$ orbitals on both the central atom and all of the pendant atoms. We can follow the same steps that we have previously to derive other molecular orbital diagrams; however, there is one important difference: we will treat each type of pendant orbital as an individual set of SALCs.
Step 1. Find the point group of the molecule and assign Cartesian coordinates so that z is the principal axis.
The BH$_3$ molecule is trigonal planar, and its point group is $D_{3h}$. The $z$ axis is colinear with the principle axis, the $C_3$ axis.
Step 2. Identify and count the pendant atoms' valence orbitals.
Each of the three pendant fluorine atoms has four valence orbitals: one $2s$ and three $2p$ orbitals. Thus, we can expect a total of twelve SALCs from the three atoms. However, we will treat each type of orbital from the F atoms as its own set; each will have its own set of coordinates, with the $y$ axis along the M-L bond. Thus we expect the following:
• One set of $2s$ SALCs: there is one 2s orbital on each of three atoms. This set will have three SALCs.
• One set of $2p_x$ SALCs: there is one $2p_x$ orbital on each of three atoms. Each of these individual orbitals is perpendicular to the M-L bond and coplanar with the molecule. This set will have three SALCs.
• One set of $2p_y$ SALCs: there is one $2p_y$ orbital on each of three atoms. Each of these individual orbitals is colinear with the M-L bond. This set will have three SALCs.
• One set of $2p_z$ SALCs: there is one $2p_z$ orbital on each of three atoms. Each of these individual orbitals is perpendicular to the M-L bond and parallel with the principal axis of the molecule. This set will have three SALCs.
Step 3. Generate the $\Gamma$'s for each set of SALCs
Use the $D_{3h}$ character table to generate one reducible representation ($\Gamma$) for each of the four sets of pendant atomic orbitals shown in Figure $2$; in this case we need four $\Gamma$ because there are four types of valence orbital (the $s, p_x, p_y, p_z$).
• For the set of $s$ orbitals: perform the operation for each class in the character table. Each $s$ orbital is assigned a value depending on whether it moves or not: assign a value of 1 if it remains in place during the operation, or 0 (zero) if it moves out of its original place.
• for each set of $p$ orbitals: Now there are phases (signs of the wavefunction), so a negative value is possible. Assign a value of 1 to each $p$ orbital if it remains in place and in phase during the operation; assign a -1 if the atom's position remains, while the phase of the orbital is inverted (the positive lobe moves to the position of the negative lobe and vice versa); assign a 0 if it moves out of its original position.
The $\Gamma$ for each type is given below:
$\begin{array}{|c|cccccc|} \hline \bf{D_{3h}} & E & 2C_3 & 3C_2 & \sigma_h & 2S_3 & 3\sigma_v \ \hline \bf{\Gamma_{2s}} & 3 & 0 & 1 & 3 & 0 & 1 \ \bf{\Gamma_{2p_x}} & 3 & 0 & -1 & 3 & 0 & -1 \ \bf{\Gamma_{2p_y}} & 3 & 0 & 1 & 3 & 0 & 1 \ \bf{\Gamma_{2p_z}} & 3 & 0 & -1 & -3 & 0 & 1 \ \hline \end{array} \nonumber$
Step 4. Break $\Gamma$'s into irreducible representations for individual SALCs
Reduce each $\Gamma$ into its component irreducible representations. Using either of the processes described previously, we find that each of the $\Gamma$ reduces to two irreducible representations under the $D_{3h}$ point group; in each case one is singly degenerate ($A$) and the other doubly degenerate ($E$). This equates to three SALCs for each $\Gamma$ (orbital type) and a total of twelve SALCs. This should give us confidence! We were in fact expecting 12 SALCs because we found that there were twelve pendant atomic orbitals to start with.
$\begin{array}{|c|cccccc|} \hline {D_{3h}} & E & 2C_3 & 3C_2 & \sigma_h & 2S_3 & 3\sigma_v \ \hline \bf{\Gamma_{2s}} & \bf 3 & \bf 0 & \bf 1 & \bf 3 & \bf 0 & \bf 1 \ A_{1}' & 1 & 1 & 1 & 1 & 1 & 1 \ E' & 2 & -1 & 0 & 2 & -1 & 0 \ \hline \end{array} \nonumber$
$\begin{array}{|c|cccccc|} \hline {D_{3h}} & E & 2C_3 & 3C_2 & \sigma_h & 2S_3 & 3\sigma_v \ \hline \bf{\Gamma_{2p_x}} &\bf 3 &\bf 0 &\bf -1 &\bf 3 &\bf 0 &\bf -1 \ A_{2}' & 1 & 1 & -1 & 1 & 1 & -1 \ E' & 2 & -1 & 0 & 2 & -1 & 0 \ \hline \end{array} \nonumber$
$\begin{array}{|c|cccccc|} \hline {D_{3h}} & E & 2C_3 & 3C_2 & \sigma_h & 2S_3 & 3\sigma_v \ \hline \bf{\Gamma_{2p_y}} &\bf 3 &\bf 0 &\bf 1 &\bf 3 &\bf 0 &\bf 1 \ A_{1}' & 1 & 1 & 1 & 1 & 1 & 1 \ E' & 2 & -1 & 0 & 2 & -1 & 0 \ \hline \end{array} \nonumber$
$\begin{array}{|c|cccccc|} \hline {D_{3h}} & E & 2C_3 & 3C_2 & \sigma_h & 2S_3 & 3\sigma_v \ \hline \bf{\Gamma_{2p_z}} &\bf 3 &\bf 0 &\bf -1 &\bf -3 &\bf 0 &\bf 1 \ A_{2}" & 1 & 1 & -1 & -1 & -1 & 1 \ E" & 2 & -1 & 0 & -2 & 1 & 0 \ \hline \end{array} \nonumber$
Step 5: Sketch the SALCs
We'll skip this step for now to make this problem simpler.
Step 6: Draw the MO diagram by combining SALCs with AO’s of like symmetry.
We now know the symmetries of different pendant orbital SALCs - and it is helpful if we also know their relative energies, and the energies of the valence orbitals on the central boron atom. We can make predictions based on periodic trends, or we can use a table of ionization energies as a guide. The symmetries of the boron $2s$ and $2p$ orbitals can be found from the $D_{3h}$ character table. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/05%3A_Molecular_Orbitals/5.04%3A_Larger_%28Polyatomic%29_Molecules/5.4.06%3A_BF.txt |
Concept Review Questions
1. What are the assumptions (axioms) of molecular orbital (MO) theory?
2. Explain qualitatively why the vectorial addition of atomic orbitals creates molecular orbitals (which underlying physical phenomenon is described by the vectorial addition?).
3. Explain qualitatively why molecular orbital theory is suitable to describe covalent bonding.
4. What are the three criteria that determine the degree of covalent interaction in MO theory?
5. Which three rules determine the degree of orbital overlap in MO theory?
6. Explain why the combination of a large, diffuse orbital and a small orbital produces only weak covalent interaction.
7. Explain why sigma interactions between atomic orbitals typically produce larger orbital overlaps than pi interactions.
8. Explain why the combination of orbitals of the same energy leads to the largest degree of covalent interaction.
9. What can be said about energy and location of a bonding and an anti-bonding molecular orbital that are made from atomic orbitals of large energy difference?
10. MO theory – even though designed for covalent bonding – can also make statements about ionic bonding. Explain why.
11. Explain the principles of SALC.
Dr. Kai Landskron (Lehigh University). If you like this textbook, please consider to make a donation to support the author's research at Lehigh University: Click Here to Donate.
Homework Problems
Section 1
Exercise 1
What will most likely lead to the smallest covalent interaction?
a) Overlap of a small and a large orbital.
b) Overlap of two small orbitals.
c) Overlap of two large orbitals.
Answer
a) Overlap of a small and a large orbital.
Exercise 2
What will most likely lead to the largest covalent interaction?
a) orbital overlap in sigma-fashion
b) orbital overlap in pi-fashion
c) orbital overlap in delta-fashion
Answer
a) orbital overlap in sigma-fashion
Exercise 3
Qualitatively construct the MO diagrams composed of
a) two 2s atom orbitals A and B of equal energy.
b) two 2s atom orbitals where the orbital energy of atom A is significantly higher than that of B. If both the bonding and the antibonding MO are filled with electrons, where will bonding and antibonding electrons primarily be located? Briefly explain your decision.
Answer
Exercise 4
Decide by “inspection” which of the following combinations of orbitals have the “right” symmetries to form molecular orbitals.
a) The 2px orbital of the first N atom and the 2py orbital of the second N atom in the molecule N2. The z axis is defined as the bond axis in N2.
b) The F 2px and the H 1s orbital in the HF molecule. The z axis is defined as the bond axis.
c) The 2pz orbital of F and the 1s orbital in the HF molecule. The z axis is defined as the bond axis.
Answer
a)
b)
c)
Exercise 5
The CH4 molecule belongs to the point group Td. You can find the character table of the point group in the internet.
a) Calculate the reducible representation for the ligand group orbitals (LGOs).
b) Calculate the irreducible representations of the ligand group orbitals (LGOs).
c) Draw a qualitative molecular orbital diagram for CH4.
Answer
Exercise 6
Which are the symmetry types of the central atom orbitals in the PCl5 molecule?
Answer
1. Determine point group of PCl5. --> D3h.
2. Decide what are the valence orbitals of the central atom: 3s, 3p
3. Look up the character table of D3h, e.g., in the internet. You will find their symmetries to be: A1' (3s), A2'' (3pz), E'(3px, 3py)
Exercise 7
For the hypothetical BrKr+ molecule: Toward which atom is the HOMO polarized? Explain briefly why.
Answer
Exercise 8
Reconstruct the MO diagram for water and NH3 (repeat what we did in class without looking at your notes (only use the respective character tables).
Answer
Water
NH3
Dr. Kai Landskron (Lehigh University). If you like this textbook, please consider to make a donation to support the author's research at Lehigh University: Click Here to Donate.
Exercise $1$
A Molecular Orbital Diagram for a diatomic molecule (two atoms) always has the same basic pattern.
• Draw a picture of the levels.
• Label each level with σ, σ*, π, π*
Answer
Exercise $2$
A Molecular Orbital Diagram for a diatomic molecule (two atoms) varies in the number of electrons. How do you populate the electrons?
Answer
• Count the valence electrons on the molecule. That's the number of valence electrons on each atom, adjusted for any charge on the molecule. (eg C22- has 10 valence electrons: 4 from each carbon -- that's 8 -- and two more for the 2- charge).
• Fill electrons into the lowest energy orbitals first.
• Pair electrons after all orbitals at the same energy level have one electron.
Exercise $3$
Construct a qualitative molecular orbital diagram for chlorine, Cl2. Compare the bond order to that seen in the Lewis structure (remember that an electron in an antibonding orbital cancels the stabilization due to bonding of an electron in a bonding orbital).
Answer
Exercise $4$
1. Construct a qualitative molecular orbital diagram for oxygen, O2.
2. Compare the bond order to that seen in the Lewis structure.
3. How else does this MO picture of oxygen compare to the Lewis structure? What do the two structures tell you about electron pairing?
4. Compounds that have all of their electrons paired are referred to as diamagnetic. Those with unpaired electrons are referred to as paramagnetic. Paramagnetic materials are attracted by a magnetic field, but diamagnetic things are not. How would you expect molecular oxygen to behave?
Answer
Exercise $5$
1. Construct a qualitative molecular orbital diagram for peroxide anion, O22-.
2. Compare the bond order to that seen in the Lewis structure.
3. How else does this MO picture of oxygen compare to the Lewis structure? What do the two structures tell you about electron pairing?
4. Based on molecular orbital pictures, how easily do you think dioxygen could be reduced to peroxide (through the addition of two electrons)?
Answer
Exercise $6$
Construct a qualitative molecular orbital diagram for diboron, B2. Do you think boron-boron bonds could form easily, based on this picture?
Answer
Exercise $7$
1. Construct a qualitative molecular orbital diagram for dicarbon, C2.
2. Compare the bond order to that seen in the Lewis structure.
3. How else does this MO picture of oxygen compare to the Lewis structure? What do the two structures tell you about electron pairing?
Answer
Exercise $8$
1. Construct a qualitative molecular orbital diagram for acetylide anion, C22-.
2. Compare the bond order to that seen in the Lewis structure.
3. How else does this MO picture of oxygen compare to the Lewis structure? What do the two structures tell you about electron pairing?
4. Based on molecular orbital pictures, how easily do you think dicarbon could be reduced to acetylide (through the addition of two electrons)?
Answer
Exercise $9$
Make drawings and notes to summarize the effect of populating antibonding orbitals.
Answer
Exercise $10$
Researchers at Johns Hopkins recently reported the formation of Na4Al2 in a pulsed arc discharge (they put a lot of electric current through a sample of sodium and aluminum; Xinxing Zhang, Ivan A. Popov, Katie A. Lundell, Haopeng Wang, Chaonan Mu, Wei Wang, Hansgeorg Schnöckel, Alexander I. Boldyrev, Kit H. Bowen, Angewandte Chemie International Edition, 2018, 57(43), 14060-14064. Copyright 2018, John Wiley & Sons. Used with permission.).
1. The compound is ionic. Explain which atoms form the cations, based on periodic trends.
2. Therefore, what atoms form the anion?
3. The anion is one molecule. What is the charge on this molecule?
4. Show how to calculate the total valence electrons in this molecular anion.
5. Draw a Lewis structure for this molecular anion.
6. Construct a diatomic molecular orbital energy level diagram for this molecule. Label the energy levels (sigma, pi, etc.) and add in the correct number of electrons.
7. Show how to calculate the bond order in the molecule.
Answer
a) Na, because Na has a lower ionization potential (and a lower electronegativity) than Al.
b) Al
c) 4-, because there are four Na+
d) total e- = 2 x 3 e- (per Al) + 4 e- (for the negative charge) = 10 e-
g) $\textrm{bond order} = \frac{( \# bonding \: e^{-} - \# antibonding \: 3^{-})}{2} = \frac{8-2}{2}= 3$
Exercise $1$
Draw the molecular orbital diagram for ($\ce{NO^-}$).
Answer
Add texts here. Do not delete this text first.
Exercise $1$
Draw and compare the molecular orbital diagrams to the Lewis structure diagrams for $\ce{O2}$, $\ce{O2^-}$, and $\ce{O2^2-}$.
Answer
The Lewis structures and MO diagrams are shown below. In general, the bond order derived from the MO diagram agrees with the Lewis structure for each species. In the case of the two ions, the number of unpaired electrons from MO theory and the Lewis structure also agrees. In the case of dioxygen, however, there is an inconsistency between the Lewis representation and the MO diagram. While the Lewis structure would lead us to believe that all electrons are paired, the MO diagram indicates paramagnetism, with two unpaired electrons. The MO diagram explains the magnetic behaviour of dioxygen.
Exercise $1$
Draw the molecular orbital diagram for hydroxide ion ($\ce{OH^-}$).
Answer
Add texts here. Do not delete this text first.
Exercise $1$
Draw the molecular orbital diagram for hydroxide fluoride ($\ce{HF}$).
Answer
Add texts here. Do not delete this text first.
Exercise $1$
Draw the molecular orbital diagram for borane ($\ce{BN}$).
Answer
Add texts here. Do not delete this text first.
BH3
Exercise $1$
Draw the molecular orbital diagram for ($\ce{FeH6}$).
Answer
Exercise $1$
Draw the molecular orbital diagram for boron trifluoride ($\ce{BF3}$).
• What is the non-bonding orbital?
• What are the HOMO and LUMO orbitals?
• Why does e' have 2 lines of energy levels compared to a1', which has only a 1-line energy level?
• What is the bond order of $\ce{BH3}$ in this MO diagram?
Answer
The calculated MO diagram is shown below.
• $a_2''$ is the lone non-bonding orbital.
• The HOMO is the $a_2''$ nonbonding orbital, the LUMO is a bonding $e'$ set of two degenerate orbitals/
• $e'$ is doubly degenerate
• The bond order is 3 | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/05%3A_Molecular_Orbitals/5.P%3A_Problems.txt |
In a very real sense, we can make an acid anything we wish. The differences between the various acid-base concepts are not concerned with which is 'right', but which is most convenient to use in a particular situation.
James E. Huheey, Ellen A. Keiter, and Richard L. Keiter
The concept of acids and bases is often associated with the movement of hydrogen ions from one molecule or ion to another. However, a host of acid-base concepts have been developed to help chemists organize and make sense of a wide range of reactions (Table 6.1).
Table 6.1. Summary of major acid-base models.
Definition Theoretical paradigm and notable features. Acid Base Illustrative sample reactions
Arrhenius
(1894)
Interested in what the substance does to the state of an aqueous solution. In particular it assesses proton donation to & removal from water using [H3O+] as a proxy. Readily accommodates the pH concept as a measure of the state of a solution.
Increases
[H3O+]
Decreases
[H3O+]
$\underset{acid}{HCl}~+~H_2O~\rightarrow~H_3O^+~+~Cl^-$
$\underset{base}{NH_3}~+~H_2O~\rightarrow~NH_4^+~+~OH^-$
Brønsted-Lowry
(1923)
Envisions acid-base reactivity in terms of the transfer of an H+ from one substance to another. Allows for conjugate acids and bases and solvent autoionization. Donates H+ Accepts H+
$\underset{acid}{HCl}~+~\underset{base}{NH_3}~\rightarrow~\underset{conj.~acid}{NH_4^+}~+~\underset{conj.~base}{Cl^-}$
$\underset{acid}{HOAc}~+~\underset{base}{NH_3}~\rightarrow~\underset{conj.~acid}{NH_4^+}~+~\underset{conj.~base}{OAc^-}$
$\underset{amphoteric}{2~H_2O}~\rightarrow~\underset{conj.~acid}{H_3O^+}~+~\underset{conj.~base}{OH^-}$
Lux-Flood
(1939-~47)
Describes reactions involving oxides and oxyanions in terms of the transfer of oxide ion (O2-). Mainly used in geochemistry, although it also can be used to describe some redox reactions. Oxide acceptor Oxide donor
$\underset{acid}{SiO_2}~+~\underset{base}{CaO}~\rightarrow~CaSiO_3$
$\underset{base}{H_2O}~+~\underset{acid}{CO}~\rightarrow~H_2~+~CO_2$
Solvent System Applies aspects of the Arrhenius , Brønsted-Lowry, and Lux-Flood acid base concepts to solvent cation & anion formation in a generalized reaction. Can be used to describe solution chemistry in nonaqueous solvent systems like BrF3. Is a solvent cation or increases the solvent cation concentration, often by receiving a lone pair- bearing group Is a solvent anion or increases the solvent anion concentration, often by donating a lone pair- bearing group
$\underset{acid}{SbF_5}~+~\underset{base}{BrF_3}~\rightarrow~\underset{conj.~base}{SbF_6^-}~+~\underset{conj.~acid}{BrF_2^+}$
$\underset{amphoteric}{2~BrF_3}~\rightarrow~\underset{conj.~acid}{BrF_2^+}~+~\underset{conj.~base}{BrF_4^-}$
Lewis (1923) Envisions acid-base reactivity in terms of electron pair donation. Encompasses the Arrhenius , Brønsted-Lowry, Lux-Flood, and Solvent System definitions and readily integrates with molecular orbital descriptions of chemical reactivity in Frontier orbital theory. Accepts an electron pair Donates an electron pair $\underset{base}{:NH_3}~+~\underset{acid}{BF_3}~\rightarrow~H_3N~+~BF_3$
Nucleophile-Electrophile Applies the Lewis concept to organic reactivity. Nucleophiles are Lewis bases which tend to react to form a bond with Lewis acid sites called electrophilic centers.
(The electrophile)
Tends to react by receiving an electron pair from a nucleophile, forming a bond in the process
(The nucleophile)
Donates an electron pair to form a bond to an electrophile
$\underset{base}{Br^-}~+~\underset{acid}{CH_3-Cl}~\rightarrow~Br-CH_3~+~Cl^-$
Usanovich
(1939)
Extends Lewis theory to include the donation and acceptance of any number of electrons, whether through the formation of an adduct or electron transfer. Accepts electrons Donates electrons
$\underset{base}{:NH_3}~+~\underset{acid}{BH_3}~\rightarrow~\underset{adduct}{H_3N-BH_3}$
$\underset{acid}{Fe^{2+}}~+~\underset{base}{Zn^0}~\rightarrow~Fe^0~+~Zn^{2+}$
Frontier Orbital (1960s) Envisions Lewis Acid-base/Electrophile-nucleophile reactions in terms of the donation and acceptance of electrons between the reactant's frontier orbitals. Specifically, the reaction is envisioned in terms of donation of the base's HOMO electrons into the acid's LUMO level. Possesses a LUMO capable of forming an occupied bonding MO on mixing with a base's HOMO. Possesses an electron-bearing HOMO capable of forming a filled bonding MO on mixing with an acid's LUMO.
base acid adduct
Some concepts involve defining acids and bases in particular ways that allow for the understanding of particular types of chemical systems. For example, the familiar Arrhenius and Brønsted acid and base concepts used in general chemistry help chemists make sense of the behavior of compounds which can transfer H+ ions among themselves, often in aqueous solution. However, the solvent system acid-base concept defines acids and bases in terms of the transfer of a lone-pair bearing group and is particularly useful for conceptualizing the reactivity of main group halides, oxides, and related compounds. Some acid-base definitions seek to encompass an extremely wide range of chemical reactions. For instance, the Lewis acid-base definition encompasses the Arrhenius , Brønsted, and solvent system definitions and has also found wide use in inorganic chemistry owing to the ease with which Lewis acid-base interactions may be described by the Frontier Orbital approach in terms of interacting molecular orbitals on the acid and base. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/06%3A_Acid-Base_and_Donor-Acceptor_Chemistry/6.01%3A_Acid-Base_Models_as_Organizing_Concepts.txt |
The Arrhenius acid-base concept defines acids and bases in terms of how they affect the amount of hydronium ions, $\ce{H_3O^{+}}$, (and by extension hydroxide ions, $\ce{OH^{-}}$) in aqueous solutions. Simply, in the Arrhenius definition an acid is a substance that increases the concentration of hydronium ions when it is dissolved in water. This typically occurs when the acid dissociates by loss of a proton to water according to the general equation:
$\ce{HA(aq) + H_2O(l) ⇌ H_3O^{+}(aq) + A^{-}(aq)} \label {6.2.1}$
where $\ce{A}$ is the deprotonated form of the acid. For example, what hydrochloric and acetic acid, $\ce{CH3CO2H}$, have in common is that both increase the amount of hydronium ion when they are dissociated in solution.
\begin{align*} \ce{HCl(aq) + H_2O(l) } &\ce{-> H_3O^+(aq) + Cl^{-}(aq)} \[4pt] \ce{CH_3CO_2H(aq) + H_2O(l)} &\ce{<=> H_3O^{+}(aq) + CH_3CO_2^{-}(aq)} \end{align*} \nonumber
In terms of the Arrhenius definition, the major difference between hydrochloric and acetic acid is that hydrochloric acid dissociates completely in solution to yield stoichiometric amounts of $\ce{H_3O^{+}}$, while acetic acid only partially dissociates. Acids like $\ce{HCl}$ that completely dissociate in water are classified as strong in the Arrhenius definition, while those like acetic acid that do not are classified as weak.
Although all weak acids incompletely dissociate, the extent of dissociation can vary widely. The relative strengths of weak Arrhenius acids is conveniently expressed in terms of the equilibrium constant for their acid dissociation reaction, $K_a$.
$K_a=\dfrac{[\ce{H^{+}}][\ce{A^{-}}]}{[\ce{HA}]} \nonumber$
The pKa values for selected weak acids are given in Table $1$.
Table $1$: Values of $K_a$, $pK_a$, $K_b$, and $pK_b$ for selected monoprotic acids.
Acid $HA$ $K_a$ $pK_a$ $A^−$ $K_b$ $pK_b$
*The number in parentheses indicates the ionization step referred to for a polyprotic acid.
sulfuric acid (2nd ionization) $HSO_4^−$ $1.0 \times 10^{−2}$ 1.99 $SO_4^{2−}$ $9.8 \times 10^{−13}$ 12.01
hydrofluoric acid $HF$ $6.3 \times 10^{−4}$ 3.20 $F^−$ $1.6 \times 10^{−11}$ 10.80
nitrous acid $HNO_2$ $5.6 \times 10^{−4}$ 3.25 $NO_2^−$ $1.8 \times 10^{−11}$ 10.75
formic acid $HCO_2H$ $1.78 \times 10^{−4}$ 3.750 $HCO_2^−$ $5.6 \times 10^{−11}$ 10.25
benzoic acid $C_6H_5CO_2H$ $6.3 \times 10^{−5}$ 4.20 $C_6H_5CO_2^−$ $1.6 \times 10^{−10}$ 9.80
acetic acid $CH_3CO_2H$ $1.7 \times 10^{−5}$ 4.76 $CH_3CO_2^−$ $5.8 \times 10^{−10}$ 9.24
pyridinium ion $C_5H_5NH^+$ $5.9 \times 10^{−6}$ 5.23 $C_5H_5N$ $1.7 \times 10^{−9}$ 8.77
hypochlorous acid $HOCl$ $4.0 \times 10^{−8}$ 7.40 $OCl^−$ $2.5 \times 10^{−7}$ 6.60
hydrocyanic acid $HCN$ $6.2 \times 10^{−10}$ 9.21 $CN^−$ $1.6 \times 10^{−5}$ 4.79
ammonium ion $NH_4^+$ $5.6 \times 10^{−10}$ 9.25 $NH_3$ $1.8 \times 10^{−5}$ 4.75
water $H_2O$ $1.0 \times 10^{−14}$ 14.00 $OH^−$ $1.00$ 0.00
acetylene $C_2H_2$ $1 \times 10^{−26}$ 26.0 $HC_2^−$ $1 \times 10^{12}$ −12.0
ammonia $NH_3$ $1 \times 10^{−35}$ 35.0 $NH_2^−$ $1 \times 10^{21}$ −21.0
As can be seen from the table the Ka values for weak acids are less than one (otherwise they would not be weak) and vary over many orders of magnitude. Consequently it is customary to tabulate acid ionization constants as pKa values:
$pK_a=-\log K_a \nonumber$
Because pKa values essentially place the Ka values on a negative base ten logarithmic scale, the stronger the weak acid, the lower its pKa. Weak acids with larger Ka values will have lower pKa values than weaker acids with smaller Ka. Moreover, each unit increase or decrease in the pKa corresponds to a tenfold increase or decrease in the corresponding Ka.
While Arrhenius acids increase the concentration of $\ce{H_3O^{+}}$ in aqueous solution, Arrhenius bases decrease $\ce{H_3O^{+}}$. Strong bases do this stoichiometrically. Most are hydroxide salts of alkali metals or quaternary ammonium salts that dissociate completely when dissolved in water:
$\ce{MOH(aq) -> M^{+}(aq) + OH^{-}(aq)} \nonumber$
This added hydroxide decreases the concentration of $H_3O^+$ by shifting the water autoionization equilibrium towards water.
$\ce{2H_2O(l) <=> H_3O^{+}(aq) + OH^{-}(aq)} \nonumber$
In contrast, most weak bases react with water to produce an equilibrium concentration of hydroxide ion according to the base dissociation reaction
$\ce{B(aq) + H_2O(l) <=> BH^{+}(aq) + OH^{-}(aq)} \nonumber$
in which $\ce{B}$ is the weak base. The ionization constant for this reaction, called the base ionization constant or $K_b$, is typically used as a measure of a weak base's strength.
Because both hydroxide and hydronium ion are products of water autoionization, the concentrations of hydronium ion and hydroxide ion in aqueous solution will vary reciprocally with one another. This means that Arrhenius acids can be recognized as substances that decrease the hydroxide concentration and Arrhenius bases as substances that increase it.
Since the Arrhenius acid-base concept is concerned about the state of the water autoionization reaction, Arrhenius acids and bases may also be recognized by their effect on the solution pH. Arrhenius acids decrease the pH and Arrhenius bases will increase it.
NOTE
To qualify as an Arrhenius acid, upon the introduction to water, the chemical must cause, either directly or otherwise:
• an increase in the aqueous hydronium concentration,
• a decrease in the aqueous hydroxide concentration, or
• a decrease in the solution pH.
Conversely, to qualify as an Arrhenius base, upon the introduction to water, the chemical must cause, either directly or otherwise:
• a decrease in the aqueous hydronium concentration,
• an increase in the aqueous hydroxide concentration, or
• an increase in the solution pH.
Because the Arrhenius acid-base model defines acids and bases in terms of their impact on the state of an aqueous solution the Arrhenius concept is unable to describe reactions in nonaqueous solvents, gases, molten liquids, and the solid state. Consequently other models should be used to describe reactions involving the transfer of $H^+$ and other fragments in nonaqueous media. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/06%3A_Acid-Base_and_Donor-Acceptor_Chemistry/6.02%3A_Arrhenius_Concept.txt |
The Brønsted-Lowry acid-base concept
The Brønsted-Lowry acid-base concept overcomes the Arrhenius system's inability to describe reactions that take place outside of aqueous solution by moving the focus away from the solution and onto the acid and base themselves. It does this by redefining acid-base reactivity as involving the transfer of a hydrogen ion, $H^+$, between an acid and a base. Specifically, a Brønsted acid is a substance that loses an $H^+$ ion by donating it to a base. This means that a Brønsted base is defined as a substance which accepts $H^+$ from an acid when it reacts.
Because the Brønsted-Lowry concept is concerned with $H^+$ ion transfer rather than the creation of a particular chemical species, it is able to handle a diverse array of acid-base concepts. In fact, from the viewpoint of the Brønsted-Lowry concept, Arrhenius acids and bases are just a special case involving hydrogen ion donation and acceptance involving water. Arrhenius acids donate $H^+$ ion to water, which acts as a Brønsted base in accepting it to give $H_3O^+$:
$\label{ 6.3.1}$
Similarly, Arrhenius bases act as Brønsted bases in accepting a hydrogen ion from the Brønsted acid water:
$\nonumber$
In this way it can be seen that Arrhenius acids and bases are defined in terms of their causing hydrogen ions to be donated to and abstracted from water, respectively, while Brønsted acids and bases are defined in terms of their ability to donate and accept hydrogen ions to and from anything.
Becasue the Brønsted-Lowry concept can handle any sort of hydrogen ion transfer it readily accommodates many reactions that Arrhenius theory cannot, including those that take place outside of water, such as the reaction between gaseous hydrochloric acid and ammonia:
$\nonumber$
The classification of acids as strong or weak usually refers to their ability to donate or abstract hydrogen ions to or from water to give $H_3O^+$ and $OH^-$, respectively, i.e., their Arrhenius acidity and basicity. However, acids and bases may be classified as strong and weak under the Brønsted-Lowry definition based on whether they completely transfer or accept hydrogen ions; it is just that in this case it is important to specify the conditions under which a given acid or base acts strong or weak. For example, acetic acid acts as a weak acid in water but is a strong acid in triethylamine, since in the latter case it completely transfers a hydrogen ion to triethylamine to give triethylammonium acetate. Alternatively, the acidity or basicity of a compound may be specified using a thermodynamic scale like the Hammett acidity.
Conjugate Acids and Bases
By redefining acids and bases in terms of hydrogen ion donation and acceptance, the Brønsted-Lowry system makes it easy to recognize that when an acid loses its hydrogen ion it becomes a substance that is capable of receiving it back again, namely, a base. Consider, for example, the base dissociation of ammonia in water. When ammonia acts as a Brønsted base and receives a hydrogen ion from water, ammonium ion and hydroxide are formed:
$\nonumber$
The ammonium ion is itself a weak acid that can undergo dissociation:
$\nonumber$
In this case ammonia and ammonium ion are acid-base conjugates. In general acids and bases that differ by a single ionizable hydrogen ion are said to be conjugates of one another.
The strengths of conjugates vary reciprocally with one another, so the stronger the acid the weaker the base and vice versa. For example, in water, acetic acid acts as a weak Brønsted acid:
$\nonumber$
and acetic acid's conjugate base, acetate, acts as a weak Brønsted base.
$\nonumber$
However, in liquid ammonia acetic acid acts as a strong Brønsted acid:
$\nonumber$
while its conjugate base, acetate, is neutral.
$\nonumber$
The reciprocal relationship between the strengths of acids and their conjugate bases has several consequences:
1. Under conditions when an acid or base acts as a weak acid or base its conjugate acts as weak as well. Conversely, when an acid or base acts as a strong acid or base its conjugate acts as a neutral species.
2. When a Brønsted acid and base react with one another, the equilibrium favors formation of the weakest acid-base pair. That is why the acid-base reaction between acetic acid and ammonia in liquid ammonia proceeded to give the weak acid ammonium ion and neutral acetate. This consequence is particularly important for understanding the behavior of acids and bases in nonaqueous solvents, as illustrated by the following example.
Example $1$
Can a solution of lithium diisopropylamide in heptane be used to form lithium cyclopentadienide? The $pK_a$ of cyclopentadiene and diisopropylamine are ~15 and 40, respectively, and the proposed reaction is as follows:
Solution:
Since cyclopentadiene is a stronger acid than diisopropylamine (the stronger the acid the lower the $pK_a$) the equilibrium will favor protonation of the diisopropylamine by cyclopentadiene. Consequently addition of a heptane solution of lithium diisopropylamide to monomeric cyclopentadiene should give lithium cyclopentadienide. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/06%3A_Acid-Base_and_Donor-Acceptor_Chemistry/6.03%3A_Brnsted-Lowry_Concept/6.3.01%3A_Brnsted-Lowry_Concept.txt |
Because the acidity of a given substance depends on the interplay between the relatively large values of its proton affinity and the energy associated with solvation of an acid's conjugate base, it can sometimes be difficult to estimate the strength of an acid in a given solvent in the absence of detailed computations. Nevertheless a variety of simple ideas may be used to roughly estimate the relative strengths of acids. These should never be substituted for a detailed consideration of solvation but can serve as useful aids when thinking about trends and designing new Brønsted acids and bases.
Four main factors should be considered when thinking about the relationship between molecular structure and Brønsted acidity and basicity
Some simple factors that it can be helpful to consider when thinking about the strength of a given acid or base are:
1. Bond strength effects
The weaker the bond to the ionizable hydrogen, the stronger the acid. Strongly bonded hydrogen ions are difficult to remove, weakly bonded ones much less so.
2. Inductive effects
Inductive effects involve the donation or withdrawal of electrons from an atom by a group connected to it through bonds. Electron donating groups increase the electron density while electron withdrawing groups decrease it. Atoms or groups that withdraw electron density away from a center increase its acidity while those which donate electrons to the center decrease its acidity. The reasons for this follow from the heterolytic bond cleavage of acid ionization:
$E-H→E:^-+ H^+ \nonumber$
• When a bond to hydrogen is more polarized away from the H (more like$^{-δ}E-H^{δ+}$) it is easier to cleave off the hydrogen ion from that E-H bond. This may be seen from how the pKa values of acetic acid and its mono-, di-, and tri-chlorinated derivatives decreases with the extent of chlorination of the methyl group.
• Polarized E-H bonds also make for stronger Brønsted acids because the resulting $E:^-$ conjugate base is more stable.
This leads to the third major factor that should be considered when thinking about acid strength.
3. Electronegativity Effects, Especially as Seen Using the Conjugate Base Principle
The more stable the acid's conjugate base, the stronger the Brønsted acid. All reactions are in theory reversible, and so when considering the propensity of an acid to donate hydrogen ions, it can be helpful to look at the reverse of hydrogen ion donation, namely, protonation of the acid's conjugate base. If deprotonation of the acid gives a very stable conjugate base then deprotonation of the acid will be more favorable.
Two factors determine the stability of an acid's conjugate base.
• Conjugate bases in which a small amount of charge is on a large atom, spread over a large number of atoms, and on electronegative atoms tend to be more stable. Conjugate bases in which a small amount of charge is spread over a large number of electronegative atoms are especially stable. That is why magic acid, a mixture of HF and $SbF_5$, is so acidic: the single negative charge on its conjugate base is spread over six F atoms and on Sb in $SbF_6^-$.
• Groups which tend to inductively polarize E-H bonds also tend to stabilize the conjugate base formed when that bond ionizes. In general, the more electronegative an atom, the better able it is to bear a negative charge. All other things being equal, weaker bases have negative charges on more electronegative atoms; stronger bases have negative charges on fewer electronegative atoms. This is apparent from how inductive effects lead to an increase in the acidity of E-H bonds as the electronegativity of the element to which the acidic hydrogen is bound increases from left to right across a row of the periodic table. This horizontal periodic trend in acidity and basicity is apparent from the homologous series below:
Horizontal periodic trend in acidity and basicity
Notice how the inductive polarization of the E-H bond, which results in greater acidity, contributes to the greater stability of the conjugate base. For the case above, look at where the negative charge ends up in each conjugate base. In the conjugate base of ethane, the negative charge is borne by a carbon atom, while on the conjugate base of methylamine and ethanol the negative charge is located on a nitrogen and an oxygen, respectively. Remember that electronegativity also increases as we move from left to right along a row of the periodic table, meaning that oxygen is the most electronegative of the three atoms, and carbon the least.
Thus, the methoxide anion is the most stable (lowest energy, least basic) of the three conjugate bases, and the ethyl carbanion anion is the least stable (highest energy, most basic). Conversely, ethanol is the strongest acid, and ethane the weakest acid.
4. Size effects on Bond Strength and Charge Delocalization
There are two classes of size effects to be considered:
1. The larger the atom to which a H is bound in an E-H bond, the weaker the bond and the stronger the acid.
2. Increased charge delocalization with increasing size. Electrostatic charges, whether positive or negative, are more stable when they are ‘spread out’ over a larger area. The greater the volume over which charge is spread in the acid's conjugate base, the more stable that base and the stronger the acid.
The impact of size effects are readily seen in the increase in acidity of the hydrogen halides, as illustrated by the vertical periodic trend in acidity and basicity below:
Vertical periodic trend in acidity and basicity
On going vertically down the halogen group from F to I the H-X bond strength decreases in the acid, making it easier to ionize, while the charge becomes more diffuse in the resultant X- ion, making the conjugate base more stable.
The increase in the acidity of the hydrogen halides down a group suggests that size effects are more important than inductive effects. In the case of the hydrogen halides, because fluorine is the most electronegative halogen element, we might expect fluoride to also be the least basic halogen ion. But in fact, it is the least stable, and the most basic! It turns out that when moving vertically in the periodic table, the size of the atom trumps its electronegativity with regard to basicity. The atomic radius of iodine is approximately twice that of fluorine, so in an iodide ion, the negative charge is spread out over a significantly larger volume.
Exercise $\PageIndex{}$
The structure of the amino acids serine and cysteine are shown below. Which do you expect will have the more acidic side chain?
Answer
Cysteine, since the cysteine side chain possesses an ionizable S-H bond while serine's side chain possesses an ionizable O-H bond. Since S is larger than O, cysteine's S-H bond will be weaker than serine's O-H bond, and the cysteine side chain's thiolate conjugate base more stable than the serine side chain's alkoxide conjugate base. In fact, the side chain $pK_a$ of cysteine is 8.3 while serine is considered to be nonionizable under physiological conditions.
Contributors and Attributions
Stephen M. Contakes (Westmont College),
* who expanded this section from a combination of
https://chem.libretexts.org/Courses/University_of_Arkansas_Little_Rock/Chem_1403%3A_General_Chemistry_2/Text/16%3A_Acids_and_Bases/16.6%3A_Molecular_Structure%2C_Bonding%2C_and_Acid-Base_Behavior
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Book%3A_Organic_Chemistry_with_a_Biological_Emphasis_(Soderberg)/Chapter_07%3A_Organic_compounds_as_acids_and_bases/7.3%3A_Structural_effects_on_acidity_and_basicity
6.3.03: The acid-base behavior of binary element hydrides is determined primarily by the ele
The compounds formed between the elements and hydrogen are called binary hydrides. All such compounds can in principle act as Brønsted acids in reactions with a suitably strong base. However, as the electronegativity decreases down a group and increases from left to right across the periodic table, the acidity of binary hydrides increases. In fact, on the left side of the periodic table the hydrides of extremely electropositive alkali and alkaline earth metals are not acidic but basic. They are perhaps best considered to be ionic salts of the hydride ion ($\ce{H^{-}}$). Consequently substances such as $\ce{NaH}$ and $\ce{CaH2}$ tend to act as Brønsted bases in their reactions.
$\ce{NaH(s) + H2O(l) → Na^{+}(aq) + OH^{-}(aq) + H2(g)} \nonumber$
$\ce{CaH2(s) + 2H2O(l) → Ca^{2+}(aq) + 2OH^{-}(aq) + 2H2(g)} \nonumber$
On the right side of the periodic table the binary hydrides of the nonmetals exhibit appreciable acidity.
$\ce{HBr(aq) + H2O(l) → H^{+}(aq) + Br^{-}(aq)} \nonumber$
For this reason the binary nonmetal hydrides are termed acidic hydrides. Nevertheless not all are equally acidic. The dilute aqueous acid ionization constants for these hydrides are given in Figure $1$. As can be seen from the constants in Figure $2$, the ability of the hydrides to transfer a hydrogen to water increases across a period and down a group.
These trends are largely due to changes in the electronegativity and size of the nonmetal atom:
1. Going across a period, the acid strength increases as there is an increase in electronegativity and the molecule gets more polar, with the hydrogen getting a larger partial positive charge. This makes it easier to heterolytically cleave the E-H bond to produce a stable anion.
$E-H → E:^- + H^+ \nonumber$
2. Going down a group, the acid strength increases because the bond strength decreases as a function of increasing size of the nonmetal, and this has a larger effect than the electronegativity. In fact HF is a weak acid because it is so small that the hydrogen-fluorine bond is so strong that it is hard to break. Remember, the weaker the bond, the stronger the acid strength. This is further illustrated in Table $1$, where the weakest bond has produced the strongest acid.
Table $1$ Acid strength as function of bond energy
Relative Acid Strength HF << HCl < HBr < HI
H–X Bond Energy (kJ/mol) 570 432 366 298
Ka 10-3 107 109 1010
Contributors and Attributions
• Stephen M. Contakes (Westmont College),
* who expanded this section from a combination of | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/06%3A_Acid-Base_and_Donor-Acceptor_Chemistry/6.03%3A_Brnsted-Lowry_Concept/6.3.02%3A_Rules_of_Thumb_for_thinking_about_the_relationship_between_Molecular_Structure_and_B.txt |
Acids with an acidity greater than that of sulfuric acid are termed superacids.
Superacids are able to dissociate completely because when they do so they give an extremely stable anion in which the residual negative charge is distributed among multiple electronegative atoms. For example, in mixtures of $\ce{HF}$ and antimony pentaflouride the dissociation of a hydrogen ion from the $\ce{HF}$ is promoted by the formation of a Lewis acid-base adduct between the $\ce{HF}$ flouride and $\ce{SbF_5}$.
This gives extremely stable anions like $\ce{SbF_6^{-}}$ and $\ce{Sb_2F_{11}^{-}}$, in which a single negative charge is delocalized among many electronegative groups and atoms; of these, $CF_3$ and F are particularly common.
Superacids are able to protonate species that would otherwise act as neutral or even acidic species in water or aprotic media. Phosphoric, nitric, carboxylic, and other ordinary acids are all protonated when dissolved in superacids:
However, superacids also protonate many species that are not usually considered to have acid base properties. Even alkanes can be protonated and, in fact, superacids are used to generate carbocations in a variety of synthetic applications. They do this by protonating alkanes to give unstable species which then decompose to carbocations in further reactions. For example, mixtures of $\ce{SbF_5}$ and $\ce{FSO_3H}$ called Magic Acid can even protonate methane, which gives a $\ce{CH_5^{+}}$ cation:
$\ce{F_5Sb + FSO_3H + CH_4 -> F_5Sb-OSO_2F^{-} + CH_5^{+}} \nonumber$
which subsequently decomposes to give an ordinary carbocation
$\ce{CH_5^{+} -> CH_3^{+} + H_2}\nonumber$
Since liquid superacids are extremely corrosive and can be costly to separate from reaction mixtures, solid superacids are used industrially instead. The main use for these solid superacids is to generate carbocations for use in hydrocarbon isomerization and alkylation chemistry. Many solid superacids consist of sulfuric acid/sulfates attached to a metal oxide surface to give structures similar to that shown below:
$]\ These structures exhibit Hammett acidity parameters in the superacid range. However, the mechanism by which these solid superacids generate carbocations isn't entirely clear since they contain both Brønsted acid sites (at OH) and Lewis acid sites (at M) that could be involved in the carbocation generation process. The Hammett Acidity Function The acid ionization constants typically used to measure weak acids' acidity are only valid in dilute aqueous solutions. A more general measure of acidity that in principle is valid for any acid is the Hammett acidity function. The Hammett acidity function, $H_o$, is analogous to the pH used to describe the acidity of aqueous solutions but instead refers to the pure acid: \[pH = -\log(A_{\ce{H^{+}}}) \tag{for dilute aqueous solutions}$
$H_o = -\log (A_{\ce{H^{+}}}) \tag{for pure acids}$
where $\ce{A_{H^{+}}}$ is the activity of $\ce{H^{+}}$, which in many dilute solutions is approximately equal to the hydrogen ion concentration (that is why the pH is often defined in terms of [H+].
At first glance it may seem that the Hammett acidity function is simply a generalization of the pH concept for use in nonaqueous solutions. This is especially so since in water the the pH and $H_o$ do refer to the same quantity. However, the hydrogen ion of the Hammett acidity function is more than a generalization of the pH concept. Its real genius lies in that $A_{H^+}$ does not necessarily represent an actual chemical species of identity $H^+$ but rather an acid's ability to protonate weak indicator bases, B, specifically via the reaction:
$\ce{H^{+} + B <=> BH^{+}}\nonumber$
This reaction gives the weak acid $BH^+$ which can ionize in the reaction that is the reverse of that above:
$\ce{BH^{+} <=> B + H^{+}}\nonumber$
The extent of this ionization will depend on $A_{H^+}$ according to
$K_{ion} = \dfrac{[\ce{BH^{+}}]}{A_{\ce{H^{+}}}[B]}\nonumber$
Taking the negative logarithm of both sides and rearranging gives the Henderson-Hasselbach equation for the indicator base, $\ce{B}$:
$-K_{ion} = -\log{A_{\ce{H^{+}}}} -\log{\dfrac{[\ce{BH^{+}}]}{[\ce{B}]}}\nonumber$
which can be rearranged to give
$pK_{ion} = H_o -\log{\dfrac{[\ce{BH^{+}}]}{[\ce{B}]}}\nonumber$
or
$H_o= pK_{ion} -\log{\dfrac{[\ce{BH^{+}}]}{[\ce{B}]}}\nonumber$
From this it is apparent that $H_o$ represents an acid's ability to donate a hydrogen ion, as measured in terms of its ability to shift the equilibrium between $\ce{B}$ and $\ce{BH^{+}}$ towards $\ce{BH^{+}}$. More negative values of $H_o$ correspond to stronger Brønsted acids with a greater hydrogen ion transfer ability while less negative ones indicate weaker Brønsted acidity.
The value of $H_o$ has been experimentally determined for a number of strong acids by measuring the ratio of $\ce{BH^{+}}$ to $\ce{B}$ using weakly basic aromatics like 2,4,6-trinitroaniline, various nitrotoluenes, and triflouromethylbenzene as the indicator base, B. Some of these Hammett acidity parameters are shown in Table $1$.
As can be seen from table $1$, sulfuric acid has a Hammett acidity of -12. Since superacids are defined as acids with greater Brønsted acidity than pure sulfuric acid this means that superacids have $(H_o<-12)$.
Table $1$: Hammett Acidity for selected strong acids.
Acid $H_o$
Sulfuric acid, $H_2SO_4$ -12
Perchloric Acid, $HClO_4$ -13.0
Triflic acid (triflouromethanesulfonic acid), $CF_3SO_3H$ -14.6
Flourosulfonic acid, $FSO_3H$ -15.6
Magic Acid, $F_5Sb---FSO_3H$ -24 to -21$^a$
Fluoroantimonic acid, $SbF_5---FH$ -24 to -21$^{a,b}$
a. Difficult to estimate reliably.
b. Depends on the $SF_5$ to $FSO_3H$ ratio.
6.3.05: Thermodynamics of Solution-Phase Brnsted Acidity and Basicity
The behavior of Brønsted-Lowry acids and bases in solution is heavily influenced by solvation
This is particularly true in aqueous systems in which the energies and entropies of hydration can be quite significant. Consider the enthalpy and entropy changes for the dissociation of a number of acids in dilute aqueous solution at 25°C as given in Table $1$.
Table $1$: Thermochemical parameters for acid dissociation of selected inorganic acids.1,2
Acid
$\Delta S^°$
(J/mol-K)
$-T \times \Delta S^°$
(kJ/mol)
$\Delta H^°$
(kJ/mol)
$\Delta G^°$
(kJ/mol)
$K_a$
(calcd. from $\Delta G^°$ )
HF -102 30 -16 14 $6.8 \times 10^{-4}$
HCl -35 10 -57 -47 $10^8$
HBr -13 4 -65 -61 $10^{10}$
HI 11 -3 -62 -65 $10^{11}$
HClO -95.4 28.4 13.9 42.3 $4 \times 10^{-8}$
HClO2 -86.6 25.8 -14.6 11.2 $1.1 \times 10^{-2}$
H3PO4 → H+ + H2PO4- -66.9 20.0 -7.9 12.1 $7.9 \times 10^{-3}$
CH3CO2H → H+ + CH3CO2- -92.4 27.5 -0.4 27.1 $1.8 \times 10^{-5}$
As can be seen from the data in Table $1$, the major driving force for dissociation of inorganic acids is the enthalpy of dissociation, which is exothermic for all of the acids listed except very weakly acidic HClO. In contrast, the entropy term usually disfavors dissociation since it is negative for all acids except weakly polar HI (for which it contributes little to the overall free energy of dissociation). The situation is exactly reversed in the case of acetic acid, $CH_3CO_2H$, for which the enthalphly of dissociation is small and the entropy is the dominant contributor to the free energy.
Further insight into the factors which govern Brønsted acid strength in aqueous solution may be gained by examining the relative contributions of heterolytic H-A bond breaking and hydration to the enthalpy of dissociation. This is done using the thermodynamic cycle depicted in Figure $1$.
As can be seen from Figure $1$, while most of the process involves solvation energies, the heterolytic bond dissociation enthalpy is also a major factor in determining the acid dissociation enthalpy. The heterolytic bond dissociation energy step is a gas phase Brønsted acid base reaction:
$\ce{HA(g) → H^{+}(g) + A^{-}(g)} \nonumber$
Consequently it will be important to consider gas phase acidity before examining the thermodynamics of acid ionization further. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/06%3A_Acid-Base_and_Donor-Acceptor_Chemistry/6.03%3A_Brnsted-Lowry_Concept/6.3.04%3A_Brnsted-Lowry_Superacids_and_the_Hammett_Acidity_Function.txt |
Proton Affinity and Gas Phase Acidity/Basicity Describe Thermodynamic Aspects of Hydrogen Ion Transfer
Defining acidity in terms of hydrogen ion donation and acceptance within the Brønsted-Lowry acid base concept allows for the understanding of acidity and basicity in a variety of liquid, solid, and gaseous media. The latter are particularly important for understanding the thermodynamics of hydrogen ion donation and acceptance since the energies of gas phase species are uninfluenced by solvation factors. These reactions are perhaps best thought of in terms of the association of a hydrogen ion and a base, $\ce{B}$.
Formally the gas phase proton affinity (PA) of $\ce{B}$ is defined as the negative of the enthalpy change for the its association with hydrogen:
$\ce{H^{+}(g) + B(g) <=> BH^{+}(g)} \label{eq1}$
where $\Delta H_{rxn}$ is the Proton Affinity of $\ce{B}$.
However, it is simpler to think about the PA in terms of the reverse reaction of Equation \ref{eq1}, which describes the dissociation of $\ce{BH^{+}}$:
$\ce{BH^{+}(g) <=> H^{+}(g) + B(g)} \nonumber$
where $\Delta H_{rxn}$ is the PA of $\ce{B}$ and is also the heterolytic bond dissociation enthalpy, from which it can be seen that the proton affinity is just the enthalpy for heterolytic cleavage of the $\ce{H-B}$ bond.
The absolute or gas phase basicity (GB) of $\ce{B}$ is the negative of the Gibb's free energy change for the same reaction:
$\ce{H^{+}(g) + B(g) <=> BH^{+}(g)} \nonumber$
where $\Delta~G_{rxn}$ is the negative of the Absolute or Gas phase basicity of $\ce{B}$.
Again, it is simpler to think about the GB of $\ce{B}$ as the free energy change upon heterolytic dissociation of the $\ce{B-H}$ bond in $\ce{BH^{+}}$:
$\ce{BH^+(g) ⇌ H^{+}(g) + B(g)} \nonumber$
where $\Delta G_{rxn}$ is the Absolute or Gas phase basicity of $\ce{B}$ and is also the free energy for heterolytic bond dissociation.
Because both PA and GB values correspond to heterolytic bond dissociation, more positive values of the PA and GB correspond to higher thermodynamic affinity between $\ce{B}$ and $\ce{H^{+}}$.
Note that the gas phase acidity (GA) is a very similar quantity to the gas phase basicity, being defined for an acid of formula $\ce{HA}$ as the Gibb's free energy change for:††
$\ce{HA(g) <=> H^{+}(g) + A^{-}(g)} \nonumber$
where $\Delta G_{rxn}$ is the Absolute or Gas phase acidity of $\ce{HA}$.
Since $\ce{A^{-}}$ and $\ce{B}$ are just both convenient shorthand notations for a Brønsted base gas phase acidity and gas phase basicity, both refer to ionization of a gas phase acid. The quantities only differ in that gas phase acidities correspond to ionization of a neutral acid, $\ce{HA}$, while gas phase basicities correspond to ionization of a monocationic base, $\ce{BH^{+}}$. For many purposes this difference may be ignored.
Gas phase proton affinities and acidities/basicities have been determined for a large number of species and are available through the NIST chemistry webbook. Selected values are given in Table $1$. As can be seen from the data in Table $1$, typical proton affinity and gas phase acidity values fall between 600 and 1750 kJ/mol, with proton affinities being slightly larger by 20-50 kJ/mol.
Table $1$: Gas phase proton affinity and acidity values for selected species. Data are taken from the NIST Chemistry Webbook. When more than one value is available, the number presented below represents the average of those reported (after rejection of outliers at the 90% confidence interval using a G-test).
Base Proton Affinity (kJ/mol) Gas Phase Acidity (kJ/mol)
hexafluorobenzene $C_6F_6$ 648
$H_2O$ 691
benzene 750
$NH_3$ 854 819
aniline 882 851
pyridine 930 898
triethylamine 982 951
$ClO_4^-$ 1238 1201
$ClO_3^-$ 1310 1284
$I^-$ 1313 1293
$Br^-$ 1349 1328
$NO_3^-$ 1368 1330
$Cl^-$ 1390 1367
$NO_2^-$ 1424 1396
$ClO_2^-$ 1488 1461
$anilide^-$ 1537 1506
$F^-$ 1555 1524
$CH_3O^-$ 1573 1570
$O^-$ 1601 1576
$H:^-$ 1675 1649
$NH_2:^-$ 1686 1656
$FCH_2:^-$ 1734 1676
$CH_3CH_2:^-$ 1758 1723
Proton Affinity and Gas Phase Acidity/Basicity Data are Consistent with Expected Acidity Trends but also Illuminate the Role of Solvation and Dissociation Entropy in Solution Phase Acid-Base Behavior
Two lessons may be derived from gas phase acidity and proton affinity data:
1. The gas phase proton affinities and acidities of neutral and anionic bases are large, positive, and strongly disfavor acid ionization. As shown in Figure 6.3.3.1, acid ionization is driven by the high solvation energy of the hydrogen ion ($\Delta H^o = -1091\, \text{kJ/mol}$) with smaller contributions from solvation of the acid's conjugate base.
2. Trends in gas phase proton affinities and acidities indicate the importance of solvation for acid-base behavior in solution. As can be seen from the data in Table $1$, water is a very weak base in the gas phase. In fact, water's proton affinity is even lower than the proton affinity of the conjugate bases of most strong acids. This means that transfer of hydrogen ion from most strong acids to water is endothermic in the gas phase. For instance, for
$\ce{HCl(g) + H_2O(g) -> H_3O^{+}(g) + Cl^{-}(g)} \quad \quad {\Delta H = +699 kJ/mol} \nonumber$
In other words, based on the thermodynamics of hydrogen ion transfer in the gas phase, most strong Brønsted acids should not act as Brønsted acids towards water. Nevertheless they do act as strong acids. This means that the solvation energy terms in the thermodynamic cycle of Figure 6.3.3.1 (possibly assisted by entropic effects) are driving hydrogen ion transfer in solution.
This is confirmed by the values for the proton affinities and hydration energies for the hydrohalic acids given in Table $2$ . For these acids the solvation energies for the hydrogen ion (-1130 kJ/mol) and anions (e.g., for Cl-, -363 kJ/mol) are much larger than the hydration energies of the neutral acids and contribute towards an overall exothermic enthalpy of acid dissociation in water.
Table $2$ Contribution of heterolytic bond dissociation and hydration (hyd) energies to the enthalpy of dissociation for the hydrohalic acids.1
Acid
Proton Affinity (PA)
$H-A(g) → H^+(g) + A^-(g)$
(kJ/mol)
$\Delta H_{hyd}~for~HX$
$HX(aq) → HX(g)$
(kJ/mol)
$\Delta~H_{hyd}~for~H^+$
$H^+(g) → H^+(aq)$
(kJ/mol)
$\Delta~H_{hyd}~for~X^-$
$X^-(g) → X^-(aq)$
(kJ/mol)
Total (acid dissociation enthalpy)
$H-A(aq) → H^+(aq) + A^-(aq)$
(kJ/mol)
HF +1555 +48 -1091 -524 -12
HCl +1390 +18 -1091 -378 -61
HBr +1349 +21 -1091 -348 -69
HI +1313 +23 -1091 -308 -63
Although solvation plays a crucial role in determining a substance's acid-base behavior in solution, an acid's strength cannot be predicted from solvation energies alone. This is because it is difficult to predict the relative importance of the large proton affinity and conjugate base hydration enthalpy terms (and to a lesser extent the acid hydration enthalpy) in a given case. This is also illustrated by the data in Table $2$. The hydrohalic acid bond dissociation energy and halide hydration enthalpy both decrease in magnitude by approximately the same amount on going from HCl to HI, so that the acid dissociation enthalpies remain approximately constant. While proton affinities and hydration enthalpies often change in tandem, it is not always the case that they change by similar magnitudes. This may be seen by comparing the data shown for HF with those for the other hydrohalic acids in Table $2$. In the case of HF, the 165 kJ/mol increase in proton affinity between HCl and HF is not compensated for by the -146 kJ/mol decrease in hydration energy on going from Cl- to F- (due to fluorine's ability to form hydrogen bonds), leading HF to exhibit an anomalously small ionization enthalpy.
How Gas phase acidity values are measured
Gas phase acidities can be conveniently determined thermochemically relative to a standard by measuring the relative affinity of a hydrogen ion for two bases, $B_1$ & $B_2$, by using mass spectrometry to follow the equilibrium of the hydrogen ion transfer reaction:
$HB_1^+ + B_2 → B_1 + HB_2^+ \nonumber$
The free energy change of this reaction may be found from the relative populations of $HB_1^+$ and $HB_2^+$ using mass spectrometry:
$\Delta G = \dfrac{[B_1][HB_2^+]}{[HB_1^+][B_2]} \nonumber$
and the enthalpies and entropies of reaction from the temperature dependence of the free energies (since $\Delta G = \Delta H - T \times \Delta S$.
Gas phase acidity values determined from hydrogen ion transfer equilibrium values are called relative acidities. In contrast, absolute acidities are those which can be traced back to the standard forms of the elements at standard temperature and pressure. Absolute acidities can be found from relative ones as long as the absolute acidity of at least one of the bases is known. Fortunately, the absolute affinities of a number of gas phase acidity standards have been determined thermochemically. For instance, the gas phase proton affinity of H-A corresponds to the sum of the following processes:
$H-A(g) ⇌ H(g) + A(g)~~~~~~~ { \Delta H=Bond~Dissociation~Energy~of~HA,~BDE_{HA}} \nonumber$
$H(g) ⇌ H^+(g) + e^-(g)~~~~~~~~~~~~~~ { \Delta H=Ionization~Energy~of~H,~IE_H}\nonumber$
$A(g) + e^-(g) ⇌ A^-(g)~~~~~~~~~~~~~~~ { \Delta H=Electron~Affinity~of~A^-,~IE_A}\nonumber$
$\rule{10 cm}{0.03 cm}$
$HA(g) ⇌ H^+(g) + A^-(g)~~~~~~~~~~~ { \Delta H_{rxn} = Proton~affinity~of~A^-}\nonumber$
so that
${ Proton~Affinity~of~A^- = Delta H_{rxn} = BDE_{HA} + IE_H - EA_A}\nonumber$
For a more detailed explanation of how gas phase acidities are found see the gas phase ion thermochemistry page at the NIST chemistry webbook. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/06%3A_Acid-Base_and_Donor-Acceptor_Chemistry/6.03%3A_Brnsted-Lowry_Concept/6.3.06%3A_Thermodynamics_of_Gas_Phase_Brnsted_Acidity_and_Basicity.txt |
Oxyacids (also known as oxoacids) are compounds of the general formula $\ce{H_{n}EO_{m}}$, where $\ce{E}$ is a nonmetal or early transition metal and the acidic hydrogens are attached directly to oxygen (not $\ce{E}$). This class of compounds includes such well-know acids as nitric acid ($\ce{HNO_2}$) and phosphoric acid, ($\ce{H_3PO_4}$).
The acidity of oxyacids follows three trends:
Trend 1: In a homologous series the acidity increases with the electronegativity of the central atom
Elements in the same group frequently form oxyacids of the same general formula. For example, chlorine, bromine, and iodine all form oxyacids of formula $\ce{HOE}$: hypochlorous, hypobromous and hypoiodous acids. In the case of these homologous oxyacids, the acidity is largely determined by the electronegativity of the central element. Central atoms that are better able to inductively pull electron density towards themselves make the oxygen-hydrogen bond that is to be ionized more polar and stabilize the conjugate base, $OE^-$. Thus the acid strength in such homologous series increases with the electronegativity of the central atom. This may be seen from the data for the hypohalous acids in Table $1$, in which the acid strength increases with the electronegativity of the halogen so that the order of acidity is:
$\ce{HClO>HBrO>HIO} \nonumber$
Table $1$: Relationship of central atom's electronegativity to acid ionization constant in the hypohalous acids.
HOX Electronegativity of X Ka
HOCl 3.0 4.0 × 10−8
HOBr 2.8 2.8 × 10−9
HOI 2.5 3.2 × 10−11
Note that the influence of central atom electronegativity on the strength of oxyacids is exactly the opposite to that observed for the binary hydrides in Table $5$, for which the acidity increased down a group, giving the order of acidity:
$\ce{HI>HBr>HCl \gg HF} \nonumber$
The reason for this is that in the hydrogen halides, the bond to be broken (H-E bond) decreased in strength down the group, while in oxyacids the bond to be broken is always an O-H bond and so varies much less in strength with the electronegativity of the central atom.
Trend 2: For oxoacids of a given central atom the acidity increases with the central element's oxidation state or, in other words, the number of oxygens bound to the central atom.
Here we are looking at the trend for acids in which there are variable numbers of oxygen bound to a given central atom. An examples is the perchloric ($\ce{ClO_4^{-}}$), chloric ($\ce{ClO_3^{-}}$), chlorous ($\ce{ClO_2^{-}}$), and hypochlous ($\ce{ClO^{-}}$) acid series. In such a series, the greater the number of oxygens, the stronger the acid. This can be explained in several ways. From the viewpoint of the acid itself the key factor is again the inductive effect, in this case involving the ability of the oxygens attached to the central atom to pull on electron density across the OH bond. This is seen from the charge density diagram for the chlorine oxoacids shown in Figure $1$, in which the partial positive charge on the acidic hydrogen increases with the number of oxygens present.
The increase in oxoacid acidity with the number of oxygens bound to the central atom may also be seen by considering the stability of the conjugate oxyanion. That the stability of the conjugate base increases with the number of oxygens may be seen from the charge distribution diagrams and Lewis bonding models for the chlorine oxyanions shown in figure $2$ . As the negative charge is spread over more oxygen atoms it becomes increasingly diffuse.
Exercise $1$
Sulfur and selenium both forms oxoacids of formula $\ce{H_2EO_3}$, where E is either S or Se. These are called sulfurous and selenous acid, respectively. Which oxoacid would you expect to be more acidic: selenous acid or sulfurous acid?
Answer
Sulfurous acid should be more acidic. Since sulfur is more electronegative than selenium sulfur will polarize OH bonds to a greater extent, making them more acidic. This prediction is borne out by a comparison of the $pK_a$ values for the acids:
Acid $pK_{a1}$ $pK_{a2}$
sulfurous acid, $H_2SO_3$ 1.85 7.2
selenous acid, $H_2SeO_3$ 2.62 8.32
Trend 3: For polyprotic oxoacids the acidity decreases as each successive proton is removed
Oxoacids with multiple O-H bonds are called polyprotic since they can donate more than one hydrogen ion. In this case hydrogen ions are removed in successive ionization reactions. Examples include phosphoric and carbonic acid:
$\ce{H_3PO_4 ⇌ H^{+} + H_2PO_4^{-}} \quad \quad pK_{a1} = 2.2 \nonumber$
$\ce{H_2PO_4^{-} ⇌ H^{+} + HPO_4^{2-}} \quad \quad pK_{a2} = 7.2 \nonumber$
$\ce{HPO_4 ⇌ H^{+} + PO_4^{3-}} \quad \quad pK_{a3} = 12.4 \nonumber$
$\ce{H_2CO_3 ⇌ H^{+} + HCO_3^{-}} \quad \quad pK_{a1} = 3.6 \nonumber$
$\ce{HCO_3^{-} ⇌ H^{+} + CO_3^{2-}} \quad \quad pK_{a1} = 10.3 \nonumber$
The dissociation constants for successive ionization constants decrease by about five orders of magnitude between successive hydrogen ions. This is reflected in Linus Pauling's rules for oxoacids and their oxyanions:
Pauling's Rules
1. The $pK_a$ for an oxyacid of general formula $\ce{E(OH)_{q}(O)_{p}}$ is given by $pK_a = 8 - 5 \times p \label{PaulingRules}$
2. As an oxoaxid undergoes successive ionizations the $pK_a$ increases by five each time.
The central theme of Pauling's Rules is that the more oxygens there are on the central atom, the more resonance structures that can be constructed for the conjugate base, which increases its stability and increases the acidity of the acid. However, as the acids successively ionize, they have fewer resonance structures. Pauling's Rules are phenomenological (i.e., not based on a theoretical framework). However, as empirical rules, they often work quite well, but it should be borne in mind that they are approximate.
Exercise $2$: How well do Pauling's rules for oxoacids work?
Calculate the theoretical $pK_a$ values for phosphoric and carbonic acid and their dissociation produces and compare the results with the experimental $pK_a$ values.
Answer
For phosphoric acid, Pauling's rules (Equation \ref{PaulingRules}) predict the $pK_a$ values quite well:
• $H_3PO_4$: $p = 3$ and $q =1$ and $pK_{a1, predicted} = 8 - 5 \times 1 = 3 \nonumber$ This is slightly greater than the observed value of 2.2.
• $H_2PO_4^-$: $pK_{a2, predicted} = pK_{a1, experimental} + 5 = 7.2 \nonumber$ This is spot on with experiment.
• $HPO_4^{2-}$: $pK_{a3, predicted} = pK_{a2, experimental} + 5 = 12.2 \nonumber$ This is slightly less than the experimental value of 12.4.
For carbonic acid Pauling's rules predict $pK_{a1}$ reasonably well, but $pK_{a2}$ less so:
• $H_2CO_3$: $p = 2$, $q =1$ and $pK_{a1, predicted} = 8 - 5 \times 1 = 3 \nonumber$ This is slightly lower than the observed value of 3.6.
• $HCO_3^-$: $pK_{a2, predicted} = pK_{a1, experimental} + 5 = 8.6 \nonumber$ This is 1.7 units less than the experimental value of 10.3.
In some cases discrepancies between experimental $pK_a$ values and those predicted by Pauling's rules suggest that structural rearrangements may be taking place upon ionization or else that the reported $pK_a$ values do not really represent the ionization in question because they do not fully account for all the equilibria taking place in solution. In the case of carbonic acid, however, the reason for the discrepancy between the predicted and experimental $pK_{a2}$ values is not entirely clear. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/06%3A_Acid-Base_and_Donor-Acceptor_Chemistry/6.03%3A_Brnsted-Lowry_Concept/6.3.07%3A_The_Acidity_of_an_Oxoacid_is_Determined_by_the_Electronegativity_and_Oxidation_State.txt |
Metal Ions Act as Acids by Polarizing Bound Water Molecules Called Aqua Ligands
Aqueous solutions of simple salts of metal ions can also be acidic, even though a metal ion cannot donate a proton directly to water to produce $H_3O^+$. Instead, a metal ion can act as a Lewis acid and interact with water, a Lewis base, by coordinating to a lone pair of electrons on the oxygen atom to form a hydrated metal ion.
A water molecule coordinated to a metal ion is more acidic than a free water molecule for two reasons. First, repulsive electrostatic interactions between the positively charged metal ion and the partially positively charged hydrogen atoms of the coordinated water molecule make it easier for the coordinated water to lose a proton.
Second, the positive charge on the $Al^{3+}$ ion attracts electron density from the oxygen atoms of the water molecules, which decreases the electron density in the $\ce{O–H}$ bonds, as shown in Figure $\PageIndex{5b}$. With less electron density between the $O$ atoms and the H atoms, the $\ce{O–H}$ bonds are weaker than in a free $H_2O$ molecule, making it easier to lose an $H^+$ ion. This is shown schematically in Figure $1$.
The acidity of a given metal ion largely depends on its charge to size ratio and electronegativity, although in some cases hardness and ligand field effects also play a role.
The magnitude of this effect depends on the following factors, of which the first two are generally considered the most important (Figure $2$):
The charge on the metal ion
A divalent ion ($\ce{M^{2+}}$) has approximately twice as strong an effect on the electron density in a coordinated water molecule as a monovalent ion ($\ce{M^{+}}$) of the same radius.
The radius of the metal ion
For metal ions with the same charge, the smaller the ion, the shorter the internuclear distance to the oxygen atom of the water molecule and the greater the effect of the metal on the electron density distribution in the water molecule.
The first two of these factors explain why most alkali metal cations exhibit little acidity while aqueous solutions of small, highly charged metal ions, such as $Al^{3+}$ and $Fe^{3+}$, are acidic:
$\ce{[Al(H2O)6]^{3+}(aq) <=> [Al(H2O)5(OH)]^{2+}(aq) + H^{+}(aq)} \label{16.36}$
The $\ce{[Al(H2O)6]^{3+}}$ ion has a $pK_a$ of 5.0, making it almost as strong an acid as acetic acid. Because of the two factors described previously, the most important parameters for predicting the effect of a metal ion on the acidity of coordinated water molecules are the charge and ionic radius of the metal ion.
A simple empirical equation for predicting the pKa of metal ions in water has been proposed by Wulfsberg:1
$pK_a = 15.14 - 88.16~pm \left( \frac{Z^2}{r} \right) \nonumber$
where $Z$ is the charge on the metal ion and $r$ its radius. As can be seen from figure $3$, in general the acidity of metal ions increases with $\frac{Z^2}{r})$. Nevertheless a number of ions have considerably lower $pK_a$ values than predicted from this correlation, suggesting that the acidity of metal cations cannot be predicted using a simple electrostatic model alone.
Although the charge-to-size ratio is the simplest and most powerful predictor of metal ion acidity in water, three additional factors also can play a role:
Electronegativity
All other things being equal, more electronegative elements are better able to withdraw electron density from a bound water ligand and consequently better at enhancing the ability of that water molecule to lose a hydrogen ion. The electrostatic model of ion acidity can be extended to account for electronegativity effects but only needs to be done so for metals with Pauling electronegativities greater than 1.5. The empirical relationship that has been proposed to account for the effect of electronegativity is:
$pK_a = 15.14 - 88.16 \left( \frac{Z^2}{r} + 0.096( \chi - 1.50) \right) \nonumber$
where $r$ is the ion radius in pm and $\chi$ is its Pauling electronegativity. As can be seen from figure $4$, the electronegativity term accounts for some of the deviations in metal ion acidity predicted from charge and size effects alone.
However, there are still large deviations between the predicted and observed pKa for a number of ions. In particular, the modified electrostatic model overestimates the pKa of Al3+ and Sn4+ and underestimates the pKa of Hg2+, Sn2+, and Tl3+. While the exact reasons for these discrepancies is not entirely clear, at least some are thought to arise from the impact of the fourth factor that determines metal ion acidity.
Hardness and Softness
Cation hardness or softness is assessed according to Pearson's Hard-Soft Acid Base Principle (HSABP). In general, soft cations are more acidic than hard cations of the same charge and radius, as may be seen from the examples in Table $1$. The greater than expected acidity of softer cations is thought to reflect the importance of covalent contributions to the metal-water bond.
Table $1$: Comparison of the $pK_a$ values at 25°C of hard and soft cations with approximately the same radius and charge. Based on Gutmann2 as reported by Burgess.3
Cation Classification Radius (pm) $pK_a1$
$K^+$ hard 1.33 14
$Ag^+$ soft 1.26 10
$Mg^{2+}$ hard 0.65 12.2
$Cu^{2+}$ soft 0.69 7.3
$Ca^{2+}$ hard 0.99 12.6
$Cd^{2+}$ soft 0.97 9.0
$Sr^{2+}$ hard 1.13 13.1
$Hg^{2+}$ soft 1.10 3.6
Ligand field effects
Ligand field effects involve bonding and antibonding interactions between the d orbitals on a transition metal and ligand orbitals. The importance of ligand field interactions on cation acidity is not well established, but ligand field interactions might influence the acidity of a hydrated metal in two cases: (a) when the ligand field stabilization of the aqua complex (metal with $\ce{H_2O}$ or $\ce{OH^{-}}$ bound) is greater or less than that in its conjugate base and (b) in complexes which undergo Jahn-Teller distortions that alter the metal's ability to polarize the OH bond. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/06%3A_Acid-Base_and_Donor-Acceptor_Chemistry/6.03%3A_Brnsted-Lowry_Concept/6.3.08%3A_High_Charge-to-Size_Ratio_Metal_Ions_Act_as_Brnsted_Acids_in_Water.txt |
The proton-centered nature of Arrhenius and Brønsted-Lowry definitions of acids and bases limits the utility of the acid-base formalism to reactions in protic solvents. A more generalizable theory is the solvent-system acid/base definition, which can be used to describe acid/base chemistry in non-protic solutions.
The solvent system acid-base concept generalizes the Arrhenius acid-base concept
Remember that the Brønsted-Lowry concept seeks to generalize Arrhenius acidity in ways that allow all hydrogen ion transfers to be thought of as acid-base reactions.
$\ce{H_2O(l) + H_2O(l) → H_3O^{+}(aq) + OH^{-}(aq)} \nonumber$
Like the Brønsted-Lowry acid-base concept, the solvent system acid-base concept is a way to generalize the Arrhenius acid-base concept by focusing on cations and anions generated in any solvent that autoionizes. These include Brønsted-Lowry type autoionizations:
\begin{align*} \ce{2NH_3(l)} &\ce{⇌ NH_4^{+} + NH_2^{-}} \[4pt] \ce{2H_2SO_4(l)} &\ce{⇌ H_3SO_4^{+} + HSO_4^{-}} \end{align*} \nonumber
However, the solvent system definition also allows for autoionizations that involve the transfer of an ion other than hydrogen. For example,
\begin{align*}\ce{2SeOCl_2(l) }&\ce{⇌ SeOCl^{+} + SeOCl_3^{-}} \[4pt] \ce{2BrF_3(l)} &\ce{ ⇌ BrF_2^{+} + BrF_4^{-}} \end{align*} \nonumber
However, the solvent system concept does not define acidity in terms of ion transfer. Rather, like the Arrhenius concept, it defines acids and bases in terms of the impact those acids and bases have on the concentrations of cations and anions in solution (Figure $1$).
Definition: Solvent System Definitions of Acids and Bases
This means that under the solvent system definition:
• an acid is the solvent cation or any substance that increases the concentration of the solvent cations normally produced by solvent autoionization.
• a base is the solvent anion or any substance that increases the concentration of the solvent anions normally produced by solvent autoionization.
Note that in the solvent system concept, salts of the solvent cation are acids and salts of the solvent anion are bases. For instance, if a salt of $\ce{BrF_4^{-}}$ such as sodium tetrafluorobromate ($\ce{NaBrF_4}$) is added to $\ce{BrF_3(l)}$, the concentration of $\ce{BrF_4^{-}}$ increases.
The Solvent System Concept
The solvent system definition collapses to the Arrhenius definition for Brønsted acids in water. For instance, $\ce{HCl}$ is an acid in water since it increases the concentration of $\ce{H_3^{+}O}$ when it dissociates:
$\ce{HCl(aq) + H_2O(l) → H_3O^{+}(aq) + Cl^{-}(aq)}\nonumber$
The solvent system broadens the Arrhenius definition by allowing for similar reactions in a variety of solvents. For instance, $\ce{HCl}$ also acts as an acid in liquid ammonia since it gives an $\ce{NH_4^{+}}$ ion on dissociation.
$\underset{\textcolor{red}{acid}}{\ce{HCl}} + \ce{NH_3(l)} → \ce{NH_4^{+}} + \ce{Cl^{-}} \nonumber$
Antimony pentafluoride acts as an acid in liquid $\ce{BrF3}$ since it abstracts a fluoride to give $\ce{BrF_2^{+}}$.
$\underset{\textcolor{red}{acid}}{\ce{SbF_5}} + \ce{BrF_3(l)} ⇌ \ce{SbF_6^{-}} + \ce{BrF_2^{+}} \nonumber$
In contrast, the fluoride ion of potassium fluoride acts as a base, since it adds to $\ce{BrF_3}$ to give $\ce{BrF_4^{-}}$.
$\underset{\textcolor{blue}{base}}{\ce{KF}} + \ce{BrF_3(l)} ⇌ \ce{K^{+}} + \ce{BrF_4^{-}} \nonumber$
The solvent system definition also allows for acid-base neutralization reactions. An example would be the reaction between $\ce{KF}$ and $\ce{SbF_5}$ in $\ce{BrF_3}$.
$\underset{\textcolor{blue}{base}}{\ce{KF}}+ \underset{\textcolor{red}{acid}}{\ce{SbF_5}} → \ce{KSbF_6}\nonumber$
although in this case it may be easier to see what is happening by writing the complete ionic form of the neutralization reaction equation.
$\ce{K^{+}} + \underset{\textcolor{blue}{base}}{\ce{F^{-}}} + \underset{\textcolor{red}{acid}}{\ce{SbF_5}} → \ce{KSbF_6}\nonumber$
The solvent system concept allows for successive ionizations just as the Brønsted-Lowry system does. Acids like sulfuric acid can be described well under the Arrhenius, Brønsted-Lowry, and solvent system definitions.
\begin{align*} \ce{2H_2SO_4} &⇌ \ce{H_3SO_4^{+} + HSO_4^{-}} \[4pt] \ce{HSO_4^{-} + H_2SO_4} &⇌ \ce{H_3SO_4^+ + SO_4^{2-}} \end{align*} \nonumber
The solvent system definition also describes the autoionization of nonprotic solvents like thionyl chloride, $\ce{SOCl_2}$:
\begin{align*} \ce{SOCl_2(l)} &⇌ \ce{ SOCl^{+} + SOCl_3^{-}} \[4pt] \ce{SOCl^{+} + SOCl_2(l)} &⇌ \ce{SO^{2+} + SOCl_3^{-}} \end{align*} \nonumber
Successive equilibria are often useful in applying the solvent system to neutralization reactions, as may be seen in Example $1$.
Example $1$
The reaction between sodium sulfite, $\ce{Na_2SO_3}$, and thionyl chloride, $\ce{SOCl_2}$, is a neutralization reaction according to the solvent system acid-base concept.
$\ce{Na_2SO_3(s) + SOCl_2(l) → 2NaCl(s) + 2SO_2(g)} \nonumber$
Write out the relevant equilibria and use them to explain the reaction in acid-base terms.
Solution
In the reaction shown the $\ce{Na^+}$ just acts as a counterion. Consequently the reaction that should be considered is
$\ce{2SO_3^{2-} + SOCl_2 → 2Cl^{-} + 2SO_2}\nonumber$
A good place to start is to consider each of the species involved under the solvent-system acid-base concept by seeing if it is possible to write out ionization (or autoionization) reactions for them.
For $\ce{SO_3^{2-}}$ these are
\begin{align*} \ce{2SO_3^{2-}} &⇌ \underset{\textcolor{red}{acid}}{\ce{SO_2}} + \underset{\textcolor{blue}{base}}{\ce{SO_4^{4-}}} \[4pt] \underset{\textcolor{red}{acid}}{\ce{SO_2}} + \underset{\textcolor{blue}{base}}{\ce{2SO_3^{2-}}} &⇌ \underset{\textcolor{red}{acid}}{\ce{SO^{2+}}} + \underset{\textcolor{blue}{base}}{\ce{SO_4^{4-}}}\end{align*} \nonumber
These equations reveal that
• $\ce{SO_3^{2-}}$ and $\ce{SO_2}$ are amphoteric since they can act as either an acid or a base
• $\ce{SO_4^{4-}}$ acts only as a base since it is the solvent anion
• $\ce{SO^{2+}}$ acts only as an acid since it is the solvent cation
Since $\ce{SO_3^{2-}}$ is amphoteric, this does not resolve the issue of whether it is acting as an acid or a base. However, notice that the product of the reaction between $\ce{SO_3^{2-}}$ and $\ce{SOCl_2}$ is $\ce{SO_2}$. This means that $\ce{SO3^{2-}}$ is acting as if it is following the pathway:
$\ce{2SO_3^{2-} ⇌ SO_2 + SO_4^{4-}} \nonumber$
The ionizations of $\ce{SOCl_2}$ were already given in the main text but are also reproduced here for convenience:
\begin{align*} \ce{SOCl_2(l)} &\ce{<=> } \ce{SOCl^{+}} + \ce{SOCl_3^{-}} \[4pt] \ce{SOCl^{+}} + \ce{SOCl_2(l)} &\ce{<=>} \ce{SO^{2+}} + \ce{SOCl_3^{-}}\end{align*} \nonumber
We could write an expression involving ionization of $\ce{2SO^{2+}}$ to $\ce{S^{4+}}$ and $\ce{SO_2}$ but that is unnecessary for solving the present problem.
The equations we have written reveal that
• $\ce{SOCl_2}$ and $\ce{SOCl^{+}}$ are amphoteric since they can act as either an acid or a base
• $\ce{SOCl_3^{-}}$ acts only as a base since it is the solvent anion
• $\ce{SO^{2+}}$ acts only as an acid since it is the solvent cation
Again, since $\ce{SOCl_2}$ is amphoteric this does not resolve the issue of whether it is acting as an acid or a base. However, notice that the product of the reaction between $\ce{SO_3^{2-}}$ and $\ce{SOCl_2}$ is $\ce{SO_2}$. The only species in a reaction pathway involving $\ce{SOCl_2}$ that can react to form $\ce{SO_2}$ is $\ce{SO^{2+}}$. This means that $\ce{SOCl_2}$ is acting as if it is following the pathway:
\begin{align*} \ce{2SOCl_2(l) &⇌ SOCl^+ + SOCl_3^{-}} \[4pt] \ce{SOCl^{+} + SOCl_2(l) &⇌ SO^{2+} + SOCl_3^{-}} \[4pt] \ce{SO^{2+} + base &⇌ SO_2} + \text{the base's conjugate acid} \end{align*} \nonumber
But what sort of base could react with $\ce{SO^{2+}}$ to give $\ce{SO_2}$? The candidates in our list of bases are $\ce{SO_4^{4-}}$, $\ce{SO_3^{2-}}$, and $\ce{SO^{2+}}$ It must be one that donates an oxide ion, such as $\ce{SO_4^{4-}}$. So in other words the $\ce{SOCl_2}$ is following the pathway:
\begin{align*} \ce{2SOCl_2(l) &⇌ SOCl^+ + SOCl_3^{-}} \[4pt] \ce{SOCl^{+} + SOCl_2(l) &⇌ SO^{2+} + SOCl_3^{-}} \[4pt] \ce{SO^{2+} + SO_4^{4-} &⇌ SO_2 + SO_3^{2-}} \end{align*} \nonumber
and $\ce{SO_3^{2-}}$ the pathway
$\ce{2SO_3^{2-} ⇌ SO_2 + SO_4^{4-}} \nonumber$
Adding these together gives the following net reaction:
$\ce{3SOCl_2(l) + SO_3^{2-} ⇌ SO_2 + 2SOCl_3^{-}} \nonumber$
At first glance this seems like it is not the correct equation. However, there is one acid-base reaction we are neglecting. The $\ce{Cl^{-}}$ product of the reaction between sodium sulfite and thionyl chloride acts as a base in thionyl chloride:
$\ce{SOCl_2(l) + Cl^- ⇌ SOCl_3^{-}} \nonumber$
So the net reaction above simply corresponds to the reaction between sodium sulfite and thionyl chloride with additional reactions between the product $\ce{Cl^{-}}$ and thionyl chloride added in. However, if a 1:1 ratio of thionyl chloride and sodium sulfate is used in the reaction, then the thionyl chloride will be consumed and the equilibrium between thionyl chloride and chloride ion shifted towards chloride, giving the desired net equation:
$\ce{2SO_3^{2-} + SOCl_2 → 2Cl^- + 2SO_2} \nonumber$
As Example $1$ illustrates, the application of the solvent system concept to the understanding of acid-base reactions in solution can be quite involved but also serves to allow chemists to apply a detailed understanding of solution chemistry to a wider variety of reactions. However, it is usually much simpler to think about solvent system neutralization reactions using the Lewis acid-base concept. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/06%3A_Acid-Base_and_Donor-Acceptor_Chemistry/6.03%3A_Brnsted-Lowry_Concept/6.3.09%3A_The_Solvent_System_Acid_Base_Concept.txt |
Amphoteric Protic Solvents Undergo Solvent Autoionization just as Water Does
The Brønsted-Lowry concept allows for an understanding of hydrogen ion transfer chemistry in amphoteric protic solvents. Amphoteric protic solvents are those which can both accept and receive hydrogen ions. From the viewpoint of the Brønsted-Lowry concept, the acid-base chemistry in these solvents is governed by autoionization equilibria analogous to water autoionization.
$\ce{2H_2O(l) ⇌ H_3O^+ + OH^{-}} \quad \quad K_w = 1.0\times10^{-14}\nonumber$
For example, sulfuric acid ionizes according to the equation:
$\ce{2H_2SO_4(l) ⇌ H_3SO_4^{+} + HSO_4^{-}} \quad \quad K = 4\times10^{-4}\nonumber$
The magnitude of the solvent autoionization constant in a given amphoteric solvent determines the amount of protonated and deprotonated* solvent present. Since sulfuric acid's autoionization constant is much larger than that of water ($K_w = 10^{-14}$), the concentrations of $\ce{H_3SO_4^{+}}$ and $\ce{HSO_4^{-}}$ present in pure sulfuric acid are ~0.02 M, much greater than the $10^{-7}$ M $H_3O^+$ and $\ce{OH^{-}}$ present in pure water.
In contrast, ammonia's autoionization constant is much less than that of water and only $10^{-14}\, M\, \ce{NH_4^{+}}$ and $\ce{NH_2^{-}}$ are present in pure ammonia.
$\ce{2NH_3(l) ⇌ NH_4^+ + NH_2^{-}} \quad \quad K = 10^{-27}\nonumber$
The Solvent Leveling Effect Limits the Strongest Acid or Base that Can Exist
The conjugate acid and base of the solvent are the strongest Brønsted acids and bases that can exist in that solvent. To see why this is the case for acids, consider the reaction between a Brønsted acid ($\ce{HA}$) and solvent ($\ce{S}$):
$\ce{HA + S ⇌ HS^{+} + A^{-}}\nonumber$
This equilibrium will favor dissociation of whichever is a stronger acid - $\ce{HA}$ or $\ce{HS^{+}}$. If the acid is stronger it will mostly dissociate to give $\ce{HS^{+}}$, while if the solvent's conjugate acid is stronger the acid will be mostly un-ionized and remain as $\ce{HA}$.**
Any acid significantly stronger than $\ce{HS^{+}}$ will act as a strong acid and effectively dissociate completely to give the solvent's conjugate acid ($\ce{HS^{+}}$). This also means that the relative acidity of acids stronger than $\ce{HS^{+}}$ cannot be distinguished in solvent $\ce{S}$. This is called the leveling effect since the solvent "levels" the behavior of acids much stronger than itself to that of complete dissociation. For example, there is no way to distinguish the acidity of strong Arrhenius acids like $\ce{HClO_4}$ and $\ce{HCl}$ in water since they both completely dissociate. However, it is possible to distinguish their relative acidities in solvents that are more weakly basic than the conjugate base of the strongest$†$ acid, since then the acids will dissociate to different extents. Such solvents are called differentiating solvents. For example, acetonitrile acts as a differentiating solvent towards $\ce{HClO_4}$ and $\ce{HCl}$. Both $\ce{HClO_4}$ and $\ce{HCl}$ partly dissociate in MeCN, with the stronger acid $\ce{HClO_4}$ dissociating to a greater extent than $\ce{HCl}$.
The leveling effect can also occur in basic solutions. The strongest Brønsted base, $\ce{B}$, that can exist in a solvent is determined by the relative acidity of the solvent, and the base's conjugate acid, $\ce{BH^{+}}$, determines whether the base will remain unprotonated and able to act as a base in that solvent. If the solvent is represented this time as $\ce{HS}$ then the relevant equilibrium is:
$\ce{B + HS ⇌ BH^{+} + S^{-}} \nonumber$
The position of this equilibrium depends on whether $\ce{B}$ or $\ce{S^{-}}$ is the stronger base. If the solvent's conjugate base, $\ce{S^{-}}$, is stronger, then the base B will remain unprotonated and available to act as a base. However, if $\ce{B}$ is a stronger base than $\ce{S^{-}}$, it will deprotonate the solvent to give $\ce{BH^{+}}$ and $\ce{S^{-}}$. In this way the strongest base that can exist in a given solvent is the solvent's conjugate base.
The relative strength of Brønsted bases can only be determined in solvents that are more weakly acidic than $\ce{BH^{+}}$; otherwise the bases will all be leveled to $\ce{S^{-}}$.
It is important to consider the leveling effect of protic solvents when performing syntheses that require the use of basic reagents. For instance, hydride and carbanion reagents (lithium aluminum hydride, Grignard reagents, alklyllithium reagents, etc.) cannot be used as nucleophiles in protic solvents like water, alcohols, or enolizable aldehydes and ketones. Since carbanions are stronger bases than these solvents' conjugate bases, they will instead act as Brønsted bases and deprotonate the solvent. For example if one adds n-butyllithium to water, the result (along with much heat and possibly a fire) will be butane and a solution of lithium hydroxide:
$\ce{Li^{+} + ^{-}:CH_2CH_2CH_2CH_3 + H_2O(l) → Li^{+}(aq) + OH^{-}(aq) + CH_3CH_2CH_2CH_3(g)} \nonumber$
Since hydride and carbanion reagents cannot be used as nucleophiles in protic solvents like water or methanol, they are commonly sold as solutions in solvents such as hexanes (for alklyllithium reagents) or tetrahydrofuran (for Grignard reagents and $\ce{LiAlH_4}$). | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/06%3A_Acid-Base_and_Donor-Acceptor_Chemistry/6.03%3A_Brnsted-Lowry_Concept/6.3.10%3A_Acid-Base_Chemistry_in_Amphoteric_Solvents_and_the_Solvent_Leveling_Effect.txt |
Non-Nucleophilic Organic Superbases Act as Strong Bases in Organic Solvents but Are Unreactive Towards Other Electrophiles
A variety of strong organic and inorganic bases are available for use in organic synthesis (alkyllithium reagents, diisopropyl amide derivatives, hydrides, and hydroxides). Some of these exhibit poor functional group tolerance owing to their ability to react with electrophilic functional groups. Hence there is considerable interest in the development of bases that can remove hydrogen ions from very weakly acidic organic substrates (i.e. like C-H bonds) without reacting with electrophilic functional groups. Classical non-nucleophilic bases used widely in organic chemistry include diisopropylethylamine (DIEA), lithium diisopropylamide (LDA), and the simple and complex ionic hydrides like \(\ce{CaH2^{+2}}\) and \(\ce{LiAlH4}\). The latter, however, are non-nucleophilic by virtue of their insolubility in organic solvents, on account of which they largely act as bases when an organic substrate interacts with the hydride at a crystal surface.
Many strong bases are called superbases in everyday usage. However, more rigorous definitions have been developed to clearly distinguish superbases from simple strong bases. Most of these use some aspect of the thermodynamics of hydrogen ion bonding by 1,8-Bis(dimethylamino)naphthalene as a delimiter between ordinary strong bases and superbases. In other words, by these criteria any base stronger than 1,8-bis(dimethylamino)naphthalene is considered a superbase. By this measure organic superbases may be defined as those having larger proton affinities than 1048 kJ/mol,1 although most workers use defined superbases as those having a \(pK_a\) greater than 1,8-bis(dimethylamino)naphthalene's value of 12.2
Non-nucleophilic Organic Superbases Use Charge Delocalization, Proton Chelation, or Both to Tightly Bind Hydrogen Ion
A variety of compounds with a high affinity for hydrogen ions in organic solvents have been specifically developed as strong non-nucleophilic organic superbases.3 Common classes include:
Napthalene-type proton sponges
Napthalene-type proton sponges are two basic groups held in close proximity to one another at the 1 and 8 positions of a napthalene ring. The compound 1,8-Bis(dimethylamino)naphthalene is perhaps the best known superbase of this type and is even sold by the trade name of Proton Sponge™. Its structure is shown below.
Proton sponge has the typical properties of an organic superbase. Unlike lithium diisopropylamindes and metal hydride reagents, 1,8-Bis(dimethylamino)naphthalene is only weakly basic in water. However, it possesses a very strong affinity for hydrogen ions in organic solvents.
In this case the high basicity of 1,8-Bis(dimethylamino)naphthalene is due to a combination of steric and electronic factors. First, it is a hydrogen chelator in that the hydrogen is held in a strong symmetric N---H---N hydrogen bond in its conjugate base. Second, the formation of that hydrogen bond relieves some of the steric strain associated with dimethyl amino groups held close to one another at the 1 and 8 positions on the naphthalene ring.
Aminidines and Guanidenes
Aminidines are nitrogen derivatives of carboxylic acid derivatives. Acyclic aminidines have \(pK_a\) values ~12, similar to that of 1,8-Bis(dimethylamino)naphthalene, although cyclic aminidines are several orders of magnitude higher. For example, DBU is estimated to exhibit a \(pK_a\) value of 24.3 in MeCN.4 Examples of aminidine superbases include the organic catalysts 1,8-Diazabicyclo(5.4.0)undec-7-ene (DBU) and 1,5-Diazabicyclo(4.3.0)non-5-ene (DBN), and vinamidinium superbases, which bind hydrogen ion at the imine nitrogen.
Of these the vinamidinium superbases are hydrogen ion chelators.
Guanidenes are nitrogen analogues of carbonate and bind hydrogen ion at a nitrogen lone pair just as aminidines do, although monomeric guanidines are slightly more basic (\(pK_a ~13\)) than analogous aminidines ( \(pK_a ~12\)). Common classes of guanidene superbases include pentalkyl and bicyclic guanidines, shown below.
The high basicity of aminidines and guanidines is derived from the ability of these systems to delocalize charge in the protonated form.
Exercise \(1\)
Use the structure of protonated aminidines and guanidines to explain why guanidines are stronger bases than their aminidine analogues.
Answer
There is greater delocalization of charge in guanidinium ions. According to the resonance picture of bonding, in guanidinium ions the positive charge is delocalized over three nitrogen atoms while in amidinium ions it is only delocalized over two.
Phosphatrane-type Superbases
These exhibit enhanced basicity at the phosphorus atom of a special type of azaphosphine called a proazaphosphatrane:
Of these the best known is Verkade's superbase, which is estimated to have a \(pK_a\) of 29 in acetonitrile,5 meaning it is considerably more basic than aminidine and guanidine type superbases. The key to the remarkable basicity of Verkade's superbase is the ability of the proazaphosphoatrane's amine to stabilize the protonated azaphosphine through the formation of a P-N bond, giving a pseudo trigonal bipyramidal phosphorous.
These interactions are so stabilizing that the transannular amine nitrogen exhibit no basicity, even when superacids like Magic Acid are added.6
Phosphazene Superbases
Phosphazene superbases, also known as Schwesinger superbases, function similarly to aminidines and guanidines in that they bind a hydrogen ion at an imine nitrogen to give a resonance-stabilized cation. It is just that in this case the imine is that of a phosphazene and the greater resonance stabilization of the resulting cation means that they exhibit greater basicity. One phosphazene superbase has been reported to have a \(pK_a\) of 42 in acetonitrile. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/06%3A_Acid-Base_and_Donor-Acceptor_Chemistry/6.03%3A_Brnsted-Lowry_Concept/6.3.11%3A_Non-nucleophilic_Brnsted-Lowry_Superbases.txt |
The Lewis acid-base concept generalizes the Brønsted and solvent system acid base concepts by describing acid-base reactions in terms of the donation and acceptance of an electron pair.
Under the Lewis definition
• Lewis acids are electron pair acceptors
• Lewis bases are electron pair donors
and in Lewis acid-base reactions, a Lewis base donates an electron pair to the Lewis acid, which accepts it. The reaction between borane, $BH_3$, and $NH_3$ is the classic example:
In this case the Lewis acid-base reaction results in the formation of a bond between $BH_3$, and $NH_3$. When the acid and base combine to form a larger unit, that unit is said to be an adduct, and the resulting bond is said to be a coordinate covalent or dative bond. Such coordinate covalent bonds are often represented by an arrow that indicates the direction of electron donation from the base to the acid. For instance, the reaction between $BH_3$ and $NH_3$ could also have been written as
The arrow notation for the coordinate covalent bond is really just a convenient formalism - a bookkeeping tool to help keep track of where the electrons came from and where they might return if the reverse reaction occurs. In actuality a coordinate covalent bond is just an ordinary covalent bond like any other.
For example, in Brønsted acid-base reactions, the hydrogen ion is an acid because it accepts an electron pair from the Brønsted base. Consequently, under the Lewis acid-base concept, Brønsted acid-base reactions involve the formation of an adduct between $H^+$ and a base.
$\nonumber$
The Lewis acid-base concept nicely explains ionization reactions involving nonaqueous solvents. For instance, the autoionization of $SOCl_2$ is an acid-base reaction between two $SOCl_2$ molecules.
$\nonumber$
Example $1$
Explain how the autoionization of $SOCl_2$ is a Lewis acid-base displacement reaction.
Solution
In the autoionization of $SOCl_2$ the pair of electrons donated comes from the S-Cl bond, and the S-Cl bond is broken to give a lone pair bearing a $Cl^-$ base and an $SOCl^+$ Lewis acid fragment. If you are having trouble seeing how this works, it can be instructive to consider the reverse of this process. It is a Lewis acid-base reaction to give a Lewis acid-base adduct:
From the autoionization reaction it is also apparent that $SOCl_2$ itself acts as a Lewis acid towards the liberated $Cl^-$. So the autoionization reaction involves a transfer of the $Cl^-$ base between the two Lewis acids - a Lewis acid-base displacement reaction.
It can be helpful to keep several distinctions in mind when using the Lewis acid-base concept to describe chemical reactions.
1. Many Lewis-Acid base reactions are displacement reactions. This is because the hydrogen ion is usually bound to something at the start of the reaction. In such cases the Lewis acid $H^+$ unit is transferred from one Lewis base to another :
$\nonumber$
Such acid-base reactions are sometimes called displacement reactions since the base group in the initial Lewis acid-base complex is displaced by the incoming Lewis base to generate another complex.
2. Substances are sometimes considered amphoteric because they exhibit Lewis acidity and basicity at different types of atomic centers. The classic example is aluminum hydroxide, $\ce{Al(OH)3}$. In water $\ce{Al(OH)3}$ can act as a Lewis acid towards OH- ion. The reaction occurs by formation of an adduct at $\ce{Al(OH)3}$'s Al3+ center:
$\ce{Al(OH)3 + OH^{-} \rightarrow Al(OH)4^{-}} \nonumber$
Notice that in addition to acting as a Lewis acid, in this reaction $Al(OH)_3$
• does not act as a Brønsted acid or base since no H+ ion transfer occurs
• but does act as an Arrhenius acid since OH- ion is consumed from solution, decreasing [OH-] and increasing [H+]
In water $\ce{Al(OH)_3}$ also acts as a Lewis base towards H+ through the lone pairs on its hydroxide ligands.
$\ce{Al(OH)3 + 3 H^{+} + 3 H2O \rightarrow Al(H2O)6^{3+}} \nonumber$
Notice that in addition to acting as a Lewis base, in this reaction $\ce{Al(OH)3}$ also acts as a
• Brønsted base, since a H+ ion is transferred onto the OH- ligand
• Arrhenius base, since H+ ion is consumed from solution, decreasing [H+]
• Lewis acid at its Al3+ center, since Al-O metal-ligand bonds are formed
Substances that rapidly undergo Lewis acid-base reactions are called nucleophiles and electrophiles.
Two factors govern the course of chemical reactions - thermodynamics and kinetics. Thermodyamics determines what possible fates of the reaction can take place while kinetics determines which among those possible fates will take place quickly under a particular set of reaction conditions. Because of this, it can be helpful to distinguish Lewis acids and bases that tend to undergo reaction quickly with one another from those which do so more slowly. For this reason, synthetic chemists use the terms electrophile and nucleophile to refer to Lewis acids and bases that react quickly:
• electrophile - Lewis acids that rapidly react with a given Lewis base or class of Lewis bases are said to be good electrophiles
• nucleophile - Lewis bases that rapidly react with a given Lewis acid or class of Lewis acids are said to be good nucleophiles
The electrophile-nucleophile concept is sometimes referred to as the Ingold-Robinson acid-base concept after two organic chemists who did much to illustrate the utility of thinking about Lewis acid-base reactions in kinetic terms.
Several distinctions that should be kept in mind when using the electrophile-nucleophile/Ingold-Robinson concept to think about chemical reactivity:
1. Whether a given Lewis acid or base is able to react rapidly depends on the reaction conditions and the substrate (substance it is reacting with). This means that in principle, the terms nucleophile and electrophile should not be used without appropriately qualifying what the substances act as a nucleophile or electrophile towards and under what conditions. Unfortunately this is rarely done. In such cases, the conditions and substrate must be inferred from the context. For instance, when you learned that the chloride ion is a good nucleophile when studying organic chemistry, what was meant is that chloride is a good nucleophile towards electrophilic carbon atoms in organic molecules.
2. Acid and base strength and electrophilicity-nucleophilicity are not exactly the same thing. In many contexts where Lewis acids and bases are classified as electrophilic and nucleophilic substances, it is common to also refer to substances as being good acids or bases on the basis of their Brønsted acidity and basicity. In these contexts, the terms acid and base typically refer to the thermodynamic propensity of a substance to give and accept hydrogen ions while the terms nucleophile and electrophile refer to a substance's kinetic propensity to react with carbon-based Lewis acids and bases, respectively.
A substance may be both a good nucleophile and a strong Brønsted base. For instance, alkyllithium carbanion reagents and alkali metal hydrides are strong bases and good nucleophiles towards electrophilic carbon atoms in organic solvents at moderate temperatures. However, this is not always the case. Under such conditions iodide ion, $I^-$, is a good nucleophile towards electrophilic carbon atoms but is a poor Brønsted base (it is the conjugate of the strong acid HI). In fact, for the halide ions the order of Brønsted basicity is the opposite of the order of nucleophilicity towards electrophilic carbon:
stronger Brønsted base: $F^-$, $Cl^-$, $Br^-$, $I^-$: weaker Brønsted base
better nucleophlie: $I^-$, $Br^-$, $Cl^-$, $F^-$: poorer nucleophile
3. Although the nucleophile-electrophile concept is most often used to describe organic reactions, it is also useful for describing inorganic reactions as well. For instance, the formation of hypochlorite ion in chlorine water possibly involves nucleophilic attack of hydroxide on an electrophilic chlorine atom in an SN2 type process:
$\nonumber$
The Usanovich Acid-base Concept
The Usanovich Acid-base concept encompasses a wider range of reactions than the Lewis acid-base concept but is perhaps too general to serve as a convenient framework for understanding reaction chemistry. The Usanovich acid-base concept was developed by the Russian chemist Mikhail Usanovich and extends the Lewis acid-base concept's definition of acids and bases as electron pair donors and acceptors even further. Within the Usanovich definition:
• Acids are anything that accepts electrons, increases cation concentrations, decreases anion concentrations, or reacts with bases
• Bases are anything that donates electrons, increases anion concentrations, decreases cation concentrations, or reacts with acids
These definitions essentially expand the Lewis definition by removing the requirement that the electrons donated or accepted be a pair and practically mean that Usanovich acid-base reactions include Lewis acid-base reactions plus oxidation-reduction reactions. In other words:
• Usanovich acids are Lewis acids plus oxidants
• Usanovich bases are Lewis bases plus reductants
Under this definition $BF_3$, $Fe^{3+}$, and $[Fe(CN)_6]^{3-}$ all act as acids, and $:NH_3$, $CN^-$, and Zn are all bases:
$\underset{\textcolor{red}{acid}}{BF_3} + \underset{\textcolor{blue}{base}}{NH_3} → F_3B-NH_3 \nonumber$
$\underset{\textcolor{red}{acid}}{Fe^{3+}} + 6 \underset{\textcolor{blue}{base}}{CN:^-} → [Fe(CN)_6]^{3-} \nonumber$
$2\underset{\textcolor{red}{acid}}{[Fe(CN)_6]^{3-}} + \underset{\textcolor{blue}{base}}{Zn} → 2[Fe(CN)_6]^{3-} + Zn^{2+} \nonumber$
However, notice that while it is easy to think about how any acids might interchangeably react with a given base (and vice versa), under the other acid-base concepts chemists have found it more difficult to think about reactions between oxidants/reductants and Lewis acids and bases. For example, it is difficult to think about how a reaction between $BF_3$ and Zn metal might occur. Would two Zn donate a pair of electrons to the $BF_3$? For this reason, most chemists find it more convenient to think about Lewis acid-base and redox reactions under separate categories rather than unite them under the Usanovich definition.
$2\underset{\textcolor{red}{oxidant}}{[Fe(CN)_6]^{3-}} + \underset{\textcolor{blue}{reductant}}{Zn} → 2[Fe(CN)_6]^{3-} + Zn^{2+} \nonumber$
Again, it can be helpful to remember Huheey's dictum that deciding between acid-base concepts is not a matter of which concept is the most correct but rather of which is more convenient for a given application.
6.04: Lewis Concept and Frontier Orbitals
Another way the Lewis-Acid base concept is widely employed for understanding chemical reactivity is through the frontier orbital approach to chemical reactions. The frontier orbital concept conceptualizes chemical bonding and reactivity in terms of the interactions between frontier orbitals on the chemical species undergoing an interaction (e.g. molecules, atoms, ions, or groups as they interact to form a bond or undergo a reaction). Frontier orbitals are those at the frontier between occupied and unoccupied. They are often taken to be the highest energy occupied and lowest energy unoccupied molecular orbitals, called the HOMO and LUMO levels. However, it can sometimes be more convenient to think about them as atomic orbitals or Valence Bond approach-derived orbitals. When developing rough qualitative frontier orbital descriptions of the orbital interactions involved in a given system, the choice of what types of orbitals to use is often a matter of what is the most informative and convenient.
In particular, the frontier orbital concept envisions a Lewis acid-base interaction as involving an interaction between some of the frontier orbitals of the Lewis acid and base, specifically the donation of electrons from the base's HOMO level into the acid's LUMO level. For example, in the frontier orbital approach, adduct formation between $NH_3$ and $BH_3$ involves the donation of electrons from ammonia's a1 HOMO into BH3's a2'' LUMO level.
As expected when two orbitals of the appropriate symmetry combine, the result of the interaction is the formation of a lower energy bonding orbital between the acid and base (Figure $1$). Since it is occupied by an electron, the net result of the interaction is the lowering of the base's lone pair energy as it interacts with the Lewis acid. In this case the interaction just follows the general pattern for Lewis-Acid base adduct formation, which is:
The frontier orbital concept illuminates the orbital interactions involved in reactions. For instance, from a frontier orbital perspective the alkyl halide substitution reaction between hydroxide and CH3Cl via the SN2 mechanism involves a Lewis acid-base interaction:
Notice how the frontier orbital approach even explains the displacement of the chloride leaving group. The donation of electrons from the hydroxide HOMO populates the antibonding CH3Cl LUMO, breaking the C-Cl bond.
With the frontier orbital approach, it becomes apparent that pericyclic reactions are Lewis acid-base reactions. For example, in the Diels-Alder reaction, the dienophile acts as a Lewis base and the diene as a Lewis acid.
The orbital interactions involved in a given reaction can include both reactants acting as both an acid and base. In these cases the HOMO of each reactant interacts with the LUMO of the other. A good example of this involves $\Pi$-type interactions between a metal ion with occupied d orbitals and a pi-acceptor ligand. The ligand acts as a base and the metal as an acid to give a M-CO single bond:
However, the metal can also act as a base towards the ligand LUMO ( $\pi ^*$ ) orbitals.
The frontier orbital concept can also be used to describe interactions between one or more singly occupied frontier orbitals. In fact, when approximate molecular orbital diagrams are being constructed, it can often be helpful to focus on one or two particularly instructive interactions. In these cases, it is sometimes convenient to think about a chemical bond as involving the formation of bonding and antibonding interactions between singly occupied frontier orbitals on different molecular fragments. For instance, when thinking about the chemistry of $CH_3Re(CO)_5$ it can sometimes be helpful to think about the C-Re bond as arising from the interaction between a singly-occupied sp3 type orbital on a CH3 group and a singly occupied d2sp3 orbital on the Re.
In introducing singly occupied orbitals, the frontier orbital concept formally handles interactions that the Lewis acid-base concept does not, since Lewis acid-base behavior is formally confined to the acceptance and donation of electron pairs. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/06%3A_Acid-Base_and_Donor-Acceptor_Chemistry/6.04%3A_Lewis_Concept_and_Frontier_Orbitals/6.4.01%3A_The_frontier_orbital_approach_considers_Lewis_acid_base_reactions_in_t.txt |
The electrons donated from a Lewis base to a Lewis acid in a Lewis acid-base reaction are donated and accepted at particular atomic centers. For instance, the formation of an adduct between ammonia and BF3 involves the donation of the lone pair on the ammonia nitrogen atom to the Lewis acid site on BF3.
Because Lewis acid-base reactions involve electron donation and acceptance at particular sites, substituent groups which alter the electron density at a site through inductively donating or withdrawing electron density will affect the Lewis acid-base properties of that site. For instance, the BF3 affinities of 4-substituted pyridines increase slightly as the substituent on the aromatic ring is changed from electron donating Me to electron withdrawing CF3.
Substituent inductive effects are much as one might predict. Since Lewis bases donate electron pairs and Lewis acids accept them:
• electron withdrawing substituents tend to decrease the Lewis basicity of basic sites while electron donating substituents increase site Lewis basicity by making them more electron rich.
• electron withdrawing substituents increase the Lewis acidity of acidic sites by making those sites more electron deficient while electron donating substituents tend to decrease Lewis acidity by making sites less electron deficient.
Nevertheless it can be difficult to predict substituent-based trends in Lewis acidity and basicity by inductive effects alone. This is because inductive effects are modest and often exists in competition with other substituent effects, such as
• Steric effects, discussed in section 6.4.7
• Hardness effects, discussed in section 6.6
• $\pi$-donation and acceptance effects, which can increase or decrease electron density at a given site as well as create an energy barrier for any structural distortions that might occur upon adduct formation.
Avoid misconceptions about the impact of$\pi$ donation and acceptance on Lewis acid-base affinity
As with $\sigma$-based induction effects, $\pi$-donation tends to increase Lewis basicity and decrease Lewis acidity while $\pi$-withdrawal tends to decrease Lewis basicity and increase Lewis acidity. However, care needs to be taken in assessing the effect of $\pi$-donation effects on Lewis acidity and basicity. For example, some textbooks claim that the Lewis acidity of boron trihalides is dominated by the reduction of boron acidity through $\pi$-donation from the halide substituents:
According to this explanation, the extent of this $\pi$-donation decreases down the halogen group as the boron-halogen bond distance decreases. This is consistent with the observed trend in Lewis acidity of the boron trihalides towards most bases, which runs counter to that suggested by inductive effects alone:
BI3 > BBr3 > BCl3 >> BF3
However, computational work suggests that this explanation is incorrect, since
• atomic size effects are important mainly for substituents in which the connected atom is in row 3+ or higher;
• atomic size effects mainly involve changes in the extent of $\sigma$-overlap. In other words, the larger halogens are less able to reduce the electron deficiency at the boron center through $\sigma$ interactions while $\pi$-interactions play little or no role.
The conjugate bases of many Brønsted superacids have electron-withdrawing substituents that make them such poor Lewis bases that they are useful as noncoordinating anions.
The nonreactivity of Brønsted superacids' conjugate bases towards hydrogen ions is often mirrored in nonreactivity towards other Lewis Acids/electrophiles, most notably metals. This makes these substances useful as inert or noncoordinating ions, although since all are reactive towards suitably electrophilic centers, they are perhaps better understood as weakly coordinating.
A number of noncoordinating anions are commonly used in synthetic and other applications. The conjugate base of perchloric acid, perchlorate, was a common noncoordinating inert anion in classical coordination chemistry and continues to be used widely in electrochemistry. In contrast, the conjugate bases of triflic acid, hexafluoroboric acid, and tetrafluoroboric acid are now more commonly used as counterions for isolating reactive cations.
Even less reactive noncoordination anions include derivatives of tetraphenylborate, particularly those with electron withdrawing substituents.
Other classes of noncoordinating ions include fluoroantimonate clusters, derivatives of the carborane anion ($CB_{11}H_{11}^-$), and fluorinated aluminum tetraalkoxides. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/06%3A_Acid-Base_and_Donor-Acceptor_Chemistry/6.04%3A_Lewis_Concept_and_Frontier_Orbitals/6.4.02%3A_All_other_things_being_equal_electron_withdrawing_groups_tend_to_make_.txt |
When a Lewis acid-base adduct is formed, electron density and negative charge are transferred from the Lewis base to the acid.
The formation of a Lewis-acid base complex involves the transfer of electron density from the base to the acid.
Weakly bound Lewis acid-base adducts in which charge is incompletely transferred are called charge transfer complexes.
In many weakly bound Lewis acid-base complexes, the transfer of electron density and, consequently, charge from the base group to the acid group is only partial:
$A~+~B~\rightleftharpoons~A^{\delta-}B^{\delta+} \nonumber$
Such Lewis acid-base adducts are commonly called charge transfer complexes (CT complexes) or donor-acceptor complexes (DA complexes). In these
• the base is called the donor (D) since it is a net donor of electrons and, consequently, their negative charge
• the acid is called the acceptor (A) since it is a net acceptor of electrons and, consequently, their negative charge
A particularly well-known class of charge transfer complexes is the iodine charge transfer complexes. In iodine charge transfer complexes the $I_2$ acts as a Lewis acid. This is possible since iodine is a Row 3+ element and so is capable of forming hypervalent complexes on reaction with a Lewis base. For example, $I_2$ reacts with $I^-$ to give the triiodide ion.
$\nonumber$
Triiodide is well-known from introductory chemistry from the bright blue color that appears when the triiodide complexes with starch to give the dark blue starch-iodide complex.
In contrast to the stable triiodide anion, iodine charge transfer complexes are only weakly associated. Complexes between iodine and amines are a well known example:
$\nonumber$
Iodine also forms weakly-associated charge transfer complexes with many solvents. For instance, iodine can weakly associate with both acetone and benzene:
Organic Charge Transfer Complexes
It is worth noting in passing that several organic charge transfer complex systems have been developed in which organic electron donors and acceptors weakly associate with each other. These are of considerable interest for use in molecular electronics applications, and as a result, a large variety of organic electron donors and acceptors have been developed. An example of an organic electron acceptor is tetracyanoethylene (TCNE):
and an example of an organic electron acceptor is tetrathiofulvalene (TTF).
Together they form a charge transfer complex:
Similar organic charge transfer systems have been investigated for use in organic photovoltaic cells. There is considerable interest in developing such organic photovoltaic devices owing to the ease with which organic materials may be fabricated from organic solutions and suspensions.
Charge transfer complexes exhibit charge transfer transitions in which absorption triggers the transfer of an electron from the donor to the acceptor.
When iodine is dissolved in solutions of donor solvents, the striking purple color of molecular iodine is replaced by a yellow-brown color. This is because charge transfer complexes like those formed by $I_2$ can absorb light in ways that neither the donor nor the acceptor can on their own. Specifically, charge transfer complexes exhibit charge transfer bands (CT bands) in their absorption spectra. In the charge transfer transition the initial partial transfer of charge from the donor Lewis base to the acceptor Lewis acid in the charge transfer complex is pushed further by photoexcitation.
$D^{\delta+}-A^{\delta-}~\overset{h\nu_{CT}}{~\longrightarrow}~D^+-A^- \nonumber$
The nature of these charge transfer transitions is seen from the orbital description of binding for iodine charge transfer complexes. When a donor-I2 complex forms, the formation of donor-I2 bonding and antibonding orbitals results in a shift in the $I2 \sigma \rightarrow \sigma*$ transition to higher energy, as a new charge transfer band is formed, associated with the excitation of an electron from the largely amine-centered amine-I2 $\sigma$ orbital to the largely I2-centered amine-I2 $\sigma *$ orbital.
Solutions of I2 as mixtures with Lewis bases such as amines and in donor solvents clearly have charge transfer bands in their absorption spectra. Several such spectra are given in Figure 6.4.2.3.
In principle, the energies of both the charge transfer and $I_2~\pi *~\rightarrow ~D-A~ \sigma *$ transitions both increase with donor strength, as shown in Figure 6.4.2.4.
As can be seen in Figure 6.4.2, the charge transfer band energy might be expected to increase as the donor HOMO increases in energy to become closer in energy to the acceptor LUMO. Although care should be taken when interpreting the solution phase spectra of I2, this expectation is borne out by a cursory and qualitative analysis of the spectra in Figure 6.4.2.3. The CT transition energy shifts towards lower wavelengths (and thus higher energy) as the highest occupied atomic orbital energy for the donor atom increases on going from acetone (oxygen, -15.85 eV) to chloroform (chlorine, -13.67eV) and finally benzene (carbon, -10.66 eV).**
Charge Transfer Bands in Transition Metal Chemistry
Charge transfer transitions are responsible for the intense color of many transition metal complexes. In these cases, however, the weak Lewis acid-base interaction involves incomplete electron donation and acceptance in a $p \pi -d \pi$ (or $p\pi-d\pi*$)-bond between a metal and ligand. The charge transfer bands in the absorption spectra of these complexes involve the transfer of electrons between the metal and ligand. In particular,
• Metal to ligand charge transfer (MLCT or CTTL) bands involve the transfer of an electron from a filled or partly filled metal d orbital to a ligand $\pi*$-type orbital.
• Ligand to metal charge transfer (LMCT or CTTM) bands involve the transfer of an electron from a filled or partly filled ligand orbital to a metal d-orbital.
• Metal to metal charge transfer bands can be observed in some bimetallic complexes. However, these are usually thought of only as an electron transfer rather than as a shift in the status of a Lewis acid-base interaction.
Because metal-ligand charge transfer bands involve intermolecular electron transfer between the metal and ligand to generate a high energy redox state, the CT excited state is both a better oxidant and reductant than the ground state. Consequently there has been intense research into the development of metal complexes whose charge transfer excited states are powerful oxidants and reductants in the expectation that they will be able to drive the photocatalytic oxidation and reduction of substrates.
* The apparent absorptivity of I2 in hexanes was calculated from the absorption spectrum of 215 µM I2 in hexanes. All other apparent absorptivities were calculated from absorption spectra of solutions that were 44 µM in I2.
** The band positions are not the CT band energies, and the HOMO energies given are atomic energy levels and do not necessarily correspond to the HOMO of the donor in solution. Because of this and other simplifications this analysis is not intended to replace a rigorous computational analysis of the factors that give rise to CT band positions. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/06%3A_Acid-Base_and_Donor-Acceptor_Chemistry/6.04%3A_Lewis_Concept_and_Frontier_Orbitals/6.4.03%3A_The_electronic_spectra_of_charge_transfer_complexes_illustrate_the_imp.txt |
A number of spectroscopic and thermodynamic scales have been developed to quantify the strength of Lewis acids and bases.1,2 For the sake of simplicity only thermodynamic scales will be described. Thermodynamic scales are based on the energetics of Lewis acid-base reactions. When substances act as a Lewis base ($\ce{B}$) they form adducts with a Lewis acid ($\ce{A}$):
$B:~+~A~⇌ ~B-A \nonumber$
Various thermodynamic parameters for this process may be taken as a measure of the strength of the Lewis base towards that acid. For instance, in the case of iodine charge transfer complexes, the equilibrium constant for adduct formation is sometimes used as an informal measure of Lewis base strength.
The thermodynamic parameters used to define Lewis acid and base strength are similar to those used to define Brønsted acidity and basicity. Just as Brønsted acidity and basicity were defined in terms of the free energy change for the dissociation and association of a hydrogen ion, Lewis acidities and basicities are defined in terms of the free energy change for adduct formation:
$B:~+~A_{reference}~ ⇌ ~B-A~~~~~~- \Delta G~=~Lewis~basicity~of~B~towards~A \nonumber$
$B:_{reference}~+~A~ ⇌ ~B-A~~~~~~- \Delta G~=~Lewis~acidity~of~A~towards~B \nonumber$
The physical meaning of Lewis acidities and basicities may be easier to grasp by considering that Lewis acidities and basicities correspond to the free energy for dissociation of acid-base adducts:
$B-A~ ⇌ ~B:~+~A_{reference}~~~~~~ \Delta G~=~Lewis~basicity~of~B~towards~A \nonumber$
$B-A~ ⇌ ~B:_{reference}~+~A~~~~~~ \Delta G~=~Lewis~acidity~of~A~towards~B \nonumber$
Since in practice it is much easier to measure the reaction enthalpies for these processes, many scales use the enthalpy change instead of the free energies. To distinguish these enthalpy changes from Lewis acidities and basicities, the enthalpy changes are called Lewis acid and base affinities. Of these, however, Lewis acid affinities are poorly characterized at present (perhaps one of you readers will help redress this). Thus the remainder of this section will focus on Lewis base affinities.
Before discussing Lewis base affinities it is worth noting that there is no universal reference scale of Lewis base strength. That is because the thermodynamics of a given Lewis acid-base interaction is contingent on a number of factors, including:
• The relative hardness of the acid and base according to Pearson's hard-soft acid base principle. Roughly, Lewis acids tend to associate more strongly with Lewis bases with similar charge densities and polarizabilities. This means that the affinity of a given base for Lewis acids is not a static parameter. Rather it differs markedly with the acid's hardness.
• Steric effects. Sterically hindered Lewis acid-base interactions will be weaker than sterically accessible ones.
• Solvent effects. In solutions, the adduct and the free acid and base pair will in general be differentially stabilized. To avoid solvent effects, some scales quantify Lewis acid and base strength in terms of gas phase adduct formation. However, it is not always possible or desirable to quantify a Lewis acid or base's strength in the gas phase, since gas phase affinities do not always allow for adequate prediction of the strength of a Lewis acid-base interaction in solution.
Because Lewis base affinity depends on hardness, steric effects, and solvent effects care should be taken when acid-base parameters are used to predict the strength of a given interaction. In particular, steric effects should be considered separately, and if predictions are made using scales that employ reference acids or bases that differ markedly in hardness from the interaction to be predicted or which correspond to unrealistic solvent conditions, they should always be taken as tentative.
6.4.05: In the boron triflouride affinity scale the enthalphy change on format
The physical meaning of Lewis acid and base affinities are defined as the negative of the enthalphy for formation of an adduct with a reference base and acid, respectively:
$B~+~A_{reference}~ ⇌ ~B-A~~~~~~~- \Delta H~=~\sf{Lewis~basicity~of~B~towards~A} \nonumber$
$B_{reference}~+~A~ ⇌ ~B-A~~~~~~- \Delta H~=~\sf{Lewis~acidity~of~A~towards~B} \nonumber$
Two common scales of Lewis base affinity involve the use of SbCl5 and BF3 as the reference base. Specifically,
• In the Guttman donor-acceptor scale, Lewis bases' donor numbers are equal to the negative enthalpy change for formation of the base's SBCl5 adduct in dilute 1,2-dichloroethane (EDC*):
$Base(EDC~solution) ~+~SbCl_5(EDC~solution)~\rightarrow~Base-SbCl_5(EDC~solution)~~~~~~- \Delta H~=~\sf{Guttman~Donor~Number} \nonumber$
• The BF3 affinity scale has largely supplanted the Guttman scale owing to the ease with which its results may be correlated with computational data (it is easier to calculate energies for adducts of BF3 than of SbF5). In the BF3 affinity scale, Lewis bases' BF3 affinities are equal to the negative enthalpy change for formation of a solution phase adduct between the base and gaseous BF3. The reaction is:
$Base(solution) ~+~BF_3(g)~\rightarrow~Base-BF_3(solution)~~~~~~- \Delta H~=~\sf{BF_3 ~Affinity}~ \nonumber$
BF3 affinity values for a wide range of organic bases were compiled by Laurence, Graton, and Gal in reference 2 and salient trends in this data will be discussed in subsequent sections. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/06%3A_Acid-Base_and_Donor-Acceptor_Chemistry/6.04%3A_Lewis_Concept_and_Frontier_Orbitals/6.4.04%3A_Substances%27_solution_phase_Lewis_Basicity_towards_a_given_acid_may_b.txt |
When a Lewis acid and base form an adduct, it shifts the bond lengths, and strengths in the acid and base also shift.
According to the frontier orbital description of Lewis acid-base interactions, adduct formation results in a net transfer of electron density from the base HOMO to the acid LUMO. This transfer
• increases the electron density between the acid (A) and base (D)
• alters the strength of the bonding within the base as the base LUMO becomes partially populated.
These changes result in structural and spectroscopic shifts on adduct formation that can be taken as a measure of the strength of the Lewis acid-base interaction. Since these shifts depend on the nature of the bonding in the acid and base, in practice structural and spectroscopic measurements of Lewis basicity are applied to a few classes of Lewis acid-base interactions, most notably hydrogen bonding and halogen bonding. Since hydrogen bonds will be discussed in section 6.5.1. the remainder of this subsection will focus on halogen bonds.
Halogen bonds are coordinate covalent bonds formed between a Lewis base and an organohalogen.
The halogens in compounds containing a singly-bonded halogen can engage in a special type of intermolecular interaction called halogen bonding, since the bonded halogens are electrophilic on the opposite side of the bond. This electrophilicity is apparent from the charge distribution in $CF_5I$ shown in Figure 4.5.1, in which the charge is distributed on the I such that the I is Lewis basic perpendicular to the C-I bond and Lewis acidic opposite to it. This region of diminished electron density on the halogen opposite the existing sigma bond is called a $\sigma$-hole.*
Schematically,
Since the halogen can act as a Lewis acid in the direction opposite the R-X bond it can form adducts with Lewis bases in that direction.
$\nonumber$
The bonds holding these adducts together are called halogen bonds, often abbreviated XB.
The halogen is said to act as a halogen bond donor when it forms a halogen bond with a base along that direction, with the base acting as the halogen bond acceptor. Consequently, the two previous schemes may also be given as:
The ability of singly-bonded halogen atoms to form halogen bonds forms the basis of many efforts to engineer structures in which halogen-containing compounds are held together into chains, sheets, or 3D structures using halogen bonds. One easy-to-see example involves 1D chains formed between p-diiodotetrafluorobenzene and p-bis(dimethylamino)benzene.3
Trends in halogen bond strength in a closely-related series of compounds may be assessed by examining how the vibrational stretching frequency of I2 and the I2 $\pi^*$ to $\sigma^*$ transition energy change upon formation of the adduct.
From the description of halogen bonding above it should be apparent that the iodine charge transfer complexes described in section 6.4.2 are held together by halogen bonds. This means that the description of the bonding in $I_2$ charge transfer complexes given in that section illustrate how vibrational and electronic transitions shift when a halogen-bonded adduct is formed between a halogen bond donor and acceptor.
The orbital interactions involved in halogen bond formation are shown in Figure 6.4.5.3. As can be seen from that figure, the formation of an adduct between $I_2$ and a Lewis base is expected to result in weakening of the I-I bond and a shift in the $I_2~ \pi^*~\rightarrow~\sigma^*$ transition energy to higher energy/shorter wavelengths (i.e., a blueshift). Further, both the weakening of the I-I bond and the $I_2~ \pi^*~\rightarrow~\sigma^*$ blueshift are expected to increase with the HOMO energy of the Lewis base.
Because the I-I bond strength and $I_2~ \pi^*~\rightarrow~\sigma^*$ blueshift increase with the Lewis basicity of the donor, the blueshift and the I-I vibrational stretching frequency are sometimes used as measures of donor's halogen bond affinity. However, the data obtained from such spectroscopic measurements should be interpreted with a degree of caution. As seen in Table 6.4.5.1, the extent to which the I-I bond is weakened (indicated by $\Delta \nu_{I-I}$) and the $I_2's~ \pi^*~\rightarrow~\sigma^*$ band is blue shifted are imperfectly correlated overall. However, a careful comparison of the data in 6.4.5.1. reveals that the I-I bond weakening and $I_2's~ \pi^*~\rightarrow~\sigma^*$ blue shift correlate well for Lewis bases of a given type. Consequently, shifts in the I-I stretching frequency and $I_2~ \pi^*~\rightarrow~\sigma^*$ absorption band are not useful as an absolute measure of Lewis basicity but can be used to rank the halogen bond affinity of a closely-related set of donors.**
Table $\sf{1}$. Spectroscopic measures of Lewis basicity for $I_2$ adducts of selected bases arranged in order of extent of blue shift. The data are excerpted from the extensive compilation given in reference 4.
Base (type) $\Delta \nu_{I-I}~cm^{-1}$ Blue shift of $I_2's~ \pi *~\rightarrow ~\sigma*$
4-methylpyridine (pyridines) 29.1 4730
pyridine (pyridines) 27.4 4560
tetrahydrothiophene (thioethers) 44.5 3640
tetrahydrofuran, THF (ethers) 6.4 2280
diethyl ether (ethers) 5.5 1950
acetone (ketone) 5.0 1850
acetophenone (ketone) 4.6 1650
acetonitrile (nitrile) 4.1 1610
hexamethylbenzene (aromatic) 10.4 1070
benzene (aromatic) 4.3 450
Notes
* Other compounds also exhibit $\sigma$-holes (and $\pi$-holes) that can be used in crystal engineering. For details see reference 2.
** for more details consult reference 4, pp. 286-309.
Contributors and Attributions
Consistent with the policy for original artwork made as part of this project, all unlabeled drawings of chemical structures are by Stephen Contakes and licensed under a Creative Commons Attribution 4.0 International License.
6.4.07: Bulky groups weaken the strength of Lewis acids and bases because they
Steric effects can influence the ability of a Lewis acid or base to form adducts by introducing:
• front strain (F-strain) whereby bulky groups make it difficult for the Lewis acid and Lewis base centers to approach and interact.
• back strain (B-strain) associated with steric interactions that do not directly impede the Lewis acid and base centers from interacting but instead occur as the acid and base rearrange upon adduct formation. For instance, when trivalent boron compounds form adducts with amines, the boron center changes from a more open trigonal geometry to a more hindered tetrahedral one:
• internal strain (I-strain) is also associated with the geometry changes incident on adduct formation. However, while B-strain involves direct steric clashes that occur on adduct formation, I-strain is the strain involved in deforming bond and torsional angles away from more stable local geometries. Thus it is more important for Lewis base centers embedded in rings or clusters.
6.4.08: Frustrated Lewis pair chemistry uses Lewis acid and base sites within
In certain cases when a Lewis acid-base adduct is sterically hindered from quantitatively forming an adduct (and perhaps even when they are so able), the Lewis acid and Lewis base can act together to heterolytically cleave chemical bonds. In such cases the pair of electrons that would otherwise be donated by the base is said to be a frustrated Lewis pair (FLP) and in performing the heterolytic cleavage the Lewis acid and base are said to exhibit frustrated Lewis pair chemistry.
The Stephan's phosphinoborane illustrates how FLP chemistry works. It reversibly cleaves dihydrogen to give a zwitterionic species containing both acidic and basic element hydride bonds.
Note that the Lewis acid and base do not need to be part of the same molecule. Hydrogen is also heterolytically cleaved by mixtures of sterically encumbered triarylphosphines and triarylboranes. This time the reaction gives phosphonium and hydroborate ions:
Given the description of FLP systems as frustrated Lewis acid-base adducts, it is natural to think about the cleavage of substrates like H2 as involving nucleophilic attack by the Lewis base while the Lewis acid acts like an electrophile in a heterolytic cleavage mechanism.
However, only some FLP systems cleave substrates through this mechanism. In others the mechanism involves homolytic cleavage by a diradical intermediate.
In either case, when substrates like \(\ce{H2}\) are cleaved by FLPs, they are generally separated into more reactive fragments (e.g., in the case of \(\ce{H2}\), \(\ce{H^{+}}\) and \(\ce{H^{-}}\)). Because of this the substrates are said to be activated to undergo further reactions. Because FLP systems can cleave and activate substrates, there is considerable interest in the development of FLP systems that can facilitate reactions of economic or environmental interest. So far FLP systems have been developed that can catalyze various hydrogenations and activate numerous substrates, including \(\ce{CO}\), \(\ce{SO2}\), and \(\ce{N2}\). | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/06%3A_Acid-Base_and_Donor-Acceptor_Chemistry/6.04%3A_Lewis_Concept_and_Frontier_Orbitals/6.4.06%3A_Lewis_base_strength_may_also_be_estimated_by_measuring_structural_or_e.txt |
As suggested by the Lewis acid-base concept's ability to describe halogen bonding in terms of the interaction between electrophilic and nucleophilic centers, the Lewis concept can be applied to the understanding of a variety of intermolecular interactions. In the next two sections the utility of Lewis Theory for understanding hydrogen bond and $\pi - \pi$ stacking interactions will be examined.
Contributors and Attributions
Stephen M. Contakes, Westmont College
6.05: Intermolecular Forces
This section sidesteps the discussion of topics related to acid-base chemistry and indulges in a brief aside about one type of intermolecular interaction that is used in host-guest chemistry. This section will first describe the concept of host-guest chemistry, then outline the types of interactions and design principles that can be used to promote host-guest interactions, and finally conclude with a discussion of $\pi-\pi$ interactions.
In Host-Guest Chemistry a large molecule or network material noncovalently binds a smaller molecule in a binding pocket.
In host-guest chemistry, a large molecule or network host uses noncovalent interactions to bind a smaller molecule guest in a binding pocket. Schematically this interaction is often represented as:
Host-guest binding is analogous to the way many biomolecule-substrate interactions occur - e.g., to how antibodies bind antigens and many enzymes bind their substrates. In this way many biomolecular systems can be said to form host-guest complexes, although that terminology is not often used in the biochemistry field. Usually when someone refers to a host-guest complex in inorganic or organic chemistry they are referring to organic or metal-organic cages that bind small organics. A simple example involves the binding of buckminsterfullerene, C60,, by calix-[5]-arenes.
In principle any type of intermolecular interaction can hold a host-guest complex together, but whichever one you choose, it helps to pay attention to a few design principles.
The types of interactions that are used to effect host-guest binding are diverse. In the fullerene-calixarene example above, the interactions will primarily involve $\pi$ stacking interactions between the aromatic rings of the calixarenes and those in the fullerene. Other host-guest systems use some combination of:
• hydrogen bonds
• halogen bonds
• ionic interactions whereby charged groups from the host attract oppositely charged groups of the guest
• ion-dipole interactions whereby charged groups on one component interact with the appropriate end of a polar group on the other.
• dipole-dipole interactions
• cation-$\pi$, X-H-$\pi$, and similar interactions in which full or partial positive charges induce a complementary dipole in a $\pi$ system.
• $\pi - \pi$ interactions
• dispersion forces
Many effective host-guest systems employ a few design principles to favor host-guest binding. These principles include:
• host preorganization - when conformationally flexible (-i.e., floppy) hosts bind a guest, the host will conform to the shape of the guest and so become more rigid. In other words, when a floppy host binds to a guest, a lot of conformational entropy is lost. This will make the $\Delta S$ term for the binding free energy more negative and tends to disfavor host-guest binding. To avoid this, effective hosts minimize the loss of conformational entropy on binding. This is usually done by limiting the conformational entropy of the host to begin with by fusing multiple binding sites into a macrocyclic ring, cage, or cluster. The more rigid these are the better.
• host-guest size and shape complementarity - this simply means that complexes form when the binding pocket or cavity of the host is complementary to that of the guest. Specifically, its size and shape both fit the guest and allow for tight interactions (i.e., the guest doesn't rattle around inside). Moreover, any charged, polar, or any-directional binding sites on the host are arranged in such a way that they interact favorably with complementary sites on the guest.
• preferencing of guest desolvation via preferential solvent-solvent interactions* - $\pi - \pi$ interactions like those which hold the calixarene-fullerene complex together are weak compared to stronger interactions like hydrogen bonds or even water-$\pi$ bonding (a type of X-H $\pi$ bonding. However, water-water hydrogen bonds and other interactions between H-bond-capable and polar systems are stronger still. This means that favorable water-water interactions must be broken when a nonpolar host or guest is dissolved in a very polar or ionic solution. Every dipole-induced dipole or X-H $\pi$ interaction that takes place is a lost opportunity for solute molecules to engage in hydrogen bonding interactions. Thus, when an aqueous solution of a nonpolar host and guest are mixed, the nonpolar parts of these molecules will tend to stick together, freeing water or other polar molecules to interact with one another as they do so.**
$\pi - \pi$ interactions occur when aromatic $\pi$ systems bind face to face with one another and involve a combination of dispersion and dipole-induced dipole interactions.
Since the electron distribution in aromatic systems is relatively easily distorted, they can engage in atypically strong dispersion and dipole-induced dipole interactions called $\pi - \pi$ stacking interactions. These interactions are so named because they occur when the planes of aromatic rings are stacked parallel to one another. This parallel stacking can occur in either a sandwich or a displaced stacking arrangement.
In either case the rings can be attracted to one another by a combination of dispersion and electrostatic interactions. The latter depend on the specific bonding parameters of the system and can be attractive or repulsive. The eclipsed arrangement of the benzene dimer is electrostatically disfavored, while attractions between the positive hydrogen substituents and the $\pi$ system of an adjacent ring are attractive in the eclipsed arrangement.
With electronegative substituents, the polarity of the ring system is reversed so that there are favorable electrostatic interactions in an eclipsed heterodimer of benzene and hexafluorobenzene.
In the heterodimer of benzene and hexafluorobenzene, at last we have an example of a $\pi$ stacking system that can more readily be described as involving a Lewis acid-base interaction. Specifically, the relatively electron-rich benzene $\pi$ system acts as a Lewis base in donating "electron pairs" to the electron-poor hexafluorobenzene $\pi$ system.
Notes
* The most prominent example of desolvation as a driving force is the hydrophobic effect. The hydrophobic effect governs the structure of soaps, micelles, biomolecules, and other amphiphilic or nonpolar systems in water. In the hydrophobic effect, the preference of an aqueous solution of a hydrophobic solute to maximize water-water interactions means that hydrophobic groups tend to stick together in solution.
** A good analogy for this effect is the magnetic marble analogy. When magnetic and ordinary marbles are mixed, the magnetic marbles will tend to stick together because they will attract one another strongly, not because the interactions between the nonmagnetic marbles are somehow disfavored.
† This is not to say that the stacking arrangements lead to attraction in every instance. There are arrangements in which the parallel rings are expected to repel. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/06%3A_Acid-Base_and_Donor-Acceptor_Chemistry/6.05%3A_Intermolecular_Forces/6.5.01%3A_Host-Guest_Chemistry_and_-_stacking_interactions.txt |
A hydrogen bond is a bonding interaction in which an E-H hydrogen bond donor unit acts as a Lewis acid in donating a hydrogen ion to a hydrogen bond acceptor, A.
Classically, hydrogen bonds (H-bonds) have been defined as a special type of intermolecular force between D-H and :A, where D & A = N, O, F, and occasionally S. The D-H is said to be the hydrogen bond donor while :A is the hydrogen bond acceptor.
Consider, for instance, a hydrogen bond between two water molecules:
An O-H bond in the water on the left serves as the hydrogen bond donor while one of the lone pairs on the oxygen of the water on the right acts as the hydrogen bond acceptor.* The strength of this classical hydrogen bonding interaction is ~12-20 kJ/mol, much greater than typical dipole-dipole interaction energies (<4 kJ/mol), and even though the hydrogen bonds in water are somewhat flexible, the O-H---O hydrogen bond axis prefers to be linear in many cases [1]. These factors suggest that hydrogen bonds are more than just a special type of dipole-dipole interaction involving hydrogen. Instead, hydrogen bonds involve a combination of factors, including:
• dipole-dipole interactions
• covalent bonding
• dispersion forces
• X-H $\pi$-interactions**
The relative importance of these interactions can differ significantly from one system to another.
In considering how to better understand hydrogen bonding, both Brønsted and Lewis acid-base concepts have been employed. In Brønsted terms, hydrogen bonding has been described as involving a partial transfer of a hydrogen ion from D to A. The Lewis definition is often more useful for understanding the role of covalency, however, owing to the ease with which Lewis theory accommodates orbital interactions through the Frontier orbital approach to chemical bonding and reactivity. Moreover, from the classical definition of hydrogen bonding as involving donation of H by a D-H unit and its acceptance by a lone pair on A, it is apparent that in some sense D-H is acting as a Lewis acid in donating an electron-deficient H to A while A is acting as a Lewis base in using its lone pair to accept it. In this respect, the hydrogen bond is perhaps better thought of as involving Lewis acid-base adduct formation between a D-H bond and A.
This more expansive definition of a hydrogen bond as involving a Lewis acid-base interaction accommodates the much wider variety of hydrogen bonding interactions that have been recognized since the 1990s. While the classical definition of hydrogen bonding accounts for much of the hydrogen bonding that occurs in water, organic, and biomolecular systems, the large number of multicenter H-bonds and the ability of C-H, P-H, and Au-H to function as hydrogen bond donors and for $\pi$ systems to function as hydrogen bond acceptors is excluded by the classical definition of a hydrogen bond. In contrast, the Lewis-based definition allows for the understanding of covalent contributions to hydrogen bonding in such systems in terms of the overlap of frontier orbitals on D, H, and A. Consider, for example, the frontier orbital interactions involved in a typical hydrogen bond between D-H and A:††
Hydrogen bonds like the one above are called asymmetric hydrogen bonds since the D-H bond is shorter and more stabilized than the A-H bond. When D and A are of equal or similar energy, however, the D-H and A-H interactions are of similar energy, and the H will be centered in between D and A, making it difficult to identify which atom is the donor and which is the acceptor. Such H-bonds are called symmetric H-bonds. An example is the H-bond in [F---H----F]-, for which the stabilizing and destabilizing frontier orbital interactions are
In general, symmetric hydrogen bonds are stronger and more stable than asymmetric ones. In fact, one approach to predicting the strength of classical hydrogen bonds called Gili's pKa Slide Rule‡ considers the pKa of the protonated form of both the neutral donor (D-H) and the neutral acceptor (A, which is then protonated to H-A+). In this approach the H-bond is considered to involve transfer of a hydronium ion between the donor and acceptor, and the pKa values of the donor and acceptor serve as proxies for D- and A's ability to draw the hydrogen ion in the H-bond to itself. When the
• donor is more basic (i.e., its pKa is higher) relatively little transfer occurs, and the hydrogen bond is better represented as D-H----A
• acceptor is more basic (i.e., its pKa is higher) the hydrogen ion is transferred from D to A, giving a hydrogen bond better represented as D----H-A
• donor and acceptor are about equally basic (have similar pKa values) the hydrogen ion is roughly shared equally between D and A in a symmetric H-bond, D--H--A
Specifically, Gili proposed the following criteria for predicting the strength of these ordinary hydrogen bonds:
Hydrogen Bond type $\Delta pK_a$
Strong (11.5-15.8 kcal/mol) 0-3
Medium Strong (6.8-11.5 kcal/mol) 3-11
Medium (4.1-6.8) kcal/mol 11-21
Medium Weak (2.5-4.1 kcal/mol) 21-31
Weak (1.1-2.5 kcal/mol) >31
Experimentally, the existence of hydrogen bonds and their approximate strength is inferred from crystallographic bond distances and angles, changes in the D-H vibrational frequency, and shifts in 1H NMR peak positions, although all such measures should be interpreted with caution.
If hydrogen bonds involve full or partial coordinate covalent bond formation between a D-H bond and an acceptor site, then the resulting adduct should exhibit evidence of bond formation. In practice such evidence is sought from three main sources:
1. Crystallographic bond distances and angles. In the structure of crystalline solids, hydrogen bonds can be distinguished from Van der Waals contacts in that
• the D----A distance is closer than that expected from the Van der Waals surfaces of D and A.
• when the H can be located, the D-H distance is longer than non-hydrogen-bonded D-H bonds in similar systems.
• for two-centered hydrogen bonds unconstrained by external factors, the D-H---A angle will approach linearity, as VSEPR theory would predict for a D:H:A unit.
2. D-H vibrational frequencies. Since hydrogen bonding involves weakening of the D-H bond and the formation of an H-A bond, in theory both the D-H and A-H vibrational modes will be found in the IR spectrum of a hydrogen-bonded system, albeit redshifted to lower wavenumbers than the unweakened D-H and A-H bonds. In practice, the shift in the D-H vibrational band to lower wavenumbers (along with a broadening and intensifying of the band) is taken as evidence for hydrogen bonding (Figure 6.5.1.1).
3. Shifts in the 1H NMR peak position of the donor hydrogen. As the hydrogen is partially transferred from D to A it becomes deshielded and so is shifted downfield, sometimes by as much as several ppm.
Care should be taken in applying these criteria since:
• The criteria work best for classical hydrogen bonded systems. Other aspects of the bonding in a given system can influence apparent bond strengths and NMR chemical shifts, particularly for intramolecular H-bonds and in nonclassical hydrogen bonding systems.
• Stronger hydrogen bonds are more likely to adhere to these criteria than weaker ones. Moreover, stronger hydrogen bonds generally exhibit greater linearity, shorter distances relative to those expected from Van der Waals radii, larger stretching frequency lowering for D-H, and greater downfield shifts in their NMR spectra. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/06%3A_Acid-Base_and_Donor-Acceptor_Chemistry/6.05%3A_Intermolecular_Forces/6.5.02%3A_Hydrogen_bonds_may_be_considered_as_a_special_type_of_Lewis_acid-base_interaction_in.txt |
Origin of the Hard-Soft Acid-Base Principle
One of the strengths of the Lewis acid-base concept is the readiness with which it illuminates the role that covalent and electrostatic interactions in acid base behavior, specifically through its ability to explain chemical interactions in terms of frontier orbitals and the interactions between charged groups as electrons are donated from a base to an acid. However, simply acknowledging the presence of such interactions does little to illuminate the degree to which each mode of explanation best explains the bonding in a given system? To what extent is a given adduct better described as held together by covalent bonds as opposed to ionic ones - e.g. better described as a molecule rather than an ion pair? Moreover, does it even matter, given that the orbitals of quantum mechanics result from the combination of electrons' wavelike behavior with their electrostatic attraction to nuclei in either case?1 These questions and more are addressed by one of the most important conceptual tools in contemporary inorganic chemistry, the Hard-Soft acid-base principle.
The Hard-Soft acid-base principle (HSAB principle) stems from the recognition that some Lewis acids and bases seem to have a natural affinity for one another.* Consider the following:
• Some metals are commonly found in nature as salts of chloride or as oxide ores while others are found in combination with sulfur. Geochemists even use the Goldschmidtt classification scheme to classify the halide and oxide formers as lithophiles and the sulfide formers as chalcophiles.
• In living systems small highly charged metals ions like Fe3+ are usually found bound to N and O atoms while larger metals with lower charges such as Zn2+ are often found attached to at least one S atom. Similarly, metals prefer to bind to one coordination site over the other when forming complexes with ambidentate ligands. The most well-known instances involve complexes of cyanate and thiocyanate, which can coordinate metals through either the N or chalcogen atom. For instance, Cu2+ and Zn2+ form N-thiocyanato complexes in species like [Cu(NCS)2(py)2] and [Zn(NCS)4]2- while their larger cogeners Au3+ and Hg2+ preferentially forms S-thiocyanato complexes, giving species like [Hg(SCN)4]2-.
Ambidentate ligands
Ambidentate ligands possess multiple coordination sites through which a metal may bind. For instance, thiocyanate may coordinate metals (M) at either the S or N to give S-thiocyanato or N-thiocyanato complexes.
• The solubility trends for the alklai metal halides and silver halides are opposite, even though both involve salts of formula M+X- (salts can be thought of as involving Lewis acid-base adduct formation between the anions and cations). Specifically, although the silver halides are all relatively insoluble in water, the very modest solubility they possess follows the order:
X = F >> Cl > Br > I (for the solubility of AgX)
In contrast, the much more ample solubility of the alkali metal halides** follows the opposite order. For example, the order for the lithium halides is
X = F << Cl < Br << I (for the solubility of LiX)
Notes
1. Albeit suitably fudged through such niceties as the Aufbau principle and other approximations.
* Despite the fruitfulness of this observation, in general it is important to reduce the potential for observer bias by checking observations like these against compounds reported in the chemical literature and databases like the Inorganic Crystal Structure and Cambridge Crystallographic Databases.
** These are very soluble in water, to the point where some solutions are perhaps better described as solutions of water in the halide.
† This can be predicted based on the relative hardness of BF3, BR3, and BH3 in the list of hard and soft acids. However, for those of you who may be confused as to why H is considered a better electron donor for the purposes of softening a Lewis acid center while alkyl groups are better electron donors for the purposes of stabilizing carbocations in organic chemistry, the dominant effect is the lower electronegativity of H relative to carbon (in CH3). The effect of electron donation due to hyperconjugation isn't as great for thermodynamically stable bases like BX3/BR3.
6.06: Hard and Soft Acids and Bases
Pearson Absolute Hardness is a useful metric of hardness based on orbital energies
The recognition that hard acids and bases possess a large HOMO-LUMO gap suggests that the gap size itself might serve as a useful index of hardness. The basis for this idea may be found by considering Pearson’s definition of absolute hardness, η.1 Pearson's absolute hardness, η, is half the second derivative of a species' energy with respect to changes in total number of electrons, $N_{e^-}$.
$\text{Pearson’s absolute hardness} = η = \dfrac{1}{2} \dfrac{d^2 E}{dN_{e^-}^2} \nonumber$
$(in~~eV)$
and, since acids' and bases’ hardness and softness are inversely related, Pearson’s absolute softness, σ, is just the inverse of hardness.
$\text{Pearson’s softness} = \dfrac{1}{ η} \nonumber$
Of these, Pearson’s absolute hardness is related to the Mulliken definition of electronegativity as the first derivative of a species' energy with respect to changes in total number of electrons.
$\text{Mulliken electronegativity} = χ = \dfrac{d E}{dN_{e^-}} \nonumber$
$(in~~eV)$
Operationally, both the Pearson hardness and Mulliken electronegativity are approximated in terms of the energies associated with unit changes in the number of electrons – i.e., in terms of ionization energies and electronegativities. Specifically,
$\text{Pearson’s absolute hardness} = η ≈ \dfrac{IE – EA}{2} \nonumber$
$\text{Mulliken electronegativity} = χ ≈ \dfrac{IE + EA}{2} \nonumber$
where $IE$ and $EA$ are in $eV$.
The connection between Pearson’s absolute hardness/softness and the HOMO-LUMO gap then follows from Koopman’s theorem, in which the ionization energy (IE) is just the opposite of the HOMO energy.
$IE = -E_{HOMO} \nonumber$
Similarly, the electron affinity (EA), defined as the opposite of the energy released on absorption of an electron, may be taken as an approximation of the LUMO energy.
$E_{LUMO} ≈ -EA \nonumber$
So Pearson’s absolute hardness is just half the HOMO-LUMO gap (band gap) size in electron volts:
$\text{Pearson's absolute hardness} = η ≈ \dfrac{E_{LUMO} – E_{HOMO}}{2} \nonumber$
where all values are given in $eV$,
and the Mulliken electronegativity is just the average of the HOMO and LUMO energies (~Fermi energy):
$Mulliken~electronegativity, χ ≈ -\dfrac{E_{LUMO} + E_{HOMO}}{2} \nonumber$
$where~all~values~are~given~in~eV$
The relationships between the Pearson absolute hardness, Mulliken electronegativity, and HOMO and LUMO energies are depicted schematically for the group 1A monocations in Figure $1$.
Values of the Pearson hardness and Mulliken electronegativity for several acids and bases are given in Table $1$. The values in the table confirm the expected trends, showing an increase in hardness with size (down a group), with increasing charge, and as substituent electronegativity increases for a series of isolobal ions.
Table $1$: Pearson Absolute Hardness and Related Parameters for Selected Acids and Bases.2 Taken from Pearson, R. G., Absolute electronegativity and hardness: application to inorganic chemistry. Inorganic Chemistry 1988, 27 (4), 734-740.
Species Ionization Energy, IE or I (eV) Electron Affinity, EA or A (eV) Mulliken Electronegativity, χ (eV) Pearson Absolute Hardness, η (eV)
Selected Acids
Group 1A monocations
Li+ 75.64 5.39 40.52 35.12
Na+ 47.29 5.14 26.21 21.08
K+ 31.63 4.34 17.99 13.64
Rb+ 27.28 4.18 15.77 11.55
Cs+ 25.1 3.89 14.5 10.6
Group 11 monocations
Cu+ 20.29 7.73 14.01 6.28
Ag+ 21.49 7.58 14.53 6.96
Au+ 20.5 9.23 14.9 5.6
Isoelectronic Row 3 Metal Cations
Na+ 47.29 5.14 26.21 21.08
Mg2+ 80.14 15.04 47.59 32.55
Al3+ 119.99 28.45 74.22 45.77
Changes with Transition Metal Ion Charge
Fe2+ 30.65 16.18 23.42 7.24
Fe3+ 54.8 30.65 42.73 12.08
Co2+ 33.50 17.06 25.28 8.22
Co3+ 51.3 22.5 42.4 8.9
Boron trihalides
BF3 15.81 -3.5 6.2 9.7
BCl3 11.60 0.33 5.97 5.64
BBr3 10.51 0.82 5.67 4.85
CO2 13.8 -3.8 5.0 8.8
CS2 10.08 0.62 5.35 5.56
Selected Bases
Group 17 monoanions (taken to be identical to the free atom values; for arguments as to why this is reasonable see Pearson, R. G., Inorg. Chem. 1988, 27 (4), 734-740.)
F- 17.42 3.40 10.41 7.01
Cl- 13.01 3.62 8.31 4.70
Br- 11.84 3.36 7.60 4.24
I- 10.45 3.06 6.76 3.70
Group 15 hydrides
NH3 10.7 -5.6 2.6 8.2
PH3 10.0 -1.9 4.1 6.0
Trimethylpnictides
NMe3 7.8 -4.8 1.5 6.3
PMe3 8.6 -3.1 2.8 5.9
AsMe3 8.7 -2.7 3.0 5.7
Group 16 hydrides
H2O 12.6 -6.4 3.1 9.5
H2S 10.5 -2.1 4.2 6.2
Phosphorous trihalides
PF3 12.3 -1.0 5.7 6.7
PC13 10.2 0.8 5.5 4.7
PBr3 9.9 1.6 5.6 4.2
Drago-Wayland Acid-Base Parameters allow for estimation of the electrostatic and covalent contributions to the enthalpy of formation of a Lewis acid-base adduct.
Although Pearson hardness values are a useful metric of acids' and bases' hardness, they cannot easily be used to estimate the Lewis acid-base interaction energy. This is not the case for the EC model developed by Drago and Wayland.3 In the Drago and Wayland's EC Model, the enthalpy of formation of an acid-base adduct, AB,
$A + B ⇌ AB ~~~~~~\Delta H_{AB~adduct} \nonumber$
from the acid (A) and base (B) can be calculated as the sum of products of electrostatic (E) and covalent (C) factors that reflect the propensity of the acid and base to engage in strong electrostatic and covalent interactions with one another:
$-\Delta H_{AB~adduct} = E_AE_B + C_AC_B \nonumber$
where EA and CA are the electrostatic and covalent parameters for the acid and EB and CB the electrostatic and covalent parameters for the base given in units of kcal½ mol. E and C parameters for selected acids and bases are given in Table $2$ .
Table $2$: E and C Parameters for selected Lewis acids and bases according to the ECW model.4,5 Taken from Vogel, G. C.; Drago, R. S., The ECW Model. Journal of Chemical Education 1996, 73 (8), 701 and Drago, R. S. A modern approach to acid-base chemistry. Journal of Chemical Education 1974, 51 (5), 300 (for BF3).
Species E (kcal½ mol) C (kcal½ mol) W (kcal mol-1) C/E
Acids
I2 0.5 2 0 4
ICl 2.92 1.66 0 0.57
H2O 1.31 0.78 0 0.59
SO2 0.51 1.56 0 3.1
HCCl3 (chloroform) 1.56 0.44 0 0.28
(CH3)3COH 1.07 0.69 0 0.65
(CF3)3COH 3.06 1.88 -0.87 0.61
2.27 1.07 0 0.47
2.23 1.03 0 0.46
2.30 1.11 0 0.48
1.38 0.68 0 0.49
BF3 1.62 9.88 6.10
B(CH3)3 2.90 3.60 0 1.2
Al(CH3)3 8.66 3.68 0 0.43
Ga(C2H5)3 6.95 1.48 0 0.21
In(CH3)3 6.60 2.15 0 0.33
Zn[N(Si(CH3)3)]2 2.75 2.32 0 0.84
Cd[N(Si(CH3)3)]2 2.50 1.83 0 0.73
Bases
NH3 2.31 2.04 0.88
CH3NH2 2.16 3.13 1.4
(CH3)2NH 1.80 4.21 2.3
(CH3)3N 1.21 5.61 4.6
1.78 3.54 2.0
1.81 3.73 2.0
1.53 2.94 1.9
CH3CN 1.64 0.71 0.43
ClCH2CN 1.67 0.33 0.20
CH3C(O)CH3 1.74 1.26 0.72
(C2H5)2O 1.80 1.63 0.91
1.64 2.18 1.3
(EtOAc) 1.63 0.95 0.58
2.35 1.31 0.56
(CH3)2S 0.25 3.75 15
(CH3CH2)2S 0.24 3.92 15
0.26 4.07 16
(CH3)2Se 0.05 4.24 83
(CH3)2SO 2.4 1.47 0.61
(CH3)3P 0.25 5.81 24
(CH3O)3P 0.13 4.83 37
(C6H5)3PO 2.59 1.67 0.64
(C6H5)3PS 0.35 3.65 10
C6H6 0.70 0.45 0.64
† These values were fixed to parameterize the rest of the E and C parameters.
Trends in the relative values of C and E for the acids and bases in Table $2$ are very roughly consistent with the trends in hardness and softness outlined earlier. However, the parameters suggest that some trends in hardness reflect changes in species' ability to engage in ionic interactions while others reflect changes in species' ability to engage in strong covalent interactions. For example, CB for the dimethylchalcogenides increases steadily from 1.5 to 4.25 kcal½ molon going from Me2O to Me2Se while EB decreases from 1.68 to 0.05 kcal½ mol, suggesting that electrostatic and covalent factors are both involved in a decrease in base hardness down group 16. However, in the case of phosphines and amines it appears that electrostatic factors are primarily responsible for the decrease in hardness down group 15. On going from Me3N to Me3P, CB only increases slightly from 5.61 to 5.81 kcal½ mol while EB decreases from 1.21 to 0.25 kcal½ mol - i.e., to 20% of the Me3N value.
However, some species tend to have very high values for both E and C, reflecting their ability to engage in strong electrostatic and covalent interactions, while others have small values for both, reflecting their relative stability as free species in solution.
As can be seen by comparing the acids and bases listed in the absolute hardness table (Table $1$) and the EC parameters table (Table $2$). The EC model has primarily been applied to organic and main group organometallic acids and bases. However, a variety of extensions have been proposed that enable its wider applicability.
The simple EC model only includes electrostatic and covalent considerations and thus ignores steric, lattice energy, and other contributions to the interaction energy. Thus it is only useful for analyzing the interaction energies of sterically unhindered adducts in which solvation energy and other contributions to the overall interaction energy are insignificant. However, additional refinements of the model attempt to extend its usefulness by accomodating various factors, such as
• Steric strain. Specifically, Hancock and Martell6 introduced a D parameter to account for any additional steric strain introduced upon adduct formation,* giving
$-\Delta H_{AB~adduct} = E_AE_B + C_AC_B - D_AD_B \nonumber$
• Charge transfer upon adduct formation. As discussed in section 6.4.2, the formation of a Lewis acid-base complex results in a net transfer of electron density from the electron donor (base) to the acceptor (acid). Drago and Wong7 extended the EC model to include that charge transfer by adding what they called receptance factors that account for the acid's ability to receive electron density (RA) and transference factors that account for the base's ability to donate electron density (TB), giving
$-\Delta H_{AB~adduct} = E_AE_B + C_AC_B + R_AT_B \nonumber$
This extension of the EC model is called the electrostatic-covalent-transfer or ECT model. Notably, it has been applied successfully to adducts involving ions, for which the RATB term can account for as much as 31% of the interaction energy.
• Any constant energy term, such as the energy needed to cleave a dimer in order to make the Lewis acid (e.g., Al2Cl6 → 2AlCl3). Drago and Vogel4 extended the EC model to accommodate these constant energy terms, which they designated W. The resulting model is called the ECW model, for which
$-∆H_{AB~adduct} = E_AE_B + C_AC_B + W \nonumber$
where
$W = W_A + W_B \nonumber$
A simple application of EC and extended EC models is that they allow the enthalpy of adduct formation to be calculated. These enthalpy calculations based on the EC model are consistent with the superior interaction energy of hard-hard and soft-soft interactions compared to hard-soft ones. Consider, for example, the transfer of I2 between BF3 and InMe3, H3N-BF3, and H3N-InMe3.
$H_3N-BF_3 + InMe_3 ⇌ BF_3 + H_3N-InMe_3 \nonumber$
Since H3N is a hard base and BF3 and InMe3 are harder and softer acids, respectively, the equilibrium is expected to favor the reactant, H3N-BF3. Assuming the equilbrium is enthalpically driven, this qualitative analysis is consistent with the expected endothermic enthalpy of the reaction, as may be seen from the calculated enthalpies of adduct formation for both H3N-BF3 and H3N-InMe3.
For H3N-BF3:
$-\Delta H_{H_3N-BF_3} = E_{BF_3}E_{NH_3} + C_{BF_3}C_{NH_3} \nonumber$
$-\Delta H_{H_3N-BF_3} = (1.62)(2.31) + (9.88)(2.04) \nonumber$
$-∆H_{H_3N-BF_3} = 3.74~kcal/mol + 20.1~kcal/mol = 23.8~kcal/mol \nonumber$
$∆H_{H_3N-BF_3} = -23.8~kcal/mol \nonumber$
For H3N-InMe3:
$-∆H_{H_3N-InMe_3} = E_{InMe_3}E_{NH_3} + C_{InMe_3}C_{NH_3} \nonumber$
$-∆H_{H_3N-InMe_3} = (6.6)(2.31) + (2.15)(2.04) \nonumber$
$-∆H_{H_3N-InMe_3} = 15.25~kcal/mol + 4.39~kcal/mol = 19.63~kcal/mol \nonumber$
$∆H_{H_3N-InMe_3} = -19.63~kcal/mol \nonumber$
So H3N-BF3 is enthalpically favored by -4.2 kcal/mol [=-23.8 kcal/mol - (-19.63 kcal/mol) according to Hess' Law].
In addition to their utility for estimating enthalpies of Lewis acid-base complex formation, EC and related models serve as a useful tool for estimating the relative importance of ionic, covalent, and steric factors in complex formation. Specifically,
• The relative contributions of ionic and covalent factors can be calculated directly, as from the E and C terms. This can provide insight into why some complexes are more stable than others. Such calculations reveal that the hard-hard adduct H3N-BF3 is favored over the hard-soft adduct H3N-InMe3 because of the strong covalent interaction holding together H3N-BF3. The covalent term accounts for 84% of the energy of the H3N-BF3 interaction and is largely lost upon formation of H3N-InMe3, for which it contributes only 22% of the interaction energy. When the electrostatic term is accounted for, it can be seen that the formation of H3N-InMe3 from H3N-BF3 is disfavored, since it would result in a loss of 15.7 kcal/mol of covalent stabilization that would be incompletely compensated for by a gain of 11.5 kcal/mol of electrostatic stabilization.
• The role of other contributions to the bonding in a Lewis acid-base complex may be estimated from the discrepancy between the experimental and EC model calculated stabilization energies. This is because the EC parameters assume sigma bonding and so any deviation between the calculated and experimental enthalpies of complex formation can be attributed to non-sigma contributions.
• This is done explicitly in extensions of the EC model. For instance, in the case of the Hancock and Martell, ECT, and ECW extensions of the EC model, the contributions of the steric, charge transfer, and constant energy factors (like dimer dissociation) are directly computed in the model.
• Comparisons of the difference between the energies calculated using ordinary EC parameters and the observed enthalpies has also been used as an estimate of the
• steric strain energy in strained adducts, which exhibit a less exothermic heat of adduct formation than expected
• $\pi$-backbonding energy in adducts which are capable of such interactions and exhibit a more exothermic heat of formation than expected.8-10
• For examples see (a) Drago, R. S., The interpretation of reactivity in chemical and biological systems with the E and C model. Coordination Chemistry Reviews 1980, 33 (3), 251-277; (b) Drago, R. S.; Bilgrien, C. J., Inductive transfer and coordination of ligands in metal—metal bonded systems. Polyhedron 1988, 7 (16), 1453-1468; (c) Drago, R. S., The question of a synergistic metal-metal interaction leading to .pi.-back-bond stabilization in dirhodium tetrabutyrate adducts. Inorganic Chemistry 1982, 21 (4), 1697-1698.
• The role of electrostatic, covalent, and other contributions to the spectroscopic behavior of Lewis acid-base complexes can be assessed similarly using specialized versions of the ECW or ECT model that allow for calculation of changes in spectroscopic parameters upon adduct formation - e.g., OH stretching frequencies. Details are beyond the scope of this text but may be found in the original literature,11 for example, Drago, R. S.; Vogel, G. C. JACS 1992, 114 (24), 9527-9532; Vogel, G. C.; Drago, R. S., J. Chem. Educ. 1996, 73 (8), 701; Drago, R. S.; Wong, N. M., J. Chem. Educ. 1996, 73 (2), 123.).
Pearson Softness Parameters have been proposed as a means of calculating the equilibrium constant for formation of a Lewis acid-base adduct, although they are more commonly used to estimate metal toxicity
Pearson also proposed a softness parameter that can be used to estimate the thermodynamics of Lewis acid-base complex formation, this time in the form of the equilibrium constant for
$A + B \overset{K}{⇌} AB \nonumber$
for which Pearson proposed
$logK = S_AS_B + \sigma_A \sigma_B \nonumber$
where SA and SB are acid and base parameters related to the strength of the acid-base interaction and $\sigma_A$ and $\sigma_B$ are acid and base parameters related to softness. This approach hasn't found as wide an acceptance in studies of chemical reactivity as the Drago-Wayland and absolute hardness parameters. However, because the toxicity of metal ions sometimes depends on their propensity to bind soft Lewis bases in living systems, the softness parameters which Pearson and Mawby subsequently proposed based on the relative energies of metal fluorine and metal iodine bonds12 have been widely used as a tool for predicting metal ion toxicity.13
Notes
* More precisely, the D parameters account for the difference in steric strain in the adduct relative to a water adduct of the Lewis acid. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/06%3A_Acid-Base_and_Donor-Acceptor_Chemistry/6.06%3A_Hard_and_Soft_Acids_and_Bases/6.6.01%3A_Quantitative_Measures_of_Hardness_Softness_and_Acid-Base_Interactions_from_a.txt |
The Hard Soft Acid Base Principle is a Conceptual Tool for Thinking About Patterns of Lewis Acid-Base Reactivity
The explanation of the trends in metal distribution, halide salt solubility, and preferred metal coordination patterns is rooted in Arland, Chatt, and Davies' observation that Lewis acids and bases could be classified into two groups based on their propensity to form stable compounds with one another (e.g. acids in a class tend to form more stable adducts with bases in the same class than they did with bases in the other).1 Arland, Chatt, and Davies somewhat boringly termed these groups class a and class b but today they are known by Ralph Pearson's name for them. Pearson called the class a acids and bases hard and class b acids and bases soft. These terms reflect how "soft" these substance's electron clouds are towards distortion or, in other words, their polarizability (Figure $1$). Pearson terms acids and bases which are relatively polarizable soft and those which are difficult to polarize hard.
Recognizing Hard and Soft Acids and Bases
Hard acids and bases come in two varieties:
1. hard acid and base sites that possess few valence electrons and for which polarization therefore involves distorting core electrons, which are difficult to distort because they are close to the nucleus and experience a high nuclear charge. The most common examples of such substances are Lewis acid hard acids towards the left of the periodic table.
2. hard acid and base sites with a high charge density (highly charged relative to size) and/or which are electron deficient. In these cases polarization involves distorting electrons that already experience strong unshielded electrostatic interactions.
Soft acids and bases also come in two varieties
1. soft acids and bases which posses many valence electrons and so are more readily polarized. In consequence, all other things being equal, soft acids and bases are more likely to be found towards the middle or right of the periodic table.
2. soft acids and bases with little charge density and/or which are relatively electron rich.
Note that the hard-soft classification should not be thought of as if all hard acids and bases are equally hard and all soft acids and bases equally soft. There is a graduation in hardness and softness and a number of intermediate acids and bases which do not fit neatly in either category. With this caveat in mind, representative hard, soft, and borderline acids are given below. Notice how they illustrate the trends just outlined.
As expected, hard acids tend to be found towards the left side of the periodic table and involve higher oxidation states and/or electron donating substituents while soft acids are more common to the right of the periodic table and involve lower oxidation states and/or electron donating substituents.
Illustrative hard, soft, and borderline bases are given below. Again, notice how these substances illustrate the general trends.
Qualitative Estimation of the Relative Hardness and Softness of Lewis Acids and Bases
As can be seen from the examples above, hard acids are relatively electron-poor and hard bases electron-rich since they have comparatively
• small frontier orbitals, reflective of their relatively small atom/ion/fragment sizes
• high (for acids) or low (for bases) oxidation states on the base atom, reflected in a large positive formal charge (for acids) or negative formal charge (for bases)
• low polarizability, due to loss or gain of substantial numbers of electrons, or the localization of
• positive charge on an electropositive element or an atom bearing electron-withdrawing substituents
• negative charge on an electronegative element or an atom bearing electron-donating substituents
In contrast to hard acids and bases, soft acids are relatively electron-rich and soft bases larger and more electron poor since they have comparatively
• large frontier orbitals, reflective of their relatively large atom/ion/fragment sizes
• low oxidation states, often resulting in small or nonexistent atomic charges
• high polarizability, as might be expected of species in which electron-electron repulsions are lower and electrons are spread over a large volume. Sometimes this is indicated by
• positive charge on an electronegative element or an atom bearing electron-donating substituents
• negative charge on an electropositive element or an atom bearing electron-withdrawing substituents
Exercise $1$
Rank the acids or bases in each set in increasing order of expected hardness.
1. Cr2+ and Cr3+
2. H+, Cs+, and Tl+
3. SCN- (acting as a base at N) and SCN- (acting as a base at S)
4. AlF3, AlH3, AlMe3
5. The side chains of the following proteinogenic amino acids
Answer
(a) Cr2+ < Cr3+ All other things being equal, hardness increases with oxidation state.
(b) Tl+ < Cs+ < H+ The order reflects Cs+ and Tl+'s larger size relative to H+ (which doesn't possess any electrons that can be polarized anyway) and that Tl+ still possesses two valence electrons while Cs+ possesses none.
(c) SCN- (acting as a base at S) < SCN- (acting as a base at N). The order reflects the greater electronegativity of N than S and N's possession of a more negative formal charge of -1.
(d) AlH3 < AlMe3 < AlF3. The hardness increases as the substituents on the Lewis acid Al center become less electron donating and more electron withdrawing (and, incidentally, harder bases) as their electronegativity increases in the order H- < CH3- < F-. Note that the order of electron donating ability for H- and CH3- is the opposite observed for carbocations, for which hyperconjugation plays a larger role.
(e) Sec < Cys < Ser. The hardness increases as the electronegativity of the Lewis base chalcogen increases on going from a selenol to a thiol to an alcohol.
The Hard-Soft Acid-Base Principle (HSAB Principle)
The Hard-Soft acid-base principle (HSAB Principle) explains patterns in Lewis acid-base reactivity in terms of a like reacts with like preference. Both thermodynamically and kinetically, hard acids prefer hard bases and soft acids soft bases. Specifically,
• Thermodynamically, hard acids form stronger acid-base complexes with hard bases while soft acids form stronger complexes with soft bases.
• Kinetically, hard acids/electrophiles react more quickly with hard bases/nucleophiles while soft acids/electrophiles react more quickly with soft bases/neucleophiles.
Applications of the HSAB principle include
1. Predicting the equilibrium or speed of Lewis acid-base metathesis and displacement reactions. In a Lewis acid-base metathesis reaction the acids and bases swap partners $\ce{A1:B1 + A2:B2 <=>[k_1, K_{eq}] A1:B2 + A2:B1} \nonumber$
For example, the equilibrium position of the metathesis reaction between $\ce{TlF}$ and $\ce{K2S}$ favors the products:
$\ce{2TlF + H2S <=>> Tl2S + 2KF} \nonumber$
consistent with the HSAB's hard-hard and soft-soft preference.
$\nonumber$
The HSAB principle also allows for prediction of the position of displacement reactions, in which a Lewis acid or base forms an adduct using a base or acid from an existing Lewis acid-base complex. In these reactions, the displacement of acid or base from the reactant complex may be thought of as a sort of metathesis reaction, one in which in the unbound acid or base switches places with one in the complex. For example, the reaction between $\ce{HI}$ and methylmercury cation
$\ce{HI + HgSCH3^{+} <=> CH3SHgI + H^{+}} \nonumber$
involves displacement of an iodide from $\ce{HI}$ to give $\ce{CH3HgI}$. The position of the equilibrium favors $\ce{CH3HgI}$ since both $\ce{CH3Hg^{+}}$ and $\ce{I^{-}}$ are soft, while $\ce{H^{+}}$ is a hard acid.
$\nonumber$
Exercise $2$
Predict the position of equilibrium for the following reaction.
$\ce{Fe2O3 + 3Ag2S <=> Fe2S3 + 3Ag2O} \nonumber$
Answer
The equilibrium will favor the reactants (K<1) since the hard-hard and soft-soft interactions in the reactants are more stable than the hard-soft interactions in the products.
Exercise $3$
Predict whether $K$ for the following equilibria will be <<1, ~1, or >>1.
1. $\ce{2HF + (CH_3Hg)_2S ⇌ 2CH_3HgF + H_2S}$
2. $\ce{Ag(NH_3)_2^+ + 2PH_3 ⇌ Ag(PH_3)_2^+ + 2NH_3}$
3. $\ce{Ag(PH_3)_2^+ + 2H_3B-SH_2 ⇌ 2H_3B-PH_3 + Ag(SH_2)_2^{+}}$
4. $\ce{H_3B-NH_3 + F_3B-SH_2 ⇌ H_3B-SH_2 + F_3B-NH_3}$
Answer
a. K< < 1 since the reactant adducts are hard-hard and soft-soft while the products involve hard-soft interactions.
b. K>>1 since the reactant complex, diamine silver(I), is a complex of a hard base, NH3, with the soft acid, Ag+, while the product is a complex of the same soft acid with the soft base phosphine.
c. K~1 since all the adducts amongst the reactants and products involve soft acids and bases.
d. K>>1 since BH3 is a softer acid than BF3, so it will form a stronger complex with the softer base H2S while the harder BF3 forms a stronger complex with the harder base NH3.
2. Predicting the relative strengths of a given set of Lewis acids or bases towards a particular substrate. Consider, for example, the relative strengths of a BH3, BMe3, and BF3 towards group 15 hydrides like NH3, PH3, and AsH3. Of the boranes listed, the hardest acid BF3 is the strongest acid towards the hard base NH3 while BH3 is the strongest towards AsH3.
Exercise $4$
Which acid will form the most stable complex with $\ce{CO}$: $\ce{BH3}$, $\ce{BF3}$, or $\ce{BMe3}$?
Answer
$\ce{BH3}$. Since $\ce{CO}$ forms complexes primarily through its carbon lone pair, it is a soft base and so will form the strongest complex with the softest Lewis acid.
Exercise $5$
When lactones react with nucleophiles, they can undergo ring opening reactions to give either an alcohol or a carboxylic acid, as shown for propiolactone below:
In the reaction above, sterically unhindered alkoxides give one product and sterically unhindered thioalkoxides the other. Explain why this is the case and predict the products of the reaction between propiolactone and the sodium salts of ethoxide and thioethoxide.
Answer
The two reaction products correspond to nucleophilic attack at the lactones' two electrophilic carbon centers. Specifically, the acid is produced by attack at the softer CI center of the CH2 directly attached to the ester oxygen and the alcohol by nucleophilic attack at the harder CIII center of the ester carbonyl.
Consequently, it is reasonable to expect that the harder base ethoxide will nucleophilically attack the harder carbonyl carbon while the softer thioethoxide will attack the softer methylene carbon.
The theoretical interpretation of the hard-soft acid-base principle is that hard-hard preferences reflect superior electrostatic stabilization while soft-soft preferences reflect superior covalent stabilization.
The hard-hard and soft-soft preferences in Lewis acid-base interactions reflect that
• The lone pair of a hard base is strongly stabilized electrostatically by a hard acid.
• The lone pair of a soft base is strongly stabilized by forming a covalent bond with a soft acid.
• The lone pair of a hard or soft base is comparatively weakly stabilized by an acid opposite to it in hardness or softness since the overall electrostatic and covalent stabilization of the adduct is comparatively weak.
To see why this is the case it is helpful to divide the contributions to the interaction energy between an acid and a base as follows:
$\nonumber$
Of the three contributions to the interaction energy, only the ionic and covalent terms directly relate to the hardness of the interacting acid and base. One approach to thinking about how hardness influences the ionic and covalent contributions is to consider the frontier orbitals involved in the acid-base interaction. This is sometimes done through the use of the Salem-Klopman equation,1,* although in the treatment which follows a more qualitative approach will be employed.
Both hard acids and bases will have comparatively low energy HOMO levels and high energy LUMO levels, with a correspondingly high HOMO-LUMO gap. In contrast, soft acids and bases will have comparatively high-energy HOMO levels and low-energy LUMO levels, giving a comparatively smaller HOMO-LUMO gap.
Given this, consider the frontier orbital interactions involved in the formation of an acid-base complex for the possible cases, as illustrated schematically below.
The large gap in energy between hard bases’ highly stabilized HOMO lone pairs and the high energy LUMO of hard acids ensures that in hard acid-hard base adducts the dominant stabilizing interaction will involve electrostatic attraction between the base lone pair and the electropositive Lewis acid center. Fortunately, since the electron clouds in hard bases are relatively dense and electron rich while hard Lewis acids are highly charged and small, these electrostatic interactions are strong.
In contrast, in soft acid-soft base adducts, the dominant stabilizing interaction will be covalent. This is because the small gap in energy between a soft base HOMO and soft acid LUMO enables the formation of a well-stabilized bonding orbital with significant electron density between the acid and base.
The orbitals' interactions between hard acids and soft bases and soft acids and hard bases are intermediate between the hard acid-hard base and soft acid-soft base cases.
This means that the adducts are stable relative to free acid and base – just not as well stabilized as in the hard acid and hard base case. In the case of hard acids and soft bases the hard acids are less able to stabilize the soft bases’ relatively diffuse electron pair electrostatically and there isn’t as much covalent stabilization as in adducts of soft acids and bases due to the hard acid’s high energy.
Notes
* Despite the fruitfulness of this observation, in general it is important to reduce the potential for observer bias by checking observations like these against compounds reported in the chemical literature and databases like the Inorganic Crystal Structure and Cambridge Crystallographic Databases.
** These are very soluble in water, to the point where some solutions are perhaps better described as solutions of water in the halide.
† This can be predicted based on the relative hardness of BF3, BR3, and BH3 in the list of hard and soft acids. However, for those of you who may be confused as to why H is considered a better electron donor for the purposes of softening a Lewis acid center while alkyl groups are better electron donors for the purposes of stabilizing carbocations in organic chemistry, the dominant effect is the lower electronegativity of H relative to carbon (in CH3). The effect of electron donation due to hyperconjugation isn't as great for thermodynamically stable bases like BX3/BR3.
†† For more on the Salem-Klopman equation see Fleming, I., Molecular orbitals and organic chemical reactions. Reference ed.; Wiley: Hoboken, N.J., 2010; pp. 138-143. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/06%3A_Acid-Base_and_Donor-Acceptor_Chemistry/6.06%3A_Hard_and_Soft_Acids_and_Bases/6.6.02%3A_Hard-Hard_and_Soft-Soft_preferences_may_be_explained_and_quantified_in_terms.txt |
Learning Objectives
• Explain the physical basis of the Hubbard and Mott models of metal-insulator transitions.
• Understand why good superconductors derive from bad metals.
• Know the structures and the periodic trends in band gaps and colors of semiconductors.
• Obtain the band gap of an intrinsic semiconductor from the temperature dependence of the conductivity.
• Predict the doping type when impurities or defects are introduced into a semiconductor.
• Correlate the band picture and Fermi level with n- or p-type doping.
• Understand the physical principles of operation of diodes, LEDs, solar cells, and FETs.
• Explain the differences in structures and electronic properties of crystalline and amorphous semiconductors.
The band model (like MO theory) is based on a one-electron model. This was an approximation we made at the very beginning of our discussion of MO theory: we used hydrogen-like (one-electron) solutions to the Schrödinger equation to give us the shapes of s, p, d, and f atomic orbitals. In a one-electron atom, these orbitals are degenerate within a given shell, and the energy differences between, e.g., 2s and 2p orbitals arise only when we consider the energy of an electron in the field of other electrons in the atom. Moving from atoms to molecules, we made linear combinations to generate one-electron molecular orbitals (and, in solids, one-electron energy bands). But as in multi-electron atoms, life is not so simple for real molecules and solids that contain many electrons. Electrons repel each other, and so their movement in molecules and in solids is correlated.
• 7.1.1: Prelude to Electronic Properties of Materials - Superconductors and Semiconductors
Correlated electron effects give rise to metal-insulator transitions that are driven by small changes in temperature, pressure, or composition, as well as to superconductivity - the passage of current with zero resistance at low temperatures. In this chapter we will develop some simple models to understand these interesting and important electronic properties of solids.
• 7.1.2: Metal-Insulator Transitions
Under experimentally accessible temperatures and pressures, Si and Ge are always semiconducting (i.e., insulating), and Pb is always metallic. Why is Sn different? The reason has to do with orbital overlap. Theory tells us in fact that any (and all) insulators should become metallic at high enough pressure, or more to the point, at high enough density. For most insulators, however, the pressures required are far beyond those that we can achieve in the laboratory.
• 7.1.3: Periodic Trends- Metals, Semiconductors, and Insulators
• 7.1.4: Semiconductors- Band Gaps, Colors, Conductivity and Doping
There are a number of places where we find semiconductors in the periodic table.
• 7.1.5: Semiconductor p-n Junctions
• 7.1.6: Diodes, LEDs and Solar Cells
Diodes are semiconductor devices that allow current to flow in only one direction. Diodes act as rectifiers in electronic circuits, and also as efficient light emitters (in LEDs) and solar cells (in photovoltaics). The basic structure of a diode is a junction between a p-type and an n-type semiconductor, called a p-n junction. Typically, diodes are made from a single semiconductor crystal into which p- and n-dopants are introduced.
• 7.1.7: Amorphous Semiconductors
Amorphous semiconductors are disordered or glassy forms of crystalline semiconductor materials with network structures involving primarily covalent bonding. Crystalline silicon, which has the diamond structure, is an ordered arrangement of fused six-membered silicon rings and the local bonding environment of the silicon atoms is tetrahedral. The silicon atoms in amorphous silicon (a-Si) are also predominantly tetrahedrally coordinated, but there is no long-range order in the structure.
• 7.1.8: Discussion Questions
• 7.1.9: Problems
• 7.1.10: References
7.01: Molecular Orbitals and Band Structure
In Chapter 6 we developed an energy band picture for metals, starting from atomic orbitals and building up the molecular orbitals of the solid metallic crystal. This treatment gave us a useful picture of how electrons behave in metals, moving at very fast speed between scattering events, and migrating in an electric field at a slow drift velocity. It also taught us that a metal is something with a partially filled band, meaning that the Fermi level cuts through one of its bands of orbitals. An insulator or a semiconductor has a similar band picture, except that the bands are either completely full or completely empty. In this case the Fermi level lies in the gap between fully occupied and unoccupied bands. We will see in this chapter that the properties of semiconductors (along with their useful electronic applications) depend on the addition of small amounts of impurities ("dopants") that change the position of the Fermi level, resulting in conduction by electrons or "holes."
Modern integrated circuits contain billions of nanoscale transistors and diodes that are essential for logic and memory functions. Both kinds of devices rely on junctions between crystalline silicon regions that contain a few parts per million of boron or phosphorus impurities.
While the band picture works well for most crystalline materials, it does not tell us the whole story of conduction in solids. That is because the band model (like MO theory) is based on a one-electron model. This was an approximation we made at the very beginning of our discussion of MO theory: we used hydrogen-like (one-electron) solutions to the Schrödinger equation to give us the shapes of s, p, d, and f atomic orbitals. In a one-electron atom, these orbitals are degenerate within a given shell, and the energy differences between, e.g., 2s and 2p orbitals arise only when we consider the energy of an electron in the field of other electrons in the atom. Moving from atoms to molecules, we made linear combinations to generate one-electron molecular orbitals (and, in solids, one-electron energy bands). But as in multi-electron atoms, life is not so simple for real molecules and solids that contain many electrons. Electrons repel each other and so their movement in molecules and in solids is correlated. While this effect is weak in a "good" metal such as sodium - where the wavefunctions are highly delocalized - it can be quite important in other materials such as transition metal oxides. Correlated electron effects give rise to metal-insulator transitions that are driven by small changes in temperature, pressure, or composition, as well as to superconductivity - the passage of current with zero resistance at low temperatures. In this chapter we will develop some simple models to understand these interesting and important electronic properties of solids. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/07%3A_The_Crystalline_Solid_State/7.01%3A_Molecular_Orbitals_and_Band_Structure/7.1.01%3A_Prelude_to_Electronic_Properties_of_Materials_-_Superconductors_and_Semiconduct.txt |
In Chapter 6 we learned that metals and insulators not only have different electrical properties but also have very different crystal structures. Metals tend to have high coordination numbers (typically 8 or 12) whereas insulators have low coordination numbers that can be rationalized as "octet" bonding arrangements. For example, in crystalline Si or Ge (diamond structure), each atom has four nearest neighbors. There are two electrons per bond, and thus each atom has eight electrons in its valence shell. Sn, the element below Ge, exists in two different forms, one (gray tin) with the diamond structure that is a brittle narrow-gap semiconductor, and the other (white tin) with a body-centered tetragonal structure that is a malleable metal. These two forms are very close in energy, and in fact metallic white tin transforms to the brittle semiconducting gray form at low temperature. Extremely cold weather in 18th century Europe caused many tin organ pipes to break and eventually turn to dust. This transformation has been called tin blight, tin disease, tin pest or tin leprosy. The dust is actually grey tin, which lacks the malleability of its metallic cousin white tin.
Under experimentally accessible temperatures and pressures, Si and Ge are always semiconducting (i.e., insulating), and Pb is always metallic. Why is Sn different? The reason has to do with orbital overlap. Theory tells us in fact that any (and all) insulators should become metallic at high enough pressure, or more to the point, at high enough density. For most insulators, however, the pressures required are far beyond those that we can achieve in the laboratory.
How can we rationalize the transition of insulators to the metallic state? Indeed, how can we understand the existence of insulators at all?
The Hubbard Model
Let's consider a chain of a large number (N) of atoms as we did in Chapter 6. For convenience, we can say that these are atoms such as H, Na, or Cs that have one valence electron. The simple band model we developed earlier suggests that the chain should be metallic, because N atoms combine to make N orbitals, and the N valence electrons only fill the band of orbitals halfway. But this conclusion doesn't depend on the density, which creates a paradox. If atoms in the chain are very far apart, we suspect that the electrons should localize on the atoms.
Electron hopping in a 1-D chain of atoms.
A solution to this problem was proposed by J. Hubbard in 1963.[1] Hubbard considered the energy required to transfer an electron from an atom to its nearest neighbor, as shown in the picture at the right. Because each atom already has one electron (with random spin), moving an electron over by one atom requires overcoming the energy of electron-electron repulsion to make a cation-anion pair. For well-separated atoms this energy (U) is given by:
$U = IP -Ea -\frac{e^{2}}{4\pi \varepsilon_{0} d}$
where IP and EA are the ionization energy and electron affinity, ε0 is the permittivity of free space, and the last term in the equation represents the coulombic attraction between the cation and the anion. For atoms such as alkali metals, U is on the order of 3–5 eV, which is much larger than the thermal energy kT. Thus we expect there to be very few anion-cation pairs at room temperature, and the chain of atoms should be insulating.
Energy vs. DOS for the chain of atoms as the density and degree of orbital overlap between atoms increases. Increasing overlap broadens the neutral atom and anion-cation states into bands, each of which has a bandwidth Δ. A transition to the metallic state occurs abruptly when Δ exceeds the Hubbard gap U.
What happens when we squeeze the atoms together? In the Hubbard model, as the distance between atoms decreases, the energies of both the neutral atom states and the anion-cation states broaden into bands, each of which has a band width Δ. The lower band can accommodate exactly N electrons (not 2N as in the MO picture we developed earlier) because each orbital can only take one electron without spin-pairing. Thus for small Δ the lower band is full and the upper band is empty. However, as we continue to compress the chain, the orbital overlap becomes so strong that Δ ≈ U. At this point, the bands overlap and some of the electrons fill the anion-cation states. The chain then becomes conducting and the material is metallic.
Some materials, such as Sn and VO2, happen to have just the right degree of orbital overlap to make the Hubbard transition occur by changing the temperature or pressure. Such materials can be very useful for electrical switching, as illustrated at the right for rutile structure VO2. Most materials are far away from the transition, either on the metallic or insulating side. An interesting periodic trend that illustrates this concept can be seen among the transition metal monoxides, MO (M = Ti, V, Cr, Mn, Fe, Co, Ni), all of which have the NaCl structure. TiO and VO are metallic, because the 3d orbitals have significant overlap in the structure. However, CrO, MnO, FeO, CoO, and NiO are all insulators, because the 3d orbitals contract (and therefore Δ < U) going across the transition metal series. In contrast, the analogous sulfides (TiS, VS,....NiS) are all metallic. The sulfides have the NiAs structure, in which all the metal atoms are eclipsed along the stacking axis (the hexagonal c-axis). The short metal-metal distances along that axis result in strong orbital overlap, making Δ > U.
Vanadium dioxide has the rutile structure, and in its undistorted form it is metallic, with one valence electron per V atom. Distortion of the lattice makes pairs of V atoms, resulting in an electronically insulating state. The metal-insulator transition can be driven reversibly by changing the temperature, pressure, or orbital occupancy. Electrical switching of this transition in VO2 is being studied for applications in high performance thin film transistors[2]
The Mott Model
A simpler, less atomistic model of the metal-insulator transition was formulated by Neville Mott.[3] The Mott model considers the behavior of an electron in a material as a function of the density of all the other valence electrons. We know for a one-electron hydrogen-like atom (H, Na, Cs, etc.) the Schrödinger equation contains a potential energy term:
$V(r) = -(\frac{e^{2}}{4 \pi \varepsilon_{0} r})$
This potential energy function gives rise to familiar ladder of allowed energy levels in the hydrogen atom. However, in a metal, this Coulomb potential must be modified to include the screening of nuclear charge by the other electrons in the solid. In this case there is a screened Coulomb potential:
$V(r) = -(\frac{e^{2}}{4 \pi \varepsilon_{0}r}) exp(-qr)$
where q, which is the inverse of the screening length, is given by:
$q^{2} = 4m_{e}^{2}(\frac{3n}{\pi})^{\frac{1}{3}}(\frac{2\pi}{h})$
Here n is the density of atoms (or valence electrons), me is the electron mass, and h is Planck's constant. At distances much larger than the screening length q-1, the electron no longer "feels" the charge on the nucleus. Mott showed that there is a critical density of electrons nc above which the valence electrons are no longer bound by individual nuclei and are free to roam the crystal. This critical density marks the transition to the metallic state, and is given by the Mott criterion:
$\mathbf{n_{c}^{\frac{1}{3}}a_{H} \approx 0.26}$
In this equation, aH is an effective Bohr radius for the valence electrons in the low-density limit, e.g. the average orbital radius of electrons in the 6s shell of a Cs atom when computing the value for Cs metal.
Solutions of lithium metal in liquid ammonia at low (top, ionic conductor) and higher (bottom, metal) Li concentration. From a video by Joshua Judkins
The important concept from the Mott model is that the metal-insulator transition depends very strongly on the density of valence electrons. This is consistent with the orbital overlap model of Hubbard, but also more general in the sense that it does not depend on a periodic structure of atoms. The Mott model is thus applicable to such diverse systems as metal atoms dissolved in liquid ammonia, metal atoms trapped in frozen gas matrices, and dopants in semiconductors.[4] In some systems, it is possible to continuously tune the density of valence electrons with rather striking results. For example, dissolving alkali metal (Li, Na, ...) in liquid ammonia (bp -33 oC) produces a blue liquid. The solvated alkali cations and negatively charged electrons impart ionic conductivity (as in a salt solution) but not electronic conductivity to the blue liquid ammonia solution. But as the concentration of electrons increases, a reflective, bronze-colored liquid phase forms that floats over the blue phase. This bronze phase is metallic and highly conducting. Eventually, with enough alkali metal added, the entire liquid is converted to the electronically conducting bronze phase.
The electrical switching of VO2 between insulating and metallic phases (see above) can also be rationalized in terms of the Mott transition. Adding more electron density (by chemical or electric field doping) increases the concentration of valence electrons, driving the phase transition to the metallic state.
Thermodynamics and phase transitions
Thermodynamically, the metal-insulator transition is a first-order phase transition. In such a transition, the structure and properties change abruptly (think of the breakfast-to-lunch transition at McDonalds - there is just no way to get pancakes after, or hamburgers before 10:30 AM![5]). Thus in the case of Sn metal, the changes in structure (from four- to eight-coordination) and in electronic conductivity (insulator to metal) occur simultaneously. As in other first order phase transitions such as ice to water to steam, there is a latent heat associated with the transition and a discontinuity in derivative properties such as the heat capacity.
A typical phase diagram for a metal-insulator transition is shown at the right for V2O3. The octahedrally coordinated V3+ ion has a d2 electron count, so there are two unpaired spins per atom, and at low temperature the spins in the lattice order antiferromagnetically. As we learned in Chapter 8, above the Néel temperature an antiferromagnet becomes a paramagnet, which is also a Mott insulator. Increasing the pressure, or doping with electrons (e.g., by substituting some d3 Cr3+ for V3+) pushes the electron density over the Mott transition, the spins pair, and the solid becomes metallic.
7.1.03: Periodic Trends- Metals Semiconductors and Insulators
As we consider periodic trends in the electronic properties of materials, it is important to review some of the key bonding trends we have learned in earlier chapters:
• Going down the periodic table, atoms in solids tend to adopt structures with higher coordination numbers.
• The second row of the periodic table is special, with strong s-p hybridization and π-bonding between atoms.
• Electrons in higher quantum shells are less strongly bound, so the energy difference between bonding and antibonding orbitals becomes smaller for heavier atoms.
We also know that most of the elements in the periodic table are metals, but the elements in the top right corner are insulating under ordinary conditions (1 atm. pressure) and tend to obey the octet rule in their compounds.
At the transition between metals and non-metals in the periodic table we encounter a crossover in electronic properties, as well as in other properties such as the acidity of the oxides (see Ch. 3). The group of elements at the border is loosely referred to as the metalloids. Several of these elements (such as C, Sn, and As) can exist as different allotropes that can be metals, insulators, or something in between.
A more rigorous delineation of the electronic properties of these elements (and of many compounds) can be made by considering their band structures and the temperature dependence of the electronic conductivity. As we have previously discussed, metals have partially filled energy bands, meaning that the Fermi level intersects a partially filled band. With increasing temperature, metals become poorer conductors because lattice vibrations (which are called phonons in the physics literature) scatter the mobile valence electrons. In contrast, semiconductors and insulators, which have filled and empty bands, become better conductors at higher temperature, since some electrons are thermally excited to the lowest empty band. The distinction between insulators and semiconductors is arbitrary, and from the point of view of metal-insulator transitions, all semiconductors are insulators. We typically call an insulator a semiconductor if its band gap (Egap) is less than about 3 eV. A semimetal is a material that has a band gap near zero, examples being single sheets of sp2-bonded carbon (graphene) and elemental Bi. Like a narrow gap semiconductor, a semimetal has higher conductivity at higher temperature. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/07%3A_The_Crystalline_Solid_State/7.01%3A_Molecular_Orbitals_and_Band_Structure/7.1.02%3A_Metal-Insulator_Transitions.txt |
Semiconductors, as we noted above, are somewhat arbitrarily defined as insulators with band gap energy < 3.0 eV (~290 kJ/mol). This cutoff is chosen because, as we will see, the conductivity of undoped semiconductors drops off exponentially with the band gap energy and at 3.0 eV it is very low. Also, materials with wider band gaps (e.g. SrTiO3, Egap = 3.2 eV) do not absorb light in the visible part of the spectrum
There are a number of places where we find semiconductors in the periodic table:
• Early transition metal oxides and nitrides, especially those with d0 electron counts such as TiO2, TaON, and WO3
• Oxides of later 3d elements such as Fe2O3, NiO, and Cu2O
• Layered transition metal chalcogenides with d0, d2 and d6 electron counts including TiS2, ZrS2, MoS2, WSe2, and PtS2
• d10 copper and sliver halides, e.g., CuI, AgBr, and AgI
• Zincblende- and wurtzite-structure compounds of the p-block elements, especially those that are isoelectronic with Si or Ge, such as GaAs and CdTe. While these are most common, there are other p-block semiconductors that are not isoelectronic and have different structures, including GaS, PbS, and Se.
A 2" wafer cut from a GaAs single crystal. GaAs, like many p-block semiconductors, has the zincblende structure.
The p-block octet semiconductors are by far the most studied and important for technological applications, and are the ones that we will discuss in detail.
Zincblende- and wurtzite-structure semiconductors have 8 valence electrons per 2 atoms. These combinations include 4-4 (Si, Ge, SiC,…), 3-5 (GaAs, AlSb, InP,…), 2-6 (CdSe, HgTe, ZnO,…), and 1-7 (AgCl, CuBr,…) semiconductors. Other variations that add up to an octet configuration are also possible, such as CuIInIIISe2, which has the chalcopyrite structure, shown at the right.
The chalcopyrite structure is adopted by ABX2 octet semiconductors such as CuIInIIISe2 and CdIISnIVP2. The unit cell is doubled relative to the parent zincblende structure because of the ordered arrangement of cations. Each anion (yellow) is coordinated by two cations of each type (blue and red).
How does the band gap energy vary with composition? There are two important trends
(1) Going down a group in the periodic table, the gap decreases:
C (diamond) > Si > Ge > α-Sn
Egap (eV): 5.4 1.1 0.7 0.0
This trend can be understood by recalling that Egap is related to the energy splitting between bonding and antibonding orbitals. This difference decreases (and bonds become weaker) as the principal quantum number increases.
(2) For isoelectronic compounds, increasing ionicity results in a larger band gap.
Ge < GaAs < ZnSe
0.7 1.4 2.8 eV
Sn < InSb < CdTe < AgI
0.0 0.2 1.6 2.8 eV
This trend can also be understood from a simple MO picture, as we discussed in Ch. 2. As the electronegativity difference Δχ increases, so does the energy difference between bonding and antibonding orbitals.
The band gap is a very important property of a semiconductor because it determines its color and conductivity. Many of the applications of semiconductors are related to band gaps:
• Narrow gap materials (HgxCd1-xTe, VO2, InSb, Bi2Te3) are used as infrared photodetectors and thermoelectrics (which convert heat to electricity).
• Wider gap materials (Si, GaAs, GaP, GaN, CdTe, CuInxGa1-xSe2) are used in electronics, light-emitting diodes, and solar cells.
Color wheel showing the colors and wavelengths of emitted light.
Semiconductor solid solutions such as GaAs1-xPx have band gaps that are intermediate between the end member compounds, in this case GaAs and GaP (both zincblende structure). Often, there is a linear relation between composition and band gap, which is referred to as Vegard's Law. This "law" is often violated in real materials, but nevertheless offers useful guidance for designing materials with specific band gaps. For example, red and orange light-emitting diodes (LED's) are made from solid solutions with compositions of GaP0.40As0.60 and GaP0.65As0.35, respectively. Increasing the mole fraction of the lighter element (P) results in a larger band gap, and thus a higher energy of emitted photons.
Colors of semiconductors
The color of absorbed and emitted light both depend on the band gap of the semiconductor. Visible light covers the range of approximately 390-700 nm, or 1.8-3.1 eV. The color of emitted light from an LED or semiconductor laser corresponds to the band gap energy and can be read off the color wheel shown at the right.
Fe2O3 powder is reddish orange because of its 2.2 eV band gap
The color of absorbed light includes the band gap energy, but also all colors of higher energy (shorter wavelength), because electrons can be excited from the valence band to a range of energies in the conduction band. Thus semiconductors with band gaps in the infrared (e.g., Si, 1.1 eV and GaAs, 1.4 eV) appear black because they absorb all colors of visible light. Wide band gap semiconductors such as TiO2 (3.0 eV) are white because they absorb only in the UV. Fe2O3 has a band gap of 2.2 eV and thus absorbs light with λ < 560 nm. It thus appears reddish-orange (the colors of light reflected from Fe2O3) because it absorbs green, blue, and violet light. Similarly, CdS (Egap = 2.6 eV) is yellow because it absorbs blue and violet light.
Electrons and holes in semiconductors
Pure (undoped) semiconductors can conduct electricity when electrons are promoted, either by heat or light, from the valence band to the conduction band. The promotion of an electron (e-) leaves behind a hole (h+) in the valence band. The hole, which is the absence of an electron in a bonding orbital, is also a mobile charge carrier, but with a positive charge. The motion of holes in the lattice can be pictured as analogous to the movement of an empty seat in a crowded theater. An empty seat in the middle of a row can move to the end of the row (to accommodate a person arriving late to the movie) if everyone moves over by one seat. Because the movement of the hole is in the opposite direction of electron movement, it acts as a positive charge carrier in an electric field.
The opposite process of excitation, which creates an electron-hole pair, is their recombination. When a conduction band electron drops down to recombine with a valence band hole, both are annihilated and energy is released. This release of energy is responsible for the emission of light in LEDs.
An electron-hole pair is created by adding heat or light energy E > Egap to a semiconductor (blue arrow). The electron-hole pair recombines to release energy equal to Egap (red arrow).
At equilibrium, the creation and annihilation of electron-hole pairs proceed at equal rates. This dynamic equilibrium is analogous to the dissociation-association equilibrium of H+ and OH- ions in water. We can write a mass action expression:
$n \times p = K_{eq} = n_{i}^{2}$
where n and p represent the number density of electrons and holes, respectively, in units of cm-3. The intrinsic carrier concentration, ni, is equal to the number density of electrons or holes in an undoped semiconductor, where n = p = ni.
Note the similarity to the equation for water autodissociation:
$[H^{+}][OH^{-}] = K_{w}$
By analogy, we will see that when we increase n (e.g., by doping), p will decrease, and vice-versa, but their product will remain constant at a given temperature.
Temperature dependence of the carrier concentration. Using the equations $K_{eq} = e^{(\frac{- \Delta G^{o}}{RT})}$ and $\Delta G^{o} = \Delta H^{o} - T \Delta S^{o}$, we can write:
$n \times p = n_{i}^{2} = e^{(\frac{\Delta S^{o}} {R})} e^{(\frac{- \Delta H^{o}}{RT})}$
The entropy change for creating electron hole pairs is given by:
$\Delta S^{o} = R ln (N_{V}) + R ln (N_{V}) = R ln (N_{C}N_{V})$
where NV and NC are the effective density of states in the valence and conduction bands, respectively.
and thus we obtain
$n_{i}^{2} = N_{C}N_{V} e^{({- \Delta H^{o}}{RT})}$
Since the volume change is negligible, $\Delta H^{o} \approx \Delta E^{o}$, and therefore $\frac {\Delta H^{o}}{R} \approx \frac{E_{gap}}{k}$, from which we obtain
$n_{i}^{2} = N_{C}N_{V} e^{(\frac{-E_{gap}}{kT})}$
and finally
$\mathbf{n= p = n_{i} = (N_{C}N_{V})^{\frac{1}{2}} e^{(\frac{-E_{gap}}{2kT})}}$
For pure Si (Egap = 1.1 eV) with N ≈ 1022/cm3, we can calculate from this equation a carrier density ni of approximately 1010/cm3 at 300 K. This is about 12 orders of magnitude lower than the valence electron density of Al, the element just to the left of Si in the periodic table. Thus we expect the conductivity of pure semiconductors to be many orders of magnitude lower than those of metals.
Conductivity of intrinsic semiconductors
The conductivity (σ) is the product of the number density of carriers (n or p), their charge (e), and their mobility (µ). Recall from Chapter 6 that µ is the ratio of the carrier drift velocity to the electric field and has units of cm2/Volt-second. Typically electrons and holes have somewhat different mobilities (µe and µh, respectively) so the conductivity is given by:
$\sigma = ne\mu_{e} + pe\mu_{h}$
For either type of charge carrier, we recall from Ch. 6 that the mobility μ is given by:
$\mu = \frac{v_{drift}}{E} = \frac{e\tau}{m}$
where e is the fundamental unit of charge, τ is the scattering time, and m is the effective mass of the charge carrier.
Taking an average of the electron and hole mobilities, and using n = p, we obtain
$\mathbf{\sigma= \sigma_{o} e^{(\frac{-E_{gap}}{2kT})}}, \: where \: \sigma_{o} = 2(N_{C}N_{V})^{\frac{1}{2}}e\mu$
By measuring the conductivity as a function of temperature, it is possible to obtain the activation energy for conduction, which is Egap/2. This kind of plot, which resembles an Arrhenius plot, is shown at the right for three different undoped semiconductors. The slope of the line in each case is -Egap/2k.
Plots of ln(σ) vs. inverse temperature for intrinsic semiconductors Ge (Egap = 0.7 eV), Si (1.1 eV) and GaAs (1.4 eV). The slope of the line is -Egap/2k.
Doping of semiconductors. Almost all applications of semiconductors involve controlled doping, which is the substitution of impurity atoms, into the lattice. Very small amounts of dopants (in the parts-per-million range) dramatically affect the conductivity of semiconductors. For this reason, very pure semiconductor materials that are carefully doped - both in terms of the concentration and spatial distribution of impurity atoms - are needed.
n- and p-type doping. In crystalline Si, each atom has four valence electrons and makes four bonds to its neighbors. This is exactly the right number of electrons to completely fill the valence band of the semiconductor. Introducing a phosphorus atom into the lattice (the positively charged atom in the figure at the right) adds an extra electron, because P has five valence electrons and only needs four to make bonds to its neighbors. The extra electron, at low temperature, is bound to the phosphorus atom in a hydrogen-like molecular orbital that is much larger than the 3s orbital of an isolated P atom because of the high dielectric constant of the semiconductor. In silicon, this "expanded" Bohr radius is about 42 Å, i.e., 80 times larger than in the hydrogen atom. The energy needed to ionize this electron – to allow it to move freely in the lattice - is only about 40–50 meV, which is not much larger the thermal energy (26 meV) at room temperature. Therefore the Fermi level lies just below the conduction band edge, and a large fraction of these extra electrons are promoted to the conduction band at room temperature, leaving behind fixed positive charges on the P atom sites. The crystal is n-doped, meaning that the majority carrier (electron) is negatively charged.
Alternatively, boron can be substituted for silicon in the lattice, resulting in p-type doping, in which the majority carrier (hole) is positively charged. Boron has only three valence electrons, and "borrows" one from the Si lattice, creating a positively charged hole that exists in a large hydrogen-like orbital around the B atom. This hole can become delocalized by promoting an electron from the valence band to fill the localized hole state. Again, this process requires only 40–50 meV, and so at room temperature a large fraction of the holes introduced by boron doping exist in delocalized valence band states. The Fermi level (the electron energy level that has a 50% probability of occupancy at zero temperature) lies just above the valence band edge in a p-type semiconductor.
n- and p-type doping of semiconductors involves substitution of electron donor atoms (light orange) or acceptor atoms (blue) into the lattice. These substitutions introduce extra electrons or holes, respectively, which are easily ionized by thermal energy to become free carriers. The Fermi level of a doped semiconductor is a few tens of mV below the conduction band (n-type) or above the valence band (p-type).
As noted above, the doping of semiconductors dramatically changes their conductivity. For example, the intrinsic carrier concentration in Si at 300 K is about 1010 cm-3. The mass action equilibrium for electrons and holes also applies to doped semiconductors, so we can write:
$n \times p = n_{i}^{2} = 10^{20} cm^{-6} \: at \: 300K$
If we substitute P for Si at the level of one part-per-million, the concentration of electrons is about 1016 cm-3, since there are approximately 1022 Si atoms/cm3 in the crystal. According to the mass action equation, if n = 1016, then p = 104 cm-3. There are three consequences of this calculation:
• The density of carriers in the doped semiconductor (1016 cm-3) is much higher than in the undoped material (~1010 cm-3), so the conductivity is also many orders of magnitude higher.
• The activation energy for conduction is only 40–50 meV, so the conductivity does not change much with temperature (unlike in the intrinsic semiconductor)
• The minority carriers (in this case holes) do not contribute to the conductivity, because their concentration is so much lower than that of the majority carrier (electrons).
Similarly, for p-type materials, the conductivity is dominated by holes, and is also much higher than that of the intrinsic semiconductor.
Chemistry of semiconductor doping. Sometimes it is not immediately obvious what kind of doping (n- or p-type) is induced by "messing up" a semiconductor crystal lattice. In addition to substitution of impurity atoms on normal lattice sites (the examples given above for Si), it is also possible to dope with vacancies - missing atoms - and with interstitials - extra atoms on sites that are not ordinarily occupied. Some simple rules are as follows:
• For substitutions, adding an atom to the right in the periodic table results in n-type doping, and an atom to the left in p-type doping.
For example, when TiO2 is doped with Nb on some of the Ti sites, or with F on O sites, the result is n-type doping. In both cases, the impurity atom has one more valence electron than the atom for which it was substituted. Similarly, substituting a small amount of Zn for Ga in GaAs, or a small amount of Li for Ni in NiO, results in p-type doping.
• Anion vacancies result in n-type doping, and cation vacancies in p-type doping.
Examples are anion vacancies in CdS1-x and WO3-x, both of which give n-type semiconductors, and copper vacancies in Cu1-xO, which gives a p-type semiconductor.
• Interstitial cations (e.g. Li) donate electrons to the lattice resulting in n-type doping. Interstitial anions are rather rare but would result in p-type doping.
Sometimes, there can be both p- and n-type dopants in the same crystal, for example B and P impurities in a Si lattice, or cation and anion vacancies in a metal oxide lattice. In this case, the two kinds of doping compensate each other, and the doping type is determined by the one that is in higher concentration. A dopant can also be present on more than one site. For example, Si can occupy both the Ga and As sites in GaAs, and the two substitutions compensate each other. Si has a slight preference for the Ga site, however, resulting in n-type doping. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/07%3A_The_Crystalline_Solid_State/7.01%3A_Molecular_Orbitals_and_Band_Structure/7.1.04%3A_Semiconductors-_Band_Gaps_Colors_Conductivity_and_Doping.txt |
Semiconductor p-n junctions are important in many kinds of electronic devices, including diodes, transistors, light-emitting diodes, and photovoltaic cells. To understand the operation of these devices, we first need to look at what happens to electrons and holes when we bring p-type and n-type semiconductors together. At the junction between the two materials, mobile electrons and holes annihilate each other, leaving behind the fixed + and - charges of the electron donor and electron acceptor dopants, respectively. For example, on the n-side of a silicon p-n junction, the positively charged dopants are P+ ions and on the p-side the negatively charged dopants are B-. The presence of these uncompensated electrical charges creates an electric field, the built-in field of the p-n junction. The region that contains these charges (and a very low density of mobile electrons or holes) is called the depletion region.
The electric field, which is created in the depletion region by electron-hole recombination, repels both the electrons (on the n-side) and holes (on the p-side) away from the junction. The concentration gradient of electrons and holes, however, tends to move them in the opposite direction by diffusion. At equilibrium, the flux of mobile carriers is zero because the field-driven migration flux is equal and opposite to the concentration-driven diffusion flux.
When p-type and n-type semiconductors are joined, electrons and holes are annihilated at the interface, leaving a depletion region that contains positively and negatively charged donor and acceptor atoms, respectively. At equilibrium, the Fermi level (EF) is uniform throughout the junction. EF lies just above the valence band on the p-type side of the junction and just below the conduction band on the n-type side.
The width of the depletion layer depends on the screening length in the semiconductor, which in turn depends on the dopant density. At high doping levels, the depletion layer is narrow (tens of nanometers across), whereas at low doping density it can be as thick as 1 µm. The depletion region is the only place where the electric field is nonzero, and the only place where the bands bend. Elsewhere in the semiconductor the field is zero and the bands are flat.
In the middle of p-n junction, the Fermi level energy, EF, is halfway between the valence band, VB, and the conduction band, CB, and the semiconductor is intrinsic (n = p = ni)
7.1.06: Diodes LEDs and Solar Cells
Diodes are semiconductor devices that allow current to flow in only one direction. Diodes act as rectifiers in electronic circuits, and also as efficient light emitters (in LEDs) and solar cells (in photovoltaics). The basic structure of a diode is a junction between a p-type and an n-type semiconductor, called a p-n junction. Typically, diodes are made from a single semiconductor crystal into which p- and n- dopants are introduced.
Closeup of a diode, showing the square-shaped semiconductor crystal (black object on left) (John Maushammer, Wikipedia, CC-BY-SA)
If the n-side of a diode is biased at positive potential and the p-side is biased negative, electrons are drawn to the n-side and holes to the p-side. This reinforces the built in potential of the p-n junction, the width of the depletion layer increases, and very little current flows. This polarization direction is referred to as "back bias." If the diode is biased the other way, carriers are driven into the junction where they recombine. The electric field is diminished, the bands are flattened, and current flows easily since the applied bias lowers the built-in potential. This is called "forward bias."
Electrons (red) and holes (white) in a forward-biased diode. (S-kei. Wikipedia, CC-BY-SA)
The figure on the left illustrates a forward-biased diode, through which current flows easily. As electrons and holes are driven into the junction (black arrows in lower left figure), they recombine (downward blue arrows), producing light and/or heat. The Fermi level in the diode is indicated as the dotted line. There is a drop in the Fermi level (equal to the applied bias) across the depletion layer. The corresponding diode i-V curve is shown on the right. The current rises exponentially with applied voltage in the forward bias direction, and there is very little leakage current under reverse bias. At very high reverse bias (typically tens of volts) diodes undergo avalanche breakdown and a large reverse current flows.
Diode i-V curve
A light-emitting diode or LED is a kind of diode that converts some of the energy of electron-hole recombination into light. This radiative recombination process always occurs in competition with non-radiative recombination, in which the energy is simply converted to heat. When light is emitted from an LED, the photon energy is equal to the bandgap energy. Because of this, LED lights have pure colors and narrow emission spectra relative to other light sources, such as incandescent and fluorescent lights. LED lights are energy-efficient and thus are typically cool to the touch.
Light-emitting diode (LED). (S-kei. Wikipedia, CC-BY-SA)
Direct-gap semiconductors such as GaAs and GaP have efficient luminescence and are also good light absorbers. In direct gap semiconductors, there is no momentum change involved in electron-hole creation or recombination. That is, the electrons and holes originate at the same value of the momentum wavevector k, which we encountered in Ch. 6. k is related to the momentum (also a vector quantity) by p = hk/2π. In a direct-gap semiconductor, the top of the valence band and the bottom of the conduction band most typically both occur at k = 0. Since the momentum of the photon is close to zero, photon absorption and emission are strongly allowed (and thus kinetically fast). Polar semiconductors such as GaAs, GaN, and CdSe are typically direct-gap materials. Indirect-gap semiconductors such as Si and Ge absorb and emit light very weakly because the valence band maximum and conduction band minimum do not occur at the same point in k-space. This means that a lattice vibration (a phonon) must also be created or annihilated in order to conserve momentum. Since this "three body" (electron, hole, phonon) process has low probability, the radiative recombination of electrons and holes is slow relative to non-radiative decay - the thermalization of electron-hole energy as lattice vibrations - in indirect-gap semiconductors. The momentum selection rule thus prevents light absorption/emission and there are no pure Si LEDs or Si-based lasers.
Prof. Shuji Nakamura holding a blue LED.
While red, orange, yellow, and green LEDs can be fabricated relatively easily from AlP-GaAs solid solutions, it was initially very difficult to fabricate blue LEDs because the best direct gap semiconductor with a bandgap in the right energy range is a nitride, GaN, which is difficult to make and to dope p-type. Working at Nichia Corporation in Japan, Shuji Nakamura succeeded in developing a manufacturable process for p-GaN, which is the basis of the blue LED. Because of the importance of this work in the development of information storage (Blu-Ray technology) and full-spectrum, energy-efficient LED lighting, Nakamura shared the 2014 Nobel Prize in Physics with Isamu Asaki and Hiroshi Amano, both of whom had made earlier contributions to the development of GaN diodes.
A Solar cell, or photovoltaic cell, converts light absorbed in a p-n junction directly to electricity by the photovoltaic effect. Photovoltaics is the field of technology and research related to the development of solar cells for conversion of solar energy to electricity. Sometimes the term solar cell is reserved for devices intended specifically to capture energy from sunlight, whereas the term photovoltaic cell is used when the light source is unspecified.
Photovoltaic effect in a semiconductor p-n junction. (S-kei. Wikipedia, CC-BY-SA)
Photocurrent in p-n junction solar cells flows in the diode reverse bias direction. In the dark, the solar cell simply acts as a diode. In the light, the photocurrent can be thought of as a constant current source, which is added to the i-V characteristic of the diode. The relationship between the dark and light current in a photovoltaic cell is shown in the diagram at the left.
Current-voltage characteristic of a solar cell in the dark and under illumination with band gap light. The short-circuit photocurrent is indicated as isc, and the open-circuit photovoltage is Vphoto. The maximum power generated by the solar cell is determined by the area of the orange box.
The built-in electric field of the p-n junction separates e- h+ pairs that are formed by absorption of bandgap light in the depletion region. The electrons flow downhill, towards the n-type side of the junction, the holes flow uphill towards the p-side. If hν ≥ Egap, light can be absorbed by promoting an electron from the valence band to the conduction band. Any excess energy is rapidly thermalized. Light with hν > Eg thus can store only Eg worth of energy in an e- h+ pair. If light is absorbed outside of depletion region, i.e., on the n- or p-side of the junction where there is no electric field, minority carriers must diffuse into the junction in order to be collected. This process occurs in competition with electron-hole recombination. Because impurity atoms and lattice defects make efficient recombination centers, semiconductors used in solar cells (especially indirect-gap materials such as Si, which must be relatively thick in order to absorb most of the solar spectrum) must be very pure. Most of the cost of silicon solar cells is associated with the process of purifying elemental silicon and growing large single crystals from the melt.
In the photodiode i-V curve above, Vphoto is typically only about 70% of the bandgap energy Egap. The photocurrent is limited by the photon flux, the recombination rate, and the re-emission of absorbed light.[6] The area of the orange rectangle indicates the power generated by the solar cell, which can be calculated as P = i x V. In good single crystal or polycrystalline solar cells made of Si, GaAs, CdTe, CuInxGa1-xSe2, or (CH3NH3)PbI3 the quantum yield (the ratio of short circuit photocurrent to photon flux) is close to unity.
The equivalent circuit of a p-n junction solar cell, which results in the "light" i-V curve shown in the figure above. The solar cell is effectively a diode with a reverse-bias current source provided by light-generated electrons and holes. The shunt resistance (Rsh) in the equivalent circuit represents parasitic electron-hole recombination. A high shunt resistance (low recombination rate) and low series resistance (Rs) are needed for high solar cell efficiency.
Solar cells have many current applications. Individual cells are used for powering small devices such as electronic calculators. Photovoltaic arrays generate a form of renewable electricity, particularly useful in situations where electrical power from the grid is unavailable such as in remote area power systems, Earth-orbiting satellites and space probes, remote radio-telephones and water pumping applications. Photovoltaic electricity is also increasingly deployed in grid-tied electrical systems.
The cost of installed photovoltaics (calculated on a per-watt basis) has dropped over the past decade at a rate of about 13% per year, and has already reached grid parity in Germany and a number of other countries.[7] Photovoltaic grid parity is anticipated in U.S. power markets in the 2020 timeframe.[8] A major driver in the progressively lower cost of photovoltaic power is the steadily increasing efficiency of solar cells, which is shown in the graphic at the right. Higher efficiency solar cells require less area to deliver the same amount of power, and this lowers the "balance of system" costs such as wiring, roof mounting, etc., which scale as the area of the solar panels. Progress towards higher efficiency reflects improved processes for making photovoltaic materials such as silicon and gallium arsenide, as well as the discovery of new materials. Silicon solar cells are a mature technology, so they are now in the flat part of the learning curve and are approaching their maximum theoretical efficiencies. Newer technologies such as organic photovoltaics, quantum dot solar cells, and lead halide perovskite cells are still in the rising part of the learning curve.
Reported timeline of solar cell energy conversion efficiencies since 1976 (National Renewable Energy Laboratory)
A field effect transistor (FET) is a transistor that uses an electric field to control the width of a conducting channel and thus the current in a semiconductor material. It is classified as unipolar transistor, in contrast to bipolar transistors.
Field effect transistors function as current amplifiers. The typical structure of Si-based FETs is one in which two n-type regions (the source and the drain) are separated by a p-type region. An oxide insulator over the p-type region separates a metal gate lead from the semiconductor. This structure is called a metal-oxide-semiconductor FET (or MOSFET). When voltage is applied between source and drain, current cannot flow because either the n-p or the p-n junction is back-biased. When a positive potential is applied to the gate, however, electrons are driven towards the gate, and locally the semiconductor is "inverted" to n-type. Then the current flows easily between the n-type source and drain through the n-channel. The current flow between the source and drain is many times larger than the current through the gate, and thus the FET can act as an amplifier. Current flow can also represent a logical "1," so FETs are also used in digital logic.
Cross section of an n-type MOSFET
In electronic devices such as microprocessors, field-effect transistors are kept in the off-state most of the time in order to minimize background current and power consumption. The FET shown above, which has n-type source and drain regions, is called an NMOS transistor. In a PMOS transistor, the source and drain regions are p-type and the gate is n-type. In CMOS (complementary metal-oxide semiconductor) integrated circuits, both NMOS and PMOS transistors are used. CMOS circuits are constructed in such a way that all PMOS transistors must have either an input from the voltage source or from another PMOS transistor. Similarly, all NMOS transistors must have either an input from ground or from another NMOS transistor. This arrangement results in low static power consumption.
Transistors are most useful in the range of gate voltage (indicated by the red circle in the figure at the left) where the source-drain current changes rapidly. In this region it is possible to effect a large change in current between source and drain when a small signal is applied to the gate. An important figure of merit for FETs is the subthreshold slope, which is the slope a plot of log(current) vs. Vgate. An ideal subthreshold slope is one decade of current per 60 mV of gate bias. Typically, a decade change in source-drain current can be achieved with a change in gate voltage of ~70 mV. The performance of FETs as switches and amplifiers is limited by the subthreshold slope, which in turn is limited by the capacitance of the gate. It is desirable to have a very high gate capacitance, which requires a thin insulating oxide, but also to have a small leakage current, which requires a thick oxide. A current challenge in the semiconductor industry is to continue to scale FETs to even smaller nanoscale dimensions while maintaining acceptable values of these parameters. This is being done by developing new gate insulator materials that have higher dielectric constants than silicon oxide and do not undergo redox reactions with silicon or with metal gate leads. Only a few known materials (such as hafnium oxynitride and hafnium silicates) currently meet these stringent requirements. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/07%3A_The_Crystalline_Solid_State/7.01%3A_Molecular_Orbitals_and_Band_Structure/7.1.05%3A_Semiconductor_p-n_Junctions.txt |
Amorphous semiconductors are disordered or glassy forms of crystalline semiconductor materials. Like non-conducting glasses, they are network structures with primarily covalent bonding. Crystalline silicon, which has the diamond structure, is an ordered arrangement of fused six-membered silicon rings, all in the "chair" conformation, as we saw in Ch. 8. The local bonding environment of the silicon atoms is tetrahedral. The silicon atoms in amorphous silicon (a-Si) are also predominantly tetrahedrally coordinated, but there is no long-range order in the structure. In addition to six-membered rings, there are five- and seven-membered rings, as well as some "dangling bond" sites in which Si atoms have only three nearest neighbors.
Schematic illustration of the structures of crystalline silicon (left), amorphous silicon (middle), and amorphous hydrogenated silicon (right)
Two of the most widely studied amorphous semiconductors are a-Si and amorphous selenium, a-Se. Si and Se can both be made in glassy form, usually by sputtering or evaporation at relatively low temperature. In a-Se, as in a-Si, locally, most of the atoms have their "normal" valence, but there are many defects and irregularities in the structure. Dangling bonds in amorphous semiconductors have orbital energies in the middle of gap, and electrons in these states are effectively non-bonding. Because these dangling bond sites are far apart from each other, there is little orbital overlap between them, and they also exist over a range of energies. Electrons in these mid-gap states are therefore localized, a phenomenon known as Anderson localizaton. Amorphous Si is insulating because electrons the Fermi level (in the middle of the gap) are not mobile in the lattice. These localized states create a mobility gap, and only electrons in states that are strongly bonding or antibonding are delocalized. Therefore, unmodified a-Si is not very useful as a semiconductor. However, by hydrogenating the material as it is formed (typically in a plasma of H atoms), the under-coordinated Si atoms are bonded to hydrogen atoms. This generates filled bonding and empty antibonding orbitals, the energies of which are outside the mobility gap. Hydrogenation thus lowers the density of states in the mobility gap. Hydrogenated amorphous silicon (a-Si:H) is insulating in the dark, but is a good photoconductor because light absorption creates electrons and holes in mobile states that are outside the mobility gap.
Energy vs. DOS for an amorphous semiconductor. Disorder and dangling bonds result in localized mid-gap states.
The photoconductivity of amorphous Se is exploited in xerography. A conductive drum coated with a-Se, which is insulating, is charged with static electricity by corona discharge from a wire. When the drum is exposed to a pattern of light and dark (the image to be duplicated), the illuminated a-Se areas become conductive and the static charge is dissipated from those parts of the drum. Carbon-containing toner particles then adhere via static charge to the areas that were not exposed to light, and are transferred and bonded to paper to make the copy. The speed of the process and the high resolution of pattern transfer depend on the very low conductivity of a-Se in the dark and its high conductivity under illumination.
Charging of amorphous Se and pattern transfer in the xerographic cycle.
Amorphous hydrogenated Si is used in inexpensive thin film solar cells. The mobility gap is about 1.7 eV, which is larger than the bandgap crystalline of Si (1.1 eV). a-Si:H is a direct-gap material, and therefore thin films are good light absorbers. a-Si:H solar cells can be vapor-deposited in large-area sheets. p+Si-a-Si:H-n+Si cells have around 10% power conversion efficiency. However amorphous Si solar cells gradually lose efficiency as they are exposed to light. The mechanism of this efficiency loss, called the Staebler-Wronski effect,[9], involves photogenerated electron-hole pairs which have sufficient energy to cause chemical changes in the material. While the exact mechanism is still unclear, it has been proposed that the energy of electron-hole recombination breaks a weak Si-Si bond, and that one of the resulting dangling bonds abstracts a H atom, leaving a passivated Si-H center and a permanent dangling bond. The effect is minimized by hydrogenating a-Si and can be partially reversed by annealing.
A calculator that runs on solar and battery power
Thin layers of amorphous silicon are used in conjunction with crystalline silicon in heterojunction intrinsic thin-layer (HIT) solar cells.[10] Because the mobility gap of a-Si is wider than the bandgap of c-Si, there is a potential energy barrier at the amorphous-crystalline interface that reflects electrons and holes away from that interface. At the p+ contact, only holes can tunnel through the barrier, whereas only electrons can tunnel through the barrier to the n+ contact. The passivation of surface defects that are sites of electron-hole recombination prevents a major loss mechanism in solar cells, increasing both the photovoltage and the photocurrent relative to conventional c-Si p-n junction cells. Panasonic and Sanyo have announced the production of HIT cells with power conversion efficiencies as high as 23%.
Layered structure of a HIT solar cell. The layers are not drawn to scale. A thick crystalline n-silicon layer is the light absorber, and photogenerated holes, which are the minority carriers, are reflected away from the aluminum back contact by the thin intrinsic a-Si layer there. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/07%3A_The_Crystalline_Solid_State/7.01%3A_Molecular_Orbitals_and_Band_Structure/7.1.07%3A_Amorphous_Semiconductors.txt |
• How is magnetic ordering in the 3d transition metals (Fe, Co, Ni) and the absence of magnetism in the elements just below them (Ru, Ir, Pd) related to the metal-insulator transition?
• Why are good metals bad superconductors and vice-versa?
• Discuss why semiconducting oxides of early transition metals such as TiO2 and Nb2O5 can be doped n-type but not p-type. Conversely, semiconducting late transition metal oxides such as NiO and Cu2O can be doped p-type but not n-type.
7.1.09: Problems
1. The structure of a high temperature superconductor containing barium, europium, copper, and oxygen is shown below. What are the coordination environments of Cu in this compound? This structure is actually closely related to perovskite, ABO3. Explain the relationship between this structure and the ideal perovskite structure.
2. VO2 can exist in insulating or metallic form, depending on temperature and pressure. Which form would be stabilized by increasing the temperature? Explain your answer.
3. Explain briefly how and why the bandgaps for octet p-block semiconductors vary (1) with the average principal quantum number, and (2) with the electronegativity difference between anion and cation.
4. Indicate the type of conduction (n or p) in the following: (a) Se-doped GaAs, (b) InAs1-x, where x << 1, (c) Li0.05Ni0.95O, (d) LixWO3, where x << 1.
5. The structure of copper indium selenide, a semiconductor used in thin film solar cells, is shown below in sections.
(a) What is the stoichiometry of the compound?
(b) What kind of doping (n or p) will occur if a small amount of iodine is substituted for selenium?
(c) What kind of doping (n or p) will occur if a small fraction of the indium sites are occupied by copper atoms?
6. Using 1 eV = 1240 nm, predict the colors of anatase TiO2 (Eg = 3.1 eV), SiC (2.0 eV), ZnSnP2 (1.7 eV), ZnGeP2 (1.9 eV), and InP (1.27 eV).
7. The conductivity of a certain intrinsic (undoped) semiconductor increases by a factor of two when the temperature is raised from 300 to 330 K. What is the bandgap (in eV)? R = 8.314 J/mol-K, 1 eV/atom = 96.52 kJ/mole.
8. Pure Ge is much more conductive than pure Si. Given their bandgaps (0.74 and 1.15 eV, respectively), estimate the ratio of their conductivities at room temperature.
9. The figure below illustrates the trends in conductivity vs. inverse temperature for Si, Ge, and As-doped Ge. Identify lines (i), (ii), and (iii) with the appropriate materials. Explain why the slope of line (i) is close to zero.
10. Sketch a silicon p-n junction, showing the depletion region, band bending, and the Fermi level in the absence of light or applied potential. In the dark, the p-n junction acts as a rectifier. (a) Which way do electrons and holes flow most easily in the dark? (b) Does the built in electric field increase or decrease under forward bias? (c) In the light, the junction acts as a photodiode. In this case, under short circuit conditions, do electrons flow in the same direction or in the opposite direction as in (a)? Explain.
7.1.10: References
1. Hubbard, J. (1963). "Electron Correlations in Narrow Energy Bands". Proceedings of the Royal Society of London 276 (1365): 238–257. doi:10.1098/rspa.1963.0204. Bibcode: 1963RSPSA.276..238H.
2. T. Mizokawa, Metal-insulator transitions: orbital control, Nature Physics 9, 612–613 (2013), doi: doi:10.1038/nphys2769
3. N. F. Mott, (1961) "The Transition to the Metallic State," Phil. Mag. 6, 287. DOI: 10.1080/14786436108243318.
4. P. P. Edwards and M. J. Sienko (1982) "The Transition to the Metallic State," Acc. Chem. Res. 15, 87-93. DOI: 10.1021/ar00075a004
5. Recent data have disproven this assertion; MacDonalds has finally responded to public opinion and is offering breakfast after 10:30 AM. But the laws of thermodynamics remain immutable and eternal
6. E. Yablonovitch, O. Miller, and S. Kurtz, "Strong Internal and External Luminescence as Solar Cells Approach the Shockley–Queisser Limit," IEEE Journal of Photovoltaics, vol. 2, no. 3, pp. 303-311, July 2012.
7. "Recent facts about photovoltaics in Germany". Fraunhofer ISE. 7 January 2015. Retrieved 17 February 2015.
8. Energy Information Administration, (November 2010). Levelized Cost of New Generation Resources in the Annual Energy Outlook 2011.
9. Staebler, D. L. and Wronski, C. R. Optically induced conductivity changes in discharge-produced hydrogenated amorphous silicon. J. Appl. Physics. 51(6), June 1980.
10. Mishima, T., Taguchi, M., Sakata, H., Maruyama, E., 2011. Development status of high efficiency HIT solar cells. Sol. Energy Mater. Sol. Cell. 95, 18–21. doi:10.1016/j.solmat.2010.04.030 | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/07%3A_The_Crystalline_Solid_State/7.01%3A_Molecular_Orbitals_and_Band_Structure/7.1.08%3A_Discussion_Questions.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.
• 7.7.1: 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.
• 7.7.2: 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.
• 7.7.3: 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.
• 7.7.4: 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.
7.07: Thermodynamics of Ionic Crystal Formation
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
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/Inorganic_Chemistry_(LibreTexts)/07%3A_The_Crystalline_Solid_State/7.07%3A_Thermodynamics_of_Ionic_Crystal_Formation/7.7.01%3A_Lattice_Energy.txt |
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/Inorganic_Chemistry_(LibreTexts)/07%3A_The_Crystalline_Solid_State/7.07%3A_Thermodynamics_of_Ionic_Crystal_Formation/7.7.02%3A_Lattice_Energy_-_The_Born-Haber_cycle.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.
7.7.04: 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/Inorganic_Chemistry_(LibreTexts)/07%3A_The_Crystalline_Solid_State/7.07%3A_Thermodynamics_of_Ionic_Crystal_Formation/7.7.03%3A_Lattice_Enthalpies_and_Born_Haber_Cycles.txt |
In addition to the Lewis, VSEPR, valence bond, molecular orbital, and various acid-base models that are important for understanding chemical structure and bonding, a number of general principles are useful for understanding trends in the properties of the main group elements. These include:
1. The Periodic Law and Periodic Table, most notably the ordering of elements with similar chemical properties by group, the metal-nonmetal distinction, and the second row uniqueness principle.
2. Periodic trends in atomic properties, most notably atomic radii, ionization energy, electron affinity, and electronegativity, on account of which there are diagonal relationships between elements in the periodic table.
3. Periodic trends in elements' redox properties, which can be helpfully illuminated with the aid of Latimer, Frost, and Pourbaix diagrams.
The remaining pages of this section will outline these principles.
8.01: General Trends in Main Group Chemistry
The chemical and physical properties of the main group compounds result from the electronic properties of their constituent atoms and ions. These atomic electronic properties are determined by the number of electrons present and the ground state orbitals they occupy. Of these electrons the chemically most important are the valence electrons and the orbitals they occupy, called valence orbitals. These valence electron counts and orbital occupancies vary systematically in a way that is reflected in the organization of the periodic table. Specifically, the number of valence electrons and the types of atomic orbitals they occupy are determined by the element’s group. This is reflected in the familiar block structure of the periodic table shown in Figure $1$.
From Figure $1$ it can be seen that the main group elements are those with nsx and ns2 npy type valence electron configurations – called s and p block elements. The exact occupancy of the s and p subshells increases with the total number of valence electrons from left to right across the periodic table as shown in Figure $2$.
From Figure $2$ it can be seen that elements within a group of the periodic table have the same configuration of s and p valence electrons. This is one reason why elements within the same group tend to have similar chemical properties.[1] Consequently, it is usually convenient to think and talk about the elements and their properties by group.
Groups may always be referred to by their group number or their first element. In addition, a number have common names that are widely used. These are given in Figure $3$.
Further insight into the properties of the elements may be gained by considering how the systematic variation in electron configuration that occurs across the periodic table's rows interacts with the systematic variations in atomic properties that occur across rows and down groups. A detailed understanding of these properties, how they vary throughout the periodic table, and why they vary in the way they do will be developed in later sections. For now, it is only necessary to know the trends in valence orbital properties, which are:
• The valence orbital size tends to increase down a group as the principle atomic number increases. This increase in orbital size corresponds to lengthening of the average electron-nuclear distance and consequently a weakening of the valence electron-nucleus attraction.
• The valence orbital energy roughly decreases across a row as the effective nuclear charge, Z*, increases. The major factor driving this decrease in energy is the increase in electron-nucleus attraction as Z* increases.
• The valence orbital size roughly decreases across a row as the effective nuclear charge, Z*, increases. This decrease in size corresponds to a shortening of the valence electron-nuclear distance and strengthening of the valence electron-nucleus attraction. This size effect reinforces the aforementioned lowering of valence orbital energies, effective nuclear charge, although in general the lowering of energy due to decreasing size is less important than the lowering of energy due to increasing charge. That is because size decreases gradually from left to right while the effective nuclear charge increases by about 2/3 of a charge unit when moving from group to group.
These trends are summarized in Figure $4$.
These valence orbital property trends affect two different classes of atomic properties. The first class includes the valence orbital size-related properties such as atomic radii and the $\pi$-bonding interaction strength. The second class contains electron-nucleus attraction-related properties comprising ionization energy, electron affinity, and electronegativity.
The consequences of the valence orbital property trends for atomic properties are as follows
1. Atomic size increases down a group as the valence orbital becomes larger (n increases) and roughly decreases across a row as the effective nuclear charge increases.
2. Ionization energy decreases down a group as the valence orbital becomes larger and roughly increases across a row as the effective nuclear charge increases.
3. Electron affinity decreases down a group as the valence atomic orbital becomes larger and roughly increases across a row as the effective nuclear charge increases.
4. Electronegativity decreases down a group as the lowest incompletely occupied atomic orbital becomes larger and roughly increases across a row as the effective nuclear charge increases.
These consequences are summarized in Figure $5$.
These variations in turn explain the trends in main group element properties that are explored in the remainder of this chapter.
[1] Another is that the valence orbital energies tend to vary relatively little down a group, especially as one moves from row three downward.
Contributors and Attributions
Stephen M. Contakes (Westmont College)
8.1.01: The Periodic Table is an Organizing Concept in Main Group Chemistry
Two factors contribute to large-scale and rough variations in the sort of structure and bonding that main group elements engage in.
1. Decreasing valence orbital energy and size on moving from the lower left to the upper right of the periodic table
2. Increasing valence shell occupancy on moving from left to right across the main group
As summarized in Figure $1$, the net effect of these trends is that
Elements in earlier groups with few valence electrons act as metals since they tend to either
• lose them to give cations with an empty valence shell
or
• share them by forming delocalized bonds in clusters and the solid state, although this tendency is less pronounced as one moves towards the top of the periodic table, since elements in earlier groups with more compact lower energy orbitals less readily ionize and more readily form strong covalent bonds (for example, consider the relative stability of alkyllithium and Mg-containing Gringard reagents compared to alkylsodium or calcium reagents)
Elements in later groups with many valence electrons act as nonmetals since they tend to either
• gain additional electrons to form anions with a stable filled valence shell
or
• form stable pairwise sigma bonds in molecular and network covalent solids, although this tendency is less pronounced as one moves down a group and the orbitals become more diffuse. As a result nonmetals exhibit an increased tendency to engage in delocalized and cluster bonding on moving down the periodic table.
These periodic trends in bonding tendencies impact the sort of structures elements form and their physical properties. This is reflected in the classical division of the elements into metals, nonmentals, and metalloids depicted in Figure $2$. As can be seen from Figure $2$, the metals in the lower left of the periodic table are separated from the nonmetals in the upper right by the metal-nonmetal line. Most of the elements adjacent to the line (which ones are chosen varies between periodic tables) exhibit properties intermediate between metals and nonmetals and are called metalloids.
The metals towards lower left of the periodic table form metallic solids held together by delocalized bonding. As a result metals
• are malleable (able to be hammered into sheets) and ductile (able to be pulled into wires), since metals' solid state lattices may be deformed by dislocation movement (slip) without significantly disrupting the bonding.
• exhibit a high thermal conductivity and high electrical conductivity. Since they have an incompletely filled valence band delocalized throughout the structure, it is easy to transmit heat and charge by moving their electrons around.
In contrast, the nonmentals towards the upper right of the periodic table form molecular compounds and network covalent solids. In general these elements tend to
• be liquids, gases, or brittle solids. When present as solids, nonmetals tend to be brittle since deformation involves rupturing covalent bonds in the case of covalent solids or the rupture of intermolecular forces in the case of molecular solids.
• have low thermal and electrical conductivity, since for molecular solids the transfer of energy or electrons via collisions or through space is much much slower and, for network covalent solids held together by strong sigma bonds, involves excitation of electrons across a reasonably large band gap. Note that this is only a tendency. Network covalent solids held together by weaker covalent bonds or $\pi$-bonds sometimes have small or nonexistent band gaps. For example, the graphite allotrope of carbon is an outstanding electrical conductor because its $\pi$-valence band and $\pi$*-conduction band overlap, leaving no band gap.
The metalloids along the metal-nonmetal line form solids that exhibit bonding patterns and properties intermediate between those of metals and nonmetals. This is because is in reality there is a graduation in element properties as one moves from the lower left to the upper right of the periodic table. Consequently, although the distinction between metals, nonmetals, and metalloids can be a helpful one, these categories should not be understood too rigidly.
The same is true of the tendency of metals and nonmetals to form ionic compounds with one another, owing to the tendency of metals to lose electrons and nonmetals to gain them. In practice, there is a graduation in ionic character based on the difference in the elements' electronegativites and the size/polarizability of the atoms involved. As a result many compounds between metals and nonmetals may be profitably thought of as involving polar covalent bonds.
Contributors and Attributions
Stephen M. Contakes (Westmont College) | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/08%3A_Chemistry_of_the_Main_Group_Elements/8.01%3A_General_Trends_in_Main_Group_Chemistry/8.1.01%3A_The_Periodic_Table_is_an_Organizing_Concept_in_Main_Group_Chemistry/8.txt |
The physical and chemical properties of main group compounds depend in part on the electronegativity of the atoms involved.
Electronegativity is somewhat loosely defined as the ability of atoms to attract electrons towards themselves in covalent bonds. When atoms that differ in electronegativity form a covalent bond, the bonding electrons are not shared equally. Instead, the bonding electrons are more strongly attracted towards the more electronegative atom, resulting in a polar bond, as shown in Scheme $\PageIndex{I}$.
Because electronegativity determines the polarity of bonds and by extension molecules, it is one of the key determinants of the chemical and physical properties of main group compounds.
A classic example of the influence of bond polarity on reactivity involves the group IV/14 element hydrides. Carbon hydrides like methane and other alkanes do not react with water, alcohols, and other Brønsted acids, while silanes, germanes, and stannanes do. For example
$\sf{CH_4 + 4HOEt \rightarrow No~~reaction} \nonumber$
$\sf{SiH_4 + 4HOEt \rightarrow Si(OEt)_4 + 4 H_2 } \nonumber$
This difference in reactivity reflects two factors. The first is a kinetic one. The ability of the larger Si atom to more readily add an additional ligand to give a relatively stable trigonal bipyramidal intermediate provides a relatively low energy pathway for the reaction, enabling it to take place quickly.
$\nonumber$
The second and much more important factor that determines the difference in reactivity is thermodynamic and has to do with the polarities of the C-H and Si-H bonds. These polarities depend on the electronegativities of carbon, silicon, hydrogen, and oxygen as depicted in Scheme $\PageIndex{I}$. As can be seen in part a of Scheme $\PageIndex{I}$, because carbon's electronegativity is greater than that of H, the hydrogens of C-H bonds are electrophilic and act as weak Brønsted acids, rendering them unreactive towards other Brønsted acids like alcohols. In contrast, as shown in part b, silicon's electronegativity is lower than that of H, so that the polarization of Si-H bonds facilitates both elimination of the Si-H and O-H as $\sf{H_2}$ and nucleophilic attack of the ethanol O at the Si center.
The Pauling scale is the most commonly used scale to rate electronegativities.
Electronegativity is most commonly quantified using the Pauling scale. The Pauling scale is based on the observation that polar bonds are generally stronger than the average of their homopolar counterparts. This is well illustrated by the bond energies of H2, Cl2, and HCl given in Table $1$. As may be seen from the table, the average of the H-H and F-F bond energy of 295.5 kJ/mol is only 52% of the H-F bond energy of 567 kJ/mol.
Table $1$. Bond energies of H2, Cl2, and HCl.
In this model the greater the bond polarity (larger $\delta^+$ and $\delta^-$ ), the larger the "excess bond strength." Consequently the "excess bond strength"may be used as an empirical measure of the difference between the electronegativities $\chi$ of the elements involved. Pauling did this using equation $2$:
$\nonumber$
where A and B are a given pair of elements.
Equation $2$ may only be used to determine differences in electronegativities. To determine absolute electronegativities, Pauling assigned the electronegativity of the most electronegative element in his scale, F, a value of 4.00, although to accomodate more accurate thermochemical data the Pauling electronegativity of F has since been adjusted to 3.98. Pauling electronegativities are given for the elements in Figure $2$.
Electronegativity generally increases across a row and decreases down a group
As can be seen from the Pauling electronegativity values in Figure $4$, within the main group electronegativity decreases down a group and roughly increases on moving from left to right across a group - i.e., electronegativity roughly increases on moving from the lower left to the upper right of the periodic table.
Two factors modify the overall pattern of increasing electronegativity on moving to the upper right of the periodic table.
The first is the aforementioned ambiguous position of hydrogen. The electronegativity of hydrogen does agree with this overall pattern if hydrogen is placed at the head of the alkali metal group. However, it is over a full unit higher than any other alkali metal and, if placed at the head of the halogen group, would be almost two units lower than that of F.
The second is the reversal of the expected decrease in electronegativity down a group for the pairs Al & Ga and Si & Ge. This is because the post transition metals have anomalously high electronegativities due to the increasing effective nuclear charge associated with the filling of the d block - i.e., a sort of d-block contraction for electronegativity. The net effect is that the post transition elements in the p block all have anomalously high electronegativities. In the cases of Al & Ga and Si & Ge this results in an increase in electronegativity down the group as one moves from row 3 to row 4. In the case of the other post-transition elements it is seen that the movement between these two rows involves a modest decrease in electronegativity.
The electronegativity trends lead to diagonal relationships in which elements just to the left of and down a column to one another have similar chemical properties.
The increase in electronegativity across a row of the periodic table is sometimes compensated for by the decrease in electronegativity down a group, so that elements so diagonally related sometimes have similar electronegativities. Some such diagonally related elements have similar physical and chemical properties, although these similarities may not be due to the effect of electronegativity alone. Since atomic size also decreases across a row and increases down a group, some similarities reflect size effects as well. In any event, the result is that there are recognized diagonal relationships of chemical and physical similarity among elements diagonal to one another in the periodic table. Some of the more prominent diagonal relationships within the main group are shown in Figure $5$.
A few observations illustrate how these diagonal relationships work.
• Li & Mg – Lithium acts like Mg in that it is the only alkali metal to form stable nitride (Li3N), oxide (Li2O), and THF soluble alkyl carbanion salts as alkaline Earth metals do. This may be seen in the similar compounds formed by Li and Mg shown in Table $2$. In contrast, the larger alkali metals form insoluble polymeric carbanion reagents, superoxides or peroxides and not oxides, and extremely unstable nitrides.
Table $2$. Similar compounds formed by the diagonally related elements Li and Mg.
Type of compound Li Mg
oxide $\sf{Li_2O}$ $\sf{MgO}$
nitride $\sf{Li_3N}$ $\sf{Mg_3N_2}$
alkyl carbanion reagent alkyllithium (RLi) Gringard Reagent (RMg+)
• Be & Al – In contrast to the basic oxides and hydroxides of most alkali metals and the heavier group 13 elements, both Be and Al form amphoteric oxides and hydroxides ($Table 3$). Like Be and the remaining alkali metals, aluminum also forms a carbide (C4-“salt”) instead of an acetylide (C22-) or allylide (C34-).
Table $3$. Amphoteric behavior of the oxides and hydroxides of beryllium and aluminum.
Reaction Be Al
Oxide hydrolysis to hydroxide $\sf{BeO + H_2O \rightarrow Be(OH)_2} \nonumber$ $\sf{Al_2O_3 + 3H_2O \rightarrow 2 Al(OH)_3} \nonumber$
Hydroxide as an Arrhenius and Brønsted Base
$\sf{Be(OH)_2 \rightarrow BeO(OH)^+ + OH^-} \nonumber$
or
$\sf{Be(OH)_2 + 2H^+ \rightarrow Be^{2+} + 2H_2O} \nonumber$
$\sf{Al(OH)_3 \rightarrow 2 AlO(OH)_2^+ + OH^-} \nonumber$
or
$\sf{Al(OH)_3 + 3H^+ \rightarrow Al^{3+} + 3H_2O} \nonumber$
Hydroxide as a Lewis Acid $\sf{Be(OH)_2 + 2OH^- \rightarrow Be(OH)_4^{2- }} \nonumber$ $\sf{Al(OH)_3 + OH^- \rightarrow 2 AlO(OH)_4^-} \nonumber$
• B & Si - Both of these elements act as semiconductors, form acidic complex polymeric covalent oxides containing interlinked EO4 tetrahedra, and their halides (BCl3 & SiCl4) act as acid halides. In contrast, the oxides of carbon (CO and CO2) are monomeric, while the oxides of the heavier cogeners of these elements (Al2O3, Ga2O3, In2O3, In2O, Tl2O and GeO2, SnO2, PbO2, and PbO) are considered ionic oxides. The halides of the heavier cogeners of Si more readily act as ionic oxides, while the binary halides of carbon (e.g., compounds like CCl4) are usually not readily hydrolyzed.
The Mulliken and Allred-Rochow Electronegativity Scales connect Electronegativity to other Atomic Properties
While the Pauling scale considers electronegativity in terms of bonding, there are a number of other electronegativity scales that define electronegativity as a function of other atomic properties. Two of these will be considered here: the Mulliken (or Mulliken-Jaffe) scale, which defines electronegativity in terms of electron affinity and ionization energy (and by extension orbital energies), and the Allred-Rochow scale, which considers electronegativity in terms of effective nuclear charges.
The Mulliken Electronegativity Scale defines electronegativity in terms of Electron Affinity and Ionization Energy
The Mulliken definition of electronegativity was previously described in connection with absolute hardness and softness as the first derivative of a species energy with respect to changes in total number of electrons, $N_{e^-}$.
$Mulliken~electronegativity = χ = \dfrac{d E}{dN_{e^-}} \nonumber$
$\sf{where~~E~~is~~in~~eV}$
Since this derivative is difficult to evaluate experimentally, the Mulliken electronegativity is typically approximated as the average of the species' ionization energy and electronegativity.
$Mulliken~~electronegativity = χ ≈ \dfrac{IE + EA}{2} \nonumber$
$\sf{where~~IE~~and~~EA~~are~~in~~eV}$
which, if the IE is approximated by Koopman's theorem as the opposite of the energy of the lowest unoccupied atomic orbital (LUAO) $\sf{(IE = -E_{HOMO})}$ and the EA as the opposite of the energy released on absorption of an electron, may be taken as an approximation of the LUMO energy.
$E_{LUMO} ≈ -EA \nonumber$
and the Mulliken electronegativity is just the average of the HOAO and LUAO energies (~Fermi energy):
$Mulliken~electronegativity, χ ≈ -\dfrac{E_{LUAO} + E_{HOAO}}{2} \nonumber$
$\sf{where~~all~~values~~are~~for~~the~~gas~~phase~~atoms~~in~~units~~of~~eV}$
The Allred-Rochow Electronegativity Scale defines electronegativity in terms of the effective nuclear charge experienced by the valence electrons.
The Allred-Rochow electronegativity scale attempts to quantify electronegativity in terms of the electrostatic force attracting the bonding electrons to the atomic center. It does this by calculating the electronegativity as the associated electrostatic force.
$\sf{\chi_{Allred-Rochow,~raw} = - \dfrac{1}{4\pi \epsilon_0} \dfrac{e^2Z*}{r_{bond}^2}} \nonumber$
These raw values have units of force (N/electron or dyne/electron) and do not correspond closely to electronegativity values reported on the Pauling Scale. Consequently, an empirical correction factor is used to adjust the scale to give electronegativities that more closely correspond to their Pauling equivalents.
$\sf{\chi_{Allred-Rochow} = 3590\dfrac{e^2Z*}{r_{bond}^2}} \nonumber$+0.744
Mulliken electronegativities so calculated are given in Table $4$.
Table $4$: Allred-Rochow Electronegativity Values. Copied from https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Physical_Properties_of_Matter/Atomic_and_Molecular_Properties/Electronegativity/Allred-Rochow_Electronegativity
H
2.20
Li
0.97
Be
1.47
B
2.01
C
2.50
N
3.07
O
3.50
F
4.10
Na
1.01
Mg
1.23
Al
1.47
Si
1.74
P
2.06
S
2.44
Cl
2.83
K
0.91
Ca
1.04
Sc
1.20
Ti
1.32
V
1.45
Cr
1.56
Mn
1.60
Fe
1.64
Co
1.70
Ni
1.75
Cu
1.75
Zn
1.66
Ga
1.82
Ge
2.02
As
2.20
Se
2.48
Br
2.74
Rb
0.89
Sr
0.99
Y
1.11
Zr
1.22
Nb
1.23
Mo
1.30
Te
1.36
Ru
1.42
Rh
1.45
Pd
1.35
Ag
1.42
Cd
1.46
In
1.49
Sn
1.72
Sb
1.82
Te
2.01
I
2.21
Cs
0.86
Ba
0.97
La
1.08
Hf
1.23
Ta
1.33
W
1.40
Re
1.46
Os
1.52
Ir
1.55
Pt
1.44
Au
1.42
Hg
1.44
Tl
1.44
Pb
1.55
Bi
1.67
Po
1.76
At
1.90
These values correlate reasonably well with Pauling electronegativities. One consequence of this good agreement is that both scales are useful for making qualitative predictions about a substance's chemical behavior. For this reason most chemists continue to use the Pauling scale. In this respect, perhaps the greatest value of the Mulliken and Allred-Rochow electronegativity scales for the bench chemist lies in how they connect electronegativity to parameters that may be used to tailor molecular structure and reactivity, namely, substituent effects and oxidation states.
The effective electronegativity of an atom varies increases with oxidation state and in the presence of electron withdrawing substituents
One of the advantages of the Allred-Rochow and Mulliken electronegativity systems over that of Pauling's is that they can consider electronegativity in the context of compounds, in which electron donating and withdrawing groups modify the ability of atomic centers to attract electrons in a bond. As was considered in the section on measures of hardness, the Mulliken scale can be used to define the electronegativity of Lewis acid centers simply by considering the HOMO and LUMO energy in place of the HOAO and LUAO energies.
$Mulliken~electronegativity~of~Lewis~acid, χ ≈ -\dfrac{E_{LUAO} + E_{HOAO}}{2} \nonumber$
$\sf{where~~all~~values~~are~~in~~units~~of~~eV}$
The Allred-Rochow scale is not usually applied directly to molecules. However, its model of electronegativity as a measure of the electrostatic force experienced by the valence electrons may be applied to atoms in molecules, specifically by accounting for the influence of electron donating and withdrawing substituents on the effective nuclear charge an electron experiences. This is not usually done in any sort of rigorous way, and in the context of a typical advanced inorganic chemistry course, the influence of substituent groups on an atom's electronegativity may be estimated in one of two ways:
1. The group electronegativity concept may be used to quantitatively estimate an atom or functional group's electronegativity. A group electronegativity is the electronegativity of an atom that has been modified by attaching it to one or more substituents. These substituents modify the atom's base electronegativity by an amount equal to their empirically-determined substituent constants so that the group electronegativity, $\chi_\sf{group}$, may be calculated.1
$\chi_\sf{group} = \chi_\sf{base} + \sigma_\sf{substituent} \nonumber$
where $\chi_\sf{base}$ and $\sigma_\sf{substituent}$ are the group's base electronegativity and the substituent constant, respectively. Because the substituent constant must be known in order to calculate group electronegativities, the group electronegativity concept has primarily been applied to organic compounds.
2. The effect of substituents on electronegativity may be qualitatively rationalized in terms of inductive effects and oxidation state. In this system
• electron donating groups decrease Z* and consequently a center's effective electronegativity; conversely, by increasing the Z* experienced by an atom's valence electrons, electron withdrawing groups increase its effective electronegativity.
• when an atomic center is oxidized, Z* and its effective electronegativity increase, while when it is reduced, Z* and its effective electronegativity decrease. This effect is commonly manifested as an increase in the acidity of polarized E-OH units as the oxidation state of the central element (E) increases, as illustrated for the inorganic oxyacids in Table $4$.
Table $4$. $pK_a$ vales for selected oxyacids illustrating the increase in Brønsted acidity with central atom oxidation state.2
Acid Central atom oxidation state $pK_a$ ($pK_{a1}$ unless otherwise noted)
H2SO3 SIV 1.9
H2SO4 SVI <0
HNO2 NIII 3.1
HNO3 NV <0
H3PO3 PIII 3.1
H3PO4 PV 1.3
HClO ClI 7.40
HClO2 ClIII 1.94
HClO3 ClV -2.7 (predicted)
HClO4 ClVII ~ -153
Contributors and Attributions
Stephen M. Contakes, Westmont College
All line drawings on this page are by Stephen Contakes and licensed under a Creative Commons Attribution 4.0 International License.
To the extent it is possible to do so, all modified figures on this page made by Stephen Contakes are licensed under a Creative Commons Attribution 4.0 International License. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/08%3A_Chemistry_of_the_Main_Group_Elements/8.01%3A_General_Trends_in_Main_Group_Chemistry/8.1.02%3A_Electronegativity_increases_and_radius_decreases_towards_the_upper_le.txt |
As with electronegativity, the ionization energy of the main group elements (Figure \(1\)) decreases down a group and increases from left to right across the periodic table.
As may be seen in Figure \(1\), ionization energy does not increase steadily from left to right across a group. Instead, there are two discontinuities. These discontinuities are summarized in Figure \(2\).
Briefly, the two discontinuities are as follows:
1. There is a decrease or meager increase in ionization energy that occurs on going from the alkaline earth metals to the boron group, which is caused by the change in the highest occupied orbital from a lower energy ns to a higher energy np subshell.
2. There is a decrease or anomalously modest increase in ionization energy on going from the pnictides to the chalcogenides caused by the presence of Coulombic repulsion in the chalcogenide np4 configuration that is not present in the pnictide np3 configuration.
These discontinuities illustrate how orbital energy and electron configuration effects should be considered when considering the reactivity of the elements.
8.1.04: As may be seen from considering element's redox diagrams main group
The existence of diagonal relationships among elements in the periodic table and the unique properties of the first and second row elements have already been discussed. Thus before beginning a survey of the descriptive chemistry of the elements, it only remains to describe the general trends associated with the redox properties of the elements and to present several conceptual and graphical tools that make it possible to rapidly grasp the general features of the redox chemistry of each element. These are the subject of the present section.
Elements on the left of the periodic table tend to act as reductants; those on the right as oxidants
The ability of the elements to act as oxidants and reductants is usually described by citing the reduction potentials associated with the reactions they undergo in solution. Thus before considering trends in the redox properties of the main group elements, this section will begin by pointing out how redox potentials work and what they represent.
Substances' standard reduction potentials describe the thermodynamic propensity of a substance to undergo reduction.
Redox potentials represent the thermodynamic potential for the substance to undergo reduction relative to the ideal standard hydrogen electrode (SHE).
$2H^+(aq) + 2e^- \rightarrow H_2(g) ~~~~ E^{\circ} = 0.00V \nonumber$
The ability of a substance to act as an oxidant or reductant is described by the relevant standard reduction potential. These correspond to reactions of the form
$\sf{\underset{reductant}{{substance}_{ox}(solv)} ~~+~~} ne^- \rightarrow \underset{oxidant}{{substance}_{red}(solv)} \nonumber$
There are three things to notice about these standard reduction potentials.
1. Standard reduction potentials always corresponds to reduction of an oxidant to give a reductant. Consequently,
• thermodynamically better oxidants have more positive standard reduction potentials
• thermodynamically better reductants correspond to the products of standard reduction reactions with more negative standard reduction potentials (Figure \{1\).
2. Since $\Delta G^{\circ} = -nFE^{\circ}$, standard reduction potentials are a measure of the thermodynamic spontaneity of a reduction relative to the oxidation of H2(g). As such, reduction potentials are useful for determining whether a reaction can occur but are not useful for determining whether a reaction actually will occur under a given set of conditions. The kinetics of the reduction should also be considered. For instance, thermodynamically nitrate is a powerful oxidant in aqueous solution:
$2NO_3^-(aq) + 12H^+ + 10e^- \rightarrow N_2(g) + 6H_2O(l) ~~~~E^{\circ} = 6.229 V! \nonumber$
However, in solution there is a large kinetic barrier for nitrate reduction, and nitrate usually functions as an inert spectator ion even when strong reductants are present.
3. Since reduction potentials are thermodynamic quantities, they depend on the stability of both their oxidized and reduced forms. For reactions involving soluble species in solution, these stabilities in turn depend on the energy of solvation of any solution phase species involved. For this reason, there is no such thing as the potential of a given substance to act as an oxidant or reductant. They should always be specified relative to a given set of conditions.
Most tabulated redox potentials correspond to oxidations and reductions taking place in aqueous solutions. Moreover, for reactions involving the gain or loss of protons (or OH-), they likely correspond to reactions taking place in 1.0 M H+ solution (or, less commonly, 1.0 M OH-). Such potentials may not correspond closely to those taking place in nonaqueous solvents or in water at other pH values. Nevertheless, with these caveats it is possible to recognize rough trends in the redox properties of the elements.
There are three trends in the redox properties of the elements
The redox properties of the elements very roughly follow three general trends:
1. The noble gases are inert and as elements tend not to act as good oxidants or reductants
Otherwise
2. As one moves towards the left of the periodic table, elements tend to act as good reductants, while those towards the right tend to act as increasingly good oxidants.
3. As one moves down a group of the periodic table, elements tend to act as either weaker oxidants or better reductants.
These trends are summarized in Figure $2$.
Since substances' ability to act as oxidants and reductants also depends on the stability of their oxidation and reduction products, these trends should be regarded as at best approximate. As can be seen from the data given in Table $1$, the alkali metals furnish an important exception to the general trend that substances become more reducing towards the lower left of the periodic table. Among the elements included,2 lithium and not cesium is the most powerful reducing agent due to the high stability of the lithium cation in aqueous solution (anomalously high hydration energy). Further, from the potentials given in Table $1$, it can also be seen that on going from K+ to Rb+ and Cs+ the reduction potentials do not decrease but instead remain approximately constant. In this case the decrease in ionization energy throughout this series is compensated for by a set of similar factors.3
Table $1$ . Standard reduction potentials of aqueous alkali metal cations.
Reduction $E^{\circ}$
Li+(aq) + e- $\rightarrow$ Li(s) -3.040
Na+(aq) + e- $\rightarrow$ Na(s) -2.713
K+(aq) + e- $\rightarrow$ K(s) -2.924
Rb+(aq) + e- $\rightarrow$ Rb(s) -2.924
Cs+(aq) + e- $\rightarrow$ Cs(s) -2.923
Not only are there exceptions to the general trends in the redox properties of the free elements, but many elements also exhibit a rich and interesting redox chemistry involving the interconversion of multiple species possessing different oxidation states. In order to make sense of these it will be helpful to make use of the schematic and graphical tools that will be developed in the next sections.
Contributors and Attributions
Stephen Contakes, Westmont College
8.02: What are the main group elements and why should anyone care about them
What are the main group elements?
The main group elements are those in the s and p blocks of the periodic table, as shown in Figure \(1\).
Why should anyone care about the main group elements?
In addition to being interesting in their own right as fundamental constituents of ordinary matter with a rich and interesting chemistry, the main group elements are of tremendous societal importance. As shown in Table \(1\), thirteen of the top 25 chemicals produced in the United States are inorganic main group compounds.
Table \(1\). Top 25 industrial chemicals produced in the US in 2002.2
This chapter's approach to the descriptive chemistry of the main group elements
The aim of this chapter is to introduce you to the descriptive chemistry of the main group elements. In general descriptive chemistry involves presenting information about the chemistry of the elements, as opposed to describing more fundamental concepts, theories, and models. Nevertheless, in presenting this information no attempt will be made to be exhaustive; instead the focus will be on describing important or characteristic features of the chemistry. Moreover, as intimated by the structures shown in Scheme \(\sf{\PageIndex{I}}\), the bonds which hold main group compounds together do not always obey the rules developed in typical general and organic chemistry courses. Consequently, as opportunity presents itself, this description of the chemistry of the elements will also be used to enlarge your understanding of what types of chemical structures are possible or even common.
Scheme \(\sf{\PageIndex{I}}\). Chlorine atoms bridge Al centers in aluminum chloride dimer at left while the structure of pentaborane nonahydride shown at right includes four B-H-B bridge bonds arranged around the open face of a cluster of five B atoms held together by cluster bonds.
The elements will be considered by group after an introductory section that reviews several concepts that help to make sense of periodic trends in the descriptive chemistry of these elements.
Contributor
Stephen Contakes, Westmont College
The unknown authors of Table \(1\) and the periodic table portion of Figure \(1\).
Unless otherwise noted, all line drawings on this page are by Stephen Contakes and licensed under a Creative Commons Attribution 4.0 International License. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/08%3A_Chemistry_of_the_Main_Group_Elements/8.01%3A_General_Trends_in_Main_Group_Chemistry/8.1.03%3A_Ionization_energy_roughly_increases_towards_the_upper_left_of_the_per.txt |
The Alkali Metals
The alkali metals comprise the elements lithium through francium in group 1 of the periodic table, as shown in Figure $\sf{1}$.
Alkali metals are powerful reductants and so do not exist as the free metal in the relatively oxidizing environment at the earth's surface. As a result they are commonly found as the +1 cations in a variety of minerals like rock salt (halite, NaCl) and natron (Na2CO3·10H2O).
Because they occur as minerals, free alkali metals are prepared by electrolytic reduction of their +1 cations. For example, in the Downs process for making sodium, NaCl is electrolyzed to Na(l) and Cl2(g).
$\sf{2~NaCl(l)~~\rightarrow~~2~Na(l)~~+~~Cl_2(g)} \nonumber$
In order for this process to occur rapidly, transport of cations to the cathode and anions to the anode must occur. To permit this, the salt to be electrolyzed is melted.1 In addition, since the alkali metal is reactive the electrolysis must take place in a specialized cell that permits separation of the metal from both the salt and any oxidation products formed at the anode. For example, to prevent the explosively exothermic reaction between the chlorine gas and molten sodium products in the Downs process, specialized Downs cells in which the cathode and anode are carefully separated are employed. A typical way Downs cell is depicted in Figure $\sf{2}$.
As can be seen from the cell in Figure $\sf{2}$, the chlorine gas formed at the anode and sodium metal formed at the cathode are kept separate by allowing Cl2 gas to bubble out of the cell and the sodium metal to collect above the cathode (since sodium's density is lower than that of molten NaCl it floats to the top of the molten sodium chloride).
Alkali metals can also be formed by chemical reduction. For example, potassium can be made by reducing potassium salts with Na, carbide, or carbon according to the following reactions.
$\sf{KCl~~+~~Na~~\rightarrow~~K~~+~~NaCl} \nonumber$
$\sf{2~KX~~+~~CaC_2~~\rightarrow~~2~K~~+~~CaX_2~~+~~C~~(X~=~F,~Cl} \nonumber$
$\sf{K_2CO_3~~+~~2~C~~\rightarrow~~2~K~~+~~3~CO} \nonumber$
Once formed, alkali metals are stored under an inert atmosphere or under hydrocarbon oil to prevent their reoxidation.
Properties
In metallic form, alkali metals possess a body centered cubic (BCC) structure and are silvery solids, as shown in Figure $\sf{3}$.
Like other metals, alkali metals are good conductors of heat and electricity, malleable, and ductile. However, compared to other metals alkali metals have a small number of valence electrons and relatively low effective nuclear charges. As a result the metallic bonds which hold solid and liquid alkali metals together are weaker than those in other metals and they melt and boil at lower temperatures, as illustrated by the melting and boiling points listed in Table $\sf{1}$. Further, as can be seen from the data in Table $\sf{1}$, both the melting and boiling points decrease down the alkali metal group. This decrease in melting and boiling points reflects a decrease in metallic bonding strength as the atomic size (and consequently average electron-nucleus distance) increases down the group.
Table $\sf{1}$. Melting and boiling points of the alkali metals and selected reference substances.3
Substance Melting Point ($^{\circ}$ C) Boiling Point ($^{\circ}$ C)
Alkali Metal
Lithium, Li 181 1347
Sodium, Na 98 883
Potassium, K 64 774
Rubidium, Rb 39 688
Cesium, Cs 28 678
Francium 27 677
Non-alkali Metal
Magnesium 649 1090
Barium 727 1845
Titanium 1660 3287
Iron, Fe 1538 2861
Copper 1083 2567
Water 0 100
benzene 6 80
The relatively low strength metallic bonding in the alkali metals is also reflected in their softness. The metals are so soft that they may be pressed into sheets and cut with an ordinary lab spatula. In fact, spatulas are commonly used to cut appropriate size portions of metal when they are used in the laboratory.
The atomic properties of alkali metals reflect the relatively high energy and large size of their ns valence orbitals. In consequence, they possess larger atomic radii and lower ionization energies than most metals, as may be seen from the data shown in Table $\sf{2}$.
Table $\sf{2}$. Selected atomic properties of the alkali metals and selected reference compounds.3,4
Substance
Atomic radius
(Angstroms)
Ionization energy
(kJ/mol)
Pauling Electronegativity
Alkali Metal
Lithium, Li 1.45 513 0.98
Sodium, Na 1.80 496 0.93
Potassium, K 2.20 419 0.82
Rubidium, Rb 2.35 403 0.82
Cesium, Cs 2.60 376 0.79
Francium. Fr not determined 400 0.7
Non-alkali Metal
Magnesium, Mg 1.50 738 1.31
Barium, Ba 2.15 503 0.89
Titanium, Ti 1.40 658 1.54
Iron, Fe 1.40 759 1.83
Copper, Cu 1.35 745 1.90
Boron, B 0.85 801 2.04
Nitrogen, N 0.65 1402 3.04
As will be explained in the next section, the small ionization energies of the alkali metals is among the factors contributing to their extreme reactivity.
Contributors and Attributions
Stephen Contakes, Westmont College
8.03: Group 1 The Alkali Metals
The chemistry of the alkali metals reflects their tendency to form +1 cations.
The alkali metals tend to form +1 cations. Cation formation is favored by the relatively low ionization energies of the free metal (which makes it easier to form the cation) and the high solvation energy of their cations (which indicates that the cation is thermodynamically stabilized in solution). The variations in ionization energy and solvation energies down the alkali metal group help explain why Li is more reducing than the other alkali metals.1 The relevant ionization energies and hydration enthalpies are given in Table $\sf{1}$. As can be seen from the data in Table $\sf{1}$, the ionization energy of the metal decreases by ~20% on going from Li to Cs while the solvation energy of the cation decreases by ~75% over the same range (with ~2/3 of that decrease taking place on going from Li+ to Na+).
Table $\sf{1}$. Selected thermodynamic properties of alkali metals and their cations.2,3
Alkali Metal
Ionization Energy
(kJ/mol)
Solvation Enthalpy
of the +1 Cation
(kJ/mol)
Lithium, Li 513 -1091
Sodium, Na 496 -515
Potassium, K 419 -405
Rubidium, Rb 403 -321
Cesium, Cs 376 -296
Francium. Fr 400 -263
One result of this greater stability of Li+ in solution relative to that of other alkali metal cations is that Li tends to be the most reducing of the alkali metals, as illustrated by the reduction potentials of the metals given in Figure $\sf{1}$. Nevertheless, as can be seen in Figure $\sf{1}$, the reduction potentials of the other alkali metals do not monotonically increase on going from Na to Cs. Instead, Na is the least reducing metal, while the reduction potentials of K, Rb, and Cs are approximately equal.
Because of their tendency to form cations, alkali metals are effective reducing agents.
Because of their tendency to form cations, alkali metals are highly reducing. All react vigorously with water to give hydroxides.
$\sf{2~M(s)~~+~~2~H_2O(l)~\rightarrow~2~M^+~~+~~2~OH^-~~+~~H_2(g)~~~~~~(M~=~= Li,~Na,~K,~Rb,~Cs)} \nonumber$
The reactions are all exothermic and often lead to combustion of the evolved H2 gas. As a result, while Li usually just fizzes as it reacts, Na and K usually result in the formation of a colored flame, while the reactions of Rb and Cs result in an explosion.
More synthetically useful is the reduction of alkyl halides by Li metal to give organolithium reagents.
$\sf{R-Cl~~+~~2~Li(s)~~\overset{THF}{\longrightarrow}~~RLi~~+~~LiCl} \nonumber$
Under conditions where the metal cation is stabilized in the absence of an oxidizing agent, it is even possible to form salts of free or solvated electrons. Small amounts of sodium may be dissolved in dry donor solvents (ammonia, tertiary amines, hexamethylphosphoramide) to give paramagnetic blue solutions containing solvated electrons.
$\sf{Na(s)~~\overset{NH_3}{\longrightarrow}~~Na(NH_3)_6^+~~+~~e^-(solv.)~~~~(at~low~Na^+~concentrations)} \nonumber$
At higher concentrations of sodium, bronze solutions containing sodide anions (e.g. M-) are formed.
$\sf{2~Na(s)~~\overset{NH_3}{\longrightarrow}~~Na(NH_3)_6^+~~+~~Na^-(solv.)~~~~(at~high~Na^+~concentrations)} \nonumber$
As either electrides or sodides, the solutions are excellent reducing agents. In some cases, they can even be used to reduce metals to negative oxidation states.
$\sf{Na(NH_3)_6^+~~+~~e^-(solv.)~~+~~Au(s)~~\longrightarrow~~Na(NH_3)_6^+~~+~~Au^-(solv.)} \nonumber$
In the presence of a catalyst the solvated electron decomposes by reducing the ammonia.
$\sf{2~e^-(solv.)~~+~~2~NH_3~~\overset{catalyst}{\longrightarrow}~~2~NH_2^-~~+~~H_2} \nonumber$
Synthetically, sodium-ammonia solutions are useful for preparing acetylides and reduced organometallic complexes.
$\sf{R-C≡C-H~~+~~Na^+(solv.)~~+~~e^-(solv.)~~\longrightarrow~~R-C≡C:^-Na^+~~+~~{\textstyle \frac{1}{2}}~H_2} \nonumber$
$\sf{Fe(CO)_5~~+~~2~Na^+(solv.)~~+~~2~e^-(solv.)~~\longrightarrow~~(Na^+)_2[Fe(CO)_5^{2-}]~~+~~{\textstyle \frac{1}{2}}~H_2} \nonumber$
Solid electride salts - i.e., in which the anion is a free electron - may even be prepared by adding a ligand that encapsulates and stabilizes the metal cation and then allowing the solvent to evaporate away from the solvated electron.
$\sf{Na(NH_3)_6^+~~+~~e^-(solv.)~~+~~encapsulating~ligand~~\underset{-6~NH_3}{\longrightarrow}~~Na(encapsulating~ligand)^+e^-(s)} \nonumber$
Most of the resulting solid electride salts are only stable at low temperatures. The electron density of the free electron "counterions" in crystalline electrides is thought to be localized in cavities within the structure.
The anion of sodium, sodide, Na-, can be formed by reducing the alkali metal under suitable conditions. Among the more remarkable alkali metal anion salts is "inverse sodium hydride." Inverse sodium hydride is formally an H+ salt of Na-, in which the H+ ion is encapsulated in an adamanzane cage (Figure $\sf{2}$). This encapsulation slows the thermodynamically favorable reduction of H+ by the sodide anion, stabilizing the inverse sodium hydride long enough to permit its isolation.
Scheme $\sf{\PageIndex{I}}$. "Inverse sodium hydride" (i.e. H+Na-).
The chemistry of the alkali metals depends on the size of the alkali metal cations.
The stability of alkali metal compounds depends on the stabilization of their +1 cations. This stability in turn depends on the size of the cations. Alkali metal cation size helps determine the structure and in some cases the identity of the compound that a given alkali metal forms with a given nonmental. An example of this involves the structures formed by the alkali metal halides. The halides of smaller alkali metal cations -- Li+, Na+, K+, and Rb+ -- crystallize in the rock salt structure with six halide ions about each ion (Figure $\sf{\PageIndex{2A}}$). In contrast, the chlorides, bromides, and iodides of the larger Cs+ ion crystallize in the CsCl structure, in which there are eight halides about the larger Cs+ ion (Figure $\sf{\PageIndex{2B}}$).
Size effects are also believed to be responsible for alkali metals' unusual tendency to form peroxides and superoxides when burned in an excess of oxygen. While most metals form oxides when burned in an excess of oxygen, Li and Na also form 2:1 salts of the larger peroxide ion (O22-) and K, Rb, and Cs 1:1 salts of the similarly large superoxide ion (O2-) (Figure $\sf{3}$).
Perhaps the most striking examples of size effects in alkali metal chemistry involve alkali metal complexes with macrocyclic ligands, particularly crown ethers and cryptands like those shown in Figure $\sf{4}$.
Crown ethers and cryptands prefer to bind alkali metal cations with sizes that match that of their binding cavity. For instance, as shown in Table $\sf{2}$, 14-crown-4 preferentially binds to Li+, 15-crown-5 preferentially binds to Na+, 18-crown-6 preferentially binds to K+, and 21-crown-7 preferentially binds to Cs+.
Table $\sf{2}$. Crown ethers preferentially bind cations with sizes that match the cavity size of their crown-shaped conformers. Cavity and cation diameters are taken from reference 10.
Crown ether
...with a cavity diameter of
(in $\mathring{\mathrm{A}}$ )
...preferentially binds
...which has a cation diameter of
(in $\mathring{\mathrm{A}}$ )
1.2-1.5 Li+ 1.36
1.7-2.2 Na+ 1.94
2.6-3.2 K+ 2.66
3.4-4.3 Cs+ 3.34
Exercise $1$
Cryptand ligands also preferentially bind alkali metal ions with sizes that match the size of their binding cavities. The structure and approximate cavity sizes of several cryptands are given in Scheme $\sf{\PageIndex{II}}$. Use the information in Scheme $\sf{\PageIndex{II}}$ and Table $\sf{2}$ along with a cation diameter of 2.98 Å for Rb+ to predict which alkali metal ion each cryptand will preferentially bind.
Scheme $\sf{\PageIndex{II}}$. Structures and approximate binding cavity diameters for several cryptands.
Answer
The cryptands might be expected to selectively bind the largest ion which fits within the cavity. These predictions are summarized in Table $\sf{3}$.
Table $\sf{3}$. Predicted selectivity of crown ethers for alkali metal cations based on the hypothesis that they will selectively bind the largest ion which fits their binding cavity.
Cryptand
...with a cavity diameter of
(Å)
...will be selective for
...with a radius of
(in Å)
2.1.1-cryptand 1.60 Li+ 1.36
2.1.1-cryptand 2.20 Na+ 1.96
2.1.1-cryptand 2.80 K+ 2.66
2.1.1-cryptand 3.60 Cs+ 3.34
These predicted selectivities match the experimental observations with one exception. It turns out that 2.1.1-cryptand preferentially binds the smaller Rb+ ion over Cs+, although the difference in binding constants is a small one (for more details see reference 12). One reason for this is the flexibility of the cryptand ring system, which allows the cryptand oxygen and nitrogen Lewis base sites to adjust position to accommodate the needs of each alkali metal. In fact, because of this flexibility, all four cryptands can bind all the alkali metals, even those that are nominally too large to fit the cryptand's optimal cavity diameter.
The ability of flexibile macrocycles like cryptands and crown ethers to change conformation when binding a Lewis acid means that the cavity size estimates given in Scheme$\sf{\PageIndex{II}}$ are of limited utility in predicting the relative binding strengths.
Contributors and Attributions
Stephen Contakes, Westmont College | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/08%3A_Chemistry_of_the_Main_Group_Elements/8.03%3A_Group_1_The_Alkali_Metals/8.3.01%3A_Alkali_Metals%27_Chemical_Properties.txt |
Hydrogen Isotopes
Elemental hydrogen exists primarily as the hydrogen-1 isotope, although hydrogen-2 and hydrogen-3 are also known. Some properties of these isotopes are given in Table \(\sf{1}\).
Table \(\sf{1}\). Atomic properties of hydrogen isotopes. Adapted slightly from https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_Chemistry_(Averill_and_Eldredge)/21%3A_Periodic_Trends_and_the_s-Block_Elements/21.2%3A_The_Chemistry_of_Hydrogen
Protium Deuterium Tritium
symbol \(\sf{^1_1H}\) \(\sf{^2_1H}\) or D \(\sf{^3_1H}\) or T
neutrons 0 1 2
mass (amu) 1.00783 2.0140 3.01605
abundance (%) 99.9885 0.0115 ~10−17
half-life (years) 12.32
boiling point of X2 (K) 20.28 23.67 25
melting point/boiling point of X2O (°C) 0.0/100.0 3.8/101.4 4.5/?
Notable features NMR active with a spin of \(\sf{1/2}\) NMR active with a spin of \(\sf{1}\)
Contributors and Attributions
Stephen Contakes, Westmont College
8.04: Hydrogen
Chemical Properties
As described in Section 8.1.1.2., hydrogen and helium are distinguished from all other elements in that their valence shell only consists of the 1s orbital. In the case of neutral atomic hydrogen, this orbital is occupied by one electron. Consequently, the chemistry of hydrogen is distinguished by stable bonding arrangements in which the 1s orbital is "filled" by one of the following:
• loss of an electron to give hydrogen ion, H+. In condensed phases these H+ ions are typically stabilized as Lewis base adducts (e.g., in species like H3O+ and NH4+)
• gain of an electron to give hydride ion, H-. This type of bonding adequately explains the behavior of many metal hydrides.
• pairwise sharing of electrons to give covalent E-H bonds that can adequately be described by Lewis theory.
• multicenter sharing of electrons in multicenter covalent bonds, such as those in hydrides that bridge two or more atoms.
• contributing electrons and orbitals to the band structure of a solid state lattice. This is common in interstitial/metallic hydrides.
The first four of these possibilities are summarized in Scheme $\sf{\PageIndex{I}}$.
Scheme $\sf{\PageIndex{I}}$. Common bonding arrangements for hydrogen. The E-H and E---E bonds in the bridging hydride represent sharing of two or more electrons among the three atoms.
Elemental Hydrogen
At room temperature and pressure, elemental hydrogen exists in the form of dihydrogen, H2. Dihydrogen is a colorless, odorless gas that finds wide industrial application.
Preparation
In the laboratory H2 may be prepared by the electrolytic or chemical reduction of water involving the half reaction.
$\sf{2H_2O + 2e^- \rightarrow H_2 + OH^-} \nonumber$
Such reductions are commonly carried out on acidic solutions since the potentials required are lower, as may be seen from the Pourbaix diagram for hydrogen given in Figure $\sf{2}$.
A common method is to add Zn to a solution of hydrochloric acid.
$\sf{2~H^+(aq)~~+~~Zn(s)~\rightarrow H_2(g)~~+~~Zn^{2+}(aq) } \nonumber$
In electrolysis the electrons come from the oxidation of water at the anode so that hydrogen production involves water splitting.
$\sf{2~H_2O(l)~~ \rightarrow~~2~H_2(g)~~+~~O_2(g)} \nonumber$
Industrially it is more common to produce hydrogen via steam reforming of methane and other hydrocarbons.
$\sf{CH_4(g)~~+~~H_2O(g)~\overset{Ni}{\longrightarrow}~CO(g)~~+~~3~H_2(g)~~~~(steam~reforming)} \nonumber$
In this process the formally C4- of methane acts as the reductant. The product of steam reforming is a mixture of CO and H2.. Similar mixtures can also be produced by the anaerobic thermal decomposition of organics and in coal gasification reactions. In all cases they are called syngas (i.e., synthesis gas) since they can be used in other industrial syntheses. Its CO component is capable of acting as a reductant, so additional hydrogen can be produced from it via the water-gas shift reaction.
$\sf{CO(g)~~+~~H_2O(g)~\rightarrow~CO_2(g)~~+~~H_2(g)~~~~(water~gas~shift)} \nonumber$
The "cracking" of hydrocarbons also serves as a source of industrially-produced hydrogen, although the alkenes so produced are perhaps even more important as the source of a majority of commodity organic chemicals.
Much of the hydrogen produced industrially is consumed in the Haber-Bosch synthesis of ammonia. The quantities involved are such that hydrogen production for this purpose has been variously estimated to account for 1-2% of global energy consumption.
$\sf{N_2(g)~~+~~3~H_2(g)~\rightarrow~2NH_3(g)~~~~(Haber-Bosch~process)} \nonumber$
Dihydrogen has also been considered for use as a fuel since its combustion is both highly exothermic and green, giving rise only to water.
$\sf{2~H_2(g)~~+~~O_2(g)~\rightarrow~2H_2O(g)} \nonumber$
One of the major obstacles to the implementation of hydrogen as a fuel is that its production by steam reforming and the water-gas shift reaction generates CO and CO2, and costs more energy than is gained from its combustion. For this reason, photocatalytic water splitting is considered one of the holy grails of energy research.
Compounds of hydrogen
Compounds of hydrogen are called hydrides whether or not they contain hydride anion. There are three main types of hydrides - ionic, covalent, and interstitial hydrides. As shown in the periodic table of hydrides given in Figure $\sf{3}$, interstitial or metallic hydrides are formed by some transition metals, while ionic hydrides are mainly formed by more electropositive metals and covalent hydrides by the nonmetals. The hydrides of Be, some metalloids, and some post-transition metals are said to be intermediate hydrides since they form network covalent structures (sometimes in addition to molecular ones) and tend to function as bases and hydride donors like the ionic hydrides. Not all of the transition metals are known to form hydrides. No hydrides are known for the transition metals of groups 7-9, which are said to constitute the hydride gap.
Ionic hydrides (a.k.a. saline hydrides)
Ionic hydrides are metal salts of the hydride anion, H-. These are formed by the alkali metals and all the alkaline earth metals except Be. These are typically prepared by direct reaction of the metal and hydrogen.
$\sf{2~M(s)~~+~~H_2(g)~~\overset{\Delta}{\longrightarrow}~~2~MH(s)~~~~~(M~=~Li,~Na,~K,~Rb)} \nonumber$
$\sf{M(s)~~+~~H_2(g)~~\overset{\Delta}{\longrightarrow}~~2~MH_2(s)~~~~~(M~=~Mg,~Ca,~Sr,~Ba)} \nonumber$
As salts of H-, ionic hydrides form ionic lattices (the NaCl structure is common for MH, the Rutile and PbI2 for MH2).
Chemically ionic hydrides act as
• reducing agents towards metal oxides. For example
$\sf{2CaH_2(s)~~+~~TiO_2 (l)~\rightarrow~2CaO(s)~~+~~Ti(s)~~+~~2~H_2(aq)} \nonumber$
• strong bases towards protic E-H bonds. All react exothermically with water to liberate hydrogen gas.
$\sf{MH(s)~~+~~H_2O(l)~\rightarrow~M^+(aq)~~+~~H_2(g)~~+~~OH^-(aq)} \nonumber$
For this reason CaH2 is widely used as a drying agent for organic solvents.
Reactive metal hydrides may also be used to deprotonate reactive C-H bonds:
$\sf{NaH(s)~~+~~CH_3C \equiv C-H(g)~\rightarrow~CH_3C \equiv C:^-Na^+~~+~~H_2(g)} \nonumber$
The H:- ion in an ionic hydride may in principle act as a nucleophile. However, in practice this application is limited to the less reactive and consequently more selective hydrides of aluminum and boron, both of which are usually classified as intermediate hydrides on account of the covalent character of their E-H bonds.
Covalent and intermediate hydrides
Covalent molecular hydrides are formed by the nonmetals, metalloids, and many post-transition metals. The chemical and physical properties they possess vary across the main group and depend somewhat on the row and whether the element hydride is electron deficient, electron rich, or electron precise. Specifically,
Electron deficient hydrides are those of Be and the group 13 elements (B, Al, Ga, In, and Tl) for which the neutral monomeric element hydride (BeH2, BH3, AlH3, GaH3, InH3, and TlH3 does not possess enough electrons to satisfy the octet rule. Thus these hydrides commonly form dimers (B, Al, Ga, In, Tl) or polymers (Be) held together by bridging E-H-E bonds (Scheme $\sf{\PageIndex{IIA}}$). These E-H-E bonds are explained as three-center two-electron bonds in valence bond theory )Scheme $\sf{\PageIndex{IIB}}$), but may also be described in terms of molecular orbitals (Note $\sf{1}$).
Scheme $\sf{\PageIndex{II}}$. (A) Bridging E-H-E bonds in Al2H6 and (B) their valence bond description in terms of overlap between the H 1s and Al sp3 orbitals.
Electron precise and electron rich hydrides are formed by C, N, O, F, and their heavier cogeners. The E-H bonds in these may be described as classical two-center two-electron E-H bonds of Lewis Theory. The electron precise and electron rich hydrides are distinguished in that the electron rich hydrides possess lone pair electrons, while electron precise hydrides do not. In other words,
• electron precise hydrides are those of the group 14 elements, and include the alkanes, alkenes, and alkynes of carbon along with SiH4, GeH4, SnH4, and PbH4, of which the hydride adducts of group 13 EH3 compounds like BH4- and AlH4- are analogues.
• electron rich hydrides are NH3, H2O,, HF and their heavier analogues (PH3, H2S, HCl, etc.). Regardless of the hydride's classification, the stability of element hydrides decreases down a group. For example, among the group 14 elements, it follows the order CH4 > SiH4 > GeH4 > SnH4 > PbH4. The same is true of compounds possessing E-E bonds so that while a vast number of alkanes are known, there are relatively few silanes, fewer germanes, and only the organic analogs of stannane are known (such as (CH3)3Sn-Sn(CH3)3).
The distinction between electron precise and electron rich hydrides is important mainly in thinking about the Lewis acid-base properties of the element hydrides. As illustrated in Scheme $\sf{\PageIndex{III}}$, electron deficient hydrides tend to function as Lewis acids and electron rich hydrides as Lewis bases and Brønsted acids.
Scheme $\sf{\PageIndex{III}}$. In the absence of extremely strong acids or bases, electron-deficient hydrides like BH3 tend to act as (A) Lewis acids in forming adducts with bases like THF, while electron rich hydrides can act as (B) Lewis bases through their lone pairs, as water does with Cu2+ when anhydrous CuSO4 is dissolved in water, or as (C) Brønsted acids, as water does when it is used to quench the alkoxide product of a nucleophilic addition reaction.
The reactivity of electron precise hydrides depends on the characteristics of E. For example, while most alkanes do not act as Lewis acids at the carbon atom, row 3 and heavier electron precise hydrides can form trigonal bipyramidal adducts.
Moreover, all element hydrides - whether electron deficient, precise, or rich - can function as a weak Brønsted acids or hydride donors depending on the polarity of the E-H bond. Hydrides in which hydrogen is bound to an electron rich and electronegative element tend to act as Brønsted acids, while those involving more electropositive elements tend to function as hydride donors, as illustrated in Scheme $\sf{\PageIndex{IV}}$.
Scheme $\sf{\PageIndex{IV}}$. (A) The electron rich hydride of chlorine acts as a Brønsted acid in forming a pyridinium complex while (B) the relatively electropositive hydride in tetrahydroaluminate is widely used as a hydride donor in organic chemistry, as illustrated by the use of lithium aluminum hydride to form alcohols from ketones.
The ability of a given element hydride to function as an acid or hydride donor may be modified by a number of factors. One of these is the solvation energy of the species formed. According to Figure $\sf{4}$, Germane (GeH4) might be expected to function as a hydride donor. However, it can be deprotonated in liquid ammonia, likely because of the large solvation energy of the resulting H+ ion.
$\sf{GeH_4~~\overset{NH_3(l)}{\longrightarrow}~~H^+(solv.)~~+~~GeH_3^-(solv.)} \nonumber$
Electronic factors that affect the stability of the element hydride's conjugate acid or base forms also play a role. For example, carbon-hydrogen bonds are normally very weakly acidic but can act as strong Brønsted acids when the resulting anion is highly stabilized or as a hydride donor when the resulting cation is highly stabilized. This is illustrated by the well-known ability of C-H bonds to function as Brønsted acids, hydride donors, or neither depending on the electron-richness of the carbon center and the stability of the resulting structure (Scheme $\sf{\PageIndex{V}}$).
Scheme $\sf{\PageIndex{V}}$. (A) enolate chemistry such as that used in the formation of acetylacetate (acac) ligands is based on the ability of C-H bonds $\sf{\alpha}$ to a carbonyl to act as acids, and (B) a C-H bond of NADH functions as a hydride donor in biochemical systems. Notice the similarity in reactivity of the C-H hydride in NADH and the Al-H in LiAlH4 shown in Scheme $\sf{\PageIndex{IV}}$.
Additional aspects of the acid-base chemistry of the element hydrides are described in 6: Acid-Base and Donor-Acceptor Chemistry, as is hydrogen's ability to form hydrogen bonds.
Interstitial hydrides (a.k.a. metallic hydrides)
In interstitial or metallic hydrides, hydrogen dissolves in a metal to form nonstoichiometric compounds (solid solutions) of formula MHn. They are called metallic hydrides since they possess the typical metallic properties of luster, hardness, and conductivity and are called interstitial hydrides because the H occupies interstices in an FCC, HCP, or BCC metal lattice, as illustrated schematically in Figure $\sf{5}$.
The process of interstitial hydride formation is reversible, and metals can dissolve varying amounts of hydrogen depending on the number of interstices available. Because of this, interstitial hydrides have been considered as storage materials for hydrogen.
Note. $\sf{1}$. A Qualitative Molecular Orbital Description of the Bonding in Diborane
The B-H-B bonds in diborane may also be explained using a molecular orbital description, as illustrated by the qualitative MO diagram in Figure $\sf{6}$.
Contributors and Attributions
Stephen Contakes, Westmont College | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/08%3A_Chemistry_of_the_Main_Group_Elements/8.04%3A_Hydrogen/8.4.01%3A_Hydrogen%27s_Chemical_Properties.txt |
The alkaline earth metals
The alkaline earth metals comprise the elements Be through Ra in group 2 of the periodic table, as show in Figure \(\sf{1}\).
Contributors and Attributions
Stephen Contakes, Westmont College
8.05: Group 2 The Alkaline Earth Metals
Like the alkali metals, the alkaline earth metals are good reductants that are only found on Earth in their +2 oxidation states. Magnesium and calcium are particularly common. Magnesium-containing chlorophylls are the photosynthetic pigments of green plants and photosynthetic algae (Scheme $\sf{1}$).
Scheme $\sf{1}$. Structure of chlorophyll a, one of the principle green pigments of plants.
Seawater is about 0.05 M in Mg2+ and 0.01 M in Ca2+, and beryllium and calcium are components of many important minerals. Examples of alkaline earth metal-bearing minerals that were known and used since antiquity include beryl, Be3Al2Si6O18; gypsum, CaSO4·2H2O; limestone, CaCO3; and lime, CaO, the latter of which were more recently replaced in construction by calcium silicate-based sand lime bricks and Portland cement containing concrete.
Despite calcium and magnesium's widespread environmental distribution and importance, the metals were only isolated in the late 18th and early 19th century after the development of electrolysis. Humphrey Davy's isolation of Mg and Ca is illustrative.
$\sf{2~MO~~\overset{electrolysis}{\longrightarrow}~~2~M(s)~~+~~O_2~~~~~~Davy, (M~=~Mg,~Ca,~1808)} \nonumber$
Today Ca, Mg, and the other alkaline earth metals are produced either by electrolysis of their chlorides or chemical reduction by carbon or aluminum.
$\sf{2~MCl_2~~\overset{electrolysis}{\longrightarrow}~~2~M(s)~~+~~Cl_2~~~~~~(M~=~Be,~Mg,~Ca,~Sr)} \nonumber$
$\sf{MgO~~+~~C~~\overset{electrolysis}{\longrightarrow}~~Mg(s)~~+~~CO} \nonumber$
$\sf{3~MO~~+~~2~Al~~\overset{electrolysis}{\longrightarrow}~~M(s)~~+~~Al_2O_3~~~~~~(M~=~Ca,~Sr,~Ra)} \nonumber$
$\sf{4~BaO~~+~~2~Al~~\overset{electrolysis}{\longrightarrow}~~Ba(s)~~+~~BaAl_2O_4} \nonumber$
Once formed, the metals are reactive towards atmospheric oxygen. Nevertheless, Be and Mg may be stored and used in air since their initial oxidation gives a thin passivating layer of the oxide that seals off the bulk metal from further oxidation. The other alkaline earth metals need to be stored under an inert atomosphere to prevent their degradation to the oxide.
Physical Properties
Like the alkali metals, the alkaline earth metals are all grey or silvery solids that crystallize in a cubic lattice, HCP for Be and Mg, FCC/CCP for Ca and Sr, and BCC for Ba and Ra. All possess the typical metallic properties of high heat and thermal conductivity. Aside from the lightest member of the group, they are comparatively low melting and soft on account of their large atomic radii, low effective nuclear charges, and just two valence electrons that contribute to metallic bonding. Thus they melt and boil at higher temperatures than the alkali metals but lower temperatures than most transition metals, as illustrated by the melting and boiling points listed in Table $\sf{1}$. As with the alkali metals, the melting and boiling points decrease down the group as the atomic size increases.
Table $\sf{1}$. Melting and boiling points of the alkaline earth metals and selected reference substances.3
Substance Melting Point ($^{\circ}$ C) Boiling Point ($^{\circ}$ C)
Alkaline Earth Metal
Beryllium, Be 1278 2468
Magnesium, Mg 649 1090
Calcium, Ca 839 1484
Strontium, Sr 769 1384
Barium, Ba 727 1845
Radium, Ra 700 1140
Non-alkaline Earth Metal
Lithium, Li 181 1347
Sodium, Na 98 883
Cesium, Cs 28 678
Titanium, Ti 1660 3287
Iron, Fe 1538 2861
Copper, Cu 1083 2567
Water 0 100
Benzene 6 80
The atomic properties of alkaline earth metals reflect the
• relatively high energy and large size of their ns orbitals
• higher effective nuclear charge experienced by electrons in their ns valence orbitals when compared to those of the corresponding alkali metals
As may be seen from the atomic radii and ionization energies given in Table $\sf{2}$, overall the radii and ionization energies of the alkaline earth metals follow the expected periodic trends. The radii increase down a group, and across a row they follow the trend
$\sf{alkali~metal~>~alkaline~earth~metal~>~transition~metals} \nonumber$
Because of their smaller size and larger atomic mass, alkaline earth metals are significantly more dense than the corresponding alkali metals (Table $\sf{2}$).
Correspondingly, the ionization energies of the alkaline earth metals decrease down a group while across a row their ionization energies follow the trend
$\sf{alkali~metal~<~alkaline~earth~metal~<~transition~metals} \nonumber$
In consequence the alkaline earth metals are good reductants and prefer to form the +2 ion, although they are not as reactive as the alkali metals. Moreover, Be possesses an anomalously small radius, high ionization energy, and large electronegativity when compared with heavier alkaline earth metals. Consequently, its chemical behavior is similar to that of boron and aluminum, as will be discussed in the next section.
Table $\sf{2}$. Selected atomic properties of the alkali metals and selected reference compounds.3,4
Substance
Atomic Radius
(Angstroms)
First Ionization Energy
(kJ/mol)
Pauling Electronegativity
Density of the Solid
(g/mL)
Alkaline Earth Metal
Beryllium, Be 1.05 899 1.57 1.85
Magnesium, Mg 1.50 738 1.31 1.74
Calcium, Ca 1.80 590 1.00 1.55
Strontium, Sr 2.00 550 0.95 2.54
Barium, Ba 2.15 503 0.89 3.59
Radium, Ra 2.15 509 0.89 5.0
Non-alkaline Earth Metal
Lithium, Li (same row as Be) 1.45 513 0.98 0.53
Sodium, Na (same row as Mg) 1.80 496 0.93 0.97
Potassium, K (same row as Ca) 2.20 419 0.82 0.86
Rubidium, Rb (same row as Sr) 2.35 403 0.82 1.53
Cesium, Cs (same row as Ba) 2.60 376 0.79 1.87
Francium. Fr (same row as Ra) not determined 400 0.7 not determined
Titanium, Ti 1.40 658 1.54 4.54
Iron, Fe 1.40 759 1.83 7.87
Copper, Cu 1.35 745 1.90 8.96
Boron, B 0.85 801 2.04 2.34
Aluminum, Al 1.25 577 1.61 2.70
Nitrogen, N 0.65 1402 3.04 1.03 (at 21 K)
As will be explained in the next section, one of the factors that contributes to the reactivity of the alkali metals is their small ionization energies.
Contributors and Attributions
Stephen Contakes, Westmont College | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/08%3A_Chemistry_of_the_Main_Group_Elements/8.05%3A_Group_2_The_Alkaline_Earth_Metals/8.5.01%3A_Preparation_and_General_Properties_of_the_Alkaline_Earth_Elements.txt |
Alkaline earth metals are good reducing agents that tend to form the +2 oxidation state.
The alkaline earth metals tend to form +2 cations. As can be seen from Figure $\sf{2}$, alkaline earth metals possess large negative M2+/0 standard reduction potentials which strongly favor the +2 cation. The potentials for reduction of Ca2+, Sr2+, and Ba2+ to the metal of ~ -3 V are even similar to those of the alkali metals.
Like the alkali metals, Ca, Sr, and Ba dissolve in liquid ammonia to give solutions containing solvated electrons, although these have not been as heavily studied as those of the alkali metals.
Unlike the alkali metals, all the alkaline earth metals react with oxygen to give oxides of formula MO, although the peroxides of the heavier alkaline earths can be formed by solution phase precipitation of the metal cation with a source of peroxide anion (O22-).
$\sf{2~M(s)~~+~~O_2~\longrightarrow~~2~MO(s)~~~~~~~~~~~~~~~~~~~~~~~~~(M~=~Be,~Mg,~Ca,~Sr,~Ba,~and~presumably~Ra)} \nonumber$
$\sf{M(OH)_2(aq)~~+~~H_2O_2(aq)~\longrightarrow~~MO_2(s)~~+~~2~H_2O~~~~~~~~~~~~~~~~~~~~~~~~(M~=~Mg,~Ca,~Sr,~Ba,~and~presumably~Ra)} \nonumber$
Calcium, strontium, barium (and presumably radium) react with water to liberate dihydrogen gas and form hydroxides.
$\sf{M(s)~~+~~2~H_2O(l)~\rightarrow~M^{2+}~~+~~2~OH^-~~+~~H_2(g)~~~~~~~~~~~~~~~~~~~~~~~~~~(M~=~Ca,~Sr,~Ba,~and~presumably~Ra)} \nonumber$
although unlike the alkali metals the reduction is slow and usually liberates hydrogen without fire or explosion.
Beryllium and magnesium do not react with water at room temperature, although they do dissolve in acid and react with steam at high temperatures and pressures to give the oxide (which can be thought of as a dehydrated hydroxide).
$\sf{M(s)~~+~~2~H^+~~\rightarrow~M^{2+}~~+~~H_2(g)~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(M~=~Be~and~Mg)} \nonumber$
$\sf{M(s)~~+~~H_2O(g)~~\overset{high~T,~P}{\longrightarrow}~MO(s)~~+~~H_2(g)~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(M~=~Be~and~Mg)} \nonumber$
Presumably Ca, Sr, Ba, and Ra would react this way as well, although due to their higher reactivity the reaction is likely to be violent.
Like other metal oxides containing low oxidation state metals, the alkaline earth oxides are basic. The oxides of the heavier alkaline earth metals react with water to give the hydroxides.
$\sf{MO~~+~~H_2O~\longrightarrow~~M(OH)_2~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(M~=~Be,~Mg,~Ca,~Sr,~Ba,~and~presumably~Ra)} \nonumber$
The hydrolysis of BeO and MgO usually requires high temperatures and pressures. Consequently their hydroxides are more commonly prepared by addition of base to a soluble salt.
$\sf{M^{2+}(aq)~~+~~2~OH^-~~\rightarrow~M(OH)_2~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(M~=~Be,~Mg, Ca,~Sr,~Ba,~and~presumably~Ra)} \nonumber$
The alkaline earth metals react directly with halogens to give the dihalides, although given the exothermicity of reactions involving the powerfully reducing alkaline earth metals with oxidizing halogens in the laboratory, it is usually safer to react the hydroxides or oxides with the appropriate hydrohalic acid.
$\sf{M(s)~~+~~X_2^-~~\rightarrow~MX_2~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(M~=~Be,~Mg, Ca,~Sr,~Ba,~and~presumably~Ra; X~=~F,~Cl,~Br,~I)} \nonumber$
$\sf{M(OH)_2~~+~~2~HX~~\rightarrow~MX_2~~+~~2~H_2O~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(M~=~Mg, Ca,~Sr,~Ba,~and~presumably~Ra; X~=~F,~Cl,~Br,~I)} \nonumber$
$\sf{MO~~+~~2~HX~~\rightarrow~MX_2~~+~~H_2O~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(M~=~Mg, Ca,~Sr,~Ba,~and~presumably~Ra; X~=~F,~Cl,~Br,~I)} \nonumber$
The heavier alkaline earth metals (Mg through Ba) also reduce hydrogen to give hydrides.
$\sf{M(s)~~+~~H_2(g)~~\rightarrow~MH_2~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(M~=~Mg, Ca,~Sr,~Ba,~and~presumably~Ra)} \nonumber$
As discussed in section 8.2.1. Hydrogen's Chemical Properties, these alkaline earth hydrides are ionic salts of hydride ion. Thus they react with water and other electrophiles.
$\sf{MH_2(s)~~+~~H_2O(l)~~\rightarrow~M(OH)_2~~+~~H_2(g)~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(M~=~Mg, Ca,~Sr,~Ba,~and~presumably~Ra)} \nonumber$
The consumption of water in this reaction forms the basis for the use of calcium hydride as a drying agent for organic solvents.
Unlike the heavier alkaline earth metals, beryllium does not react directly with hydrogen, and the resulting hydride, though still nucleophilic, acts as a polar covalent hydride and hydrolyzes relatively slowly.
Divalent Alkaline Earth cations polarize anions
The classic example of alkaline earth cations' ability to polarize anions involves the decomposition of the metal carbonates. Alkali metal carbonates and nitrates thermally decompose to release CO2 and a mixture of NO2 and O2, respectively.
$\sf{MCO_3(s)~~\overset{\Delta}{\longrightarrow}~MO(s)~~+~~CO_2(g)} \nonumber$
$\sf{M(NO_3)_2(s)~~\overset{\Delta}{\longrightarrow}~MO(s)~~+~~2~NO_2(g)~~+~~O_2(g)} \nonumber$
Complex factors govern decomposition of the nitrates but, as shown in Table $\sf{1}$, the decomposition of the alkaline earth carbonates shows that on going down the group, carbonate decomposition requires increasingly higher temperatures.
Table $\sf{1}$. Decomposition temperatures of alkaline earth carbonates
Carbonate
Midpoint of decomposition range2
(K)
BeCO3 not reported; unstable at room temperature (298 K)
MgCO3 903
CaCO3 953
SrCO3 1178
BaCO3 1316
The typical explanation for this trend involves the mechanism of carbonate decomposition depicted in Scheme $\sf{\PageIndex{II}}$. As the alkaline earth metal cation becomes larger on going from Be to Ba, its ability to polarize the carbonate anion is lessened, making it more difficult to form the oxide.
Scheme $\sf{\PageIndex{I}}$. Models explaining how alkaline earth metal cations facilitate carbonate decomposition. (A) Ionic model in which the negative charge buildup is stabilized by interaction between the dication and one of the carbonate oxygens. (B) Semi-covalent representation of the same interaction, now depicted as a M=O bond (which should not be taken to imply that such a bond actually exists).
Exercise $1$
Alkaline earth metal sulfates undergo decomposition reactions similar to those of the carbonates and nitrates.
1. Write a decomposition reaction that involves the liberation of a single molecular gas from the sulfate to give an oxide.
2. Rank the alkaline earth metal sulfates in order of increasing decomposition temperature.
Answer
1. The sulfates decompose by liberating SO3 according to the reaction
$\sf{MSO_4(s)~~\overset{\Delta}{\longrightarrow}~MO(s)~~+~~SO_3(g)} \nonumber$
Note that another possible decomposition mode is
$\sf{MSO_4(s)~~\overset{\Delta}{\longrightarrow}~MO(s)~~+~~ {\textstyle \frac{1}{2}}~O_2(g)~~+~~SO_2(g)} \nonumber$
However, that reaction is not the one asked for by the prompt since it involves the liberation of two different molecular gases (O2 and SO2).
2. The expected decomposition order of the alkaline earth metal sulfates along with known decomposition temperatures is
$\sf{\underset{580~^{\circ}C}{BeSO_4}~<~\underset{895~^{\circ}C}{MgSO_4}~<~\underset{1149~^{\circ}C}{CaSO_4}~<~\underset{1374~^{\circ}C}{SrSO_4}~<~BaSO_4~<~RaSO_4} \nonumber$
Beryllium, and to a lesser extent magnesium, form polar highly covalent compounds.
The chemistry of magnesium and beryllium demonstrates the danger of drawing an overly rigorous distinction between elements as metals, nonmetals, and metalloids. This is because both Be and Mg can form compounds with considerable covalent character and, as might be expected from their relative paucity of electrons, they share much in common with the electron deficient row 13 metalloids B and Al.
In terms of Mg, the influence of covalency is evident from the structures of the Grignard reagents that Mg forms on reaction between the metal and alkyl halides.
$\sf{Mg(s)~~+~~R-X~~\overset{THF, catalytic~I_2}{\longrightarrow}~R-Mg-X} \nonumber$
The reaction is dependent on the presence of Lewis base donor ethers like Et2O or THF, which coordinate the Mg2+, completing its coordination sphere and giving tetrahedral complexes like that depicted in Scheme $\sf{\PageIndex{II}}$.
Scheme $\sf{\PageIndex{II}}$. Structure of monomeric "RMgX" Grignard reagent in THF solution.
Like molecular compounds, Grignard reagents undergo ligand substitution reactions in solution according to the Schlenk equilibrium.
$\nonumber$
They also form clusters in which the halogen lone pairs are used to bridge multiple Mg centers, as illustrated by the complex in Scheme $\sf{\PageIndex{III}}$.
Scheme $\sf{\PageIndex{II}}$. (A) Halogens like Cl can bridge multiple metal centers via their lone pairs, allowing for the formation of species like (B) "(RMgCl)2(MgCl2)2". Redrawn from references 4 and 5.
The extent of covalency is even greater in the case of beryllium, which with a radius of only 113 Å and valence electrons that experience a Slater effective nuclear charge of +1.95 atomic charge units, has considerable ability to polarize nearby Lewis bases. As a result, beryllium tends to form polar covalent bonds rather than ionic ones.
Because beryllium only possesses two valence electrons, its compounds also tend to be electron deficient and bridging Be-X-Be bonds are common. Thus in liquid ammonia, Be forms species with bridging Be-N-Be bonds like the tetrameric cluster shown in Scheme $\sf{\PageIndex{III}}$.
Scheme $\sf{\PageIndex{III}}$. Tetrameric Be cluster formed in liquid ammonia. Redrawn from reference 6.
One particularly remarkable structure is that of basic beryllium acetate in which a central oxygen bridges four Be atoms, as shown in Scheme $\sf{\PageIndex{IV}}$ along with that of Be4(NO3)6O, which is thought to possess an anlogous structure.
Scheme $\sf{\PageIndex{IV}}$. (A) Structure of basic beryllium acetate, Be4(OAc)6O, (B) in which the central OBe4 tetrahedron is circumscribed to make it easier to see that the structure consists of a OBe46+ tetrahedron in which the Be---Be edges are linked by bridging acetate ligands. Zinc forms an analogous structure. (C) The structure of Be4(NO3)6O is postulated to be analogous to that of basic beryllium acetate, with bridging nitrate ligands taking the place of the acetates.
Beryllium hydride even forms an adduct with two BH3 to give a structure in which a central Be is linked to the terminal BH2 groups by 3-center-2-electron Be-H-B bonds, as shown in Scheme $\sf{\PageIndex{V}}$.7
Scheme $\sf{\PageIndex{V}}$. Formation of an adduct between BeH2 and two BH3 (in the form of B2H6).
Contributors and Attributions
Stephen Contakes, Westmont College | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/08%3A_Chemistry_of_the_Main_Group_Elements/8.05%3A_Group_2_The_Alkaline_Earth_Metals/8.5.02%3A_Alkaline_Earth_Metals%27_Chemical_Properties.txt |
The group 13 elements comprise the elements Boron through Nihonium
The group 13 elements comprise the elements Boron through Nihonium in group 13 of the periodic table, as shown in Figure \(\sf{1}\).
The group 13 elements are chemically diverse, comprising elements:
• Boron, B, an electron deficient 2nd row element sometimes classified as a nonmetal and occasionally as a metalloid;
• Aluminum, Al, an electron deficient 3rd row element sometimes classified as a metal and sometimes as a metalloid;
• Gallium (Ga), Indium (In), Thallium (Tl), and Nihonium (Nh, marked under its old symbol, Uut, in the Figures on this page), post transition metals that exhibit the inert pair effect to varying degrees.
The group 13 elements include post-transition metals.
The term post-transition metals refers to those elements that are metals following the transition metals. As with the metalloid concept there is no universal consensus as to what exactly is a post transition metal. Fortunately, in practice it is less important to precisely define what is and is not a post transition metal than to understand the reason why it might be helpful to classify an element as a post-transition metal.
The main features of the post transition metals are that they are relatively electron rich and electronegative compared to what is classically thought of as a metal. Roughly, this translates into relatively lower melting points (since more antibonding levels in the band structure tend to be occupied), increased preference for covalency, and greater brittleness than other metals (due to the resulting directional bonding). Their electron richness means that they tend to form soft cations.
Several systems are used to classify elements as belonging to the post transition metals. The main ones include:
1. Metals which follow the d-block. By this definition only the metals in groups 13 and higher and Rows 3 and higher which form relatively soft and electron rich cations and exhibit significant covalency in their bonding are included. However, if this scheme is adopted too rigidly Al is excluded since it technically doesn't follow the d-block (and has an unfilled (n-1)d subshell) and the metalloids are excluded, even though many of them also form relatively soft and electron rich cations with filled (n-1)d subshells. Another disadvantage of this system is that it entangles the issue of which elements should be classified as post-transition metals with the thorny issue of which elements should be classified as metals vs. metalloids.
2. Metals and metalloids of the p-block. This system has the advantage of emphasizing the interesting and unique properties of the metals and metalloids of p-block as well as continuities in those properties through the p-block. Consequently, it will be used here in the sections which follow. However, it has the disadvantage of excluding metals like Zn, Cd, and Hg, which form many compounds in which the metal has a (n-1)d10 configuration.
3. Metals which follow the transition elements in the sense of forming ions with a completely full (n-1)d valence shell sometimes along with Al and the p-block metalloids. This definition adds Zn, Cd, and Hg (and sometimes Cu, Ag, and Au) since they form ions with an (n-1)d 10 valence electron configuration such as Zn2+, Cd2+ and Hg2+(and Cu+, Ag+, and Au+). Since this chapter only considers the p-block elements, for the purposes of this chapter this system is functionally identical to system 2.
Whichever classification scheme one uses, it is often more helpful to think of the classification of elements as post-transition metals as a way to emphasize similarities in the chemical properties of a set of elements rather than as a way of emphasizing how post-transition metals differ from other metals. The post-transition metals are not the only ones which form soft cations or compounds better described as being held together by covalent bonds. As the previous sections made clear, even alkali metals form anions under the right circumstances, and many compounds of metals are better described in terms of covalency than ionic interactions. This was already evident in the chemistry of the alkaline earth metals Be and Mg discussed in the previous section. In subsequent chapters the bonding and reactivity in coordination complexes and organometallic compounds will largely be described in covalent terms.
Contributors and Attributions
Stephen Contakes, Westmont College
8.06: Group 13 (and a note on the post-transition metals)
As elements, Boron acts like an electron deficient nonmetal and the other group 13 elements like metals that exhibit varying degrees of covalency.
Selected properties of the Group 13 elements are given in Table $\sf{1}$.
Table $\sf{1}$. Selected Properties of Group 13 Elements other than Nihonium. For more information about how this data was sourced see note 1.
Element Elemental Form
Melting Point
($\sf{^{\circ}C}$)
Pauling Electronegativity
Ionization Energy
(kJ/mol)
Electron Affinity
(kJ/mol)
Atomic Radius
(Å)
Covalent Radius
(Å)
Boron, B
Network Covalent Solid characterized by the sharing of electrons in boron clusters, primarily linked icosahedra, like those depicted schematically below.
2180 2.04 801 27 1.92 0.84
Aluminum or Aluminium,2Al
FCC Metal
660 1.61 577 42 1.84 1.24
Gallium, Ga
Metal with a nonstandard crystal structure that may be described as packed Ga2 units, which are shown linked by the thicker pink lines in the structure below.
30 1.81 579 41 1.87 1.23
Indium, In
Distorted close packed metal
157 1.78 558 36 1.93 1.42
Thallium, Tl
HCP Metal
304 1.8 589 28.9 1.96 1.44
From the data given in Table $\sf{1}$ several things are apparent:
1. There is a discontinuity between the properties of boron and the group 13 metals in rows 3 and higher.
2. Boron forms covalent compounds characterized by multicenter bonding and cluster formation; the other elements are metals that exhibit varying degrees of covalency.
3. Trends in the properties of the row 3+ group 13 metals are far from simple, likely in part because
• The atomic properties of Ga and In are influenced by the d-block contraction, giving both a relatively high electronegativity, which when combined with their small radii promote stronger covalent interactions.
• The atomic properties of Tl are also influenced by the lanthanide contraction, resulting in an increase in electronegativity and ionization energy relative to In.
4. The interplay between changing covalent and metallic bonding interactions on going from Al to Tl results in a decrease in melting point from Al to Ga followed by an increase in melting point on going from Ga to Tl.
The group 13 elements are prepared by reduction of halides or oxides or, in the case of boron, thermal decomposition of the hydride.
Boron can be prepared by reduction of its halides or high temperature thermal decomposition of its hydrides.
$\sf{2~BCl_3~~+~~3~H_2~~\longrightarrow~~2~B~~+~~6~HCl} \nonumber$
$\sf{B_2H_6~~\overset{\Delta}{\longrightarrow}~~2~B~~+~~3~H_2} \nonumber$
The group 13 metals are made by electrolytic reduction of the cations in thermally accessible molten salts. The best known example is the production of aluminum in the Hall-Héroult process.
$\sf{2~Al_2O_3~~+~~3~C(s,~graphite)~~\overset{electrolysis, 940-980^{\circ}C, KF, Na_2AlF_6}{\longrightarrow}~~4~Al(s)~~+~~3~CO_2(g)} \nonumber$
Boron forms covalent bonds and engages in cluster and/or bridge bonding when there are too few electrons to form enough pairwise bonds to satisfy an octet.
Boron possesses only three valence electrons and can be stablilized by sharing five more. It does this in three main ways.
1. Boron forms ordinary covalent bonds and/or gains electrons to give stable molecules or ions. For example, while the boron of monomeric BH3 may not have access to enough valence electrons to achieve a stable octet, it can satisfy its octet by forming an adduct with a Lewis base. A well known example is the borohydride anion, BH4-.
$\sf{BH_3~~+~~H^-~~\longrightarrow~~BH_4^-} \nonumber$
A large number of boron compounds are of this sort. These include
a. The borates and boric acid derivatives which comprise a few ppm of the Earth's crust. These contain borate anions similar to the representative set shown in Scheme $\sf{\PageIndex{II}}$. Of these, sodium borate or borox is commonly used in household detergents. More precisely, there is a mixture of boraxes, all of which are of formula [Na(H2O)x]2H2B4O9 (often written as Na2B4O7·xH2O) and differ in the number of waters of hydration. All contain the tetraborate anion, B4O5(OH)42-, which has the structure shown at right in Scheme $\sf{\PageIndex{II}}$..
Scheme $\sf{\PageIndex{II}}$. The tetraborate anion.
As illustrated by the structures depicted in Scheme $\sf{\PageIndex{II}}$, the boron in borates is found in BO4 tetrahedra and trigonal planar BO3 units linked together in various ways.
The borates illlustrate the acid-base properties of these conventional boron compounds. Borate anions may be considered as derived from hydrolysis of boric oxide to boric acid followed by acid dissociation.
$\sf{\underset{boric~oxide}{B_2O_3}~~+~~3~H_2O~~\longrightarrow~~\underset{boric~acid}{2~B(OH)_3}} \nonumber$
$\sf{B(OH)_3~~\rightleftharpoons~~H^+~~+~~B(OH)_2O^-} \nonumber$
$\sf{B(OH)_3~~\rightleftharpoons~~3~H^+~~+~~BO_3^{3-}} \nonumber$
In this respect boron oxide, like all metal oxides, is capable of acting as a Brønsted acid, albeit a very weak one. Consequently the borate anions, as the conjugate bases of weak acids, are Brønsted bases.
The trigonal planar boron sites in species like boric acid illustrate the ability of trigonal planar boron to act as a Lewis acid, specifically by forming a Lewis base adduct.
$\sf{B(OH)_3~~+~~OH^-~~\rightleftharpoons~~B(OH)_4^-} \nonumber$
Notice that the three coordinate boron centers in Scheme $\sf{\PageIndex{II}}$ are not drawn as possessing an octet, even though it would be possible to draw resonance forms which satisfy boron's octet by drawing resonance structures in which there are B=O bonds. This is because it is unclear that such resonance structures contribute significantly to the bonding in trigonal planar boron centers. To see why, it is helpful to consider the boron trihalides.
b. The boron trihalides. Structurally, boron trihalides like BCl3 possess a trigonal planar structure like that in the BO3 units of boric acid and metal borates. What is notable about the structure is that the B-X bonds are shorter than those in analogous tetrahedral structures, as illustrated in Scheme $\sf{\PageIndex{III}}$.
Scheme $\sf{\PageIndex{III}}$. Increase in the B-Cl bond length as the boron center goes from trigonal pyramidal to tetrahedral coordination on formation of an adduct with ammonia. Reproduced from a drawing by Ben Mills based on reference 6.
c. Borazine and the polymorphs of boron nitride. One class of compounds in which multiple bonds involving boron are believed to play some role are borazine and the hexagonal polymorph of boron nitride, BN.
Borazine, B3,N3H6, is isoelectronic and isostructural to benzene, C6H6, and has very similar properties. Both have a planar hexagonal molecular structure, are nonpolar, and exist as relatively low boiling liquids at room temperature. The existence of delocalized $\pi$ bonding in the borazine ring may be inferred in that the B-N bond distances of 1.44 Å in borazine is less than the ~1.6 Å B-N distances typical for B-N single bonds in tetrahedral systems8 and close to the 1.42 Å C-C bond distance in benzene. While in benzene the electron delocalization is spread evenly around the carbon ring system, in borazine the greater electronegativity of N ensures that the electron density above and below the borazine is more heavily concentrated near the more electronegative N atoms. This is represented by the resonance and orbital descriptions of $\pi$ bonding depicted in Scheme $\sf{\PageIndex{IV}}$. As a result of the greater electron density on N and the resulting electron deficiency on B, the B and N atoms of borazine retain considerable Lewis acid and base character, respectively.
Scheme $\sf{\PageIndex{IV}}$. (A) Resonance representation of the bonding in borazine and (B) representation of the lowest energy $\pi$ bonding MO in borazine showing the greater electron density on the more electronegative N atoms.
Although the layered hexagonal structure of boron nitride is the most stable, BN also forms two polymorphs in which the B and N atoms are tetrahedrally coordinated; these are analogous to the cubic diamond and Lonsdaleite forms of carbon and are given in Figures $\sf{\PageIndex{3B}}$ and $\sf{\PageIndex{3C}}$. They may be thought of as consisting of layers of linked interlocking B3N3 chairs. The two forms differ in terms of whether the layers are arranged so as to be linked in the form of chair or boat conformation B3N3 rings.
2. Boron can achieve stable bonding configurations by sharing electrons via bridge bonds. These are common in boron hydrides. For instance, another way the B of BH3 can achieve an octet is to form two-center three-electron bridge bonds, as in diborane.
$\nonumber$
Valence bond and molecular orbital descriptions of the "two-electron three-center" bridge bonds in these molecules have already been given in 8.2.1. Hydrogen's Chemical Properties.
Those who wish to review the MO description of bonding are invited to consider the derivation of the MOs of diborane given in Example $1$.
Exercise $1$. The molecular orbitals of Diborane
Use the projection operator method to derive MOs for diborane. The structure of diborane is shown below, along with the atom labels and coordinate system which will be used in the answer key.
Answer
Diborane has D2h symmetry, under which the following sets of orbitals will transform into one another and should be considered as a set:
H1s,a, H1s,b, H1s,c, H1s,d
H1s,f, H1s,e
B1s,1, B1s,2
Bpx,1, Bpx,2
Bpy,1, Bpy,2
Bpz,1, Bpz,2
Using the projection operator method with the following generator orbitals gives:
Generator orbital E
C2
(z)
C2
(y)
C2
(x)
i
σ
(xy)
σ
(xz)
σ
(yz)
Set 1: H1s,a H1s,a H1s,c H1s,d H1s,b H1s,d H1s,b H1s,a H1s,c
Set 2: H1s,e H1s,e H1s,f H1s,e H1s,f H1s,f H1s,e H1s,f H1s,e
Set 3: B1s,1 B1s,1 B1s,2 B1s,2 B1s,1 B1s,2 B1s,1 B1s,1 B1s,2
Set 4: Bpx,1 Bpx,1 -Bpx,2 -Bpx,2 Bpx,1 -Bpx,2 Bpx,1 Bpx,1 -Bpx,2
Set 5: Bpy,1 Bpy,1 -Bpy,2 Bpy,2 -Bpy,1 -Bpy,2 Bpy,1 -Bpy,1 Bpy,2
Set 6: Bpz,1 Bpz,1 Bpz,2 -Bpz,2 -Bpz,1 -Bpz,2 -Bpz,1 Bpz,1 Bpz,2
The use of these projections to generate orbitals is straightforward but a lot of work.
For Set 1, the terminal H 1s orbitals, taking the 1s orbital on Ha as the generator orbital:
Ψ(Ag)? = [1(H1sa) + (1)( H1sc) + (1)(H1sd) + (1)( H1sb) + (1)( H1sd) + (1)( H1sb) +(1)(H1sa) + (1)( H1sc)]
Ψ (Ag) = 2H1sa + 2H1sb + 2H1sc + 2H1sd So, there will be one ag orbital (w/ all 1s in phase)
Ψ(B1g)? = [1(H1sa) + (1)( H1sc) + (-1)(H1sd) + (-1)( H1sb) + (1)( H1sd) + (1)( H1sb) +(-1)(H1sa) + (-1)( H1sc)]
Ψ (B1g) = 0H1sa + 0H1sb + 0H1sc + 0H1sd No b1g orbital
Ψ(B2g)? = [1(H1sa) + (-1)( H1sc) + (1)(H1sd) + (-1)( H1sb) + (1)( H1sd) + (-1)( H1sb) +(1)(H1sa) + (-1)( H1sc)]
Ψ (B2g) = 2H1sa - 2H1sb - 2H1sc + 2H1sd So, there will be one b2g orbital
Ψ(B3g)? = [1(H1sa) + (-1)( H1sc) + (-1)(H1sd) + (1)( H1sb) + (1)( H1sd) + (-1)( H1sb) +(-1)(H1sa) + (-1)( H1sc)]
Ψ (B3g) = = 0H1sa + 0H1sb + 0H1sc + 0H1sd No b3g orbital
Ψ(Au)? = [1(H1sa) + (1)( H1sc) + (1)(H1sd) + (1)( H1sb) + (-1)( H1sd) + (-1)( H1sb) +(-1)(H1sa) + (-1)( H1sc)]
Ψ (Au) = 0H1sa + 0H1sb + 0H1sc + 0H1sd So, there will be no au orbital
Ψ(B1u)? = [1(H1sa) + (1)( H1sc) + (-1)(H1sd) + (-1)( H1sb) + (-1)( H1sd) + (-1)( H1sb) +(1)(H1sa) + (1)( H1sc)]
Ψ (B1u) = 2H1sa - 2H1sb + 2H1sc - 2H1sd So, there will be one b1u orbital
Ψ(B2u)? = [1(H1sa) + (-1)( H1sc) + (1)(H1sd) + (-1)( H1sb) + (-1)( H1sd) + (1)( H1sb) +(-1)(H1sa) + (1)( H1sc)]
Ψ (B2u) = 0H1sa + 0H1sb + 0H1sc + 0H1sd So, there will be no b2u orbital
Ψ(B3u)? = [1(H1sa) + (-1)( H1sc) + (-1)(H1sd) + (1)( H1sb) + (-1)( H1sd) + (1)( H1sb) +(1)(H1sa) + (-1)( H1sc)]
Ψ (B3u) = 2H1sa + 2H1sb - 2H1sc - 2H1sd So, there will be one b3u orbital
So the first set gives the following terminal hydrogen 1s group orbitals:
For Set 2, The H 1s orbitals on the bridging H, taking the 1s orbital on He as the generator orbital.
Ψ(Ag)? = [1(H1se) + (1)( H1sf) + (1)(H1se) + (1)( H1sf) + (1)( H1sf) + (1)( H1se) +(1)(H1sf) + (1)( H1se)]
Ψ (Ag) = 4H1se + 4H1sf So, there will be one ag orbital (w/ both 1s in phase)
Ψ(B1g)? = [1(H1se) + (1)( H1sf) + (-1)(H1se) + (-1)( H1sf) + (1)( H1sf) + (1)( H1se) +(-1)(H1sf) + (-1)( H1se)
Ψ (B1g) = 0H1se + 0H1sf No b1g orbital
Ψ(B2g)? = [1(H1se) + (-1)( H1sf) + (1)(H1se) + (-1)( H1sf) + (1)( H1sf) + (-1)( H1se) +(1)(H1sf) + (-1)( H1se)]
Ψ (B2g) = 0H1se + 0H1sf So, there will be no b2g orbital
Ψ(B2g)? = [1(H1se) + (-1)( H1sf) + (-1)(H1se) + (1)( H1sf) + (1)( H1sf) + (-1)( H1se) +(-1)(H1sf) + (1)( H1se)]
Ψ (B3g) = = 0H1sa + 0H1sb + 0H1sc + 0H1sd No b3g orbital
Ψ(Au)? = [1(H1se) + (1)( H1sf) + (1)(H1se) + (1)( H1sf) + (-1)( H1sf) + (-1)( H1se) +(-1)(H1sf) + (-1)( H1se)]
Ψ (Au) = 0H1se + 0H1sf So, there will be no au orbital
Ψ(B1u)? = [1(H1se) + (1)( H1sf) + (-1)(H1se) + (-1)( H1sf) + (-1)( H1sf) + (-1)( H1se) +(1)(H1sf) + (1)( H1se)]
Ψ (B1u) = 0H1se + 0H1sf So, there will be no b1u orbital
Ψ(B2u)? = [1(H1se) + (-1)( H1sf) + (1)(H1se) + (-1)( H1sf) + (-1)( H1sf) + (1)( H1se) +(-1)(H1sf) + (1)( H1se)]
Ψ (B2u) = 4H1se - 4H1sf So, there will be a b2u orbital
Ψ(B3u)? = [1(H1se) + (-1)( H1sf) + (-1)(H1se) + (1)( H1sf) + (-1)( H1sf) + (1)( H1se) +(1)(H1sf) + (-1)( H1se)]
Ψ (B3u) = 0H1se + 0H1sf So, there will be no b3u orbital
So the second set gives the following bridging hydrogen 1s group orbitals:
For Set 3, The B 2s orbitals, taking the 2s orbital on B1 as the generator orbital:
Ψ(Ag)? = [1(Bs1) + (1)( Bs2) + (1)( Bs2) + (1)( Bs1) + (1)( Bs2) + (1)( Bs1) +(1)( Bs1) + (1)( Bs2)]
Ψ (Ag) = 4Bs,1 + 4Bs,2 So, there will be one ag orbital (w/ the B 1s in phase)
Ψ(B1g)? = [1(Bs1) + (1)( Bs2) + (-1)( Bs2) + (-1)( Bs1) + (1)( Bs2) + (1)( Bs1) +(-1)( Bs1) + (-1)( Bs2)]
Ψ (B1g) = 0Bs,1 + 0Bs,2 No b1g orbital
Ψ(B2g)? = [1(Bs1) + (-1)( Bs2) + (1)( Bs2) + (-1)( Bs1) + (1)( Bs2) + (-1)( Bs1) +(1)( Bs1) + (-1)( Bs2)]
Ψ (B2g) = 0Bs,1 + 0Bs,2 No b2g orbital
Ψ(B2g)? = [1(Bs1) + (-1)( Bs2) + (-1)( Bs2) + (1)( Bs1) + (1)( Bs2) + (-1)( Bs1) +(-1)( Bs1) + (1)( Bs2)]
Ψ (B3g) = 0Bs,1 + 0Bs,2 No b3g orbital
Ψ(Au)? = [1(Bs1) + (1)( Bs2) + (1)( Bs2) + (1)( Bs1) + (-1)( Bs2) + (-1)( Bs1) +(-1)( Bs1) + (-1)( Bs2)]
Ψ (Au) = 0Bs,1 + 0Bs,2 So, there will be no au orbital
Ψ(B1u)? = [1(Bs1) + (1)( Bs2) + (-1)( Bs2) + (-1)( Bs1) + (-1)( Bs2) + (-1)( Bs1) +(1)( Bs1) + (1)( Bs2)]
Ψ (B1u) = 0Bs1 + 0Bs2 So, there will be no b1u orbital
Ψ(B2u)? = [1(Bs1) + (-1)( Bs2) + (1)( Bs2) + (-1)( Bs1) + (-1)( Bs2) + (1)( Bs1) +(-1)( Bs1) + (1)( Bs2)]
Ψ (B2u) = 0Bs1 + 0Bs2 So, there will be no b2u orbital
Ψ(B3u)? = [1(Bs1) + (-1)( Bs2) + (-1)( Bs2) + (1)( Bs1) + (-1)( Bs2) + (1)( Bs1) +(1)( Bs1) + (-1)( Bs2)]
Ψ (B3u) = 4Bs1 - 4Bs2 So, there will be one b3u orbital
The boron 2s group orbitals are:
For Set 4: The B 2px orbitals, using the 2px orbital on B1 as the generator orbital:
Ψ(Ag)? = [1(Bpx,1) + (1)(-Bpx2) + (1)(-Bpx2) + (1)( Bpx1) + (1)( -Bpx2) + (1)(Bpx1) +(1)(Bpx1) + (1)(-Bpx2)]
Ψ (Ag) = 4B1px1 - 4B2px1 So, there will be one ag orbital (w/ the B 2px antiphase)
Ψ(B1g)? = [1(Bpx,1) + (1)(-Bpx2) + (-1)(-Bpx2) + (-1)( Bpx1) + (1)( -Bpx2) + (1)(Bpx1) +(-1)(Bpx1) + (-1)(-Bpx2)]
Ψ (B1g) = 0B1px1 + 0B1px2 No b1g orbital
Ψ(B2g)? = [1(Bpx,1) + (-1)(-Bpx2) + (1)(-Bpx2) + (-1)( Bpx1) + (1)( -Bpx2) + (-1)(Bpx1) +(1)(Bpx1) + (-1)(-Bpx2)]
Ψ (B2g) = 0B1px1 + 0B1px2 No b2g orbital
Ψ(B3g)? = [1(Bpx,1) + (-1)(-Bpx2) + (-1)(-Bpx2) + (1)( Bpx1) + (1)( -Bpx2) + (-1)(Bpx1) +(-1)(Bpx1) + (1)(-Bpx2)]
Ψ (B3g) = 0B1px1 + 0B1px2 No b3g orbital
Ψ(Au)? = [1(Bpx,1) + (1)(-Bpx2) + (1)(-Bpx2) + (1)( Bpx1) + (-1)( -Bpx2) + (-1)(Bpx1) +(-1)(Bpx1) + (-1)(-Bpx2)]
Ψ (Au) = 0B1px1 + 0B1px2 So, there will be no au orbital
Ψ(B1u)? = [1(B1s1) + (1)( B1s2) + (-1)( B1s2) + (-1)( B1s1) + (-1)( B1s2) + (-1)( B1s1) +(1)( B1s1) + (1)( B1s2)]
Ψ (B1u) = 0B1px1 + 0B1px2 So, there will be no b1u orbital
Ψ(B2u)? = [1(Bpx,1) + (-1)(-Bpx2) + (1)(-Bpx2) + (-1)( Bpx1) + (-1)( -Bpx2) + (1)(Bpx1) +(-1)(Bpx1) + (1)(-Bpx2)]
Ψ (B2u) = 0B1px1 + 0B1px2 So, there will be no b2u orbital
Ψ(B3u)? = [1(Bpx,1) + (-1)(-Bpx2) + (-1)(-Bpx2) + (1)( Bpx1) + (-1)( -Bpx2) + (1)(Bpx1) +(1)(Bpx1) + (-1)(-Bpx2)]
Ψ (B3u) = 4Bpx1 + 4Bpx2 So, there will be one b3u orbital
So the B 2px group orbitals are
For Set 5: The B 2py orbitals, taking 2py on B1 as the generator orbital:
Ψ(Ag)? = [1(Bpy1) + (1)(-Bpy2) + (1)(Bpy2) + (1)(-Bpy1) + (1)(-Bpy2) + (1)(Bpy1) +(1)(-Bpy1) + (1)(Bpy2)]
Ψ (Ag) = 0 Bpy1 + 0Bpy2 So, there will be no ag orbital
Ψ(B1g)? = [1(Bpy1) + (1)(-Bpy2) + (-1)(Bpy2) + (-1)(-Bpy1) + (1)(-Bpy2) + (1)(Bpy1) +(-1)(-Bpy1) + (-1)(Bpy2)]
Ψ (B1g) = 2Bpy1 - 2Bpy2 One b1g orbital
Ψ(B2g)? = [1(Bpy1) + (-1)(-Bpy2) + (1)(Bpy2) + (-1)(-Bpy1) + (1)(-Bpy2) + (-1)(Bpy1) +(1)(-Bpy1) + (-1)(Bpy2)]
Ψ (B2g) = 0Bpy1 + 0Bpy2 ; No b2g orbital
Ψ(B2g)? = [1(Bpy1) + (-1)(-Bpy2) + (-1)(Bpy2) + (1)(-Bpy1) + (1)(-Bpy2) + (-1)(Bpy1) +(-1)(-Bpy1) + (1)(Bpy2)]
Ψ (B2g) = 0Bpy1 + 0Bpy2 ; No b3g orbital
Ψ(Au)? = [1(Bpy1) + (1)(-Bpy2) + (1)(Bpy2) + (1)(-Bpy1) + (-1)(-Bpy2) + (-1)(Bpy1) +(-1)(-Bpy1) + (-1)(Bpy2)]
Ψ (Au) = 0Bpy1 + 0Bpy2 ; So, there will be no au orbital
Ψ(B1u)? = [1(Bpy1) + (1)(-Bpy2) + (-1)(Bpy2) + (-1)(-Bpy1) + (-1)(-Bpy2) + (-1)(Bpy1) +(1)(-Bpy1) + (1)(Bpy2)]
Ψ (B1u) = 0Bpy1 + 0Bpy2 ; So, there will be no b1u orbital
Ψ(B2u)? =[1(Bpy1) + (-1)(-Bpy2) + (1)(Bpy2) + (-1)(-Bpy1) + (-1)(-Bpy2) + (1)(Bpy1) +(-1)(-Bpy1) + (1)(Bpy2)]
Ψ (B2u) = 4Bpy1 + 4Bpy2; So, there will be a b2u orbital
Ψ(B3u)? = [1(Bpy1) + (-1)(-Bpy2) + (-1)(Bpy2) + (1)(-Bpy1) + (-1)(-Bpy2) + (1)(Bpy1) +(1)(-Bpy1) + (-1)(Bpy2)]
Ψ (B3u) = 0Bpy1 + 0Bpy2 ; So, there will be no b3u orbital
So the B 2py group orbitals are
For Set 6: The B 2pz orbitals, taking the 2pz orbital of B1 as the generator orbital.
Ψ(Ag)? = [1(Bpz1) + (1)(Bpz2) + (1)(-Bpz2) + (1)(-Bpz1) + (1)(-Bpz2) + (1)(-Bpz1) +(1)(Bpz1) + (1)(Bpz2)]
Ψ (Ag) = 0 Bpz1 + 0 Bpz2 So, there will be an ag orbital
Ψ(B1g)? = [1(Bpz1) + (1)(Bpz2) + (-1)(-Bpz2) + (-1)(-Bpz1) + (1)(-Bpz2) + (1)(-Bpz1) +(-1)(Bpz1) + (-1)(Bpz2)]
Ψ (B1g) = 0 Bpz1 + 0Bpz2 So, there will be no b1g orbital
Ψ(B2g)? = [1(Bpz1) + (-1)(Bpz2) + (1)(-Bpz2) + (-1)(-Bpz1) + (1)(-Bpz2) + (-1)(-Bpz1) +(1)(Bpz1) + (-1)(Bpz2)]
Ψ (B2g) = 4 Bpz1 - 4Bpz2 So, there will be a b2g orbital
Ψ(B3g)? = [1(Bpz1) + (-1)(Bpz2) + (-1)(-Bpz2) + (1)(-Bpz1) + (1)(-Bpz2) + (-1)(-Bpz1) +(-1)(Bpz1) + (1)(Bpz2)]
Ψ (B3g) = 0 Bpz1 + 0Bpz2 So, there will be no b3g orbital
Ψ(Au)? = [1(Bpz1) + (1)(Bpz2) + (1)(-Bpz2) + (1)(-Bpz1) + (-1)(-Bpz2) + (-1)(-Bpz1) +(-1)(Bpz1) + (-1)(Bpz2)]
Ψ (Au) = 0 Bpz1 + 0Bpz2 So, there will be no au orbital
Ψ(B1u)? = [1(Bpz1) + (1)(Bpz2) + (-1)(-Bpz2) + (-1)(-Bpz1) + (-1)(-Bpz2) + (-1)(-Bpz1) +(1)(Bpz1) + (1)(Bpz2)]
Ψ (B1u) = 4 Bpz1 + 4Bpz2 So, there will be a b1u orbital
Ψ(B2u)? = [1(Bpz1) + (-1)(Bpz2) + (1)(-Bpz2) + (-1)(-Bpz1) + (-1)(-Bpz2) + (1)(-Bpz1) +(-1)(Bpz1) + (1)(Bpz2)]Ψ (B2u) = 0 Bpz1 + 0Bpz2 So, there will be no b2u orbital
Ψ(B3u)? = [1(Bpz1) + (-1)(Bpz2) + (-1)(-Bpz2) + (1)(-Bpz1) + (-1)(-Bpz2) + (1)(-Bpz1) +(1)(Bpz1) + (-1)(Bpz2)]
Ψ (B3u) = 0 Bpz1 + 0Bpz2 So, there will be no b3u orbital
So the B 2pz group orbitals are
Finally, to construct MOs for the diborane group, orbitals of the same symmetry should be allowed to mix. The exact way in which this will occur should ideally be predicted by performing a quantum chemical calculation. However, an approximate diagram can be qualitatively estimated from the atomic orbital energies (which are conveniently available in a periodic table of atomic orbital energies). The result is given in 8.2.1. Hydrogen's Chemical Properties, reproduced for convenience below.
3. Share electrons among multiple boron and other atoms in a cluster. Such clusters are very common among the boron hydrides (boranes) and their derivatives. Examples are given in Scheme $\sf{\PageIndex{V}}$.
Scheme $\sf{\PageIndex{V}}$. Examples of electron-deficient boron clusters. These consist of a cluster H-B: groups in which the B atoms are held together by sharing electrons among the atoms of the cluster. The B-B and B---B bonds in these images are drawn to help define the cluster shape, in which the B atoms tend to occupy the vertices of a deltahedron.
Many aspects of the bonding in clusters will be explained in more detail in Section 15.4 Cluster Compounds, which also presents Wade's rules and the Polyhedral Skeletal Electron Pair Theory which may be used to rationalize their stability. This section instead focuses on outlining some features of borane chemistry.
Electron-deficient clusters are distinguished based on whether their shape is that of a complete deltahedron - a type of polyhedron in which all the faces are equilateral triangles. In closo-clusters the vertex atoms comprise a complete deltahedron while in nido-, arachno-, and hypho- clusters, one, two, and three of the deltahedron's vertices are unoccupied, respectively. For instance, in Scheme $\sf{\PageIndex{V}}$, tetrahedral B4H42- and octahedral B6H62- are closo- while square pyramidal B5H9 is nido- since it corresponds to an octahedron with a missing vertex.
Notice also that there are bridging B-H-B bonds along the edges of the open vertices in the nido- and arachno- clusters of Scheme $\sf{\PageIndex{IV}}$. This is a common feature of such clusters, although these bonds are not always present since the hydrogens can often be deprotonated to give stable anionic clusters.
$\sf{B_5H_9~~+~~MH~~\overset{low~T}{\longrightarrow}~~M^+~~+~~B_5H_8^-~~H_2~~~~~~~~~~(M~=~alkali~metal)} \nonumber$
Many boranes are anionic. The most iconic is icosahedral closo-dodecaborate, B12H122-, the structure of which is shown in Figure $\sf{3}$.
Typically, the higher boranes are formed by pyrolytic removal of H2, H- , and/or H+ units from BH3 or BH4-. This process may be conceptualized as giving rise to BH, BH-, and BH2 units which coalesce into clusters. The process may be seen in part by considering the preparation of closo-dodecaborate. First the diborane is pyrolyzed to give arachno-decaborane(14), B10H14, in a process in which the loss of B-H and B-H-B bonds is compensated for by the formation of cluster bonds in which multiple B atoms share the bonding electrons. The resulting arachno-decaborane(14), B10H14, has the icosahedral shape of dodecaborate.
$\nonumber$
The two missing vertices in the arachno- cluster are added by allowing the cluster to react with additional BH3 units, this time in the form of the Et3N-BH3 adduct.
$\nonumber$
Finally, it should be noted that elements other than boron can form clusters and that, consequently, boron forms mixed clusters with a variety of elements. Some examples include the carboranes depicted in Scheme $\sf{\PageIndex{VI}}$.
Scheme $\sf{\PageIndex{VI}}$. Some carborane clusters, including the ortho-, meta-, and para-"carborane", C2B10H12.
Note $\sf1$: Conjuncto Boranes
In addition to ordinary closo-, nido-, arachno-, and hypho- boranes, there are conjuncto-boranes, which consist of two clusters joined via a B-B linkage, at a vertex, edge, or face. An example of such a cluster is shown in Scheme $\sf{\PageIndex{VI}}$.
Scheme $\sf{\PageIndex{VII}}$. A face-sharing conjuncto- borane.
Note $\sf2$: Boron in Medicinal Chemistry: Neutron Capture Therapy
Boranes and their derivatives are the subject of ongoing research into effective neutron capture therapies for the treatment of tumors of the head and neck. The basis of these therapies is the nuclear chemistry of its 10B isotope, which comprises 19.6% of naturally occurring boron (the remaining 80.4% is 11B). Boron-10 can absorb neutrons to give excited 11B, which undergoes radiative decay via alpha emission:
$\sf{Neutron ~capture:~~~~\ce{^{10}_5B~~+~~^1_1n~~\longrightarrow~~[^{11}_5B]^{*}}} \nonumber$
$\sf{alpha~emission:~~~~\ce{[^{11}_5B]^{*}~~\longrightarrow~~^7_3Li~~+~~^4_2He}} \nonumber$
This sequence of reactions in medical applications to treat tumors is called neutron capture therapy. The basic principles of neutron capture therapy are summarized in Figure $\sf{4}$.
As can be seen from Figure $\sf{4}$, the process involves irradiating the boron-infused tumor with neutrons. These neutrons are poorly absorbed by tissues comprised primarily of the big six elements of biochemistry (C, H, N, O, S, and P) and so minimally damage normal tissue. However, in boron-infused tumor cells, $\ce{[^{11}_5B]^{*}}$ produced by neutron capture degrades to give high energy $\sf{^7_3Li}$ and $\sf{^4_2He}$ that damage the surrounding cancerous tissue.
Effective neutron capture therapy requires the use of a boron reagent that is otherwise nontoxic and can be selectively delivered to cancer cells in therapeutically-effective concentrations. This has been a considerable research challenge and the development of such reagents is a subject of ongoing research. Many of the compounds tried so far are borane or carborane amino or nucleic acid analogues or conjugates of borane and various biomolecules, although borane-infused liposomes (liposomes are preferentially taken up and retained by tumors) have recently been shown effective against certain types of tumors in mice.
Contributors and Attributions
Stephen Contakes, Westmont College | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/08%3A_Chemistry_of_the_Main_Group_Elements/8.06%3A_Group_13_%28and_a_note_on_the_post-transition_metals%29/8.6.01%3A_Properties_of_the_Group_13_Elements_and_Boron_Chemis.txt |
The group 13 metals also form conventional compounds as well as M-X-M type bridge bonds and metal-like clusters.
The group 13 metals form a wide variety of compounds that may be explained in terms of ionic or pairwise covalent bonding. These oxides can generally be explained as ionic. Thallium's main oxide, Tl2O, crystallizes in the CdI2 lattice, and consists of layers of edge-linked octahedra, as shown in Figure $\sf{\PageIndex{1A}}$, while the most stable form of Al2O3 , corundum, consists of layers of edge-linked AlO6 octahedra vertex-linked to adjacent layers as shown in Figure $\sf{\PageIndex{1B}}$.
Nevertheless, although it is possible to explain the structural chemistry of Al2O3 as involving an ionic lattice, Al does form covalently-bonded oxyanions just as boron does. A particularly well known example is the Al6O1818- anion, which contains AlO4 tetrahedra covalently vertex-linked into an Al6O6 ring as shown in Scheme $\sf{\PageIndex{I}}$. The calcium salt of this anion is found in tricalcium aluminate cements, which as the name suggests are formed by mixing 3 equivalents of CaO with one of Al2O3.
Scheme $\sf{\PageIndex{I}}$. Two views of the Al6O1818- anion.
The aluminum trihalides are effective Lewis acids. This property serves as the basis for their use as a catalyst in Friedel-Crafts alkylations and acylations. In these reactions, AlCl3 acts as a Lewis base to abstract a chloride ion from an alkyl or acyl halide, generating a carbon-based cation that can add to an aromatic ring in an electrophilic aromatic substitution reaction (Scheme $\sf{\PageIndex{II}}$).
Scheme $\sf{\PageIndex{II}}$. AlCl3 acts as a Lewis acid catalyst in facilitating Friedel Crafts acylation.
Trialkyl group 13 compounds also exist as monomers and dimers, although the monomer-dimer preference increasingly shifts towards monomer on going from Al to In. In general, GaR3 and InR3 species exist as monomers, while for Al the monomer and dimer are often in equilibrium (Scheme $\sf{\PageIndex{II}}$) for smaller alkyl groups, shifting towards monomer with bulky ones.
Scheme $\sf{\PageIndex{II}}$. Dimer-monomer equilibrium of trimethyl aluminum.
As with their boron analogues, trialkyl aluminum and trialkylgallium compounds are effective Lewis acids, reacting with Lewis bases at the group 13 element to give tetrahedral species.
The low electronegativity of the group 13 metals also means that the alkyl groups of group 13 trialkyl complexes are nucleophilic and basic. This reactivity is employed in the commercial use of GaR3 species for the chemical vapor deposition of semiconductor-grade GaAs via reaction of GaR3 with AsH3.
$\sf{Ga(CH_3)_3~~+~~AsH_3~~\longrightarrow~~GaAs(s)~~+~~3~CH_4} \nonumber$
Compounds containing Al-Al and Ga-Ga single and multiple bonds can be prepared, although Al≡Al bonds have so far only been demonstrated in the gas phase.1 In the case of Ga, multiple bonds can be formed by reduction of compounds of type RGaX2 by using extremely bulky R groups to prevent dimerization of any Ga=Ga or Ga≡Ga species formed. The nature of the multiply-bonded species formed this way have been the subject of much theoretical investigation. The 1997 report of the synthesis of a gallyne2 in particular set off a storm of controversy about the nature of its Ga-Ga bond, leading to competing claims that the bond should be understood as involving single, triple, or even double bond interactions. Some of the options are depicted in Figure $\sf{\PageIndex{3A}}$. Although no clear consensus has yet emerged, recent experimental and theoretical work3 has pointed to the description of bonding as involving a $\sigma$ bond, a conventional $\pi$ bond, and a Klinkhammer-type slipped $\pi$ bond, as shown in Figure $\sf{3}$.
Aluminum and gallium are also known to form atomic clusters. There are two types:
1. Naked metal clusters
Examples include anions like Tl77- , Ga117- , and Tl1311- formed by reaction of the metal or metal halide with a powerful reductant (usually an alkali or similar highly reducing metal) and "magic clusters" containing unusually stable species like Al7+ and Al13- formed by laser sputtering of aluminum (i.e., irradiating an Al surface with a laser and watching what sort of clusters come off). Among these, Al13- is so stable that it can even be prepared by solution phase reduction of dendrimer-encapsulated AlCl3 by ketyl radicals. The chemical behavior of the Al13- anion may be rationalized by thinking about it as analogous to a "halide," specifically of the superatom Al13. Superatoms like Al13 are clusters that possess well-defined energy levels and directional valence orbitals just as atoms do; consequently, they exhibit atom-like reactivity in tending to gain, lose, or share electrons to fill a shell of low energy orbitals. In the case of Al11, its reactivity mimics that of the halogens and so it is described as a superhalogen.
Examples of naked metal cluster structures are given in Figure $\sf{4}$.
As may be seen from the structures in Figure $\sf{4}$, these clusters adopt shapes similar to those of the boranes, although as is the case with Tl1311- and Al13- sometimes one of the metal atoms resides in the cluster interior. As might be expected, the stability of many of these clusters is analogous to that of the boranes in being explicable in terms of Wade's rules, although given that Tl1311- and Al13- are isostructural and both stable while differing in their valence electron counts by ten electrons it is clear that this is not always the case.
The existence of multiple stable electron counts in group 13 superatom clusters like Al13- has been explained by the jellium model of cluster stability, which treats clusters like Al13- as consisting of an "ionic core" consisting of the nuclei and their core electrons (Al3+)13 surrounded by the "jelly" of valence electrons. In the case of Al13 clusters, this model gives stable sets of energy levels allowing for configurations containing the "magic numbers" of 2, 8, 18, 20, 34, or 40 electrons. In this model, the 40 valence electrons in Al13- correspond to the last of these stable states. Thus Tl1311- and Al13- represent different regimes of stability, one consistent with a Wade-type cluster similar to the boranes explicable in terms of molecular orbitals formed by combinations of atomic orbitals and the other one explicable in terms of the orbitals given by treating Al13 as a superatom in the jellium model.
1. Metalloid clusters
Metalloid clusters consist of a small particle of metal surrounded by ligands. Some such Al and Ga clusters even contain upwards of 69 atoms. Examples include clusters with formulas [Al69{N(SiMe3)2}18]3- and [Al77N(SiMe3)2}20]2-. The structure of a relatively small example is given in Figure $\sf{5}$.
Although the group 13 metals have varying tendencies to form covalent bonds, they tend to act chemically as metals.
Some representative features of these elements illustrative of their chemistry are summarized in Table $\sf{1}$.
Table $\sf{1}$. Illustrative properties of the group 13 metals.9,10
Group 13 Metal Natural Source
Oxides
(major product of reaction of the element with O2 is given in bold)
Illustrative Stable Mononuclear Halides Redox Behavior in Acidic Solution
Aluminum, Al
Aluminosilicates like
• feldspars of formula MAlSi3O8 or M'Al2Si2O8 where M = alkali metal and M' = alkaline earth metal
• clays like kaolinite, Al2Si2O5(OH)4
• Bauxites containing Al2O3·xH2O
Al2O3 AlX3 (X = F, Cl, Br, I)
Gallium, Ga
As impurities in Aluminum and Fe/Zn ores
• Bauxites, Al2O3·xH2O
• Sphalerites, ZnS
Small amounts in Gallite, CuGaS2
Ga2O3
Ga2O
Network covalent GaF3
Dimeric Ga2X6 (X = Cl, Br, I)
GaIGaIII2Cl7
GaIGaIIIX4 (X = Cl, Br, I)
Indium, In
As impurities in Zn and Cu/Fe ores
• Sphalerites, ZnS
• Chalcopyrite, CuFeS2
In2O3
InX3 (X = F, Cl, Br, I)
InI3(InIIIX6) (X = Cl, Br)
InIInIIIX4 (X = Br, I)
InX (X = Cl, Br, I)
Thallium, Tl
As impurities in sulfide-rich ores
• CuFeS2, CuS, Cu2S
• PbS
• ZnS
small amounts in minerals like TlCu7Se4, TlPbAs5S9, others
Tl2O3
Tl2O
TlF3
TlX (X = F, Cl, Br, I)
thallium(I) triiodide, TlI(I3)
As may be seen from the data in Table $\sf{1}$, on moving down group 13 from Al to Tl
1. Geologic occurrence of the elements in oxide ores (Al, Ga) becomes increasingly replaced by a preference for occurrence in sulfides and selenides (Ga, In, Tl). This is consistent with increasing softness on moving down group 13.
2. There is an increasing preference for the +1 oxidation state. This may be seen from
• the standard potentials for reduction of M3+ and M+ to the metal. These show a decrease in the stability of the +3 ion on going down the group. In the case of Tl, Tl3+(aq) is even less stable than Tl+(aq).
• an increase in prevalence of the +1 oxidation state in the monomeric halides. The aluminum monohalides, AlX, are highly unstable; the bromide and chloride of GaI stabilizable (but perhaps not monomeric); the InI halides stable only for the less oxidizing halogens Cl, Br, and I; and for thallium, TlF3 is the only stable Tl3+ halide.
This increasing preference for an oxidation state two lower than maximum valence is not restricted to the group 13 elements. It is a common feature of post transition element chemistry that these elements can act as if their ns2 valence electrons are inert, on account of which this tendency has been referred to somewhat misleadingly as the inert pair effect.
Post-transition metals exhibit the inert pair effect, in which they act as if their ns2 valence electrons do not contribute to bonding.
Metals and metalloids of the p block commonly possess two stable oxidation states, with one corresponding to the loss of all their ns and np valence electrons and one the loss of two fewer electrons. This is evident from the well-known ions and oxidation states observed for the group 13, 14, and 15 metals and metalloids shown in Scheme $\sf{\PageIndex{III}}$.
Scheme $\sf{\PageIndex{III}}$. Common oxidation states of group 13, 14, and 15 metals and metalloids.
Moreover, there is an increasing preference for the lower oxidation state on going down a group in the periodic table so that
• The higher (n+) oxidation state is favored for lighter elements so that the order of preference is
Al3+ > Ga3+ > In3+ > Tl3+
• The lower (n - 2)+ oxidation state is favored for heavier elements
Al+ < Ga+ < In+ < Tl+
This is why the most stable oxide and chloride of Al are Al2O3 and AlCl3 while the most stable oxides and Cl of Tl are Tl2O and TlCl.
The classical explanation for the existence of this behavior is to postulate that the heavier elements' ns2 valence electrons are chemically inert - i.e., an inert pair. For this reason the observation that many post transition elements have stable n+ and (n-2)+ oxidation states and that there is an increasing preference for the lower oxidation state on moving down a group has been termed the inert pair effect.
The inert pair effect is due to the decrease in bond energies down a group.
However, the term inert pair effect is a misnomer. There are two reasons for this
1. The ns electrons do not become significantly more inert (a.k.a. lower in energy) as one descends the groups of the periodic table. A cursory look at the valence s and p orbital energies of the main group elements reveals that it is simply not the case that the ns orbitals become lower in energy on going down a group of the periodic table. As may be seen from the energies given in Figure $\sf{6}$, the energy of the ns orbitals is generally lowest for period 1 and 2 elements, after which the only overall trend that may be noted is that the ns orbital energies of the rows 3-6 elements generally do not differ by more than 20% across the entire range.
2. The inert pair effect is due to the decrease in bond energies as bond lengths increase down a group of the periodic table. To see how this leads to the appearance of ns electron inertness as one descends the periodic table, consider that the process of forming an E-X bond generally involves
• endergonic oxidation of the metal
• exergonic formation of a M-X bond
Consider the interplay between these two energies in the case of forming the monohalides and trihalides of the group 13 elements. The cost of oxidation of the metal reflects a balance between the ionization energies given in Table $\sf{2}$ and bond energies given in Table $\sf{3}$.15
Table $\sf{2}$. Ionization energy costs for formation of M+ and M3+. Recalculated from the similar table of ionization energies at en.Wikipedia.org/wiki/Inert_pair_effect.
Process B Al Ga In Tl
M $\rightarrow$ M+
IE1 (kJ/mol)
800 577 578 558 589
M $\rightarrow$ M3+
IE1 + IE2 + IE3 (kJ/mol)
6886 5137 5520 5082 5438
Inert pair oxidation:
M+ $\rightarrow$ M3+
IE2 + IE3 (kJ/mol)
6,086 4,560 4,942 4,524 4,849
Table $\sf{3}$. M-Cl homolytic bond energies in kJ/mol.15
Bond B-Cl Al-Cl Ga-Cl In-Cl Tl-Cl
Typical E-Cl Bond Dissociation energy (kJ/mol) 536 494 481 439 372.8
As may be seen from Table $\sf{3}$, the energy cost for ionization of the "inert pair" is greatest for boron, drops by ~25% on going to Al, and then remains approximately constant for the remaining members of the group. In contrast, the compensating bond dissociation energies drop more slowly on going from B to Tl, with successive ~10% and ~20% decreases on going from Ga to In and In to Tl. Because bond energies drop on going from Al to Tl while ionization energies do not, it is thought that the inert pair effect isn't due to the inertness of the ns electrons but rather due to the weakening of bond energies. When the stabilization energy due to M-E bond formation is no longer enough to pay the cost needed to oxidize away the ns electrons, the lower oxidation state will be more stable than the higher one, just as it would if the ns electrons really were inert.
The inert pair of electrons exerts structural effects.
Regardless of the origin of the inert pair effect, the inert pair of electrons influences the structure of compounds in the (n-2)+ oxidation state. Electron pairs that do this by altering the observed geometry of compounds are said to be stereochemically active inert pairs. The lone pairs of SnIICl2 are the classic example. As shown in Figure $\sf{\PageIndex{7A}}$, gas phase SnIICl2 exhibits an AX2E VSEPR geometry, while SnIICl2 in its solid state, hydrates, and salts all exhibit AX3E geometries, consistent with the influence of a lone pair of electrons.
Sometimes the coordination geometry of an element is exactly that which would be expected if the inert pair were not present. In such cases the inert pair is said to be stereochemically inert. In general, when the element's coordination number is less than six, the inert pairs will be stereochemically active, while for coordination numbers of six or more the pair may be stereochemically active or stereochemically inert.
Even when an inert pair appears stereochemically inert, it does not mean that it exerts no effect on the structure of the compound. It may be that the pair is simply fluxional and only appears to be inert because it is rapidly adjusting its position among possible locations. For instance, in the structure of BiI3 shown in Figure $\sf{\PageIndex{7B}}$, the lone pair is likely interchanging between the six faces of the octahedron defining the coordination sphere, as is the case for XeF6 (Figure $\sf{8}$).
Contributors and Attributions
Stephen Contakes, Westmont College | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/08%3A_Chemistry_of_the_Main_Group_Elements/8.06%3A_Group_13_%28and_a_note_on_the_post-transition_metals%29/8.6.02%3A_Heavier_Elements_of_Group_13_and_the_Inert_Pair_Effe.txt |
The group 14 elements comprise the elements carbon (C) through flerovium (Fl) in group 14 of the periodic table, as shown in Figure \(\sf{1}\).
Contributors and Attributions
Stephen Contakes, Westmont College
8.07: Group 14
As an element, carbon's ability to form strong multiple bonds enables it to form many allotropes, while the remaining elements in this group are network covalent or metallic solids.
Selected properties of the Group 14 elements are given in Table $\sf{1}$.
Table $1$: Selected Properties of Group 14 Elements other than Flerovium. For more information about how this data was sourced see note 1.
Property Carbon Silicon Germanium Tin Lead
*The configuration shown does not include filled d and f subshells.
The values cited are for six-coordinate +4 ions in the most common oxidation state, except for C4+ and Si4+, for which values for the four-coordinate ion are estimated.
X is Cl, Br, or I. Reaction with F2 gives the tetrafluorides (EF4) for all group 14 elements, where E represents any group 14 element.
atomic symbol C Si Ge Sn Pb
atomic number 6 14 32 50 82
atomic mass (amu) 12.01 28.09 72.64 118.71 207.2
valence electron configuration* 2s22p2 3s23p2 4s24p2 5s25p2 6s26p2
melting point/boiling point (°C) 4489 (at 10.3 MPa)/3825 1414/3265 938/2833 232/2586 327/1749
density (g/cm3) at 25°C 2.2 (graphite), 3.51 (diamond) 2.33 5.32 7.27(white) 11.30
atomic radius (Å) 1.70 2.10 2.11 2.17 2.02
covalent radius (Å) 0.75 1.14 1.20 1.40 1.45
first ionization energy (kJ/mol) 1086 787 762 709 716
electron affinity (kJ/mol) −122 −134 −119 −107 −35
Pauling electronegativity 2.55 1.90 2.01 1.96 1.8
product of reaction with O2 CO2, CO SiO2 GeO2 SnO2 PbO
type of oxide acidic (CO2) acidic neutral (CO) amphoteric amphoteric amphoteric
product of reaction with N2 none Si3N4 none Sn3N4 none
product of reaction with X2 CX4 SiX4 GeX4 SnX4 PbX2
product of reaction with H2 CH4 none none none none
As may be seen from the data in Table $\sf{1}$, consistent with general atomic property trends, atomic radii increase down the group while electron affinities, ionization energies, and electronegativities generally decrease. The trends are not always monotonic, however, likely due to the impact of the d-block contraction on the atomic properties of Ge and also the lanthanide and actinide contractions on those of Sn and Pb.
From the common elemental forms of the group 14 elements in Table $\sf{1}$ two trends are apparent:
1. In keeping with the second row uniqueness principle, carbon forms a wider variety of allotropes than the other group 14 elements. Specifically, carbon's ability to form strong $\pi$ bonds enables it to form delocalized sheet-like structures in the form of graphite and graphene (planar sheets) and nanotubes and fullerenes, which may be considered as sheets wrapped over to form tubes and ball-like structures, respectively.
2. There is a decreasing tendency towards covalency down the group. While C, Si, and Ge form network covalent structures and Pb a metallic one, tin forms both network covalent and metallic structures, which differ so slightly in their stability that the metallic form interconverts to the network covalent one above 13$^{\circ}$C.
Carbon's ability to form strong multiple bonds enables it to form a host of molecular and network covalent allotropes
Carbon's ability to form strong multiple bonds enables it to form a host of molecular and network covalent allotropes. Some of these allotropes have been known since ancient times; some are a focus of much active research. The major classes of allotropes include:
1. $\alpha$- and $\beta$-Graphite. The $\alpha$ form of graphite is the most stable carbon allotrope. It is formed naturally through the decomposition of carbonaceous material in sediments but may also be made synthetically by vaporizing the Si out of SiC.
$\sf{SiC~~\overset{\Delta}{\longrightarrow}~~~C(s)~~+~~Si(g)} \nonumber$
The structure of $\alpha$-graphite consists of hexagonal layers of interconnected rings of graphene sheets, the structure of which is shown in Scheme $\sf{\PageIndex{I}}$.
Scheme $\sf{\PageIndex{I}}$. One resonance structure depiction of part of a graphene sheet.
All the carbons in the graphene sheet are trigonal planar and connected by a network of covalent bonds that may be adequately described as involving sigma overlaps between sp2 -orbitals and a delocalized $\pi$ bond network. This delocalized $\pi$ bonding system involves a half-filled valence band of $\pi$ type orbitals and is responsible for graphite's good electrical conductivity.
In terms of the impact of the delocalized $pi$ bonds on graphite's structure, it ensures that the graphene sheets are planar with delocalized $\pi$ clouds above and below the carbon plane. Because of this, the sheets may be stacked together in various arrangements in which the carbons in adjacent layers may be eclipsed or staggered with respect to one another, as shown in Figure $\sf{1}$. The stable $\alpha$-graphite form possesses staggered layers alternating an AB pattern, while in the slightly less stable $\beta$-graphite, the layers are staggered in an ABC arrangement (Figure $\sf{\PageIndex{1b}}$).
Since the forces between the graphene layers do not vary much as the sheets move past one another, it is relatively easy for layer slip to occur; graphite thus functions effectively as a lubricant. In addition, it is possible to exfoliate individual graphene sheets using scotch tape and chemical exfoliation methods.
2. Graphene. Individual graphene sheets like that depicted in Scheme $\sf{\PageIndex{I}}$ have been the subject of intense experimental and theoretical investigation as a 2D nanomaterial. Owing to the strong carbon-carbon bonds that hold the sheet together, defect-free graphene is the strongest material known, albeit a brittle one; it also has a band structure that makes it a good conductor of electricity and leads to unusual optical properties. Among the latter, even a single graphene layer is visible to the naked eye, as shown in Figure $\sf{2}$.
3. Diamond and Lonsdaleite. When subjected to extremes of heat and pressure, graphite converts to its more dense diamond allotrope, in which all the carbons are tetrahedral and held together by sigma bonds, as shown in Figure $\sf{3}$. In this structure, all the carbons are tetrahedral and connected in chair conformation rings. As might be expected from such a tightly interconnected structure, diamond is extremely hard, although analogous structures which also possess chair form rings, specifically Wurtzite-like hexagonal BN and its all-carbon analogue Lonsdaleite, are even harder. As might be expected from diamond's all-$\sigma$ structure, it is a poor electrical conductor, and the 1.544 Å C-C bond length in diamond is longer than the 1.415 Å C-C bond length in graphite (1.415 Å).
4. Fullerenes. Since the late 1980s discovery of the sixty-carbon European football (soccer ball) shaped buckminsterfullerene shown in Figure $\sf{\PageIndex{4A}}$, it has been recognized that such molecules have long been present in soot, carbon black, and related materials. Produced today by electrolytic or laser ablation or pyrolysis, a large number of such structures have been produced, some of which are shown in Figure $\sf{\PageIndex{4B}}$.
As shown in Figure $\sf{6}$, the overall linear chains of connected carbon atoms within a graphene sheet can be classified as taking on either a zigzag or armchair arrangement. Nanotubes then are classified as armchair or zigzag based on the arrangement of the bonds perpendicular to the tube axis (Figure $\sf{7}$, top). Alternatively, armchair nanotubes may be envisioned as formed by connecting the zigzag path ends of a graphene sheet in which the direction of the zigzag path is oriented along the nanotube axis, while in zigzag nanotubes the connected zigzag paths are angled relative to the helix axis (Figure $\sf{7}$, bottom).
In ordinary carbon nanotubes the armchair or zigzag path around the tube forms a closed loop. Such carbon nanotubes are relatively linear. In helical nanotubes like that shown in Figure $\sf{7}$, the armchair or zigzag paths form a helical spiral around the nanotube surface instead. In order for this to take place, the carbon rings cannot be perfectly hexagonal but are instead twisted, and to relieve the resulting strain helical nanotubes take on a coiled shape.
Depending on the size of the nanotube, it is sometimes even possible to nest multiple nanotubes inside one another, to "bud" fullerenes off the side of a nanotube, or even to fit fullerenes inside to give a "carbon peapod". Some of the possibilities are depicted in Figure $\sf{8}$.
The chemical properties of fullerenes and nanotubes reflect their structure in the following ways:
1. They possess a delocalized $\pi$ structure and so can undergo facile oxidation and reduction reactions. For instance, buckminsterfullerene can be electrochemically reduced in six successive one-electron steps to give C606-. Similar reductions can occur by reacting a fullerene with an active metal. In many cases this simply results in a salt of a fulleride anion, but in some cases the metal goes inside the fullerene cage to give an endohedral fullerene, designated M@C60. A model of such a fullerene is given in Figure $\sf{9}$.
2. The carbons in these systems are all quaternary and so generally only undergo addition reactions, not substitutions.
3. While the planar sheets of graphite contain trigonal planar carbons, the carbons in fullerenes and nanotubes must be slightly bent into a pyramidal shape in order to accommodate the concavity of those structures. This results in considerable steric strain that can be relieved when addition occurs at a carbon to render it tetrahedral. Because of this, fullerenes undergo a variety of nucleophilic and cycloaddition reactions, primarily at the C=C bonds joining two six-membered rings.
Readers desiring a more thorough presentation of the chemistry of fullerenes are invited to read this summary of carbon nanomaterials by Andrew Barron of Rice University.
6. Vitreous carbon. Glass-like carbon, commonly referred to as glassy or vitreous carbon, is believed to consist of a mixture of concave graphene-like sheets, similar to those in nanotubes and fullerenes but not completely wrapped to form closed balls or tubes. Such forms of carbon are commonly used in electrode materials in electrochemistry (e.g., glassy carbon electrodes).
7. High-carbon content amorphous materials. These are technically not a single form or a pure allotrope but rather a family of high-carbon content materials that include amorphous carbon, carbon black, soot, pyrolyzed coal, coal, and charcoal. These contain varying amounts of hydrocarbons (particularly polycyclic aromatic hydrocarbons) and other organic material. Many of these materials contain small crystallites of diamond and graphite-like regions interconnected by an amorphous matrix of carbon-rich material. Some representative structures are given in Figure $\sf{10}$.
8. Graphyne - a possible future carbon allotrope? The discovery of fullerene and carbon nanotube structures held together exclusively by interconnected C-C and C=C bonds raises the issue of whether it might be possible to construct forms of carbon that include or are even primarily comprised of C≡C bonds. These materials, called graphynes, by analogy with the double bonded structure of graphene, are predicted to be theoretically stable but as yet have been neither observed in nature nor made experimentally. A selection of proposed graphyne structures is given in Figure $\sf{11}$.
Note: A few analytically notable group 14 isotopes
A number of group 14 isotopes find use in analytical applications. These include:
13C, a spin 1/2 NMR active isotope comprising 1% of natural carbon. It is widely used in carbon-13 NMR to structurally characterize organic and organometallic compounds. For more information on carbon 13 NMR and its use in organic chemistry see this page.
14C, an unstable isotope that occurs at ~ 1 parts per trillion in the atmosphere and undergoes radioactive decay with a half-life of 5730 years. Since the level of 14C in organisms is approximately equal to atmospheric levels at their time of death and decays thereafter, the ratio of 14C to 12C in a biologically-derived sample may be used to estimate the age of objects ~60,000 years old or less. For more information on radiocarbon dating see this chem libre texts page.
29Si, a spin 1/2 NMR active isotope comprising ~5% of naturally occurring silicon. Silicon-29 NMR is useful as a structural tool for characterizing silicon compounds ranging from glasses to organosilanes. Although there is not yet a Chem Libre texts page describing the technique, a brief introduction may be found here.
73Ge, a spin 9/2 nucleus, and spin 1/2 115Sn, 117Sn, 119Sn, and 207Pb are also NMR active isotopes that can be used to structurally characterize compounds of these elements. Their use is beyond the scope of this introductory text but for those interested, more information about the use of these nuclei in NMR is available for Ge, Sn, and Pb. In addition 73Ge is Mössbauer active, although relatively few examples of its use in the characterization of Ge in Ge-containing materials exist.
Contributors and Attributions
Stephen Contakes, Westmont College | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/08%3A_Chemistry_of_the_Main_Group_Elements/8.07%3A_Group_14/8.7.01%3A_The_Group_14_Elements_and_the_many_Allotropes_of_Carbon.txt |
Reactions and Compounds of Carbon
Carbon is the building block of all organic compounds, including biomolecules, fuels, pharmaceuticals, and plastics, whereas inorganic compounds of carbon include metal carbonates, which are found in substances as diverse as fertilizers and antacid tablets, halides, oxides, carbides, and carboranes. Like boron in group 13, the chemistry of carbon differs sufficiently from that of its heavier congeners to merit a separate discussion.
The structures of the allotropes of carbon—diamond, graphite, fullerenes, and nanotubes—are distinct, but they all contain simple electron-pair bonds (Figure 7.18). Although it was originally believed that fullerenes were a new form of carbon that could be prepared only in the laboratory, fullerenes have been found in certain types of meteorites. Another possible allotrope of carbon has also been detected in impact fragments of a carbon-rich meteorite; it appears to consist of long chains of carbon atoms linked by alternating single and triple bonds, (–C≡C–C≡C–)n. Carbon nanotubes (“buckytubes”) are being studied as potential building blocks for ultramicroscale detectors and molecular computers and as tethers for space stations. They are currently used in electronic devices, such as the electrically conducting tips of miniature electron guns for flat-panel displays in portable computers.
Although all the carbon tetrahalides (CX4) are known, they are generally not obtained by the direct reaction of carbon with the elemental halogens (X2) but by indirect methods such as the following reaction, where X is Cl or Br:
$CH_{4(g)} + 4X_{2(g)} \rightarrow CX_{4(l,s)} + 4HX_{(g)} \label{$7$}$
The carbon tetrahalides all have the tetrahedral geometry predicted by the valence-shell electron-pair repulsion (VSEPR) model, as shown for CF4 and CI4. Their stability decreases rapidly as the halogen increases in size because of poor orbital overlap and increased crowding. Because the C–F bond is about 25% stronger than a C–H bond, fluorocarbons are thermally and chemically more stable than the corresponding hydrocarbons, while having a similar hydrophobic character. A polymer of tetrafluoroethylene (F2C=CF2), analogous to polyethylene, is the nonstick Teflon lining found on many cooking pans, and similar compounds are used to make fabrics stain resistant (such as Scotch-Gard) or waterproof but breathable (such as Gore-Tex).
The stability of the carbon tetrahalides decreases with increasing size of the halogen due to increasingly poor orbital overlap and crowding.
Carbon reacts with oxygen to form either CO or CO2, depending on the stoichiometry. Carbon monoxide is a colorless, odorless, and poisonous gas that reacts with the iron in hemoglobin to form an Fe–CO unit, which prevents hemoglobin from binding, transporting, and releasing oxygen in the blood (see Figure 23.26). In the laboratory, carbon monoxide can be prepared on a small scale by dehydrating formic acid with concentrated sulfuric acid:
$\mathrm{HCO_2H(l)}\xrightarrow{\mathrm{H_2SO_4(l)}}\mathrm{CO(g)}+\mathrm{H_3O^+(aq)}+\mathrm{HSO^-_4}\label{$8$}$
Carbon monoxide also reacts with the halogens to form the oxohalides (COX2). Probably the best known of these is phosgene (Cl2C=O), which is highly poisonous and was used as a chemical weapon during World War I:
$\mathrm{CO(g)}+\mathrm{Cl_2(g)}\xrightarrow{\Delta}\textrm{Cl}_2\textrm{C=O(g)}\label{$9$}$
Despite its toxicity, phosgene is an important industrial chemical that is prepared on a large scale, primarily in the manufacture of polyurethanes.
Carbon dioxide can be prepared on a small scale by reacting almost any metal carbonate or bicarbonate salt with a strong acid. As is typical of a nonmetal oxide, CO2 reacts with water to form acidic solutions containing carbonic acid (H2CO3). In contrast to its reactions with oxygen, reacting carbon with sulfur at high temperatures produces only carbon disulfide (CS2):
$\mathrm{C(s)}+\mathrm{2S(g)}\xrightarrow{\Delta}\mathrm{CS_2(g)}\label{$1$0}$
The selenium analog CSe2 is also known. Both have the linear structure predicted by the VSEPR model, and both are vile smelling (and in the case of CSe2, highly toxic), volatile liquids. The sulfur and selenium analogues of carbon monoxide, CS and CSe, are unstable because the C≡Y bonds (Y is S or Se) are much weaker than the C≡O bond due to poorer π orbital overlap.
$\pi$ bonds between carbon and the heavier chalcogenides are weak due to poor orbital overlap.
Binary compounds of carbon with less electronegative elements are called carbides. The chemical and physical properties of carbides depend strongly on the identity of the second element, resulting in three general classes: ionic carbides, interstitial carbides, and covalent carbides. The reaction of carbon at high temperatures with electropositive metals such as those of groups 1 and 2 and aluminum produces ionic carbides, which contain discrete metal cations and carbon anions. The identity of the anions depends on the size of the second element. For example, smaller elements such as beryllium and aluminum give methides such as Be2C and Al4C3, which formally contain the C4− ion derived from methane (CH4) by losing all four H atoms as protons. In contrast, larger metals such as sodium and calcium give carbides with stoichiometries of Na2C2 and CaC2. Because these carbides contain the C4− ion, which is derived from acetylene (HC≡CH) by losing both H atoms as protons, they are more properly called acetylides. Reacting ionic carbides with dilute aqueous acid results in protonation of the anions to give the parent hydrocarbons: CH4 or C2H2. For many years, miners’ lamps used the reaction of calcium carbide with water to produce a steady supply of acetylene, which was ignited to provide a portable lantern.
The reaction of carbon with most transition metals at high temperatures produces interstitial carbides. Due to the less electropositive nature of the transition metals, these carbides contain covalent metal–carbon interactions, which result in different properties: most interstitial carbides are good conductors of electricity, have high melting points, and are among the hardest substances known. Interstitial carbides exhibit a variety of nominal compositions, and they are often nonstoichiometric compounds whose carbon content can vary over a wide range. Among the most important are tungsten carbide (WC), which is used industrially in high-speed cutting tools, and cementite (Fe3C), which is a major component of steel.
Elements with an electronegativity similar to that of carbon form covalent carbides, such as silicon carbide (SiC; Equation $\ref{Eq1}$) and boron carbide (B4C). These substances are extremely hard, have high melting points, and are chemically inert. For example, silicon carbide is highly resistant to chemical attack at temperatures as high as 1600°C. Because it also maintains its strength at high temperatures, silicon carbide is used in heating elements for electric furnaces and in variable-temperature resistors.
Carbides formed from group 1 and 2 elements are ionic. Transition metals form interstitial carbides with covalent metal–carbon interactions, and covalent carbides are chemically inert.
Example $1$
For each reaction, explain why the given product forms.
1. CO(g) + Cl2(g) → Cl2C=O(g)
2. CO(g) + BF3(g) → F3B:C≡O(g)
3. Sr(s) + 2C(s) $\xrightarrow{\Delta}$ SrC2(s)
Given: balanced chemical equations
Asked for: why the given products form
Strategy:
Classify the type of reaction. Using periodic trends in atomic properties, thermodynamics, and kinetics, explain why the observed reaction products form.
Solution
1. Because the carbon in CO is in an intermediate oxidation state (+2), CO can be either a reductant or an oxidant; it is also a Lewis base. The other reactant (Cl2) is an oxidant, so we expect a redox reaction to occur in which the carbon of CO is further oxidized. Because Cl2 is a two-electron oxidant and the carbon atom of CO can be oxidized by two electrons to the +4 oxidation state, the product is phosgene (Cl2C=O).
2. Unlike Cl2, BF3 is not a good oxidant, even though it contains boron in its highest oxidation state (+3). Nor can BF3 behave like a reductant. Like any other species with only six valence electrons, however, it is certainly a Lewis acid. Hence an acid–base reaction is the most likely alternative, especially because we know that CO can use the lone pair of electrons on carbon to act as a Lewis base. The most probable reaction is therefore the formation of a Lewis acid–base adduct.
3. Typically, both reactants behave like reductants. Unless one of them can also behave like an oxidant, no reaction will occur. We know that Sr is an active metal because it lies far to the left in the periodic table and that it is more electropositive than carbon. Carbon is a nonmetal with a significantly higher electronegativity; it is therefore more likely to accept electrons in a redox reaction. We conclude, therefore, that Sr will be oxidized, and C will be reduced. Carbon forms ionic carbides with active metals, so the reaction will produce a species formally containing either C4− or C22. Those that contain C4− usually involve small, highly charged metal ions, so Sr2+ will produce the acetylide (SrC2) instead.
Exercise $1$
Predict the products of the reactions and write a balanced chemical equation for each reaction.
1. C(s) + excess O2(g) $\xrightarrow{\Delta}$
2. C(s) + H2O(l) →
3. NaHCO3(s) + H2SO4(aq) →
Answer
1. C(s) + excess O2(g) $\xrightarrow{\Delta}$ CO2(g)
2. C(s) + H2O(l) → no reaction
3. NaHCO3(s) + H2SO4(aq) → CO2(g) + NaHSO4(aq) + H2O(l)
Reactions and Compounds of the Heavier Group 14 Elements
Although silicon, germanium, tin, and lead in their +4 oxidation states often form binary compounds with the same stoichiometry as carbon, the structures and properties of these compounds are usually significantly different from those of the carbon analogues. Silicon and germanium are both semiconductors with structures analogous to diamond. Tin has two common allotropes: white (β) tin has a metallic lattice and metallic properties, whereas gray (α) tin has a diamond-like structure and is a semiconductor. The metallic β form is stable above 13.2°C, and the nonmetallic α form is stable below 13.2°C. Lead is the only group 14 element that is metallic in both structure and properties under all conditions.
Video $1$: Time lapse tin pest reaction.
Based on its position in the periodic table, we expect silicon to be amphoteric. In fact, it dissolves in strong aqueous base to produce hydrogen gas and solutions of silicates, but the only aqueous acid that it reacts with is hydrofluoric acid, presumably due to the formation of the stable SiF62 ion. Germanium is more metallic in its behavior than silicon. For example, it dissolves in hot oxidizing acids, such as HNO3 and H2SO4, but in the absence of an oxidant, it does not dissolve in aqueous base. Although tin has an even more metallic character than germanium, lead is the only element in the group that behaves purely as a metal. Acids do not readily attack it because the solid acquires a thin protective outer layer of a Pb2+ salt, such as PbSO4.
All group 14 dichlorides are known, and their stability increases dramatically as the atomic number of the central atom increases. Thus CCl2 is dichlorocarbene, a highly reactive, short-lived intermediate that can be made in solution but cannot be isolated in pure form using standard techniques; SiCl2 can be isolated at very low temperatures, but it decomposes rapidly above −150°C, and GeCl2 is relatively stable at temperatures below 20°C. In contrast, SnCl2 is a polymeric solid that is indefinitely stable at room temperature, whereas PbCl2 is an insoluble crystalline solid with a structure similar to that of SnCl2.
The stability of the group 14 dichlorides increases dramatically from carbon to lead.
Although the first four elements of group 14 form tetrahalides (MX4) with all the halogens, only fluorine is able to oxidize lead to the +4 oxidation state, giving PbF4. The tetrahalides of silicon and germanium react rapidly with water to give amphoteric oxides (where M is Si or Ge):
$MX_{4(s,l)} + 2H_2O_{(l)} \rightarrow MO_{2(s)} + 4HX_{(aq)} \label{$1$1}$
In contrast, the tetrahalides of tin and lead react with water to give hydrated metal ions. Because of the stability of its +2 oxidation state, lead reacts with oxygen or sulfur to form PbO or PbS, respectively, whereas heating the other group 14 elements with excess O2 or S8 gives the corresponding dioxides or disulfides, respectively. The dioxides of the group 14 elements become increasingly basic as we go down the group.
The dioxides of the group 14 elements become increasingly basic down the group.
Because the Si–O bond is even stronger than the C–O bond (~452 kJ/mol versus ~358 kJ/mol), silicon has a strong affinity for oxygen. The relative strengths of the C–O and Si–O bonds contradict the generalization that bond strengths decrease as the bonded atoms become larger. This is because we have thus far assumed that a formal single bond between two atoms can always be described in terms of a single pair of shared electrons. In the case of Si–O bonds, however, the presence of relatively low-energy, empty d orbitals on Si and nonbonding electron pairs in the p or spn hybrid orbitals of O results in a partial π bond (Figure $3$). Due to its partial π double bond character, the Si–O bond is significantly stronger and shorter than would otherwise be expected. A similar interaction with oxygen is also an important feature of the chemistry of the elements that follow silicon in the third period (P, S, and Cl). Because the Si–O bond is unusually strong, silicon–oxygen compounds dominate the chemistry of silicon.
Because silicon–oxygen bonds are unusually strong, silicon–oxygen compounds dominate the chemistry of silicon.
Compounds with anions that contain only silicon and oxygen are called silicates, whose basic building block is the SiO44 unit:
The number of oxygen atoms shared between silicon atoms and the way in which the units are linked vary considerably in different silicates. Converting one of the oxygen atoms from terminal to bridging generates chains of silicates, while converting two oxygen atoms from terminal to bridging generates double chains. In contrast, converting three or four oxygens to bridging generates a variety of complex layered and three-dimensional structures, respectively.
In a large and important class of materials called aluminosilicates, some of the Si atoms are replaced by Al atoms to give aluminosilicates such as zeolites, whose three-dimensional framework structures have large cavities connected by smaller tunnels (Figure $4$). Because the cations in zeolites are readily exchanged, zeolites are used in laundry detergents as water-softening agents: the more loosely bound Na+ ions inside the zeolite cavities are displaced by the more highly charged Mg2+ and Ca2+ ions present in hard water, which bind more tightly. Zeolites are also used as catalysts and for water purification.
Silicon and germanium react with nitrogen at high temperature to form nitrides (M3N4):
$3Si_{(l)} + 2N_{2(g)} \rightarrow Si_3N_{4(s)} \label{$1$2}$
Silicon nitride has properties that make it suitable for high-temperature engineering applications: it is strong, very hard, and chemically inert, and it retains these properties to temperatures of about 1000°C.
Because of the diagonal relationship between boron and silicon, metal silicides and metal borides exhibit many similarities. Although metal silicides have structures that are as complex as those of the metal borides and carbides, few silicides are structurally similar to the corresponding borides due to the significantly larger size of Si (atomic radius 111 pm versus 87 pm for B). Silicides of active metals, such as Mg2Si, are ionic compounds that contain the Si4 ion. They react with aqueous acid to form silicon hydrides such as SiH4:
$Mg_2Si_{(s)} + 4H^+_{(aq)} \rightarrow 2Mg^{2+}_{(aq)} + SiH_{4(g)} \label{$1$3}$
Unlike carbon, catenated silicon hydrides become thermodynamically less stable as the chain lengthens. Thus straight-chain and branched silanes (analogous to alkanes) are known up to only n = 10; the germanium analogues (germanes) are known up to n = 9. In contrast, the only known hydride of tin is SnH4, and it slowly decomposes to elemental Sn and H2 at room temperature. The simplest lead hydride (PbH4) is so unstable that chemists are not even certain it exists. Because E=E and E≡E bonds become weaker with increasing atomic number (where E is any group 14 element), simple silicon, germanium, and tin analogues of alkenes, alkynes, and aromatic hydrocarbons are either unstable (Si=Si and Ge=Ge) or unknown. Silicon-based life-forms are therefore likely to be found only in science fiction.
The stability of group 14 hydrides decreases down the group, and E=E and E≡E bonds become weaker.
The only important organic derivatives of lead are compounds such as tetraethyllead [(CH3CH2)4Pb]. Because the Pb–C bond is weak, these compounds decompose at relatively low temperatures to produce alkyl radicals (R·), which can be used to control the rate of combustion reactions. For 60 yr, hundreds of thousands of tons of lead were burned annually in automobile engines, producing a mist of lead oxide particles along the highways that constituted a potentially serious public health problem. (Example 6 in Section 22.3 examines this problem.) The use of catalytic converters reduced the amount of carbon monoxide, nitrogen oxides, and hydrocarbons released into the atmosphere through automobile exhausts, but it did nothing to decrease lead emissions. Because lead poisons catalytic converters, however, its use as a gasoline additive has been banned in most of the world.
Compounds that contain Si–C and Si–O bonds are stable and important. High-molecular-mass polymers called silicones contain an (Si–O–)n backbone with organic groups attached to Si (Figure $5$). The properties of silicones are determined by the chain length, the type of organic group, and the extent of cross-linking between the chains. Without cross-linking, silicones are waxes or oils, but cross-linking can produce flexible materials used in sealants, gaskets, car polishes, lubricants, and even elastic materials, such as the plastic substance known as Silly Putty.
Example $2$
For each reaction, explain why the given products form.
1. Pb(s) + Cl2(g) → PbCl2(s)
2. Mg2Si(s) + 4H2O(l) → SiH4(g) + 2Mg(OH)2(s)
3. GeO2(s) + 4OH(aq) → GeO44(aq) + 2H2O(l)
Given: balanced chemical equations
Asked for: why the given products form
Strategy:
Classify the type of reaction. Using periodic trends in atomic properties, thermodynamics, and kinetics, explain why the observed reaction products form.
Solution
1. Lead is a metal, and chlorine is a nonmetal that is a strong oxidant. Thus we can expect a redox reaction to occur in which the metal acts as a reductant. Although lead can form compounds in the +2 and +4 oxidation states, Pb4+ is a potent oxidant (the inert-pair effect). Because lead prefers the +2 oxidation state and chlorine is a weaker oxidant than fluorine, we expect PbCl2 to be the product.
2. This is the reaction of water with a metal silicide, which formally contains the Si4 ion. Water can act as either an acid or a base. Because the other compound is a base, we expect an acid–base reaction to occur in which water acts as an acid. Because Mg2Si contains Si in its lowest possible oxidation state, however, an oxidation–reduction reaction is also a possibility. But water is a relatively weak oxidant, so an acid–base reaction is more likely. The acid (H2O) transfers a proton to the base (Si4), which can accept four protons to form SiH4. Proton transfer from water produces the OH ion, which will combine with Mg2+ to give magnesium hydroxide.
3. We expect germanium dioxide (GeO2) to be amphoteric because of the position of germanium in the periodic table. It should dissolve in strong aqueous base to give an anionic species analogous to silicate.
Exercise $2$
Predict the products of the reactions and write a balanced chemical equation for each reaction.
1. PbO2(s) $\xrightarrow{\Delta}$
2. GeCl4(s) + H2O(l) →
3. Sn(s) + HCl(aq) →
Answer
1. $\mathrm{PbO_2(s)}\xrightarrow{\Delta}\mathrm{PbO(s)}+\frac{1}{2}\mathrm{O_2(g)}$
2. GeCl4(s) + 2H2O(l) → GeO2(s) + 4HCl(aq)
3. Sn(s) + 2HCl(aq) → Sn2+(aq) + H2(g) + 2Cl(aq)
Summary
The group 14 elements show the greatest diversity in chemical behavior of any group; covalent bond strengths decease with increasing atomic size, and ionization energies are greater than expected, increasing from C to Pb. The group 14 elements show the greatest range of chemical behavior of any group in the periodic table. Because the covalent bond strength decreases with increasing atomic size and greater-than-expected ionization energies due to an increase in Zeff, the stability of the +2 oxidation state increases from carbon to lead. The tendency to form multiple bonds and to catenate decreases as the atomic number increases. The stability of the carbon tetrahalides decreases as the halogen increases in size because of poor orbital overlap and steric crowding. Carbon forms three kinds of carbides with less electronegative elements: ionic carbides, which contain metal cations and C4− (methide) or C22 (acetylide) anions; interstitial carbides, which are characterized by covalent metal–carbon interactions and are among the hardest substances known; and covalent carbides, which have three-dimensional covalent network structures that make them extremely hard, high melting, and chemically inert. Consistent with periodic trends, metallic behavior increases down the group. Silicon has a tremendous affinity for oxygen because of partial Si–O π bonding. Dioxides of the group 14 elements become increasingly basic down the group and their metallic character increases. Silicates contain anions that consist of only silicon and oxygen. Aluminosilicates are formed by replacing some of the Si atoms in silicates by Al atoms; aluminosilicates with three-dimensional framework structures are called zeolites. Nitrides formed by reacting silicon or germanium with nitrogen are strong, hard, and chemically inert. The hydrides become thermodynamically less stable down the group. Moreover, as atomic size increases, multiple bonds between or to the group 14 elements become weaker. Silicones, which contain an Si–O backbone and Si–C bonds, are high-molecular-mass polymers whose properties depend on their compositions.
• Anonymous | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/08%3A_Chemistry_of_the_Main_Group_Elements/8.07%3A_Group_14/8.7.02%3A_Inorganic_Compounds_of_the_Group_14_Elements.txt |
Carbon is the fourth most abundant element in the known universe but not nearly as common on the earth, despite the fact that living organisms contain significant amounts of the element. Common carbon compounds in the environment include the gases carbon dioxide ($CO_2$) and methane ($CH_4$).
Inorganic Chemistry of Carbon
Inorganic carbon is carbon extracted from ores and minerals, as opposed to organic carbon found in nature through plants and living things. Some examples of inorganic carbon are carbon oxides such as carbon monoxide and carbon dioxide; polyatomic ions, cyanide, cyanate, thiocyanate, carbonate and carbide. Carbon is an element that is unique to itself. Carbon forms strong single, double and triple bonds; therefore, it would take more energy to break these bonds than if carbon were to bond to another element.
For carbon monoxide the reaction is as follows:
$2C_{(s)} + O_2 \rightarrow 2CO_{(g)} \;\;\;\; \Delta H = -110.52\; kJ/mol \;CO \nonumber$
$C_{(s)} + O_2 \rightarrow CO_{2(g)} \;\;\;\; \Delta H = -393.51\; kJ/mol\; CO_2 \nonumber$
CO and CO2 are both gases. CO has no odor or taste and can be fatal to living organisms if exposed at even very small amounts (about a thousandth of a gram). This is because CO will bind to the hemoglobin that carries oxygen in the blood. CO2 will not become fatal unless living organisms are exposed to larger amounts of it, about 15%. CO2 influences the atmosphere and affects the temperature through the greenhouse gas effect. As heat is trapped in the atmosphere by CO2 gases, the Earth's temperature increases. The main source for CO2 in our atmosphere, amongst many, is volcanoes.
Allotropes
Carbon exists in several forms called allotropes. Diamond is one form with a very strong crystal lattice, known as a precious gem from the most ancient records. Another allotrope is graphite, in which the carbon atoms are arranged in planes which are loosely attracted to one another (hence its use as a lubricant). The recently discovered fullerenes are yet another form of carbon.
• Inorganic carbon may come in the form of diamond as a transparent, isotropic crystal. It is the hardest naturally occurring material on this earth. Diamond has four valence electrons, and when each electron bonds with another carbon it creates an sp3-hybridized atom. The boiling point of diamond is 4827°C.
• Unlike diamond, graphite is opaque, soft, dull and hexagonal. Graphite can be used as a conductor (electrodes) or even in pencils. Graphite consists of planes of sp2 hybridized carbon atoms in which each carbon is attached to three other carbons.
• Fullerenes are carbon cages with the formula $C_{2}n$ where $n > 13$. The most abundant fullerene is the spherical C60. Fullerene may contain atoms or molecules inside the cage (endohedral fullerenes) or covalently attached outside (exhedral or adduct fullerenes). The discovery of fullerenes is accredited to Richard Smalley and his team in 1985 at Rice University, who found it after photoablating the surface of graphite with a laser.
Applications
Carbon has a very high melting and boiling point and rapidly combines with oxygen at elevated temperatures. In small amounts it is an excellent hardener for iron, yielding the various steel alloys upon which so much of modern construction depends. An important (but rare) radioactive isotope of carbon, C-14, is used to date ancient objects of organic origin. It has a half-life of 5730 years, but there is only 1 atom of C-14 for every 1012 atoms of C-12 (the usual isotope of carbon). | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/08%3A_Chemistry_of_the_Main_Group_Elements/8.07%3A_Group_14/8.7.03%3A_Chemistry_of_Carbon_%28Z6%29.txt |
Silicon, the second most abundant element on earth, is an essential part of the mineral world. Its stable tetrahedral configuration makes it incredibly versatile, and it is used in various ways in our everyday lives. Found in everything from spaceships to synthetic body parts, silicon can be found all around us, and sometimes even in us.
Introduction
The name for silicon is taken from the Latin silex which means "flint". The element is second only to oxygen in abundance in the earth's crust and was discovered by Berzelius in 1824. The most common compound of silicon, \(SiO_2\), is the most abundant chemical compound in the earth's crust; we know it better as common beach sand.
Properties
Silicon is a crystalline semi-metal or metalloid. One of its forms is shiny, grey, and very brittle (it will shatter when struck with a hammer). It is a group 14 element in the same periodic group as carbon, but chemically behaves distinctly from all of its group counterparts. Silicon shares the bonding versatility of carbon, with its four valence electrons, but is otherwise a relatively inert element. However, under special conditions, silicon can be made to be a good deal more reactive. Silicon exhibits metalloid properties, is able to expand its valence shell, and is able to be transformed into a semiconductor, distinguishing it from its periodic group members.
Table 1: Properties of Silicon
Symbol Si
Atomic Number 14
Group 14 (Carbon Family)
Electron Configuration [Ne]3s23p2
Atomic Weight 28.0855 g
Density 2.57 g/mL
Melting Point 1414oC
Boiling Point 3265oC
Oxidation States 4, 3, 2, 1, -1, -2, -3, -4
Electronegativity 1.90
Stable Isotopes 28Si, 29Si, 30Si
Where Silicon is Found
27.6% of the Earth's crust is made up of silicon. Although it is so abundant, it is not usually found in its pure state, but rather its dioxide and hydrates. \(SiO_2\) is silicon's only stable oxide and is found in many crystalline varieties. Its purest form is quartz, but it is also found as jasper and opal. Silicon can also be found in feldspar, micas, olivines, pyroxenes and even in water (Figure 1). In another allotropic form, silicon is a brown amorphous powder most familiar in "dirty" beach sand. The crystalline form of silicon is the foundation of the semiconductor age.
Silicates
Silicon is most commonly found in silicate compounds. Silica is the one stable oxide of silicon, and has the empirical formula SiO2. Silica is not a silicon atom with two double bonds to two oxygen atoms. Silica is composed of one silicon atom with four single bonds to four oxygen molecules (Figure 2).
Silica, i.e. silicon dioxide, takes on this molecular form, instead of carbon dioxide's characteristic shape, because silicon's 3p orbitals make it more energetically favorable to create four single bonds with each oxygen rather than make two double bonds with each oxygen atom. This leads to silicates linking together in -Si-O-Si-O- networks called silicates. The empirical form of silica is SiO2 because, with respect to the net average of the silicate, each silicon atom has two oxygen atoms.
The tetrahedral SiO44- complex (see Figure 3), the core unit of silicates, can bind together in a variety of ways, creating a wide array of minerals. Silicon is an integral component in minerals, just as Carbon is an essential component of organic compounds.
Nesosilicates
In nesosilicates, the silicate tetrahedral does not share any oxygen molecules with other silicate tetrahedrals, and instead balances out its charge with other metals. The core structure of nesosilicate is simply a lone tetrahedral silica unit (Figure 4). The empirical formula for the core structure of a nesosilicate is SiO44-.
Nesosilicates make up the outer fringes of a group of minerals known as olivines.
Sorosilicates
In sorosilicates, two silicate tetrahedrals join together by sharing an oxygen atom at one of their corners. The core structure of a sorosilicate is a pair of silica tetrahedra. (see Figure 5)
The mineral thortveitie is an example of a sorosilicate complex.
Cyclosilicates
In cyclosilicates, three or more silica tetrahedrals share two corners of an oxygen atom. The core structure of a cyclosilicate is a closed ring of silica tetrahedra. (see Figure 6)
The mineral beryl is an example of a cyclosilicate complex.
Inosilicates
Inosilicates are complexes where each tetrahedral shares two corners with another silica tetrahedral to create a single chain (see Figure 7) or three corners to create a double chain (Figure 8). The core structure of an inosilicate is either an infinite single or double chain of silica tetrahedrals.
The mineral group pyroxenes are examples of single chain inosilicates.
Figure 8: The core of a double chain inosilicate
The mineral amphibole is an example of a double chain inosilicate.
Phyllosilicates
Phyllosilicates are silica complexes where each tetradedral shares three corners and creates a sheet of silicon and oxygen. (see Figure 9) The core complex of a phyllosilicate is an infinite sheet of connected silica tetrahedrals.
The mineral talc is an example of a phyllosilicate complex.
Tectosilicates
Tectosilicates are three-dimensional silicate structures. The core structure of a tectosilicate is an infinite network of connected silica tetrahedrals. (see Figure 10)
The mineral quartz is an example of a tectosilicate complex.
Although many silica complexes form network covalent solids, quartz is a particularly good example of a network solid. Silicates in general share the properties of covalent solids, and this affiliated array of properties makes them very useful in modern day industry.
Silanes
Silanes are silicon-hydrogen compounds. Carbon-hydrogen compounds form the backbone of the living world with seemingly endless chains of hydrocarbons. With the same valence configuration, and thus the same chemical versatility, silicon could conceivably play a role of similar organic importance. But silicon does not play an integral role in our day to day biology. One principal reason underlies this.
Like hydrocarbons, silanes progressively grow in size as additional silicon atoms are added. But there is a very quick end to this trend. The largest silane has a maximum of six silicon atoms (see Figure 11).
Hexasilane is the largest possible silane because Si-Si bonds are not particularly strong. In fact, silanes are rather prone to decomposition. Silanes are particularly prone to decomposition via reaction with oxygen. Silanes also have a tendency to swap out their hydrogens for other elements and become organosilanes. (see Figure 12)
Silanes have a variety of industrial and medical uses. Among other things, silanes are used as water repellents and sealants.
Silicones
Silicones are a synthetic silicon compound; they are not found in nature. When specific silanes are made to undergo a specific reaction, they are turned into silicone, a very special silicon complex. Silicone is a polymer and is prized for its versatility, temperature durability, low volatility, general chemical resistance and thermal stability. Silicone has a unique chemical structure, but it shares some core structural elements with both silicates and silanes. (See Figure 13)
Silicone polymers are used for a huge array of things. Among numerous other things, breast implants are made out of silicone.
Silicon Halides
Silicon has a tendency to react readily with halogens. The general formula depicting this is SiX4, where X represents any halogen. Silicon can also expand its valence shell, and the laboratory preparation of [SiF6]2- is a definitive example of this. However, it is unlikely that silicon could create such a complex with any halogen other than fluorine, because six of the larger halogen ions cannot physically fit around the central silicon atom.
Silicon halides are synthesized to purify silicon complexes. Silicon halides can easily be made to give up their silicon via specific chemical reactions that result in the formation of pure silicon.
Applications
Silicon is a vital component of modern day industry. Its abundance makes it all the more useful. Silicon can be found in products ranging from concrete to computer chips.
Electronics
The high tech sector's adoption of the title Silicon Valley underscores the importance of silicon in modern day technology. Pure silicon, that is, essentially pure silicon, has the unique ability of being able to discretely control the number and charge of the current that passes through it. This makes silicon play a role of utmost importance in devices such as transistors, solar cells, integrated circuits, microprocessors, and semiconductor devices, where such current control is a necessity for proper performance. Semiconductors exemplify silicon's use in contemporary technology.
Semiconductors
Semiconductors are unique materials that have neither the electrical conductivity of a conductor nor that of an insulator. Semiconductors lie somewhere in between these two classes, giving them a very useful property. Semiconductors are able to manipulate electric current. They are used to rectify, amplify, and switch electrical signals and are thus integral components of modern day electronics.
Semiconductors can be made out of a variety of materials, but the majority of semiconductors are made out of silicon. But semiconductors are not made out of silicates, or silanes, or silicones; they are made out of pure silicon, that is, essentially pure silicon crystal. Like carbon, silicon can make a diamond-like crystal. This structure is called a silicon lattice. (see Figure 15) Silicon is perfect for making this lattice structure because its four valence electrons allow it to perfectly bond to four of its silicon neighbors.
However, this silicon lattice is essentially an insulator, as there are no free electrons for any charge movement, and is therefore not a semiconductor. This crystalline structure is turned into a semiconductor when it is doped. Doping refers to a process by which impurities are introduced into ultra-pure silicon, thereby changing its electrical properties and turning it into a semiconductor. Doping turns pure silicon into a semiconductor by adding or removing a very, very small number of electrons, thereby making it neither an insulator nor a conductor, but a semiconductor with limited charge conduction. Subtle manipulation of pure silicon lattices via doping generates the wide variety of semiconductors that modern day electrical technology requires.
Semiconductors are made out of silicon for two fundamental reasons. Silicon has the properties needed to make semiconductors, and silicon is the second most abundant element on earth.
Glasses
Glass is another silicon derivate that is widely utilized by modern day society. If sand, a silica deposit, is mixed with sodium and calcium carbonate at temperatures near 1500 degrees Celsius, when the resulting product cools, glass forms. Glass is a particularly interesting state of silicon. Glass is unique because it represents a solid non-crystalline form of silicon. The tetrahedral silica elements bind together, but in no fundamental pattern behind the bonding. (see Figure 16)
The end result of this unique chemical structure is the often brittle, typically optically transparent material known as glass. This silica complex can be found virtually anywhere human civilization is found.
Glass can be tainted by adding chemical impurities to the basal silica structure. (see Figure 17) The addition of even a little Fe2O3 to pure silica glass gives the resultant mixed glass a distinctive green color.
Figure 17: Non-crystalline silica with unknown impurities
Fiber Optics
Modern fiber optic cables must relay data via undistorted light signals over vast distances. To undertake this task, fiber optic cables must be made of special ultra-high purity glass. The secret behind this ultra-high purity glass is ultra pure silica. To make fiber optic cables meet operational standards, the impurity levels in the silica of these fiber optic cables has been reduced to parts per billion. This level of purity allows for the vast communications network that our society has come to take for granted.
Ceramics
Silicon plays an integral role in the construction industry. Silicon, specifically silica, is a primary ingredient in building components such as bricks, cement, ceramics, and tiles.
Additionally, silicates, especially quartz, are very thermodynamically stable. This translates to silicon ceramics having high heat tolerance. This property makes silicon ceramics particularily useful for things ranging from space ship hulls to engine components. (see Figure 18)
Polymers
Silicone polymers represent another facet of silicon's usefulness. They are generally characterized by their flexibility, resistance to chemical attack, impermeability to water, and their ability to retain their properties at both high and low temperatures. This array of properties makes silicone polymers very useful. They are used in insulation, cookware, high temperature lubricants, medical equipment, sealants, adhesives, and even as an alternative to plastic in toys.
Production
As silicon is not normally found in its pure state, it must be chemically extracted from its naturally occurring compounds. The most prevalent form of naturally occurring silicon is silica. Silica is a strongly bonded compound, and it requires a good deal of energy to extract the silicon out of the silica complex. The principal means of extraction is via a chemical reaction at a very high temperature.
The synthesis of silicon is fundamentally a two-step process. First, use a powerful furnace to heat up the silica to temperatures over 1900 degrees celsius, and second, add carbon. At temperatures over 1900 degrees celsius, carbon will reduce the silica compound to pure silicon.
Purification
For some silicon applications, the purity of freshly produced silicon is not satisfactory. To meet the demand for high purity silicon, techniques have been devised to further refine the purity of extracted silicon.
Purification of silicon essentially involves taking synthesized silicon, turning it into a silicon compound that can be easily distilled, and then breaking up this newly formed silicon compound to yield an ultra pure silicon product. There are several distinct purification methods available, but most chemical forms of purification involve both silane and silicon halide complexes.
Trivia
• Silicon is the eighth most abundant element in the universe.
• Silicon was first identified in 1787 but first discovered as an element in 1824.
• Silicon is an important element in the metabolism of plants, but not very important in the metabolism of animals.
• Silicon is harmless to ingest and inject into the body but it is harmful to inhale.
• Silicosis is the name of the disease associated with inhaling too much of the silicon compound silica. It primarily afflicts construction workers.
• Silica is a major chemical component of asbestos.
Problems
Highlight area next to "Ans" to see answer
How many oxides does silicon have, and what are they?
Ans: 1 oxide O2
How does a silicate tetrahedral balance its charge if not bonded with another silicate?
Ans: By bonding to positively charged metals.
Carbon is to organic compounds as silica is to:
Ans: minerals
How big is the largest silicon-hydrogen compound?
Ans: The largest silane is hexasilane, with six silicon atoms and fourteen hydrogens.
Why is silicon important to computers?
Ans: It is used to make semiconductors.
Contributors and Attributions
• Thomas Bottyan (2010), Christina Rabago (2008)
8.7.04: Chemistry of Silicon (Z14)
Silicates are some of the most abundant minerals on Earth. They are some of the most common of the raw material that takes over 75% of the Earth's crust. A majority of igneous rocks and sedimentary rocks are made of silicate minerals. The most common type of silicate is (SiO4)4-.
There are many different types of silicates. Most of them have a general chemical formula of XxYy(ZzOo)Ww.
• X = +1 or +2 cations
• Y = +2, +3, or +4 cations
• Z = + 3 or +4 cations
• O = oxygen
• W = usually OH-, F-, or Cl-
• x, y, z, o, w = subscript numbers
Some of the subcategories of silicates are the following:
• Nesosilicates
• Sorosilicates
• Cyclosilicates
• Inosilicates
• Phyllosilicates
• Tectosilicates
Nesosilicates
Nesosilicates are made up of units of independent tetrahedrals. Some of the minerals that contain nesosilicates are olivine, garnet, zircon, kyanite, topaz, and staurolite. Olivine is important in the processes of igneous rock forming. It has a general formula of (Mg, Fe)2SiO4. Garnet belongs to the isomorphic group, where it often occurs as dodecahedron crystals such as pyrope, almandine, and grossularite. It is usually found in metamorphic rocks, and is known for being the January birthstone. Zircon, on the other hand, is marketed as gemstone and is oxidized to produce gemstones that are similar to diamonds known as cubic zirconia. Kyanite is a part of a polymorphic group (Al2OSiO4).
Ex. (SiO4)-4 or (Si3O12)-12
Sorosilicates
Sorosilicate is made up of two tetrahedrals shared by an oxygen. Some of the minerals that are classified as sorosilicates are hemimorphite, epidote, and allanite. Hemimorphite is usually found as bladed crystals. Epidote belongs to the isomorphic group, which is important in forming minerals. Lastly, allanite has a metamict structure that is usually black with no cleavage.
ex. (Si2O7)-6
Cyclosilicates
Cyclosilicates are made up of closed ring units of tetrahedrals sharing two oxygen atoms. They are known for their hardness and consist of a variety of gemstones. They also have poor cleavage. Some minerals that are classified as cyclosilicates are beryl, cordierite, and tourmaline. The gemstones that are classified as beryl include emerald (deep green), aquamarine (greenish-blue), and morganite (red). Tourmalines also have a variety of gemstones, which include rubellite (red-pink) and indicolite (dark blue). As for cordierite, it often show dichroism, meaning that it shows different colors at different concentrations.
Ex. (Si6O18)-12
Inosilicates
Inosilicates are made up of continuous double chain units of tetrahedrals, each sharing 2 and 3 oxygens. They include the pyroxene group, which are single chain minerals without hydroxides, and the amphibole group, which are double chains with hydroxides. The pyroxene group has two directional 90 degree cleavages. Some examples are enstatite-ferrosilite, diopside-hedenbergite, augite, and spodumene. As for the amphibole group, it has two directional cleavages at 124-56 degrees. Some examples are tremolite-actinolite and hornblende. Both of these groups are rock-forming minerals.
Ex. (SiO3)-2 or (Si2O6)-4
Phyllosilicates
Phyllosilicates comprise continuous sheet units of tetrahedrals, each sharing 3 oxygen atoms. They include the clay and mica minerals, which are rock-forming minerals. The clay group is made of hydrous aluminum layered silicates. Some examples are kaolinite and talc. On the other hand, the mica group consists of thin sheets and a multitude of ionic substitutions of Al3+ and Si4+. Some examples are muscovite (light color), biotite (black or dark colored), and lepidolite (pink colored and a source of lithium). The serpentine group also belongs to the phyllosilicates. Some examples are serpentine and crysotile.
Ex. (Si2O5)-2 or (Al Si3O10)-5
Tectosilicates
Tectosilicates consist of a continuous framework of tetrahedrals, each sharing all 4 oxygen atoms. Their structure has a great amount of Al-Si substitution. Some of the groups that are classified as tectosilicates are the SiO2 polymorphic group, the K-feldspar polymorphic group, the feldspathoid group, and the zeolite group. The SiO2 polymorphic group has a variety of quartz, such as smoky quartz, amethyst, and jasper. Some minerals in the K-felspar polymorphic group include orthoclase and microcline. Microcline has 1 Pb2+ ion replacing every 2 K1+ ions, showing an omission solid solution and causing a blue-green color in the mineral. The felspathoid group minerals are similar to feldspars but only have two-thirds of the amount of silica; they form a silica-deficient magma. Some examples are leucite and sodalite. Lastly, the zeolite group has hydrous silicates with ionic exchange and absorption properties that can act as water softeners by exchanging Na1+ ions for Ca2+ ions in solution. For example:
$Na_2Al_2Si_3O_{10}-2H_2O \rightarrow CaAl_2Si_3O_{10}-2H_2O. \nonumber$
Ex. (SiO2), (AlSiO4)-1, (Al2Si2O8)-2, or (Al2Si4O12)-2 | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/08%3A_Chemistry_of_the_Main_Group_Elements/8.07%3A_Group_14/8.7.04%3A_Chemistry_of_Silicon_%28Z14%29/8.7.4.01%3A_Silicates.txt |
Learning Objectives
• Predict some chemical reactions for a set of conditions.
• Describe some chemical and physical properties of $\ce{C}$ and $\ce{Si}$.
Group 14 Elements C, Si, Ge, Sn, Pb
Group 14 elements play more important roles in our lives and our civilization than elements of any other group. Thus, every educated person should know something about them.
• Carbon - The element of organic chemistry and life.
• Silicon - The element of information technology.
• Germanium - The element of transistor bases.
• Tin and lead - Elements known to alchemists.
Chemistry of Carbon
Carbon exists as diamond, graphite, fullerenes, and charcoal. Their structures are interesting; so are their properties. You probably know a lot about diamond and graphite, but the fullerenes were discovered after 1970, and this discovery has opened a door for a lot of interesting research. Read about them in books, magazines and journals. You might find yourself working with them some day.
Among the fullerenes, one of the most common "molecules" has 60 $\ce C$ atoms, and is represented by $\ce{C60}$; a diagram is shown here. If you connect the 60 carbon atoms with bonds, the structure looks like a cage with 5- and 6- member rings. Synthesis, bonding, symmetry and stability of the cagelike fullerenes have already attracted a lot of attention, and their properties are even more fascinating.
Regarding carbon compounds, you already know something about $\ce{CO2}$ and $\ce{CO}$, including their roles in the environment. The hard carbides such as $\ce{Fe3C}$, $\ce{WC}$, and $\ce{TiC}$ are more interesting to material scientists and engineers for their application in cutting tools. The calcium carbide $\ce{CaC2}$ produced by reducing $\ce{CaO}$ by carbon was a valuable commodity at one time due to its reaction with water to give acetylene gas:
$\ce{CaC_2 + 2 H_2O \rightarrow Ca(OH)_2 + C_2H_{2\large{(g)}}}$
Acetylene is an important industrial gas for the manufacture of polymers.
Silicon
Do you know that:
• $\ce{Si}$ comprises 27.7 % of the Earth's crust?
• silicates, $\ce{SiO2}$ based minerals, are everywhere?
• $\ce{Si}$ is an essential element for bone growth?
• $\ce{Si}$ crystals are the bases of computer chips?
• the structure of $\ce{Si}$ is the same as that of diamond, and this feature is important for computer chips?
• How we convert $\ce{SiO2}$ into $\ce{Si}$ element?
Silicates
By weight, silicon is the most abundant element in the Earth's crust. It usually exists in the form of an oxide, $\ce{SiO2}$. This formula does not do justice to so many different materials we call silicates, but these substances are indeed $\ce{SiO2}$. Some of the minerals contain impurities.
In pure form, $\ce{SiO2}$ is quartz. Small particles of quartz are sand. They are hard. In the structure at the atomic level, every silicon atom is bonded to 4 oxygen atoms, and every oxygen is bonded to two $\ce{Si}$ atoms. The four $\ce{Si-O}$ bonds point towards the corners of a tetrahedron, as do the $\ce{C-C}$ bonds in the diamond structure. When an impurity is present, the quartz may be colored. Due to various arrangements of the $\ce{Si-O-Si}$ bonds, the same substance appears in many forms.
A basic unit of silicate structures is $\ce{SiO4^4-}$. The gemstone zircon has a formula $\ce{ZrSiO4}$, and olivine has a chemical formula of $\ce{(MgFe)2SiO4}$. Two $\ce{SiO4^4-}$ units combine to give the pyrosilicate unit $\ce{Si2O7^6-}$, and it appears in akermanite, $\ce{Ca2MgSi2O7}$. When the number of units increases, the tetrahedral units combine to form rings, chains, layers and 3-dimensional networks. Thus, the structure and classification of silicate is a major part of minerals. (This site has some interesting pictures.)
Silicon and Silane
Elemental silicon can be obtained from reduction of silicates. The reduction of sand, $\ce{SiO2}$ by carbon at 3300 K in the reaction,
$\ce{SiO_2 + 2 C \rightarrow Si_{\large{(l)}} + 2 CO} \;\;\; \textrm{at 3300 K}$
gives liquid silicon. The silicon so obtained is usually not pure, and for the computer industry, the element must be purified. Crystal growth and silicon fabrication dominated the industry in the 1980s and 1990s, and perhaps into the next century, and the production of the element is only the beginning of the process.
If a more reactive element, $\ce{Mg}$, is used in the reduction, $\ce{Mg2Si}$ is formed,
$\ce{SiO_2 + 4 Mg \rightarrow Mg_2Si + 2 MgO}$
$\ce{Mg2Si}$ is a compound, and it reacts with water to form silane.
Silane, $\ce{SiH4}$, can be produced by reacting $\ce{Mg2Si}$ with acids
$\ce{Mg_2Si + 4 H_2O \rightarrow 2 Mg(OH)_2 + SiH_4}$
and $\ce{SiH4}$ is ignited when it contacts air, being much more reactive than methane,
$\ce{SiH_4 + O_2 \rightarrow SiO_2 + H_2O}$
In a basic solution, $\ce{SiH4}$ reacts with water to give $\ce{SiO(OH)3-}$,
$\ce{SiH_4 + OH^- + 3H_2O \rightarrow SiO(OH)_3^- + 4 H_2}$
Silicon Halides
Silicon tetrafluoride is formed when glass ($\ce{SiO2}$) is exposed to $\ce{HF}$, and when $\ce{Si}$ reacts with $\ce{F2}$,
$\ce{SiO2 + 4 HF_{\large{(aq)}} \rightarrow 2 H2O + SiF_{4\large{(g)}}}$
$\ce{Si + 2 F2 \rightarrow SiF_{4\large{(g)}}}$
When chlorine passes through hot sand ($\ce{SiO2}$) and carbon, $\ce{SiCl4}$ is produced,
$\ce{SiO2 + C + 2 Cl_2 \rightarrow 2CO + SiCl_{4\large{(g)}}}$
$\ce{SiCl4}$ and $\ce{SiF4}$ react with water to give silicic acid,
$\ce{SiCl4 + 4H2O \rightarrow 4 HCl + Si(OH)_{4\large{(aq)}}}$,
$\ce{SiF4 + 4H2O \rightarrow 4 HF + Si(OH)_{4\large{(aq)}}}$.
Silicone Polymers
Silicones are polymers with the general formula $\mathrm{(R_2SiO_2)_{\large n}}$ or $\mathrm{(RSiO_3)_{\large n}}$, ($\ce{R}$ = $\ce{CH3}$, $\ce{C2H5}$, $\ce{C6H5}$, etc). The chain is held together by $\ce{Si-O-Si}$ bonds. A simple one is $\mathrm{((CH_3)_2SiO_2)_{\large n}}$,
Figure 2: Chemical structure of the silicone polydimethylsiloxane (PDMS).
Of course, the 4 bonds around the $\ce{Si}$ atoms point to the corners of a tetrahedron. These siloxane polymers are widely used as sealants, adhesives, additives, flame retardants, and lubricants. They have a wide application in industries. Depending on the organic group attached to silicon, the inorganic polymer silicones have been an important class of materials.
Questions
1. Which one lists the group 14 elements in order of increasing atomic weight?
1. $\textrm{B Al Ga In Tl}$
2. $\textrm{C Si Ge Sn Pb}$
3. $\textrm{N P As Sb Bi}$
4. $\textrm{O S Se Te Po}$
5. $\textrm{F Cl Br I At}$
2. Which allotrope of carbon is the hardest: diamond, graphite, or fullerenes?
3. What compound of carbon reacts with water to give acetylene gas?
4. When you want to extract silicon element, what do you use to reduce the sand: $\ce{SiO2}$, $\ce{C}$ or $\ce{Mg}$?
5. Which is stable towards air: methane or silane?
6. Give the name of polymers whose chains are held together by $\ce{Si-O-Si}$ bonds.
7. In the crystal structure of $\ce{Si}$, how many other $\ce{Si}$ atoms are connected to a particular $\ce{Si}$ atom?
Solutions
1. Answer b
Hint...
Knowing the groups of elements enables us to correlate their chemical properties. Each list of choices is a group of elements on the period table.
a = 3A, b = 14, c = 5A, d = 6A, e = 7A.
2. Answer diamond
Hint...
Diamond is the hardest thing in the world. Fullerenes are large molecules consisting of 40 to hundreds of carbon atoms, with $\mathrm{C_{60}}$ being the most common.
3. Answer . . .$\ce{CaC2}$
Hint...
The reaction to produce acetylene gas is
$\ce{CaC2 + H2O \rightarrow C2H2 + Ca(OH)2}$
Acetylene is still an important industrial gas, as raw material for polymers.
4. Answer ...$\ce{C}$
Hint...
Carbon or coke is used for silicon metal, because $\ce{Mg2Si}$ is formed if $\ce{Mg}$ is used.
5. Answer ... methane is stable
Hint...
Methane is the major component of natural gas, and it will not react with air until ignited, whereas silane ignites explosively as soon as it contacts air.
6. Answer ...silicones
Hint...
Silicon polymers are an important class of materials invented not too long ago.
7. Answer ... 4
Hint...
Silicon and diamond have the same crystal structure. The edge of unit cells of $\ce{Si}$ is larger than that of diamond. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/08%3A_Chemistry_of_the_Main_Group_Elements/8.07%3A_Group_14/8.7.04%3A_Chemistry_of_Silicon_%28Z14%29/8.7.4.02%3A_Silicon_and_Group_14_Elements.txt |
Germanium, categorized as a metalloid in group 14, the carbon family, has five naturally occurring isotopes. Germanium, abundant in the Earth's crust, has been said to improve the immune system of cancer patients. It is also used in transistors, but its most important use is in fiber-optic systems and infrared optics.
Introduction
The metalloid was one of the elements predicted by Mendeleev in 1871 (ekasilicon) to fill out his periodic table and was discovered in 1886 by Winkler. In a mine near Freiberg, Saxony, a new mineral was found in 1885. First the mineral was called argyrodite, but later, when Clemens Winkler examined this mineral he discovered that it was similar to antimony. At first he wanted to name it neptunium, but because this name was already taken he named it germanium in honor of his fatherland Germany.
The position of where germanium should be placed on the periodic table was under discussion during the time due to its similarities to arsenic and antimony. Due to Mendeleev's prediction of ekasilicon, germanium's place on the periodic table was confirmed because of the similar properties predicted and similar properties deduced from examining the mineral.
Characteristics
Like silicon, germanium is used in the manufacture of semi-conductor devices. Unlike silicon, it is rather rare (only about 1 part in 10 million parts in the earth's crust). The physical and chemical properties of germanium closely parallel those of silicon. Germanium (Ge) has an atomic number of 32. It is grayish-white, lustrous, hard, and has similar chemical properties to tin and silicon. Germanium is brittle and silvery-white under standard conditions. Under these conditions, germanium is known as α-germanium, which has a diamond cubic crystal structure. When germanium is under pressure above 120 kilobars, it has a different allotrope known as β-germanium. Germanium is one of the few substances like water that expands when it solidifies.
Atomic Number: 32
Atomic Symbol: Ge
Atomic Weight: 72.59
Atomic Radius: 122.5 pm
Series: Metalloid
Group: 14
Density: 5.353 g/cm3
Specific Heat: 0.32J/gK
Electronic Configuration: [Ar] 4s23d104p2
Melting Point: 938.25 C
Boiling Point: 2833 C
Common Oxidation States: 4,2
Germanium, a semiconductor, is the first metallic metal to become a superconductor in the presence of a strong electromagnetic field.
Isotopes
The five naturally occurring isotopes of germanium have atomic masses of 70, 72, 73, 74, and 76. Germanium 76 is slightly radioactive and is the least common. Germanium 74 is the most common isotope, having the greatest natural abundance of the five. When it is bombarded with alpha particles, Germanium 72 generates stable Se 77.
Chemical Properties
At a temperature of 250 °C, germanium slowly oxidizes to \(GeO_2\). Germanium dissolves slowly in concentrated sulfuric acid, and is insoluble in diluted acids and alkalis. It will react violently with molten alkalis to produce [GeO3]2-. The common oxidation states of germanium are +4 and +2. Under rare conditions, germanium also occurs in oxidation states of +3, +1, and -4.
There are two forms of germanium oxides: germanium dioxide (\(GeO_2\)) and germanium monoxide (\(GeO\)). By roasting germanium sulfide, germanium dioxide can be obtained. As for germanium monoxide, it can be obtained by the high temperature reaction of germanium dioxide and germanium metal. Germanium dioxide has the unusual property of a refractive index for light but transparency to infrared light.
Applications and Uses
Like silicon, germanium is used in the manufacture of semi-conductor devices. Unlike silicon, it is rather rare (only about 1 part in 10 million parts in the earth's crust). Although it forms a compound, germanium dioxide, just like silicon, it is generally extracted from the by-products of zinc refining.
Germanium is used mainly for fiber-optic systems and infrared optics. It is also used for polymerization catalysts, electronics, and as phosphors, for metallurgy, and for chemotherapy. Because germanium dioxide has a high index of refraction and low optical dispersion, it is useful for wide-angle camera lenses. As stated before, it is an important semi-conductor, so it is used in transistors. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/08%3A_Chemistry_of_the_Main_Group_Elements/8.07%3A_Group_14/8.7.05%3A_Chemistry_of_Germanium_%28Z32%29.txt |
Mentioned in the Hebrew scriptures, tin is of ancient origins. Tin is an element in Group 14 (the carbon family) and has mainly metallic properties. Tin has atomic number 50 and an atomic mass of 118.710 atomic mass units.
Introduction
Mentioned in the Hebrew scriptures, tin is of ancient origins. Early metal smiths were quick to learn that mixing copper with tin created a more durable metal (bronze), and it is principally for its alloys that tin is valued today. Named after the Etruscan god Tinia, the chemical symbol for tin is taken from the Latin stannum. The metal is silvery white and very soft when pure. It has the look of freshly cut aluminum, but the feel of lead.
Polished tin is slightly bluish. It has been used for many years in the coating of steel cans for food because it is more resistant to corrosion than iron. It forms a number of useful low-melting alloys (solders) which are used to connect electrical circuits. Bending a bar of tin produces a characteristic squealing sound called "tin cry". Tin shares chemical similarities with germanium and lead. Tin mining began in Australia in 1872, and today tin is used extensively in industry and commerce.
Table 1: Basic Properties of tin
Color white with blueish tinge
Hardness softer than gold, harder than lead
Atomic Radius 140 pm
Density
5.77g/cm3
Melting Point 232 degrees Celsius
Boiling Point 2623 degrees Celsius
Electrical Conductivity about 1/7th that of silver
Electrode Potential >0.192V
First Ionization Energy 709 kJ/mol
Ionic Radius 93 pm
Reactions of Tin
Hydrogen Tin not affected
Nitrogen Tin absorbs it instead of hydrogen in electric discharge
Argon No sign of a combination of tin with argon
Fluorine Does not react with tin at low temperatures, but at 100 degrees Celsius they form stannic fluoride. Perhaps one of the most familiar of tin compounds, $SnF_2$, tin(II) fluoride, goes by the trade name of fluoristan and is found in some fluoride toothpastes.
Chlorine Acts on tin at room temperature
Bromine Acts on tin at room temperature
Sulfur Unites directly with tin when heated
Selenium Reacts vigorously with tin
Tellurium Reacts vigorously with tin
Nitrogen Forms a compound by direct union with tin
Arsenic Reacts with tin under heat and light
Antimony Is dissolved by molten tin
Reaction of tin with oxygen
When heated in oxygen, tin produces stannic oxide:
$Sn_{(s)} + O_{2(g)} \rightarrow SnO_{2(s)} \nonumber$
Reaction of tin with water (steam)
$Sn_{(s)} + 2H_2O_{(g)} \rightarrow SnO_{2(s)} + 2H_{2(g)} \nonumber$
Isotopes
There are 10 known stable isotopes of tin, the most of any elements on the periodic table. This high number of stable isotopes could be attributed to the fact that the atomic number of $\ce{^{50}Sn}$ is a 'magic number' in nuclear physics.
Table 4: Isotopes of tin
Isotope % Natural Abundance
112 amu 0.95%
116 amu 14.24%
117 amu 7.57%
118 amu 24.01%
119 amu 8.58%
120 amu 32.97%
122 amu 4.71%
124 amu 5.98%
Allotropes of Tin
Tin has 3 allotropes: alpha, beta and gamma tin. Alpha tin is the most unstable form. Beta tin is the most commonly found allotrope of tin, and gamma tin only exists at very high temperatures.
Oxidation States of Tin
Tin, although it is found in Group 14 of the periodic table, is consistent with the trend found in Group 13, where the lower oxidation state is favored farther down a group. Tin can exist in two oxidation states, +2 and +4, but it displays a tendency to exist in the +4 oxidation state.
Common Compounds of Tin
Tin forms two main oxides, SnO and SnO2 (amphoteric).
Electron Configuration of Tin
Tin has a ground state electron configuration of 1s22s22p63s23p64s23d104p65s24d105p2 and can form covalent tin (II) compounds with its two unpaired p-electrons. In the three dimensional figure below, the first and most inner electron shell is represented by blue electrons, the second electron shell made up of eight electrons is represented by red electrons, the third shell containing eighteen electrons is represented with green electrons, and the next outer shell again contains eighteen electrons and is represented in purple.
Uses of Tin
Early metal smiths were quick to learn that mixing copper with tin created a more durable metal (bronze) and it is principally for its alloys that tin is valued today. Nearly half of the tin metal produced is used in solders, which are low melting point alloys used to join wires. Solders are important in electrician work and plumbing. Tin is also used as a coating for lead, zinc, and steel to prevent corrosion. Tin cans are widely used for storing foods; the first tin can was used in London in 1812.
Questions
Find the oxidation state of tin in the following compounds:
a. SnCl2 Answer:2
b. SnO2 Answer:4
Write an equation for the reaction of tin with water. Under what conditions does this reaction take place?
Answer: Sn(s) + 2H2O(g) → SnO2(s) + 2H2(g) Reaction takes place if water is heated to a high temperature to form steam.
Which of these reactions take place?
a. tin with oxygen Answer: YES
b. tin with hydrogen Answer: NO
c. tin with argon Answer: NO
d. tin with chlorine Answer: YES
Arrange the following in order of increasing atomic radius: Sn, K, Ag, C, Pb.
Answer: C<Sn<Pb<Ag<K
Arrange the following in order of decreasing ionization energy: Sn, Si, Pb, I, In.
Answer: Si> I > Sn > In > Pb
Contributors and Attributions
• Taylor Hughes, (UCSB)
8.7.7.02: Lead Plumbate
Known to the ancients, lead takes its name from the Anglo-Saxon word for the metal, and its symbol comes from the Latin plumbum (from which we get the modern word "plumber" since old plumbing was done with lead pipes).
Although lead is not very common in the earth's crust, what is there is readily available and easy to refine. Its chief use today is in lead-acid storage batteries such as those used in automobiles. In pure form it is too soft to be used for much else. Lead has a blue-white color when first cut but quickly dulls on exposure to air, forming Pb2O, one of the few lead(I) compounds. Most stable lead compounds contain lead in oxidation states of +2 or +4.
Various isotopes of lead come at the end of the natural decay series of elements like uranium, thorium and actinium. These are Pb-206, Pb-207 and Pb-208.
Contributors and Attributions
Stephen R. Marsden
8.7.07: Chemistry of Lead (Z82)
Lead plumbate, also called red lead, minium or Mennige (in German), is a mineral showing colors from light red to brown/yellow tints. As a pure chemical it shows a vivid red. Minium is rare and occurs in lead mineral deposits that have been subjected to severe oxidizing conditions. It also occurs as a result of mine fires. It is most often associated with galena, cerussite, massicot, litharge, native lead, wulfenite and mimetite.
Lead plumbate is obtained by heating lead monoxide ($PbO$) to 450-480°C in air:
$3 PbO + 1/2 O_2 \rightarrow Pb_3O_4 \nonumber$
or by oxidative annealing of lead white:
$3 Pb_2CO_3(OH)_2 + O_2 \rightarrow 2 Pb_3O_4 + 3 CO_2 + 3 H_2O \nonumber$
Lead plumbate decomposes into lead monoxide and oxygen above 550°C.
$Pb_3O_4$ can be seen formally as a lead(II)plumbate(IV), $Pb_2[PbO_4]$, or $2PbO\cdot PbO_2$. In nitric acid, the lead(II) oxide reacts forming lead nitrate, while the insoluble lead(IV) oxide is left unchanged:
$Pb_3O_4 + 4 HNO_3 \rightarrow 2 Pb(NO_3)_2 + PbO_2 + 2 H_2O \nonumber$
Lead plumbate is virtually insoluble in water. However, it dissolves in hydrochloric acid (which is present in the stomach), and is therefore toxic when ingested. Lead plumbate (in a mixture with linseed oil or other organic adhesives) has been used as an anti-corrosion paint for iron. It forms insoluble iron(II) and iron(III) plumbates when brought into contact with iron oxides and with elementary iron. However, its use as a protective undercoat paint is limited due to its toxicity.
Lead plumbate was used as a red pigment in ancient and medieval periods for paintings and the production of illuminated manuscripts (the term miniature is connected to the name of the substance). | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/08%3A_Chemistry_of_the_Main_Group_Elements/8.07%3A_Group_14/8.7.06%3A_Chemistry_of_Tin_%28Z50%29.txt |
The nitrogen family includes the following compounds: nitrogen (N), phosphorus (P), arsenic (As), antimony (Sb), and bismuth (Bi). All Group 15 elements have the electron configuration ns2np3 in their outer shell, where n is the principal quantum number.
• 8.9.1: General Properties and Reactions
The nitrogen family includes the following compounds: nitrogen (N), phosphorus (P), arsenic (As), antimony (Sb), and bismuth (Bi). All Group 15 elements have the electron configuration ns2np3 in their outer shell, where n is equal to the principal quantum number. The nitrogen family is located in the p-block in Group 15, as shown below.
• 8.9.2: Chemistry of Nitrogen (Z=7)
Nitrogen is present in almost all proteins and plays important roles in both biochemical applications and industrial applications. Nitrogen forms strong bonds because of its ability to form a triple bond with its self, and other elements. Thus, there is a lot of energy in the compounds of nitrogen. Before 100 years ago, little was known about nitrogen. Now, nitrogen is commonly used to preserve food, and as a fertilizer.
• 8.9.3: Chemistry of Phosphorus (Z=15)
Phosphorus (P) is an essential part of life as we know it. Without the phosphates in biological molecules such as ATP, ADP and DNA, we would not be alive. Phosphorus compounds can also be found in the minerals in our bones and teeth. It is a necessary part of our diet. In fact, we consume it in nearly all of the foods we eat. Phosphorus is quite reactive. This quality of the element makes it an ideal ingredient for matches because it is so flammable.
• 8.9.4: Chemistry of Arsenic (Z=33)
Arsenic is situated in the 33rd spot on the periodic table, right next to Germanium and Selenium. Arsenic has been known for a very long time and the person who may have first isolated it is not known but credit generally is given to Albertus Magnus in about the year 1250. The element, which is classified as a metalloid, is named from the Latin arsenicum and Greek arsenikon which are both names for a pigment, yellow orpiment.
• 8.9.5: Chemistry of Antimony (Z=51)
Antimony and its compounds have been known for centuries. Scientific study of the element began during the early 17th century, much of the important work being done by Nicolas Lemery. The name of the element comes from the Greek anti + monos for "not alone", while the modern symbol is rooted in the Latin-derived name of the common ore, stibnite.
• 8.9.6: Chemistry of Bismuth (Z=83)
Bismuth, the heaviest non-radioactive naturally occurring element, was isolated by Basil Valentine in 1450. It is a hard, brittle metal with an unusually low melting point (271oC). Alloys of bismuth with other low-melting metals such as tin and lead have even lower melting points and are used in electrical solders, fuse elements and automatic fire sprinkler heads.
• 8.9.7: Chemistry of Moscovium (Z=115)
In studies announced jointly by the Joint Institute for Nuclear Research in Dubna, Russia, and the Lawrence Livermore National Laboratory in the U.S., four atoms of element 113 were produced in 2004 via decay of element 115 after the fusion of Ca-48 and Am-243.
Thumbnail: White and red phosphorus. (CC-SA-BY 3.0; Peter Krimbacher).
8.09: The Nitrogen Family
The nitrogen family includes the following compounds: nitrogen (N), phosphorus (P), arsenic (As), antimony (Sb), and bismuth (Bi). All Group 15 elements have the electron configuration ns2np3 in their outer shell, where n is equal to the principal quantum number. The nitrogen family is located in the p-block in Group 15.
All Group 15 elements tend to follow the general periodic trends:
• Electronegativity (the atom's ability to attract electrons) decreases down the group.
• Ionization energy (the amount of energy required to remove an electron from the atom in its gas phase) decreases down the group.
• Atomic radii increase in size down the group.
• Electron affinity (the ability of the atom to accept an electron) decreases down the group.
• Melting point (amount of energy required to break bonds to change a solid phase substance to a liquid phase substance) increases down the group.
• Boiling point (amount of energy required to break bonds to change a liquid phase substance to a gas) increases down the group.
• Metallic character increases down the group.
Element/Symbol Atomic Number Mass Electron Configuration Covalent Radius (pm) Electronegativity First Ionizaton Energy (kJ/mol) Common Physical Form(s)
Properties of Group 15 Elements
Nitrogen (N) 7 14.01 1s2 2s2 2p3 75 3.0 1402 Colorless Gas
Phosphorus (P) 15 30.97 [Ne]3s2 3p3 110 2.1 1012 White Solid / Red Solid
Arsenic (As) 33 74.92 [Ar] 3d10 4s2 4p3 121 2.0 947 Yellow Solid / Gray Solid
Antimony (Sb) 51 121.76 [Kr] 4d10 5s2 5p3 140 1.9 834 Yellow Solid / Silver-White Metallic Solid
Bismuth (Bi) 83 208.98 [Xe] 4f14 5d10 6s2 6p3 155 1.9 703 Pink-White Metallic Solid
Nitrogen
Nitrogen was discovered in 1770 by Scheele and Priestley. This nonmetallic element has no color, taste, or odor, and is present in nature as a noncombustible gas. When compared with the rest of Group 15, nitrogen has the highest electronegativity, which makes it the most nonmetallic of the group. The common oxidation states of nitrogen are +5, +3, and -3. Nitrogen makes up about 0.002% of the earth's crust; however, it constitutes 78% of the earth’s atmosphere by volume. Nitrogen has also been discovered in the atmospheres of Venus and Mars. Venus has a 3.5% nitrogen volume in its atmosphere and Mars has a 2.7% nitrogen volume in its atmosphere. Nitrogen is found naturally in animal and plant proteins and in fossilized remains of ancient plant life. Important nitrogen-containing minerals are niter, KNO3, and soda niter, NaNO3, which are found in desert regions and are important components of fertilizers. Before the process of converting nitrogen into ammonia was discovered, sources of nitrogen were limited. One of the processes of converting nitrogen to ammonia, the Haber-Bosch process, is very important for the production of nitrogen. Nitrogen has very little solubility in liquids. N2 does not have any allotropes.
The unusually stable N2(g) nitrogen gas is the source from which all nitrogen compounds are ultimately derived. N2(g) is stable due to its electronic structure: the bond between the two nitrogen atoms of N2 is a triple covalent bond, which is strong and hard to break. The enthalpy change associated with breaking the bonds in N2 is highly endothermic: ΔH = +945.4 kJ. Nitrogen gas is used as a refrigerant, metal treatment, and pressurized gas for oil recovery. Additionally, the Gibbs energies of formation of nitrogen compounds show that their formations are nonspontaneous, and the following process does not occur at normal temperatures:
$1/2N_{2(g)} + 1/2O_2 \rightarrow NO_{(g)} \nonumber$
with
$ΔG_f= +86.55\;kJ\, mol^{-1} \nonumber$
The oxides and oxyacids of nitrogen include nitrous oxide (N2O), nitrogen oxide (NO), and nitrogen dioxide (NO2). Nitrous oxide, also called “laughing gas,” is used in dental work, child birth, and to increase the speed of cars. Nitrogen oxide is found in smog and neurotransmitters. Hydrazine, N2H4, is a poisonous, colorless liquid that explodes in air. Hydrazine is a good reducing agent, and methyl hydrazine is commonly used as a rocket fuel.
Phosphorus
Phosphorus is a nonmetallic element. The most common oxidation state of phosphorus is -3. Phosphorus is the eleventh most abundant element, making up 0.11% of the earth's crust. The main source of phosphorus compounds is phosphorus rocks. Phosphorous is not found pure in nature, but in the form of apatite ores. These include compounds such as fluorapatite (Ca5(PO4)3F), which in fluoridated water is used to strengthen teeth, and hydroxylapatite (Ca10(OH)2(PO4)6), a major component of tooth enamel and bone material. Phosphorus exhibits allotropic forms: the most common forms at room temperature are white phosphorus and red phosphorus. White phosphorus is a white, waxy solid that can be cut with a knife. It forms a tetrahedral molecule, P4. White phosphorus is toxic, while red phosphorus is nontoxic. Red phosphorous forms when white phosphorous is heated to 573 Kelvin and not exposed to air. Red phosphorus is less reactive than white phosphorus. Red phosphorus has a chain-like polymeric structure, and is more stable. Both white and red phosphorus are incendiary and have been used to make match tips, although the use of white phosphorus has been largely discontinued due to toxicity. Phosphorus has many applications: phosphorus trichloride (PCl3) is used in soaps, detergents, plastics, synthetic rubber nylon, motor oils, insecticides and herbicides; phosphoric acid, H3PO4, is used in fertilizers; phosphorus is also prevalent in the food industry, used in baking powders, instant cereals, cheese, the curing of ham, and in the tartness of soft drinks.
Arsenic
Arsenic is a highly poisonous metalloid. Because it is a metalloid, arsenic has a high density, moderate thermal conductivity, and a limited ability to conduct electricity. The oxidation states of arsenic are +5, +3, +2, +1, and -3. The three allotropic forms of arsenic are yellow, black, and gray; the gray allotrope is the most common. Compounds of arsenic are used in insecticides, weed killers, and alloys. The oxide of arsenic is amphoteric, meaning it can act as both an acid and a base. Arsenic is mainly obtained by heating arsenic-containing sulfides. The chemical formula for this process is given below:
$FeAsS_{(s)} \rightarrow FeS_{(s)} + As_{(g)} \nonumber$
The $As_{(g)}$ deposits as $As_{(s)}$, which can then be used to make other compounds. Arsenic can also be obtained by the reduction of arsenic(III) oxide with $CO_{(g)}$.
Antimony
Antimony is also a metalloid. The oxidation states of antimony are +3, -3, and +5. Antimony exhibits allotropy; the most stable allotrope is the metallic form, which is similar in properties to arsenic: high density, moderate thermal conductivity, and a limited ability to conduct electricity. The oxide of antimony is antimony (III) oxide, which is amphoteric, meaning it can act as both an acid and a base. Antimony is obtained mainly from its sulfide ores, and it vaporizes at low temperatures. Along with arsenic, antimony is commonly used in alloys. Arsenic, antimony, and lead form an alloy with desirable properties for electrodes in lead-acid batteries. Arsenic and antimony are also used to produce semiconductor materials such as GaAs, GaSb, and InSb in electronic devices.
Bismuth
Bismuth is a metallic element. The oxidation states of bismuth are +3 and +5. Bismuth is a poor metal (one with significant covalent character) that is similar to both arsenic and antimony. Bismuth is commonly used in cosmetic products and medicine. Out of the group, bismuth has the lowest electronegativity and ionization energy, which means that it is more likely to lose an electron than the rest of the Group 15 elements. This is why bismuth is the most metallic of Group 15. Bismuth is also a poor electrical conductor. The oxide of bismuth is bismuth(III) oxide; it acts as a base, as expected for a metal oxide. Bismuth is obtained as a by-product of the refining of other metals, allowing other metals to recycle their by-products into bismuth.
Problems
1. How much of the earth's crust is made up of nitrogen?
2. How much of earth's crust is not made up of phosphorus?
3. What kind of bond does N2 have?
4. How are red phosphorus and white phosphorus related to each other?
5. What is the electron configuration of arsenic?
6. Does bismuth have metallic properties or nonmetallic properties?
7. True or False: nitrogen and phosphorus are metals.
8. Which Group 15 element has the greatest atomic radius?
9. Which Group 15 element is the strongest reducing agent?
10. True or False: bismuth exhibits the most metallic character.
11. What is the most common physical form of nitrogen?
12. What is the process in which nitrogen can convert into ammonia?
13. Which element has the highest first ionization energy?
14. Complete and balance the following reaction: _N2(g) + _H2(g) -> ____
15. What is the common oxidation state of all Group 15 elements?
Answers
1. 0.002% of the earth's crust is made up of nitrogen.
2. The earth's crust is made up of 0.11% phosphorus, so 99.89% of the earth's crust is not made up of phosphorus.
3. N2 has a triple covalent bond that is strong and hard to break.
4. Red phosphorus and white phosphorus are both allotropes of phosphorus. Red phosphorus comes from white phosphorus when it is heated to about 573 Kelvin.
5. [Ar] 3d10 4s2 4p3
6. Bismuth has metallic properties.
7. False; both nitrogen and phosphorus are nonmetals.
8. According to periodic trends, bismuth has the greatest atomic radius.
9. According to periodic trends, bismuth is the strongest reducing agent since it has an electronegativity value of 1.9, which is the same as antimony, but it has a lower ionization energy of 703 kJ/mol, which means it is more likely to get oxidized.
10. True; bismuth is the only metallic element of Group 15.
11. The most common physical form of nitrogen is a colorless gas.
12. The process of converting nitrogen into ammonia is known as the Haber-Bosch process.
13. According to periodic trends, nitrogen would have the highest first ionization energy, which means that it does not want to lose an electron the most.
14. N2(g) + 3H2(g) -> 2NH3(g)
15. The common oxidation state for all Group 15 elements is -3.
Contributors and Attributions
• Kirenjot Grewal (UCD), Connie Sou (UCD)
8.9.01: General Properties and Reactions
Summary of Nitrogen Group (Group VA) Trends:
• Nitrogen forms very strong multiple pπ-pπ bonds to itself and neighboring elements belonging to the same row, e.g., CN-, N2, NO+. Compounds of P, As, and Bi with multiple bonds may be obtained if large groups are introduced into the molecules, e.g., P2R2 and As2R2. Similar multiply bonded compounds of Sb are not known.
• The coordination numbers increase down the group. For nitrogen, 3 and 4 coordination predominate. Phosphorous and arsenic in addition form octahedral complexes, and higher coordination numbers are observed for Sb and Bi.
• The elements become increasingly metallic down the column in their chemical and physical properties. Down the group the oxo cations, e.g., SbO+ and BiO+, become more prevalent.
• Phosphorous and arsenic oxides are acidic, antimony oxide is amphoteric, and that of bismuth is basic.
• For nitrogen the N3- ion is well established in the ionic nitrides of the electropositive elements. The anionic derivatives of the heavier elements frequently retain element-element bonds, e.g., Sb42-, Bi42-, Sb73-, and As113-.
• Donor/Lewis base ability: N < P > As > Sb > Bi
• Steric effects: N > P ~ As > Sb > Bi and increase with bulk of substituents: PR3 > PR2H > PRH2 > PH3
• π-acidity: R3P > R3As > R3Sb > R3Bi
• Lewis acidity of the +5 fluorides: PF5 > AsF5 > SbF5
• In the +5 oxidation state, PF5 is not readily hydrolyzed, AsF5 hydrolyses, SbF5 reacts vigorously with water. BiF5 reacts explosively with water. In the +3 oxidation state, NF3 is unreactive, PF3 reacts only with OH- not OH2, AsF3 and SbF3 are soluble in water, and BiF3 is insoluble in water but soluble in inorganic acids.
N P As Sb Bi
EF5 unknown stable stable stable stable
ECl5 unknown stable known but unstable stable unknown
EBr5 unknown stable unknown stable unknown
EI5 unknown stable unknown stable unknown
EF3 stable stable stable stable stable
ECl3 known but unstable stable stable stable stable
EBr3 known but unstable stable stable stable stable
EI3 known but unstable stable stable stable stable
• The + 5 oxidation state halides are unknown for N, well defined for P, only stable for As as fluoride, well defined for the fluoride and chloride of antimony, and only known for the fluoride for Bi.
• For both oxidation states the stability order is F > Cl . Br >I as anticipated by the ordering of the mean enthalpies, i.e., fluoride forms the strongest bonds and iodine the weakest.
• The oxides show a similar trend. BiV and NV oxides and oxoacids are strongly oxidizing, whereas PV oxides and oxoacids are very stable and AsV and SbV oxides and oxoacids are mildly oxidizing. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/08%3A_Chemistry_of_the_Main_Group_Elements/8.09%3A_The_Nitrogen_Family/8.9.01%3A_General_Properties_and_Reactions/8.9.1.01%3A_Nitrogen_Group_%28Group_5%29_Trends.txt |
Nitrogen is present in almost all proteins and plays important roles in both biochemical applications and industrial applications. Nitrogen forms strong bonds because of its ability to form a triple bond with itself and other elements. Thus, there is a lot of energy in the compounds of nitrogen. Before 100 years ago, little was known about nitrogen. Now, nitrogen is commonly used to preserve food and as a fertilizer.
Introduction
Nitrogen is found to have either 3 or 5 valence electrons and lies at the top of Group 15 on the periodic table. It can have either 3 or 5 valence electrons because it can bond in the outer 2p and 2s orbitals. Molecular nitrogen ($N_2$) is not reactive at standard temperature and pressure and is a colorless and odorless gas.
Nitrogen is a non-metal element that occurs most abundantly in the atmosphere; nitrogen gas (N2) comprises 78.1% of the volume of the Earth’s air. It only appears in 0.002% of the earth's crust by mass. Compounds of nitrogen are found in foods, explosives, poisons, and fertilizers. Nitrogen is found in DNA in the form of nitrogenous bases as well as in neurotransmitters. It is one of the most produced industrial gases, and is produced commercially as a gas and a liquid.
Table 1: General Properties of Nitrogen
Name and Symbol Nitrogen, N
Category non-metal
Atomic Weight 14.0067
Group 15
Electron Configuration 1s2 2s2 2p3
Valence Electrons 2, 5
Phase Gas
History
Nitrogen, which makes up about 78% of our atmosphere, is a colorless, odorless, tasteless and chemically unreactive gas at room temperature. Its name is derived from the Greek nitron + genes for soda forming. For many years during the 1500's and 1600's, scientists hinted that there was another gas in the atmosphere besides carbon dioxide and oxygen. It was not until the 1700's that scientists could prove there was in fact another gas that took up mass in the atmosphere of the Earth.
Nitrogen was discovered in 1772 by Daniel Rutherford (and independently by others such as Priestly and Cavendish); Rutherford was able to remove oxygen and carbon dioxide from a contained tube full of air. He showed that there was residual gas that did not support combustion as did oxygen or carbon dioxide. While his experiment was the one that proved that nitrogen existed, other experiments were also going on in London, where they called the substance "burnt" or "dephlogisticated air".
Nitrogen is the fourth most abundant element in humans, and it is more abundant in the known universe than carbon or silicon. Most commercially produced nitrogen gas is recovered from liquefied air. Of that amount, the majority is used to manufacture ammonia ($NH_3$) via the Haber process. Much is also converted to nitric acid ($HNO_3$).
Isotopes
Nitrogen has two naturally occurring isotopes, nitrogen-14 and nitrogen-15, which can be separated with chemical exchanges or thermal diffusion. Nitrogen also has isotopes with masses of 12, 13, 16, and 17, but they are radioactive.
• Nitrogen 14 is the most abundant form of nitrogen and makes up more than 99% of all nitrogen found on Earth. It is a stable compound and is non-radioactive. Nitrogen-14 has the most practical uses, and is found in agricultural practices, food preservation, biochemicals, and biomedical research. It is found in abundance in the atmosphere and among many living organisms. It has 5 valence electrons and is not a good electrical conductor.
• Nitrogen-15 is the other stable form of nitrogen. It is often used in medical research and preservation. The element is non-radioactive and therefore can also be sometimes used in agricultural practices. Nitrogen-15 is also used in brain research, specifically nuclear magnetic resonance spectroscopy (NMR), because unlike nitrogen-14 (nuclear spin of 1), it has a nuclear spin of 1/2, which has benefits when it comes to MRI research and NMR observations. Lastly, nitrogen-15 can be used as a label or in some proteins in biology. Scientists mainly use this compound for research purposes and have not yet seen its full potential for uses in brain research.
Compounds
The two most common compounds of nitrogen are potassium nitrate (KNO3) and sodium nitrate (NaNO3). These two compounds are formed by decomposing organic matter that has potassium or sodium present and are often found in fertilizers and byproducts of industrial waste. Most nitrogen compounds have a positive Gibbs free energy (i.e., reactions are not spontaneous).
The dinitrogen molecule ($N_2$) is an "unusually stable" compound, particularly because nitrogen forms a triple bond with itself. This triple bond is difficult to break. For dinitrogen to follow the octet rule, it must have a triple bond. Nitrogen has a total of 5 valence electrons, so doubling that, we would have a total of 10 valence electrons with two nitrogen atoms. The octet requires an atom to have 8 total electrons in order to have a full valence shell, therefore it needs to have a triple bond. The compound is also very inert, since it has a triple bond. Triple bonds are very hard to break, so they keep their full valence shell instead of reacting with other compounds or atoms. Think of it this way: each triple bond is like a rubber band; with three rubber bands, the nitrogen atoms are very attracted to each other.
Nitrides
Nitrides are compounds of nitrogen with a less electronegative atom; in other words they are compounds with atoms that have a less full valence shell. These compounds form with lithium and Group 2 metals. Nitrides usually have an oxidation state of -3.
$3Mg + N_2 \rightarrow Mg_3N_2 \label{1}$
When mixed with water, nitride will form ammonia; the nitride ion acts as a very strong base.
$N^{3-} + 3H_2O_{(l)} \rightarrow NH_3 + 3OH^-_{(aq)} \label{2}$
When nitrogen forms compounds with other atoms, it primarily forms covalent bonds. These are normally formed with other metals and look like: MN, M3N, and M4N. These compounds are typically hard, inert, and have high melting points because of nitrogen's ability to form triple covalent bonds.
Ammonium Ions
Nitrogen goes through fixation by reaction with hydrogen gas over a catalyst. This process is used to produce ammonia. As mentioned earlier, this process allows us to use nitrogen as a fertilizer because it breaks down the strong triple bond held by N2. The famous Haber-Bosch process for synthesis of ammonia looks like this:
$N_2 + 3H_2 \rightarrow 2NH_3 \label{3}$
Ammonia is a base and is also used in typical acid-base reactions.
$2NH_{3(aq)} + H_2SO_4 \rightarrow (NH_4)_2SO_{4(aq)} \label{4}$
Nitride ions are very strong bases, especially in aqueous solutions.
Oxides of Nitrogen
Nitrogen uses a variety of different oxidation numbers from +1 to +5 for oxide compounds. Almost all the oxides that form are gasses, and exist at 25 degrees Celsius. Oxides of nitrogen are acidic and easily attach protons.
$N_2O_5 + H_2O \rightarrow 2HNO_{3 (aq)} \label{5}$
The oxides play a large role in living organisms. They can be useful, yet dangerous.
• Dinitrogen monoxide (N2O) is an anesthetic used by dentists, also known as laughing gas.
• Nitrogen dioxide (NO2) is harmful. It binds to hemoglobin molecules, not allowing the molecule to release oxygen throughout the body. It is released from cars and is very toxic.
• Nitrate (NO3-) is a polyatomic ion.
• The more unstable nitrogen oxides allow for space travel.
Hydrides
Hydrides of nitrogen include ammonia (NH3) and hydrazine (N2H4).
• In aqueous solution, ammonia forms the ammonium ion, which we described above, and it has special amphiprotic properties.
• Hydrazine is commonly used as rocket fuel.
Applications of Nitrogen
• Nitrogen provides a blanketing for our atmosphere and is used for the production of chemicals and electronic compartments.
• Nitrogen is used as fertilizer in agriculture to promote growth.
• Pressurized gas for oil production.
• Refrigerant (such as freezing food fast).
• Explosives.
• Metal treatment/protectant via exposure to nitrogen instead of oxygen.
Problems
• Complete and balance the following equations
N2 + ___H2→ ___NH_
H2N2O2 → ?
2NH3 + CO2 → ?
__Mg + N2 → Mg_N_
N2H5 + H2O → ?
• What are the different isotopes of nitrogen?
• List the oxidation states of various nitrogen oxides: N2O, NO, N2O3, N2O4, N2O5
• List the different elements that nitrogen will react with to make it basic or acidic....
• List some uses of nitrogen
Answers
• Complete and balance the following equations
N2 + 3H2→ 2NH3(Haber process)
H2N2O2 → 2HNO
2NH3 + CO2 → (NH2)2CO + H2O
2Mg + 3N2 → Mg3N2
N2H5 + H2O → N2+ H+ + H2O
• What are the different isotopes of nitrogen?
Stable forms include nitrogen-14 and nitrogen-15.
• List the oxidation states of various nitrogen oxides: +1, +2, +3, +4, +5 respectively.
• List the different elements that nitrogen will react with to make it basic or acidic :Nitride ion is a strong base when reacted with water; ammonia is generally a weak acid.
• Uses of nitrogen include anesthetic, refrigerant, and metal protector.
Contributors
• Adam Wandell (UC Davis) | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/08%3A_Chemistry_of_the_Main_Group_Elements/8.09%3A_The_Nitrogen_Family/8.9.02%3A_Chemistry_of_Nitrogen_%28Z7%29.txt |
Learning Objectives
• Compare properties of Group 15 elements.
• Explain the major application of phosphate.
• Describe the equilibria of the ionization of phosphoric acid.
Phosphorus (P) is an essential part of life as we know it. Without the phosphates in biological molecules such as ATP, ADP and DNA, we would not be alive. Phosphorus compounds can also be found in the minerals in our bones and teeth. It is a necessary part of our diet. In fact, we consume it in nearly all of the foods we eat. Phosphorus is quite reactive. This quality of the element makes it an ideal ingredient for matches because it is so flammable. Phosphorus is a vital element for plants and that is why we put phosphates in our fertilizer to help them maximize their growth.
Introduction
Phosphorus plays a big role in our existence but it can also be dangerous. When fertilizers containing phosphorus enter the water, it produces rapid algae growth. This can lead to eutrophication of lakes and rivers; i.e., the ecosystem has an increase of chemical nutrients and this can led to negative environmental effects. With all the excess phosphorus, plants grow rapidly then die, causing a lack of oxygen in the water and an overall reduction of water quality. It is thus necessary to remove excess phosphorus from our wastewater. The process of removing the phosphorus is done chemically by reacting the phosphorus with compounds such as ferric chloride, ferric sulfate, and aluminum sulfate or aluminum chlorohydrate. Phosphorus, when combined with aluminum or iron, becomes an insoluble salt. The solubility equilibrium constants of $FePO_4$ and $AlPO_4$ are 1.3x10-22 and 5.8x10-19, respectively. With solubilitys this low, the resulting precipitates can then be filtered out.
Another example of the dangers of phosphorus is in the production of matches. The flammable nature and cheap manufacturing of white phosphorus made it possible to easily make matches around the turn of the 20th century. However, white phosphorus is highly toxic. Many workers in match factories developed brain damage and a disease called "phosphorus necrosis of the jaw" from exposure to toxic phosphorus vapors. Excess phosphorus accumulation caused their bone tissue to die and rot away. For this reason, we now use red phosphorus or phosphorus sesquisulfide in "safety" matches.
Discovery of Phosphorus
Named from the Greek word phosphoros ("bringer of light"), elemental phosphorus is not found in its elemental form because this form is quite reactive. Because of this factor it took a long period of time for it to be "discovered". The first recorded isolation of phosphorus was by alchemist Hennig Brand in 1669, involving about 60 pails of urine. After letting a large amount of urine putrefy for a long time, Brand distilled the liquid to a paste, heated the paste, discarded the salt formed, and put the remaining substance under cold water to form solid white phosphorus. Brand's process was not very efficient; the salt he discarded actually contained most of the phosphorus. Nevertheless, he obtained some pure, elemental phosphorus for his efforts. Others of the time improved the efficiency of the process by adding sand, but still continued to discard the salt. Later, phosphorus was manufactured from bone ash. Currently, the process for manufacturing phosphorus does not involve large amounts of putrefied urine or bone ash. Instead, manufacturers use calcium phosphate and coke (Emsley).
Allotropes of Phosphorus
Phosphorus is a nonmetal, solid at room temperature, and a poor conductor of heat and electricity. Phosphorus occurs in at least 10 allotropic forms, the most common (and reactive) of which is so-called white (or yellow) phosphorus, which looks like a waxy solid or plastic. It is very reactive and will spontaneously inflame in air, so it is stored under water. The other common form of phosphorus is red phosphorus, which is much less reactive and is one of the components on the striking surface of a match book. Red phosphorus can be converted to white phosphorus by careful heating.
White phosphorus consists of $\ce{P4}$ molecules, whereas the crystal structure of red phosphorus has a complicated network of bonding. White phosphorus has to be stored in water to prevent natural combustion, but red phosphorus is stable in air.
When burned, red phosphorus also forms the same oxides as those obtained in the burning of white phosphosrus, $\ce{P4O6}$ when air supply is limited, and $\ce{P4O10}$ when sufficient air is present.
Diphosphorus (P2)
Diphosphorus ($P_2$) is the gaseous form of phosphorus that is thermodynamically stable above 1200 °C and until 2000 °C. It can be generated by heating white phosphorus (see below) to 1100 K and is very reactive with a bond-dissociation energy (117 kcal/mol or 490 kJ/mol) half that of dinitrogen ($N_2$).
White Phosphorus (P4)
White phosphorus (P4) has a tetrahedral structure. It is soft and waxy, but insoluble in water. Its glow occurs as a result of its vapors slowly being oxidized by the air. It is so thermodynamically unstable that it combusts in air. It was once used in fireworks and the U.S. military still uses it in incendiary bombs.
This Youtube video link shows various experiments with white phosphorus, which help show the physical and chemical properties of it. It also shows white phosphorus combusting with air.
Red Phosphorus and Violet Phosphorus (Polymeric)
Red Phosphorus has more atoms linked together in a network than white phosphorus does, which makes it much more stable. It is not quite as flammable, but given enough energy it still reacts with air. For this reason, we now use red phosphorus in matches.
Violet phosphorus is obtained from heating and crystallizing red phosphorus in a certain way. The phosphorus forms pentagonal "tubes".
Black Phosphorus (Polymeric)
Black phosphorus is the most stable form; the atoms are linked together in puckered sheets, like graphite. Because of these structural similarities black phosphorus is also flaky like graphite and possesses other similar properties.
Isotopes of Phosphorus
There are many isotopes of phosphorus, only one of which is stable (31P). The rest of the isotopes are radioactive with generally very short half-lives, which vary from a few nanoseconds to a few seconds. Two of the radioactive phosphorus isotopes have longer half-lives: 32P has a half-life of 14 days and 33P has a half-life of 25 days. These half-lives are long enough to be useful for analysis, and for this reason the isotopes can be used to mark DNA.
32P played an important role in the 1952 Hershey-Chase Experiment. In this experiment, Alfred Hershey and Martha Chase used radioactive isotopes of phosphorus and sulfur to determine that DNA was genetic material and not proteins. Sulfur can be found in proteins but not DNA, and phosphorus can be found in DNA but not proteins. This made phosphorus and sulfur effective markers of DNA and protein, respectively. The experiment was set up as follows: Hershey and Chase grew one sample of a virus in the presence of radioactive 35S and another sample of a virus in the presence of 32P. Then, they allowed both samples to infect bacteria. They blended the 35S and the 32P samples separately and centrifuged the two samples. Centrifuging separated the genetic material from the non-genetic material. The genetic material penetrated the solid that contained the bacterial cells at the bottom of the tube while the non-genetic material remained in the liquid. By analyzing their radioactive markers, Hershey and Chase found that the 32P remained with the bacteria, and the 35S remained in the supernatant liquid. These results were confirmed by further tests involving the radioactive phosphorus..
Phosphorus and Life
We get most elements from nature in the form of minerals. In nature, phosphorus exists in the form of phosphates. Rocks containing phosphate are fluoroapatite ($\ce{3Ca3(PO4)2.CaF2}$), chloroapaptite, ($\ce{3Ca3(PO4)2. CaCl2}$), and hydroxyapatite ($\ce{3Ca3(PO4)2. Ca(OH)2}$). These minerals are very similar to the bones and teeth. The arrangements of atoms and ions of bones and teeth are similar to those of the phosphate-containing rocks. In fact, when the $\ce{OH-}$ ions of the teeth are replaced by $\ce{F-}$, the teeth resist decay. This discovery led to a series of social and economical issues.
Nitrogen, phosphorus and potassium are key ingredients for plants, and their contents are key in all forms of fertilizers. From an industrial and economical viewpoint, phosphorus-containing compounds are important commodities. Thus, the chemistry of phosphorus has academic, commercial and industrial interests.
Chemistry of Phosphorus
As a member of the Nitrogen Family, Group 15 on the Periodic Table, phosphorus has 5 valence shell electrons available for bonding. Its valence shell configuration is 3s23p3. Phosphorus forms mostly covalent bonds. Any phosphorus rock can be used for the production of elemental phosphorus. Crushed phosphate rocks and sand ($\ce{SiO2}$) react at 1700 K to give phosphorus oxide, $\ce{P4O10}$:
$\ce{2 Ca3(PO4)2 + 6 SiO2 \rightarrow P4O10 + 6 CaSiO3} \label{1}$
$\ce{P4O10}$ can be reduced by carbon:
$\ce{P4O10 + 10 C \rightarrow P4 + 10 CO}. \label{2}$
Waxy solids of white phosphorus are molecular crystals consisting of $\ce{P4}$ molecules. They have an interesting property in that they undergo spontaneous combustion in air:
$\ce{P4 + 5 O2 \rightarrow P4O10} \label{3}$
The structure of $\ce{P4}$ can be understood by thinking of the electronic configuration (s2 p3) of $\ce{P}$ in bond formation. Sharing three electrons with other $\ce{P}$ atoms gives rise to the 6 $\ce{P-P}$ bonds, leaving a lone pair occupying the 4th position in a distorted tetrahedron.
When burned with insufficient oxygen, $\ce{P4O6}$ is formed:
$\ce{P4 + 3 O2 \rightarrow P4O6} \label{4}$
Into each of the $\ce{P-P}$ bonds, an $\ce{O}$ atom is inserted.
Burning phosphorus with excess oxygen results in the formation of $\ce{P4O10}$. An additional $\ce{O}$ atom is attached to the $\ce{P}$ directly:
$\ce{P4 + 5 O2 \rightarrow P4O10} \label{5}$
Thus, the oxides $\ce{P4O6}$ and $\ce{P4O10}$ share interesting features. Oxides of phosphorus, $\ce{P4O10}$, dissolve in water to give phosphoric acid,
$\ce{P4O10 + 6 H2O \rightarrow 4 H3PO4} \label{6}$
Phosphoric acid is a polyprotic acid, and it ionizes in three stages:
$\ce{H3PO4 \rightleftharpoons H+ + H2PO4-} \label{7a}$
$\ce{H2PO4- \rightleftharpoons H+ + HPO4^2-} \label{7b}$
$\ce{HPO4^2- \rightarrow H+ + PO4^3-} \label{7c}$
Phosphoric Acid
Phosphoric acid is a polyprotic acid, which makes it an ideal buffer. It gets harder and harder to separate the hydrogen from the phosphate, making the pKa values increase in basicity: 2.12, 7.21, and 12.67. The conjugate bases H2PO4-, HPO42-, and PO43- can be mixed to form buffer solutions.
Reaction Dissociation Constant
Table 1: Ionization constants for the successive deprotonation of phosphoric acid states
$H_3PO_4 + H_2O \rightarrow H_3O^+ + H_2PO^{4-}$ Ka1=7.5x10-3
$H_2PO_4^{-} + H_2O \rightarrow H_3O^+ + HPO_4^{2-}$ Ka2=6.2x10-8
$H_2PO_4^{-} + H_2O \rightarrow H_3O^+ + PO_4^{3-}$ Ka3=2.14x10-13
Overall: $H_3PO_4 + 3H_2O \rightarrow 3 H_3O^+ + PO_4^{3-}$
Past and Present Uses of Phosphorus
Commercially, phosphorus compounds are used in the manufacture of phosphoric acid ($H_3PO_4$) (found in soft drinks and used in fertilizer compounding). Other compounds find applications in fireworks and, of course, phosphorescent compounds which glow in the dark. Phosphorus compounds are currently used in foods, toothpaste, baking soda, matches, pesticides, nerve gases, and fertilizers. Phosphoric acid is not only used in buffer solutions; it is also a key ingredient of Coca Cola and other sodas! Phosphorus compounds were once used in detergents as a water softener until they raised concerns about pollution and eutrophication. Pure phosphorus was once prescribed as a medicine and an aphrodisiac until doctors realized it was poisonous (Emsley).
Questions
1. About 85% of the total industrial output of phosphoric acid is used
a. in the detergent industry
b. to produce buffer solutions
c. in the paint industry
d. to produce superphosphate fertilizers
e. in the manufacture of plastics
2. What is the product when phosphorus pentoxide $\ce{P4O10}$ reacts with water? Give the formula of the product.
3. What is the phosphorus-containing product when $\ce{PCl3}$ reacts with water? Give the formula.
Solutions
1. Answer... d
The middle number, (for example, 6-5-8) specifies the percentage of phosphorus compound in a fertilizer. Phosphorus is an important element for plant life.
2. Answer $\ce{H3PO4}$
$\ce{P4O10 + 6 H2O \rightarrow 4 H3PO4} \nonumber$
3. Answer $\ce{H3PO3}$
$\ce{PCl3 + 3 H2O \rightarrow H3PO3 + 3 HCl} \nonumber$
This is a weaker acid than $\ce{H3PO4}$.
Contributors
• Aimee Kindel (UCD), Kirenjot Grewal (UCD), Tiffany Lui (UCD)
• Chung (Peter) Chieh (Professor Emeritus, Chemistry @ University of Waterloo)
8.9.04: Chemistry of Arsenic (Z33)
Contributors and Attributions
• Freddie Chak, Jonathan Molina, Tiffany Lui (University of California, Davis)
Stephen R. Marsden
8.9.05: Chemistry of Antimony (Z51)
Antimony and its compounds have been known for centuries. Scientific study of the element began during the early 17th century, much of the important work being done by Nicolas Lemery. The name of the element comes from the Greek anti + monos for "not alone", while the modern symbol is rooted in the Latin-derived name of the common ore, stibnite. Antimony is a hard, brittle metalloid which is alloyed with other metals to increase hardness. It is also used in some semi-conductor devices. The recovery of elemental antimony parallels that of arsenic: the sulfide ore (stibnite) is roasted in air and then heated with carbon.
Antimony trisulfide
Antimony trisulfide, $Sb_2S_3$, is a sulfide mineral commonly called stibnite or antimonite. Antimony trisulfide exists as a gray/black crystalline solid (orthorhombic crystals) and an amorphous red-orange powder. It turns black due to oxidation by air. Antimony trisulfide is the most important source for antimony. It is insoluble in water and melts at 550°C. The chemical symbol of antimony (Sb) is derived from stibnite.
Amorphous (red to yellow-orange) antimony trisulfide can be prepared by treating an antimony trichloride solution with hydrogen sulfide:
$2 SbCl_3 + 3 H_2S \rightarrow Sb_2S_3 + 6 HCl \nonumber$
When antimony trisulfide is melted with iron at approximately 600°C, the following reaction yields elementary antimony:
$Sb_2S_3 + 3 Fe \rightarrow 2 Sb + 3 FeS \nonumber$
$Sb_2S_3$ is used as a pigment, in pyrotechnics (glitter and fountain mixtures) and on safety matches. In combination with antimony oxides it is also used as a yellow pigment in glass and porcelain. Antimony trisulfide photoconductors are used in vidicons for CCTV.
Contributors and Attributions
Stephen R. Marsden
8.9.06: Chemistry of Bismuth (Z83)
Bismuth, the heaviest non-radioactive naturally occurring element, was isolated by Basil Valentine in 1450. It is a hard, brittle metal with an unusually low melting point (271oC). Alloys of bismuth with other low-melting metals such as tin and lead have even lower melting points and are used in electrical solders, fuse elements and automatic fire sprinkler heads.
The metal can be found in nature, often combined with copper or lead ores, but can also be extracted from bismuth(III) oxide by roasting with carbon. Compounds of bismuth are used in pigments for oil paintings and one is in a popular pink preparation for the treatment of common stomach upset.
Contributors and Attributions
Stephen R. Marsden
8.9.07: Chemistry of Moscovium (Z115)
In studies announced jointly by the Joint Institute for Nuclear Research in Dubna, Russia, and the Lawrence Livermore National Laboratory in the U.S., four atoms of element 113 were produced in 2004 via decay of element 115 after the fusion of Ca-48 and Am-243.
Contributors and Attributions
• Stephen R. Marsden | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/08%3A_Chemistry_of_the_Main_Group_Elements/8.09%3A_The_Nitrogen_Family/8.9.03%3A_Chemistry_of_Phosphorus_%28Z15%29.txt |
The oxygen family, also called the chalcogens, consists of the elements found in Group 16 of the periodic table and is considered among the main group elements. It consists of the elements oxygen, sulfur, selenium, tellurium and polonium. These can be found in nature in both free and combined states.
• 8.11.1: General Properties and Reactions
The oxygen family, also called the chalcogens, consists of the elements found in Group 16 of the periodic table and is considered among the main group elements. It consists of the elements oxygen, sulfur, selenium, tellurium and polonium. These can be found in nature in both free and combined states. The group 16 elements are intimately related to life.
• 8.11.2: Chemistry of Oxygen (Z=8)
Oxygen is an element that is widely known by the general public because of the large role it plays in sustaining life. Without oxygen, animals would be unable to breathe and would consequently die. Oxygen is not only important to supporting life, but plays an important role in many other chemical reactions. Oxygen is the most common element in the earth's crust and makes up about 20% of the air we breathe.
• 8.11.3: Chemistry of Sulfur (Z=16)
Sulfur is a chemical element that is represented with the chemical symbol "S" and the atomic number 16 on the periodic table. Because it is 0.0384% of the Earth's crust, sulfur is the seventeenth most abundant element following strontium. Sulfur also takes on many forms, which include elemental sulfur, organo-sulfur compounds in oil and coal, H2S(g) in natural gas, and mineral sulfides and sulfates.
• 8.11.4: Chemistry of Selenium (Z=34)
Element number 34, selenium, was discovered by Swedish chemist Jons Jacob Berzelius in 1817. Selenium is a non-metal and can be compared chemically to its other non-metal counterparts found in Group 16: The Oxygen Family, such as sulfur and tellurium.
• 8.11.5: Chemistry of Tellurium (Z=52)
Discovered by von Reichenstein in 1782, tellurium is a brittle metalloid that is relatively rare. It is named from the Latin tellus for "earth". Tellurium can be alloyed with some metals to increase their machinability and is a basic ingredient in the manufacture of blasting caps. Elemental tellurium is occasionally found in nature but is more often recovered from various gold ores.
• 8.11.6: Chemistry of Polonium (Z=84)
Polonium was discovered in 1898 by Marie Curie and named for her native country of Poland. The discovery was made by extraction of the remaining radioactive components of pitchblende following the removal of uranium. There is only about 10-6 g per ton of ore! Current production for research purposes involves the synthesis of the element in the lab rather than its recovery from minerals. This is accomplished by producing Bi-210 from the abundant Bi-209.
• 8.11.7: Chemistry of Livermorium (Z=116)
In May of 2012 the IUPAC approved the name "Livermorium" (symbol Lv) for element 116. The new name honors the Lawrence Livermore National Laboratory (1952). A group of researchers of this Laboratory with the heavy element research group of the Flerov Laboratory of Nuclear Reactions took part in the work carried out in Dubna on the synthesis of superheavy elements including element 116.
Thumbnail: A sample of sulfur a member of the oxygen group of elements. (Public Domain; Ben Mills).
8.11: The Oxygen Family (The Chalcogens)
The oxygen family, also called the chalcogens, consists of the elements found in Group 16 of the periodic table and is considered among the main group elements. It consists of the elements oxygen, sulfur, selenium, tellurium and polonium. These can be found in nature in both free and combined states. The group 16 elements are intimately related to life. We need oxygen all the time throughout our lives. Did you know that sulfur is also one of the essential elements of life? It is responsible for some of the protein structures in all living organisms. Many industries utilize sulfur, but emission of sulfur compounds is often seen more as a problem than the natural phenomenon. The metallic properties of Group 16 elements increase as the atomic number increases. The element polonium has no stable isotopes, and the isotope with mass number 209 has the longest half life of 103 years.
Properties of oxygen are very different from those of other elements of the group, but they all have 2 elections in the outer s orbital, and 4 electrons in the p orbitals, usually written as s2p4.
The electron configurations for each element are given below:
• Oxygen: 1s2 2s2 2p4
• Sulfur: 1s2 2s2p6 3s2p4
• Selenium: 1s2 2s2p6 3s2p6d10 4s2p4
• Tellurium: 1s2 2s2p6 3s2p6d10 4s2p6d10 5s2p4
• Polonium: 1s2 2s2p6 3s2p6d10 4s2p6d10f14 5s2p6d10 6s2p4
Example \(1\): Polonium
Polonium can be written as [Xe] 6s2 4f14 5d10 6p4
The trends of the properties in this group are interesting. Knowing the trend allows us to predict their reactions with other elements. Most trends are true for all groups of elements, and the group trends are due mostly to the size of the atoms and number of electrons per atom. The trends are described below:
1. The metallic properties increase in the order oxygen, sulfur, selenium, tellurium, or polonium. Polonium is essentially a metal. It was discovered by M. Curie, who named it after her native country Poland.
2. Electronegativity, ionization energy (or ionization potential IP), and electron affinity decrease for the group as atomic weight increases.
3. The atomic radii and melting points increase.
4. Oxygen differs from sulfur in chemical properties due to its small size. The differences between \(\ce{O}\) and \(\ce{S}\) are more than the differences between other members.
Metallic character increases down the group, with tellurium classified as a metalloid and polonium as a metal. Melting point, boiling point, density, atomic radius, and ionic radius all increase down the group. Ionization energy decreases down the group. The most common oxidation state is -2; however, sulfur can also exist at a +4 and +6 state, and +2, +4, and +6 oxidation states are possible for Se, Te, and Po.
Table \(1\): Select properties of Group 16 elements
Oxygen Sulfur Selenium Tellurium Polonium
Boiling Pt (°C) -182.962 444.674 685 989.9 962
Ionization Energy (kJ/mol) 1314 1000 941 869 812
Ionic Radius (pm) 140 184 198 221
Oxygen
Oxygen is a gas at room temperature and 1 atm, and is colorless, odorless, and tasteless. It is the most abundant element by mass in both the Earth's crust and the human body. It is second to nitrogen as the most abundant element in the atmosphere. There are many commercial uses for oxygen gas, which is typically obtained through fractional distillation. It is used in the manufacture of iron, steel, and other chemicals. It is also used in water treatment, as an oxidizer in rocket fuel, for medicinal purposes, and in petroleum refining.
Oxygen has two allotropes, O2 and O3. In general, O2 (or dioxygen) is the form referred to when talking about the elemental or molecular form because it is the most common form of the element. The O2 bond is very strong, and oxygen can also form strong bonds with other elements. However, compounds that contain oxygen are considered to be more thermodynamically stable than O2.
The latter allotrope, ozone, is a pale-blue poisonous gas with a strong odor. It is a very good oxidizing agent, stronger than dioxygen, and can be used as a substitute for chlorine in purifying drinking water without giving the water an odd taste. However, because of its unstable nature it disappears and leaves the water unprotected from bacteria. Ozone at very high altitudes in the atmosphere is responsible for protecting the Earth's surface from ultraviolet radiation; however, at lower altitudes it becomes a major component of smog.
Oxygen's primary oxidation states are -2, -1, 0, and -1/2 (in O2-), but -2 is the most common. Typically, compounds with oxygen in this oxidation state are called oxides. When oxygen reacts with metals, it forms oxides that are mostly ionic in nature. These can dissolve in water and react to form hydroxides; they are therefore called basic anhydrides or basic oxides. Nonmetal oxides, which form covalent bonds, are simple molecules with low melting and boiling points.
Compounds with oxygen in an oxidation state of -1 are referred to as peroxides. Examples of this type of compound include \(Na_2O_2\) and \(BaO_2\). When oxygen has an oxidation state of -1/2, as in \(O_2^-\), the compound is called a superoxide.
Oxygen is rarely the central atom in a structure and can never bond with more than 4 elements due to its small size and its inability to create an expanded valence shell. Oxygen reacts with hydrogen to form water, which is extensively hydrogen-bonded, has a large dipole moment, and is considered a universal solvent.
There are a wide variety of oxygen-containing compounds, both organic and inorganic: oxides, peroxides and superoxides, alcohols, phenols, ethers, and carbonyl-containing compounds such as aldehydes, ketones, esters, amides, carbonates, carbamates, carboxylic acids and anhydrides.
Sulfur
Sulfur is a solid at room temperature and 1 atm pressure. It is usually yellow, tasteless, and nearly odorless. It is the sixteenth most abundant element in Earth's crust. It exists naturally in a variety of forms, including elemental sulfur, sulfides, sulfates, and organosulfur compounds. Since the 1890s, sulfur has been mined using the Frasch process, which is useful for recovering sulfur from deposits that are under water or quicksand. Sulfur produced from this process is used in a variety of ways including in vulcanizing rubber and as fungicide to protect grapes and strawberries.
Sulfur is unique in its ability to form a wide range of allotropes, more than any other element in the periodic table. The most common state is the solid S8 ring, as this is the most thermodynamically stable form at room temperature. Sulfur exists in the gaseous form in five different forms (S, S2, S4, S6, and S8). In order for sulfur to convert between these compounds, sufficient heat must be supplied.
Two common oxides of sulfur are sulfur dioxide (SO2) and sulfur trioxide (SO3). Sulfur dioxide is formed when sulfur is combusted in air, producing a toxic gas with a strong odor. These two compounds are used in the production of sulfuric acid, which is used in a variety of reactions. Sulfuric acid is one of the top manufactured chemicals in the United States, and is primarily used in the manufacture of fertilizers.
Sulfur also exhibits a wide range of oxidation states, with values ranging from -2 to +6. It is often the central ion in a compound and can easily bond with up to 6 atoms. In the presence of hydrogen it forms the compound hydrogen sulfide, H2S, a poisonous gas incapable of forming hydrogen bonds and with a very small dipole moment. Hydrogen sulfide can easily be recognized by its strong odor that is similar to that of rotten eggs, but this smell can only be detected at low, nontoxic concentrations. This reaction with hydrogen epitomizes how differently oxygen and sulfur act despite their common valence electron configuration and common nonmetallic properties.
A variety of sulfur-containing compounds exist, many of them organic. The prefix thio- in front of the name of an oxygen-containing compound means that the oxygen atom has been substituted with a sulfur atom. General categories of sulfur-containing compounds include thiols (mercaptans), thiophenols, organic sulfides (thioethers), disulfides, thiocarbonyls, thioesters, sulfoxides, sulfonyls, sulfamides, sulfonic acids, sulfonates, and sulfates.
Selenium
Selenium appears as a red or black amorphous solid, or a red or grey crystalline structure; the latter is the most stable. Selenium has properties very similar to those of sulfur; however, it is more metallic, though it is still classified as a nonmetal. It acts as a semiconductor and therefore is often used in the manufacture of rectifiers, which are devices that convert alternating currents to direct currents. Selenium is also photoconductive, which means that in the presence of light the electrical conductivity of selenium increases. It is also used in the drums of laser printers and copiers. In addition, it has found increased use now that lead has been removed from plumbing brasses.
It is rare to find selenium in its elemental form in nature; it must typically be removed through a refining process, usually involving copper. It is often found in soils and in plant tissues that have bioaccumulated the element. In large doses, the element is toxic; however, many animals require it as an essential micronutrient. Selenium atoms are found in the enzyme glutathione peroxidase, which destroys lipid-damaging peroxides. In the human body it is an essential cofactor in maintaining the function of the thyroid gland. In addition, some research has shown a correlation between selenium-deficient soils and an increased risk of contracting the HIV/AIDS virus.
Tellurium
Tellurium is the metalloid of the oxygen family, with a silvery white color and a metallic luster similar to that of tin at room temperature. Like selenium, it is also displays photoconductivity. Tellurium is an extremely rare element, and is most commonly found as a telluride of gold. It is often used in metallurgy in combination with copper, lead, and iron. In addition, it is used in solar panels and memory chips for computers. It is not toxic or carcinogenic; however, when humans are exposed to too much of it they develop a garlic-like smell on their breaths.
Polonium
Polonium is a very rare, radioactive metal. There are 33 different isotopes of the element and all of the isotopes are radioactive. It exists in a variety of states, and has two metallic allotropes. It dissolves easily into dilute acids. Polonium does not exist in nature in compounds, but it can form synthetic compounds in the laboratory. It is used as an alloy with beryllium to act as a neutron source for nuclear weapons.
Polonium is a highly toxic element. The radiation it emits makes it very dangerous to handle. It can be immediately lethal when applied at the correct dosage, or cause cancer if chronic exposure to the radiation occurs. Methods to treat humans who have been contaminated with polonium are still being researched, and it has been shown that chelation agents could possibly be used to decontaminate humans.
Problems
1. What properties increase down the oxygen family?
2. What element can form the most allotropes in the periodic table?
3. What is photoconductivity and which elements display this property?
4. Ozone (\(O_3\)) is a contributor to smog: True or False
5. How many electrons do elements of the oxygen family have in their outermost shell?
6. What does the term "peroxide" refer to?
7. How many elements in the oxygen family are metals, and which one(s)?
8. What is the most common oxidation state for elements in the oxygen family?
9. What is the most abundant element by mass in the Earth's crust and in the human body?
Solutions
1. Melting point, boiling point, density, atomic radius, and ionic radius.
2. Sulfur.
3. Photoconductivity is when the electrical conductivity of an element increases in the presence of light. Both selenium and tellurium display this property.
4. True.
5. Six.
6. A compound that contains oxygen in the oxidation state of -1.
7. 1; Polonium.
8. -2.
9. Oxygen. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/08%3A_Chemistry_of_the_Main_Group_Elements/8.11%3A_The_Oxygen_Family_%28The_Chalcogens%29/8.11.01%3A_General_Properties_and_Reactions.txt |
Oxygen is an element that is widely known by the general public because of the large role it plays in sustaining life. Without oxygen, animals would be unable to breathe and would consequently die. Oxygen not only is important to supporting life, but also plays an important role in many other chemical reactions. Oxygen is the most common element in the earth's crust and makes up about 20% of the air we breathe. Historically the discovery of oxygen as an element essential for combustion stands at the heart of the phlogiston controversy (see below).
The Origin and History
Oxygen is found in the group 16 elements and is considered a chalcogen. Named from the Greek oxys + genes, "acid-former", oxygen was discovered in 1772 by Scheele and independently by Priestly in 1774. Oxygen was given its name by the French scientist, Antoine Lavoisier.
Scheele discovered oxygen through an experiment which involved burning manganese oxide. Scheele came to find that the hot manganese oxide produced a gas which he called "fire air". He also came to find that when this gas was able to come into contact with charcoal, it produced beautiful bright sparks. All of the other elements produced the same gas. Although Scheele discovered oxygen, he did not publish his work until three years after another chemist, Joseph Priestly, discovered oxygen. Joseph Priestly, an English chemist, repeated Scheele's experiment in 1774 using a slightly different setup. Priestly used a 12 in burning glass and aimed the sunlight directly towards the compound that he was testing, mercuric oxide. As a result, he was able to "discover better air" that was shown to expand a mouse's lifetime to four times as long and caused a flame to burn with higher intensity. Despite these experiments, both chemists were not able to pinpoint exactly what this element was. It was not until 1775 that Antoine Lavoisier, a French chemist, was able to recognize this unknown gas as an element.
Our atmosphere currently contains about 21% of free oxygen. Oxygen is produced in various ways. The process of photochemical dissociation in which water molecules are broken up by ultraviolet rays produces about 1-2% of our oxygen. Another process that produces oxygen is photosynthesis which is performed by plants and photosynthetic bacteria. Photosynthesis occurs through the following general reaction:
$\ce{CO2 + H2O + h\nu \rightarrow} \text{organic compounds} \ce{+ O2} \nonumber$
The Dangers of Phlogiston
Phlogiston theory is the outdated belief that a fire-like element called phlogiston is contained within combustible bodies and released during combustion. The name comes from the Ancient Greek φλογιστόν phlogistón (burning up), from φλόξ phlóx (flame). It was first stated in 1667 by Johann Joachim Becher, and then put together more formally by Georg Ernst Stahl. The theory attempted to explain burning processes such as combustion and rusting, which are now collectively known as oxidation.
Properties
• Element number: 8
• Atomic weight 15.9994
• Color: gas form- colorless, liquid- pale blue
• Melting point: 54.36K
• Boiling point: 90.2 K
• Density: .001429
• 21% of earth's atmosphere
• Third most abundant element in the universe
• Most abundant element in Earth's crust at 45.4%
• 3 Stable isotopes
• Ionization energy: 13.618 eV
• Oxygen is easily reduced and is a great oxidizing agent making it readily reactive with other elements
Magnetic Properties of Oxygen
Oxygen (O2) is paramagnetic. An oxygen molecule has six valence electrons, so the O2 molecule has 12 valence electrons with the electron configuration shown below:
As shown, there are two unpaired electrons, which causes O2 to be paramagnetic. There are also eight valence electrons in the bonding orbitals and four in antibonding orbitals, which makes the bond order 2. This accounts for the double covalent bond that is present in O2.
Video $1$: A chemical demonstration of the paramagnetism of molecular oxygen, as shown by the attraction of liquid oxygen to magnets.
As shown in Video $1$, since molecular oxygen ($O_2$) has unpaired electrons, it is paramagnetic and is attracted to the magnet. In contrast, molecular nitrogen ($N_2$) has no unpaired electrons and is not attracted to the magnet.
General Chemistry of Oxygen
Oxygen normally has an oxidation state of -2, but is capable of having oxidation states of -2, -1, -1/2, 0, +1, and +2. The oxidation states of oxides, peroxides and superoxides are as follows:
• Oxides: O-2 ,
• peroxides: O2-2 ,
• superoxide: O2-1.
Oxygen does not react with itself, nitrogen, or water under normal conditions. Oxygen does, however, dissolve in water at 20 degrees Celsius and 1 atmosphere. Oxygen also does not normally react with bases or acids. Group 1 metals (alkaline metals) are very reactive with oxygen and must be stored away from oxygen in order to prevent them from becoming oxidized. The metals at the bottom of the group are more reactive than those at the top. The reactions of a few of these metals are explored in more detail below.
Lithium: Reacts with oxygen to form white lithium oxide in the reaction below.
$\ce{4Li + O_2 \rightarrow 2Li_2O} \label{1}$
Sodium: Reacts with oxygen to form a white mixture of sodium oxide and sodium peroxide. The reactions are shown below.
• Sodium oxide: $\ce{4Na + O_2 \rightarrow 2Na_2O} \label{2}$
• Sodium peroxide: $\ce{2Na + O_2 \rightarrow Na_2O_2} \label{3}$
Potassium: Reacts with oxygen to form a mixture of potassium peroxide and potassium superoxide. The reactions are shown below.
• Potassium peroxide: $\ce{2K + O_2 \rightarrow 2K_2O_2} \label{4}$
• Potassium superoxide: $\ce{K + O_2 \rightarrow KO_2} \label{5}$
Rubidium and Cesium: Both metals react to produce superoxides through the same process as that of the potassium superoxide reaction.
The oxides of these metals form metal hydroxides when they react with water. These metal hydroxides make the solution basic or alkaline, hence the name alkaline metals.
Group 2 metals (alkaline earth metals) react with oxygen through the process of burning to form metal oxides but there are a few exceptions.
Beryllium is very difficult to burn because it has a layer of beryllium oxide on its surface which prevents further interaction with oxygen. Strontium and barium react with oxygen to form peroxides. The reaction of barium and oxygen is shown below, and the reaction with strontium would be the same.
$\ce{Ba(s) + O2 (g) \rightarrow BaO2 (s) }\label{6}$
Group 13 reacts with oxygen in order to form oxides and hydroxides that are of the form $X_2O_3$ and $X(OH)_3$. The variable X represents the various group 13 elements. As you go down the group, the oxides and hydroxides get increasingly basic.
Group 14 elements react with oxygen to form oxides. The oxides formed at the top of the group are more acidic than those at the bottom of the group. Oxygen reacts with silicon and carbon to form silicon dioxide and carbon dioxide. Carbon is also able to react with oxygen to form carbon monoxide, which is slightly acidic. Germanium, tin, and lead react with oxygen to form monoxides and dioxides that are amphoteric, which means that they react with both acids and bases.
Group 15 elements react with oxygen to form oxides. The most important are listed below.
• Nitrogen: N2O, NO, N2O3, N2O4, N2O5
• Phosphorus: P4O6, P4O8, P2O5
• Arsenic: As2O3, As2O5
• Antimony: Sb2O3, Sb2O5
• Bismuth: Bi2O3, Bi2O5
Group 16 elements react with oxygen to form various oxides. Some of the oxides are listed below.
• Sulfur: SO, SO2, SO3, S2O7
• Selenium: SeO2, SeO3
• Tellurium: TeO, TeO2, TeO3
• Polonium: PoO, PoO2, PoO3
Group 17 elements (halogens) fluorine, chlorine, bromine, and iodine react with oxygen to form oxides. Fluorine forms two oxides with oxygen: F2O and F2O2. Both fluorine oxides are called oxygen fluorides because fluorine is the more electronegative element. One of the fluorine reactions is shown below.
$\ce{O2 (g) + F2 (g) \rightarrow F2O2 (g)} \label{7}$
Group 18: Some would assume that the Noble Gases would not react with oxygen. However, xenon does react with oxygen to form $\ce{XeO_3}$ and $\ce{XeO_4}$. The ionization energy of xenon is low enough for the electronegative oxygen atom to "steal away" electrons. Unfortunately, $\ce{XeO_3}$ is HIGHLY unstable, and it has been known to spontaneously detonate in a clean, dry environment.
Transition metals react with oxygen to form metal oxides. However, gold, silver, and platinum do not react with oxygen. A few reactions involving transition metals are shown below:
$2Sn_{(s)} + O_{2(g)} \rightarrow 2SnO_{(s)} \label{8}$
$4Fe_{(s)} + 3O_{2(g)} \rightarrow 2Fe_2O_{3(s)} \label{9A}$
$4Al_{(s)} + 3O_{2(g)} \rightarrow 2Al_2O_{3(s)} \label{9B}$
Reaction of Oxides
We will be discussing metal oxides of the form $X_2O$. The variable $X$ represents any metal that is able to bond to oxygen to form an oxide.
• Reaction with water: The oxides react with water to form a metal hydroxide.
$X_2O + H_2O \rightarrow 2XOH \nonumber$
• Reaction with dilute acids: The oxides react with dilute acids to form a salt and water.
$X_2O + 2HCl \rightarrow 2XCl + H_2O \nonumber$
Reactions of Peroxides
The peroxides we will be discussing are of the form $X_2O_2$. The variable $X$ represents any metal that can form a peroxide with oxygen.
Reaction with water: If the temperature of the reaction is kept constant despite the fact that the reaction is exothermic, then the reaction proceeds as follows:
$X_2O_2+ 2H_2O \rightarrow 2XOH + H_2O_2 \nonumber$
If the reaction is not carried out at a constant temperature, then the reaction of the peroxide and water will result in decomposition of the hydrogen peroxide that is produced into water and oxygen.
Reaction with dilute acid: This reaction is more exothermic than that with water. The heat produced causes the hydrogen peroxide to decompose to water and oxygen. The reaction is shown below.
$X_2O_2 + 2HCl \rightarrow 2XCl + H_2O_2 \nonumber$
$2H_2O_2 \rightarrow 2H_2O + O_2 \nonumber$
Reaction of Superoxides
The superoxides we will be talking about are of the form $XO_2$, with $X$ representing any metal that forms a superoxide when reacting with oxygen.
Reaction with water: The superoxide and water react in a very exothermic reaction that is shown below. The heat that is produced in forming the hydrogen peroxide will cause the hydrogen peroxide to decompose to water and oxygen.
$2XO_2 + 2H_2O \rightarrow 2XOH + H_2O_2 + O_2 \nonumber$
Reaction with dilute acids: The superoxide and dilute acid react in a very exothermic reaction that is shown below. The heat produced will cause the hydrogen peroxide to decompose to water and oxygen.
$2XO_2 + 2HCl \rightarrow 2XCl + H_2O_2 + O_2 \nonumber$
Allotropes of Oxygen
There are two allotropes of oxygen; dioxygen (O2) and trioxygen (O3) which is called ozone. The reaction of converting dioxygen into ozone is very endothermic, causing it to occur rarely and only in the lower atmosphere. The reaction is shown below:
$3O_{2 (g)} \rightarrow 2O_{3 (g)} \;\;\; ΔH^o= +285 \;kJ \nonumber$
Ozone is unstable and quickly decomposes back to oxygen but is a great oxidizing agent.
Miscellaneous Reactions
Reaction with Alkanes: The most common reactions that involve alkanes occur with oxygen. Alkanes are able to burn and it is the process of oxidizing the hydrocarbons that makes them important as fuels. An example of an alkane reaction is the reaction of octane with oxygen as shown below.
C8H18(l) + 25/2 O2(g) → 8CO2(g) + 9H2O(l) ∆Ho = -5.48 X 103 kJ
Reaction with ammonia: Oxygen is able to react with ammonia to produce dinitrogen (N2) and water (H2O) through the reaction shown below.
$4NH_3 + 3O_2 \rightarrow 2N_2 + 6H_2O \nonumber$
Reaction with Nitrogen Oxide: Oxygen is able to react with nitrogen oxide in order to produce nitrogen dioxide through the reaction shown below.
$NO + O_2 \rightarrow NO_2 \nonumber$
Problems
1. Is oxygen reactive with noble gases?
2. Which transition metals does oxygen not react with?
3. What is produced when an oxide reacts with water?
4. Is oxygen reactive with alkali metals? Why are the alkali metals named that way?
5. If oxygen is reactive with alkali metals, are oxides, peroxides or superoxides produced?
Solutions
1. No, noble gases are unreactive with oxygen.
2. Oxygen is mostly unreactive with gold and platinum.
3. When an oxide reacts with water, a metal hydroxide is produced.
4. Oxygen is very reactive with alkali metals. Alkali metals are given the name alkali because the oxides of these metals react with water to form a metal hydroxide that is basic or alkaline.
5. Lithium produces an oxide, sodium produces a peroxide, and potassium, cesium, and rubidium produce superoxides.
Contributors and Attributions
• Phillip Ball (UCD), Katharine Williams (UCD)
8.11.02: Chemistry of Oxygen (Z8)
Ozone is an allotropic form of oxygen. Its molecular formula is O3 and molar mass is 48 g mol-1.
Occurrence of ozone
Schonbein (1840) concluded that Van Marum's observations in 1785 of a peculiar smell, when an electric discharge was passed through oxygen (or air), was in fact a new gas. He named it ozone, which is derived from a Greek word ozoaterr meaning smell. In 1860, Soret assigned the molecular formula O3. Ozone occurs in small amounts, in the upper layer of the atmosphere, where it is formed due to the action of ultraviolet rays on the oxygen of the air. It is also present in seawater where it is formed due to the reaction of fluorine with water.
In the structure of ozone, the bond length of 127.8 pm is intermediate between a single bond (bond length 148 pm) and a double bond (bond length 110 pm). Ozone is, therefore, considered to be a resonance hybrid of the following canonical forms:
Uses of ozone
• For air purification in crowded places like cinema halls and tunnel railways. Due to its strong oxidizing power it also destroys the foul smell in slaughter houses.
• In sterilizing drinking water by oxidizing all germs and bacteria.
• For preservation of meat in cold storage.
• For bleaching delicate fabrics such as silk, ivory, oils, starch and wax.
• It helps to locate a double bond in any unsaturated organic compound by ozonolysis.
Preparation of ozone
When a silent electric discharge is passed through dry oxygen, ozone is formed. Oxygen is never converted into ozone completely and we always obtain a mixture of oxygen and ozone. This mixture is called ozonized oxygen.
$\ce{3O_2 ->[\text{electric}][\text{discharge}] 2O_3} \;\;\; \Delta{H} = +284.5\;kJ/mol \nonumber$
The reaction is initiated by a sparkless or silent electric discharge, to produce less heat, as ozone is prone to decomposing back into oxygen with an increase in temperature (Le Chatelier's Principle). Hence, ozone is prepared in a specially designed apparatus called an ozonizer to facilitate the above conditions. An 'Ozonizer' is the apparatus used to prepare ozone by the passage of silent electrical discharge. Two types of ozonizers are commonly used: the Siemen's and the Brodie Ozonizers.
Siemen's Ozonizer
Seimen's ozonizer consists of two co-axial glass tubes fused together. Tin foil is used to coat the inner side of the inner tube and the outer side of the outer tube. The inner and outer tin coatings are connected to the terminals of an induction coil, which produces a current at high voltage. A slow current of pure and dry oxygen is passed through the annular space. On subjecting oxygen to silent electrical discharge, ozonized oxygen containing 10-15% ozone is formed. By taking the following precautions, the yield of ozone can be increased in the ozonized oxygen:
• Only pure and dry oxygen should be used.
• The ozonizer should be perfectly dry.
• A fairly low temperature (around 0°C) should be maintained.
• There should be no sparking.
Brodie's ozonizer
In principle, this ozonizer is like the Siemen's ozonizer, but dilute sulfuric acid replaces the tin foil. Two carbon electrodes are dipped in the acid and connected to an induction coil. A current of dry oxygen is passed through the space between the tubes. Ozonized oxygen containing about 5% $O_3$ comes out at the other end. If the apparatus is kept cool, the proportion of ozone may go up 20-25%.
Problems
1. How is ozone formed in the upper atmosphere?
Solution
1. Ozone is formed in the upper atmosphere in two steps:
i) Photodissociation of oxygen by ultraviolet radiation of wavelength less than 240 nm.
$\ce{O_2 ->[h\nu] O^{\boldsymbol{\cdot}} + O^{\boldsymbol{\cdot}} } \nonumber$
Oxygen atoms are really diradicals with two unpaired p-orbital electrons, but are represented with a single electron here.
ii) Combination of highly reactive oxygen atoms with oxygen molecules.
$\ce{O_2 + O^{\boldsymbol{\cdot}} \rightarrow O_3 } \nonumber$
Contributors and Attributions
Binod Shrestha (University of Lorraine)
8.11.2.01: Ozone
Contrary to a common misconception, ozone is not in the form of a thick layer surrounding the atmosphere. Equally untrue is another misconception that there is a hole in this ozone layer. To understand how ozone spreads in the atmosphere, one needs to know the structure and composition of the atmosphere.
Depletion of Ozone
Ozone in the ionosphere, mesosphere, and stratosphere is being depleted. The concentration of ozone is gradually being reduced. As the content of ozone is highest in the ionosphere with the air itself being very thin, the depletion is negligible in the ionosphere. But in the mesosphere and stratosphere, the air is thicker and the ozone content is less. The depletion of ozone is of a higher order in these layers. The so-called hole in the ozone layer simply means that above some continents (specifically Antarctica, Asia, and parts of South America), the mesosphere and stratosphere have lost their original level of ozone content.
The depletion of the ozone layer is a global phenomenon in terms of both cause and effect. The geographical limits of countries are not barriers to either dispersal of gases in the layers of the atmosphere or depletion of gases. The causes for depletion may arise in any country. The effects (in terms of depletion) may arise in any other country. The effects (in terms of ozone depletion) need not be exactly above the country causing the depletion.
Causes of Depletion
It is now established that chlorofluorocarbon (CFC) chemicals evolved from various refrigerants, coolants, and propellants are the primary reason for the depletion of the ozone. CFCs are a group of chlorine-bearing gases of low specific gravity. They rise to the stratosphere and mesosphere. Due to ionizing solar radiation in these layers (which is the primary reason for production of ozone), fresh chlorine gas is produced from CFCs. This nascent chlorine gas has the capacity to react with ozone and bring down the level of ozone substantially.
Destruction of ozone in the ozone layer
The ozone in the ozone layer protects the earth from harmful ultraviolet radiation by trapping this radiation. The ozone layer has been thinning gradually and poses potential health hazards for the future. The thinning of the ozone layer has been attributed to the presence of chlorofluorohydrocarbons (e.g., $\ce{CFCl3}$ and $\ce{CF2Cl2 }$) in the atmosphere. These chemicals have been used as aerosol propellants and cooling mixtures in refrigerators. How these chemicals affect ozone concentration is illustrated with the equations given below.
$\ce{CF_2Cl_2 + h\nu \rightarrow CF_2Cl^{\cdot} + Cl^{\cdot} }\label{i}$
Chlorine atoms, being highly reactive, react with ozone ($\ce{O3}$):
$\ce{Cl^{\cdot} + O_3 \rightarrow ClO^{\cdot} + O_2}\label{ii}$
The monoxide of chlorine further reacts with another molecule of $\ce{O3}$:
$\ce{ClO^{\cdot} + O_3 \rightarrow Cl^{\cdot} + 2O_2 } \label{iii}$
The chlorine atom so obtained reacts with another ozone molecule. Hence, steps \ref{ii} and \ref{iii} are repeated again and again, and lead to the depletion of the concentration of ozone.
Contributors and Attributions
Template:SHRESTHA | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/08%3A_Chemistry_of_the_Main_Group_Elements/8.11%3A_The_Oxygen_Family_%28The_Chalcogens%29/8.11.02%3A_Chemistry_of_Oxygen_%28Z8%29/8.11.2.1.02%3A_Ozone.txt |
Learning Objectives
• Describe the chemistry of the oxygen group.
• Give the trend of various properties.
• Remember the names of Group 16 elements.
• Explain the Frasch process.
• Describe properties and applications of $\mathrm{H_2SO_4}$.
• Explain properties and applications of $\mathrm{H_2S}$.
Sulfur is a chemical element that is represented with the chemical symbol "S" and the atomic number 16 on the periodic table. Because it is 0.0384% of the Earth's crust, sulfur is the seventeenth most abundant element following strontium. Sulfur also takes on many forms, which include elemental sulfur, organo-sulfur compounds in oil and coal, H2S(g) in natural gas, and mineral sulfides and sulfates. This element is extracted by using the Frasch process (discussed below), a method where superheated water and compressed air are used to draw liquid sulfur to the surface. Offshore sites, Texas, and Louisiana are the primary sites that yield extensive amounts of elemental sulfur. However, elemental sulfur can also be produced by reducing H2S, commonly found in oil and natural gas. For the most part, though, sulfur is used to produce SO2(g) and H2SO4.
Known from ancient times (mentioned in the Hebrew scriptures as brimstone) sulfur was classified as an element in 1777 by Lavoisier. Pure sulfur is tasteless and odorless with a light yellow color. Samples of sulfur often encountered in the lab have a noticeable odor. Sulfur is the tenth most abundant element in the known universe.
Sulfur at a Glance
Atomic Number 16
Atomic Symbol S
Atomic Weight 32.07 grams per mole
Structure orthorhombic
Phase at room temperature solid
Classification nonmetal
Physical Properties of Sulfur
Sulfur has an atomic weight of 32.066 grams per mole and is part of group 16, the oxygen family. It is a nonmetal and has a specific heat of 0.706 J g-1 oC-1. The electron affinity is 200 kJ mol-1 and the electronegativity is 2.58 (unitless). Sulfur is typically found as a light-yellow, opaque, and brittle solid in large amounts of small orthorhombic crystals. Not only does sulfur have twice the density of water, but it is also insoluble in water. On the other hand, sulfur is highly soluble in carbon disulfide and slightly soluble in many common solvents. Sulfur can also vary in color and blackens upon boiling due to carbonaceous impurities. Even as little as 0.05% of carbonaceous matter darkens sulfur significantly.
Most sulfur is recovered directly as the element from underground deposits by injecting super-heated water and piping out molten sulfur (sulfur melts at 112o C). Compared to other elements, sulfur has the most allotropes. While the S8 ring is the most common allotrope, there are 6 other structures with up to 20 sulfur atoms per ring.
• Under appropriate conditions, sulfur vapor can contain $S$, $S_2$, $S_4$, $S_6$, and $S_8$.
• At room temperature, rhombic sulfur (Sα) is a stable solid comprising cyclic $S_8$ molecules.
• At 95.5 °C, rhombic sulfur becomes monoclinic sulfur (Sβ). The crystal structure found in monoclinic sulfur differs from that of rhombic sulfur. Monoclinic sulfur is also made up of $S_8$molecules.
• Monoclinic sulfur becomes liquid sulfur (Sλ) at 119 °C. Liquid sulfur is a straw-colored liquid made up of $S_8$ molecules and other cyclic molecules containing a range of six to twenty atoms.
• At 160 oC, this becomes a dark, viscous liquid called liquid sulfur (Sμ). The molecules are still made up of eight sulfur atoms, but the molecule opens up and transforms from a circle into a long spiral-chain molecule.
• At 180 °C, the chain length and viscosity reach their maximum. Chains break and viscosity decreases at temperatures that exceed 180 °C.
• Sulfur vapor is produced when liquid boils at 445 °C. In the vapor that is produced, $S_8$ molecules dominate, but as the vapor continues to heat up, the molecules break up into smaller groups of sulfur atoms.
• To produce plastic sulfur, S is poured into cold water. Plastic sulfur is rubberlike and is made up of long, spiral-chain molecules. If plastic sulfur sits for long, it will reconvert to rhombic sulfur.
While oxygen has fewer allotropes than sulfur, including $\ce{O}$, $\ce{O_2}$, $\ce{O_3}$, $\ce{O_4}$, $\ce{O_8}$, metallic $\ce{O}$ (and four other solid phases), many of these actually have a corresponding sulfur variant. However, sulfur has more tendency to catenate (the linkage of atoms of the same element into longer chains). Here are the values of the single and double bond enthalpies:
$\begin{array}{c|r} \ce {O-O} & \ce{142\ kJ/mol} \ \ce {S–S} & \ce{268\ kJ/mol} \ \ce {O=O} & \ce{499\ kJ/mol} \ \ce {S=S} & \ce{352\ kJ/mol} \ \end{array} \nonumber$
This means that $\ce{O=O}$ is stronger than $\ce{S=S}$, while $\ce{O–O}$ is weaker than $\ce{S–S}$. So, in sulfur, single bonds are favored and catenation is easier than in oxygen compounds. It seems that the reason for the weaker $\ce{S=S}$ double bonds has its roots in the size of the atom: it's harder for the two atoms to come to a small enough distance, so that the $p$ orbital overlap is small and the $\pi$ bond is weak. This is attested by looking down the periodic table: $\ce{Se=Se}$ has an even weaker bond enthalpy of $\ce{272 kJ/mol}$.
What happens when the solid sulfur melts? The $\ce{S8}$ molecules break up. When suddenly cooled, long chain molecules are formed in the plastic sulfur which behave like rubber. Plastic sulfur transforms into rhombic sulfur over time.
Compounds
Reading the following reactions, figure out and notice the change of the oxidation state of $\ce{S}$ in the reactants and products. Common oxidation states of sulfur are -2, 0, +4, and +6. Sulfur (brimstone, stone that burns) reacts with $\ce{O2}$ giving a blue flame (Figure $1$):
$\ce{S + O_2 \rightarrow SO_2} \nonumber$
$\ce{SO2}$ is produced whenever a metal sulfide is oxidized. It is recovered and oxidized further to give $\mathrm{SO_3}$, for production of $\mathrm{H_2SO_4}$. $\mathrm{SO_2}$ reacts with $\mathrm{H_2S}$ to form $\mathrm{H_2O}$ and $\ce{S}$.
$\mathrm{2 SO_2 + O_2 \rightleftharpoons 2 SO_3} \nonumber$
$\mathrm{SO_3 + H_2O \rightleftharpoons H_2SO_4} \;\;(\text{a valuable commodity}) \nonumber$
$\mathrm{SO_3 + H_2SO_4 \rightleftharpoons H_2S_2O_7} \;\;\; (\text{pyrosulfuric acid}) \nonumber$
Sulfur reacts with sulfite ions in solution to form thiosulfate,
$\ce{S + SO_3^{2-} -> S_2O_3^{2-}} \nonumber$
but the reaction is reversed in an acidic solution.
Oxides
There are many different stable sulfur oxides, but the two that are commonly found are sulfur dioxide and sulfur trioxide. Sulfur dioxide is a commonly found oxide of sulfur. It is a colorless, pungent, and nonflammable gas. It has a density of 2.8 kg/m3 and a melting point of -72.5 oC. Because organic materials are more soluble in $SO_2$ than in water, the liquid form is a good solvent. $SO_2$ is primarily used to produce $SO_3$. The direct combustion of sulfur and the roasting of metal sulfides yield $SO_2$ via the contact process:
$\underbrace{S(s) + O_2(g) \rightarrow SO_2(g)}_{\text{Direct combustion}} \nonumber$
$\underbrace{2 ZnS(s) + 3 O_2(g) \rightarrow 2 ZnO(s) + 2 SO_2(g)}_{\text{Roasting of metal sulfides}} \nonumber$
Sulfur trioxide is another one of the commonly found oxides of sulfur. It is a colorless liquid with a melting point of 16.9 oC and a density of kg/m3. $SO_3$ is used to produce sulfuric acid. $SO_2$ is used in the synthesis of $SO_3$:
$\underbrace{2 SO_2 (g) + O_2(g) \rightleftharpoons 2 SO_3(g)}_{\text{Exothermic, reversible reaction}} \nonumber$
This reaction needs a catalyst to be completed in a reasonable amount of time with $V_2O_5$ being the catalyst most commonly used.
Hydrogen Sulfide H2S
• Hydrogen sulfide, $\ce{H2S}$, is a diprotic acid. The equilibria below, $\mathrm{H_2S \rightleftharpoons HS^- + H^+} \nonumber$ $\mathrm{HS^- \rightleftharpoons S^{2-} + H^+} \nonumber$ have been discussed in connection with Polyprotic Acids.
Other Sulfur-containing Compounds
Perhaps the most significant compound of sulfur used in modern industrialized societies is sulfuric acid ($H_2SO_4$). Sulfur dioxide ($SO_2$) finds practical applications in bleaching and refrigeration but it is also a nuisance gas resulting from the burning of sulfurous coals. Sulfur dioxide gas then reacts with the water vapor in the air to produce a weak acid, sulfurous acid ($H_2SO_3$), which contributes to the acid rain problem.
• Sulfuric acid, H2SO4, is produced by reacting $SO_3$ with water. However, this often leads to pollution problems. SO3(g) is reacted with 98% H2SO4 in towers full of ceramic material to produce H2S2O7 or oleum. Water is circulated in the tower to maintain the correct concentration and the acid is diluted with water at the end in order to produce the correct concentration. Pure sulfuric acid has no color and odor, and it is an oily, hygroscopic liquid. However, sulfuric acid vapor produces heavy, white smoke and a suffocating odor.
• Dilute sulfuric acid, H2SO4(aq), reacts with metals and acts as a strong acid in common chemical reactions. It is used to produce H2(g) and liberate CO2(g) and can neutralize strong bases.
• Concentrated sulfuric acid, H2SO4 (conc.), has a strong affinity for water. In some cases, it removes H and O atoms. Concentrated sulfuric acid is also a good oxidizing agent and reacts with some metals.
$C_{12}H_{22}O_{11}(s) \rightarrow 12 C(s) + 11 H_2O(l) \nonumber$
(Concentrated sulfuric acid used in forward reaction to remove H and O atoms.)
Applications of Sulfuric Acid
• as a strong acid for making $\ce{HCl}$ and $\mathrm{HNO_3}$.
• as an oxidizing agent for metals.
• as a dehydrating agent.
• for manufacture of fertilizer and other commodities.
• Sulfurous acid (H2SO3) is produced when $SO_2$(g) reacts with water. It cannot be isolated in its pure form; however, it forms salts as sulfites. Sulfites can act as both reducing agents and oxidizing agents.
O2(g) + 2 SO32-(aq) $\rightarrow$ 2 SO42- (aq) (Reducing agent)
2 H2S(g) + 2 H+(aq) + SO32-(aq) $\rightarrow$ 3 H2O(l) + 3 S(s) (Oxidizing agent)
H2SO3 is a diprotic acid that acts as a weak acid in both steps, and H2SO4 is also a diprotic acid but acts as a strong acid in the first step and a weak acid in the second step. Acids like NaHSO3 and NaHSO4 are called acid salts because they are the product of the first step of these diprotic acids.
Boiling elemental sulfur in a solution of sodium sulfite yields thiosulfate. Not only are thiosulfates important in photographic processing, but they are also common analytical reagents used with iodine (like in the following two reactions).
$2 Cu^{2+}_{(aq)} + 5 I^-_{(aq)} \rightarrow 2 CuI_{(s)} + I^-_{3(aq)} \nonumber$
$I^-_{3(aq)} + 2 S_2O^{2-}_{3(aq)} \rightarrow 3 I^-_{(aq)} + S_4O^{2-}_{6(aq)} \nonumber$
with excess triiodide ion titrated with Na2S2O3(aq).
Other than sulfuric acid, perhaps the most familiar compound of sulfur in the chemistry lab is the foul-smelling hydrogen sulfide gas, $H_2S$, which smells like rotten eggs.
• Sulfur halides are compounds formed between sulfur and the halogens. Common compounds include SF2, S2F2, SF4, and SF6. While SF4 is a powerful fluorinating agent, SF6 is a colorless, odorless, unreactive gas. Compounds formed by sulfur and chloride include S2Cl2, SCl4, and SCl2. SCl2 is a red, bad-smelling liquid that is utilized to produce mustard gas ($S(CH_2CH_2Cl)_2$).
$SCl_2 + 2CH_2CH_2 \rightarrow S(CH_2CH_2Cl)_2 \nonumber$
Production -The Frasch Process
Sulfur can be mined by the Frasch process. This process has made sulfur a high purity (up to 99.9 percent pure) chemical commodity in large quantities. Most sulfur-containing minerals are metal sulfides, and the best known is perhaps pyrite ($\mathrm{FeS_2}$, known as fool's gold because of its golden color). The most common sulfate-containing mineral is gypsum, $\mathrm{CaSO_4 \cdot 2H_2O}$, also known as plaster of Paris.
The Frasch process is based on the fact that sulfur has a comparatively low melting point. The process forces (99.5% pure) sulfur out by using hot water and air. In this process, superheated water is forced down the outermost of three concentric pipes. Compressed air is pumped down the center tube, and a mixture of elemental sulfur, hot water, and air comes up the middle pipe. Sulfur is melted with superheated water (at 170 °C under high pressure) and forced to the surface of the earth as a slurry.
Sulfur is mostly used for the production of sulfuric acid, $\ce{H2SO4}$. Most sulfur mined by the Frasch process is used in industry for the manufacture of sulfuric acid. Sulfuric acid, the most abundantly produced chemical in the United States, is manufactured by the contact process. Most (about 70%) of the sulfuric acid produced in the world is used in the fertilizer industry. Sulfuric acid can act as a strong acid, a dehydrating agent, and an oxidizing agent. Its applications use these properties. Sulfur is an essential element of life in sulfur-containing proteins.
Applications
Sulfur has many practical applications. As a fungicide, sulfur is used to counteract apple scab in organically farmed apple production. Other crops that utilize sulfur fungicides include grapes, strawberries, and many vegetables. In general, sulfur is effective against mildew diseases and black spot. Sulfur can also be used as an organic insecticide. Sulfites are frequently used to bleach paper and preserve dried fruit.
The vulcanization of rubber includes the use of sulfur as well. Cellophane and rayon are produced with carbon disulfide, a product of sulfur and methane. Sulfur compounds can also be found in detergents, acne treatments, and agrichemicals. Magnesium sulfate (epsom salt) has many uses, ranging from bath additives to exfoliants. Sulfur is being increasingly used as a fertilizer as well. Because standard sulfur is hydrophobic, it is covered with a surfactant by bacteria before oxidation can occur. Sulfur is therefore a slow-release fertilizer. Lastly, sulfur functions as a light-generating medium in sulfur lamps.
Concentrated sulfuric acid was once one of the most produced chemicals in the United States; the majority of the H2SO4 that is now produced is used in fertilizer. It is also used in oil refining, production of titanium dioxide, and in emergency power supplies and car batteries. The mineral gypsum, or calcium sulfate dihydrate, is used in making plaster of Paris. Over one million tons of aluminum sulfate is produced each year in the United States by reacting H2SO4 and Al2O3. This compound is important in water purification. Copper sulfate is used in electroplating. Sulfites are used in the paper making industry because they produce a substance that coats the cellulose in the wood and frees the fibers of the wood for treatment.
Emissions and the Environment
Particles, SO2(g), and H2SO4 mist are the components of industrial smog. Because power plants burn coal or high-sulfur fuel oils, SO2(g) is released into the air. When catalyzed on the surfaces of airborne particles, SO2 can be oxidized to SO3. A reaction with NO2 works as well as shown in the following reaction:
$SO_{2(g)} + NO_{2(g)} \rightarrow SO_{3(g)} + NO_{(g)} \nonumber$
H2SO4 mist is then produced after SO3 reacts with water vapor in the air. If H2SO4 reacts with airborne NH3, (NH4)2SO4 is produced. When SO2(g) and H2SO4 reach levels that exceed 0.10 ppm, they are potentially harmful. By removing sulfur from fuels and controlling emissions, acid rain and industrial smog can be kept under control. Processes like fluidized bed combustion have been presented to remove SO2 from smokestack gases.
• Dhawale, S.W. "Thiosulfate: An interesting sulfur oxoanion that is useful in both medicine and industry--but is implicated in corrosion." J. Chem. Educ. 1993, 70, 12.
• Lebowitz, Samuel H. "A demonstration working model of the Frasch process for mining sulfur." J. Chem. Educ. 1931, 8, 1630.
• Nagel, Miriam C. "Herman Frasch, sulfur king (PROFILES)." J. Chem. Educ. 1981, 58, 60.
• Riethmiller, Steven. "Charles H. Winston and Confederate Sulfuric Acid." J. Chem. Educ. 1995 72 575.
• Sharma, B. D. "Allotropes and polymorphs." J. Chem. Educ. 1987, 64, 404.
• Silverstein, Todd P.; Zhang, Yi. "Sugar Dehydration without Sulfuric Acid: No More Choking Fumes in the Classroom!" J. Chem. Educ. 1998 75 748.
• Tykodi, R. J. "In praise of thiosulfate." J. Chem. Educ. 1990, 67, 146.
• Thomas Jefferson National Accelerator Facility - Office of Science Education."It's Elemental-The Element Sulfur." Jefferson Lab.
• Sulfur's Electron Shell
Problems
1. Draw a diagram that summarizes the allotropy of sulfur. Use symbols, arrows, and numbers.
2. Direct combustion of sulfur is the only method for producing SO2(g). True or False.
3. Sulfites are not oxidizing agents. They are good reducing agents. True or False.
4. Give the reaction for the production of sulfur trioxide.
5. Choose the incorrect statement.
1. Sulfur produces cellophane and rayon.
2. Standard sulfur is hydrophobic.
3. SO2 can oxidize to SO3
4. Sulfur influences the development of acid rain and industrial smog.
5. All of the above are correct.
6. Which reaction is responsible for the destruction of limestone and marble statues and buildings?
1. $\ce{CaCO3 \rightarrow CaO + CO2}$
2. $\ce{SO2 + H2O \rightarrow H2SO3}$
3. $\ce{BaO + CO2 \rightarrow BaCO3 \rightarrow BaSO3}$ upon reaction with $\ce{SO2}$
4. $\ce{CaCO3 + H2O \rightarrow Ca(OH)2 + CO2}$
5. $\ce{CaCO3 + SO2 \rightarrow CaSO3 + CO2 \rightarrow CaSO4}$ upon oxidation
7. Give the formula of thiosulfate ion.
8. What is the oxidation state of $\ce{S}$ in $\ce{SF6}$, $\ce{H2SO4}$, $\ce{NaHSO4}$, $\ce{SO4^2-}$, and $\ce{SO3}$?
9. What is the phase of sulfur at 298 K? Enter the type of crystals.
10. Give the name of the process by which sulfur is forced out of the ground using hot water and air.
Solutions
1. The diagram may be drawn in any way. However, the symbols (S2), (S4), (S6), (S?), and (S8(g)) must be included. The temperatures should be written next to the arrows.
2. False
3. False
4. $2 SO_{2(g)} + O_{2(g)} \rightarrow 2 SO_{3(g)}$
5. A
6. e.
Consider...
$\ce{SO2}$ in $\ce{H2SO3}$ is the acid in acid rain, which attacks $\ce{CaCO3}$, marble. $\ce{SO2}$ reduces pigments in organic matter.
7. $\ce{S2O3^2-}$
Consider...
Sulfate is $\ce{SO4^2-}$; replacement of an $\ce{O}$ by an $\ce{S}$ gives thiosulfate $\ce{S2O3^2-}$. The two $\ce{S}$ in $\ce{S2O3^2-}$ have different oxidation states: one is +6, the other is (-2), average +2.
8. 6
Consider...
Oxidation state for $\ce{S}$ in $\ce{H2SO3}$, $\ce{SO3^2-}$, $\ce{SO2}$, etc. is 4. The oxidation state of $\ce{S}$ is the same for all in the list.
9. rhombic sulfur
Consider...
The term rhombic describes a type of crystal. Monoclinic sulfur is meta stable at 298 K.
10. Frasch process
Consider...
The Frasch process is used to mine elemental sulfur. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/08%3A_Chemistry_of_the_Main_Group_Elements/8.11%3A_The_Oxygen_Family_%28The_Chalcogens%29/8.11.03%3A_Chemistry_of_Sulfur_%28Z16%29.txt |
Element number 34, selenium, was discovered by Swedish chemist Jons Jacob Berzelius in 1817. Selenium is a non-metal and can be compared chemically to its other non-metal counterparts found in Group 16: The Oxygen Family, such as sulfur and tellurium.
Properties
Chemical Symbol: Se
Atomic Number: 34
Atomic Weight: 78.96
Electron Configuration: [Ar] 4s23d104p4
Melting Point: 493.65 K
Boiling Point: 958 K
Electronegativity: 2.55 (Pauling)
Oxidation States: Se-2, Se+6, Se+4
Ionization Energies: First: 941 kJ/mol
Second: 2045 kJ/mol
Third: 2973.7 kJ/mol
History
Selenium was discovered by Berzelius in 1818. It is named for the Greek word for "moon", selene. The discovery of selenium was an important finding, but at the same time seemingly accidental. Fellow scientist Martin Klaproth discovered a contamination of sulfuric acid creating a red colored product which he believed to be due to the element tellurium. However, Berzelius went on to further analyze the impurity and came to the conclusion that it was an unknown element that shared properties similar to those of tellurium. Based on the Greek word “selene,” meaning moon, Jons Berzelius decided to call the newly found element selenium.
Allotropes and Physical Properties
Selenium can exist in multiple allotropes that are essentially different molecular forms of an element with varying physical properties. For example, one allotrope of selenium can be seen as an amphorous (“without crystalline shape”) red powder. Selenium also takes a crystalline hexagonal structure, forming a metallic gray allotrope which is known to be stable. The most thermodynamically stable allotrope of selenium is trigonal selenium, which also appears as a gray solid. Most selenium is recovered from the electrolytic copper refining process. This is usually in the form of the red allotrope.
Selenium is mostly noted for its important chemical properties, especially those dealing with electricity. Unlike sulfur, selenium is a semiconductor, meaning that it conducts some electricity, but not as well as conductors. Selenium is a photoconductor, which means it has the ability to change light energy into electrical energy. Not only is selenium able to convert light energy into electrical energy, but it also displays the property of photoconductivity. Photoconductivity is the idea that the electrical conductivity of selenium increases due to the presence of light -- or, in other words, it becomes a better photoconductor as light intensity increases.
Isotopes
Isotopes of an element are atoms that have the same atomic numbers but a different number of neutrons (different mass numbers) in their nuclei. Selenium is known to have over 20 different isotopes; however, only 5 of them are stable. The five stable isotopes of selenium are 74Se, 76Se, 77Se, 78Se, 80Se.
Uses
Due to selenium’s property of photoconductivity, it is known to be used in photocells, exposure meters in photography, and also in solar cells. Selenium can also be seen in the products of plain-paper photocopiers, laser printers and photographic toners. Besides its uses in the electronic industry, selenium is also popular in the glass-making industry. When selenium is added to glass, it is able to negate the color of other elements found in the glass and essentially decolorizes it. Selenium is also able to create a ruby-red colored glass when added. The element can also be used in the production of alloys and is an additive to stainless steel.
Health Hazards
Selenium, a trace element, is important in the diet and health of both plants and animals, but can be only taken in very small amounts. Exposure to an excess amount of selenium is known to be toxic and causes health problems. With an upper intake level of 400 micrograms per day that can be tolerated, too much selenium can lead to selenosis and may result in health problems and even death. Compounds of selenium are also known to be carcinogenic.
Chemical Reactivity
Reaction with hydrogen
Selenium forms hydrogen selenide, H2Se, a colorless flammable gas, when reacted with hydrogen.
Reaction with oxygen
Selenium burns in air, displaying a blue flame, and forms solid selenium dioxide.
$Se_{8(s)} + 8O_{2(g)} \rightarrow 8SeO_{2(s)} \nonumber$
Selenium is also known to form selenium trioxide, SeO3.
Reaction with halides
Selenium reacts with fluorine, F2, and burns to form selenium hexafluoride.
$Se_{8(s)} + 24F_{2(g)} \rightarrow 8SeF_{6(l)} \nonumber$
Selenium also reacts with chlorine and bromine to form diselenium dichloride, $Se_2Cl_2$ and diselenium dibromide, $Se_2Br_2$.
$Se_8 + 4Cl_2 \rightarrow 4Se_2Cl_{2(l)} \nonumber$
$Se_8 + 4Br_2 \rightarrow 4Se_2Br_{2(l)} \nonumber$
Selenium also forms $SeF_4$, $SeCl_2$ and $SeCl_4$.
Selenides
Selenium reacts with metals to form selenides. Example: Aluminum selenide
$3 Se_8 + 16 Al \rightarrow 8 Al_2Se_3 \nonumber$
Selenites
Selenium reacts to form salts called selenites, e.g., silver selenite (Ag2SeO3) and sodium selenite (Na2SeO3).
Problems
1. Describe selenium’s property of photoconductivity.
2. Does selenium react with hydrogen? If so, what compound is produced?
3. Describe selenium’s purpose as a trace element.
4. What are some common uses for selenium?
5. Does selenium react with oxygen?
Solutions
1. Selenium’s ability to change light energy into electrical energy increases as light intensity increases.
2. Yes, selenium reacts with hydrogen and forms hydrogen selenide, H2Se.
3. Selenium is important to the health of plants and animals, but is only safe in small amounts. Too much selenium can be toxic and cause serious health problems.
4. Selenium is used in the glass-making industry and also in electronics. It is used in photo cells, solar cells, photocopiers, laser printers and also photographic toners.
5. Selenium burns in air and forms selenium dioxide. It is also able to form selenium trioxide.
Reference
1. Minaev, V. S., S. P. Timoshenkov, and V. V. Kalugin. "Structural and Phase Transformations in Condensed Selenium." Journal of Optoelectronics and Advanced Materials, volume 7, number 4, 2005, pp. 1717–1741.
2. Mary Elvira Weeks and Henry M. Leicester. Discovery of the Elements, 7th edition. Easton, PA: Journal of Chemical Education, 1968.
3. Petrucci, Ralph H. General Chemistry. 9th ed. Upper Saddle River: Prentice Hall, 2007
Contributors and Attributions
• David Jin (UCD) | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/08%3A_Chemistry_of_the_Main_Group_Elements/8.11%3A_The_Oxygen_Family_%28The_Chalcogens%29/8.11.04%3A_Chemistry_of_Selenium_%28Z34%29.txt |
Discovered by von Reichenstein in 1782, tellurium is a brittle metalloid that is relatively rare. It is named from the Latin tellus for "earth". Tellurium can be alloyed with some metals to increase their machinability and is a basic ingredient in the manufacture of blasting caps. Elemental tellurium is occasionally found in nature but is more often recovered from various gold ores, all containing $AuTe_2$.
History
Tellurium was discovered in a gold ore from the mines in Zlatna, near present day Sibiu, Transylvania. The ore was known as "Faczebajer weißes blättriges Golderz" (white leafy gold ore from Faczebaja) or antimonalischer Goldkies (antimonic gold pyrite). In 1782, while serving as the Austrian chief inspector of mines in Transylvania, Franz-Joseph Müller von Reichenstein concluded that a certain ore did not contain antimony, but that it contained bismuth sulfide. However, the following year, he reported that this was erroneous and that the ore contained mostly gold and an unknown metal very similar to antimony. After 3 years of testing, Müller determined the specific gravity of the mineral and noted the radish-like odor of the white smoke, which passed off when the new metal was heated. In 1789, another Hungarian scientist, Pál Kitaibel, also discovered the element independently in an ore from Deutsch-Pilsen which had been regarded as argentiferous molybdenite, but later he gave the credit to Müller. In 1798, it was named by Martin Heinrich Klaproth, who earlier isolated it from the mineral calaverite.
Properties
Tellurium is a semimetallic, lustrous, crystalline, brittle, silver-white element. It is usually available as a dark grey powder and has metal and non-metal properties. Te forms many compounds corresponding to those of sulfur and selenium. When burned in the air, tellurium has a greenish-blue flame and forms tellurium dioxide as a result. Tellurium is unaffected by water or hydrochloric acid, but dissolves in nitric acid. It has an atomic mass of 127.6 g/mol-1 and a density of 6.24 g-cm-3. Its boiling point is 450 degrees Celsius and its melting point is 1390 °C.
Source and Abundance
There are eight naturally occurring isotopes of tellurium, of which three are radioactive. Tellurium is among the rarest stable solid elements in the Earth's crust. At 0.005 ppm, it is comparable to platinum in abundance. However, tellurium is far more abundant in the wider universe. Tellurium was originally and is most commonly found in gold tellurides. However, the largest source for modern production of tellurium is as a byproduct of blister copper refinement. The treatment of 500 tons of copper ore results in 0.45 kg of tellurium. Tellurium can also be found in lead deposits. Other tellurium sources, known as subeconomic deposits because the cost of abstraction outweighs the yield in tellurium, are lower-grade copper and some coal.
Originally, the copper tellurium ore is treated with sodium bicarbonate and elemental oxygen to produce a tellurium oxide salt, copper oxide, and carbon dioxide:
$Cu_2Te + Na_2CO_3 + 2O_2 \rightarrow 2CuO + Na_2TeO_3 + CO_2 \nonumber$
Then, the sodium tellurium oxide is treated with sulfuric acid to precipitate out tellurium dioxide, which can be treated with aqueous sodium hydroxide to reduce to pure tellurium and oxygen gas:
$TeO_2 + 2NaOH \rightarrow Na_2TeO_3 + H_2O \rightarrow Te + 2NaOH + O_2 \nonumber$
Industrial and Commercial Use
Tellurium has many unique industrial and commercial uses that improve product quality and quality-of-life. Many of the technologies that utilize tellurium have important uses for the energy industry, the military, and health industries. Tellurium is used to color glass and ceramics and can improve the machining quality of metal products. When added to copper alloys, tellurium makes the alloy more ductile, whereas it can prevent corrosion in lead products. Tellurium is an important component of infrared detectors used by the military as well as x-ray detectors used by a variety of fields including medicine, science, and security. In addition, tellurium-based catalysts are used to produce higher-quality rubber. CdTe films are one of the highest efficiency photovoltaics, metals that convert sunlight directly into electrical power, at 11-13% efficiency and are, therefore, widely used in solar panels. CdTe is a thin-film semiconductor that absorbs sunlight.
Tellurium can be replaced by other elements in some of its uses. For many metallurgical uses, selenium, bismuth, or lead are effective substitutes. Both selenium and sulfur can replace tellurium in rubber production. Technologies based on tellurium have global impacts. As a photovoltaic, CdTe is the second most utilized solar cell in the world, soon said to surpass crystalline silicon and become the first. According to the US military, tellurium-based infrared detectors are the reason that the military has such an advantage at night, an advantage which, in turn, has an effect on global and domestic politics.
Environmental Impacts
Tellurium extraction, as a byproduct of copper refinement, shares environmental impacts associated with copper mining and extraction. While a generally safe process, the removal of copper from other impurities in the ore can lead to leaching of various hazardous sediments. In addition, the mining of copper tends to lead to reduced water flow and quality, disruption of soils and erosion of riverbanks, and reduction of air quality.
Resource Limitations v. Demand
About 215-220 tons of tellurium are mined across the globe every year. In 2006, the US produced 40% of the global production, Peru produced 30%, Japan produced 20%, and Canada produced 10% of the world's tellurium supply (since the chart can't be any bigger). The leading countries in production are the United States with 50 tons per year, Japan with 40 tons per year, Canada with 16 tons per year, and Peru with 7 tons per year (year 2009). When pure, tellurium costs \$24 per 100 grams. Because tellurium is about as rare as platinum on earth, the United States Department of Energy expects a supply shortfall by the year 2025, despite the always improving extraction methods. As demand increases to provide the tellurium needed for solar panels and other such things, supply will continue to decrease and thus the price will skyrocket. This will cause waves in the sustainable energy movement as well as military practices and modern medicine.
8.11.06: Chemistry of Polonium (Z84)
Polonium was discovered in 1898 by Marie Curie and named for her native country of Poland. The discovery was made by extraction of the remaining radioactive components of pitchblende following the removal of uranium. There is only about 10-6 g per ton of ore! Current production for research purposes involves the synthesis of the element in the lab rather than its recovery from minerals. This is accomplished by producing Bi-210 from the abundant Bi-209. The new isotope of bismuth is then allowed to decay naturally into Po-210. The sample pictured above is actually a thin film of polonium on stainless steel.
Although radioactive, polonium has a few commercial uses. You can buy your own sample of polonium at a photography store. It is part of the special anti-static brushes for dusting off negatives and prints.
Contributors and Attributions
Stephen R. Marsden
8.11.07: Chemistry of Livermorium (Z116)
In May of 2012 the IUPAC approved the name "Livermorium" (symbol Lv) for element 116. The new name honors the Lawrence Livermore National Laboratory (1952). A group of researchers of this laboratory with the heavy element research group of the Flerov Laboratory of Nuclear Reactions took part in the work carried out in Dubna on the synthesis of superheavy elements including element 116. Atoms of Lv were formed by the reaction of Cm-248 with Ca-48. The resulting Lv-292 decays by alpha emission to Fl-288.
Contributors and Attributions
Stephen R. Marsden | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/08%3A_Chemistry_of_the_Main_Group_Elements/8.11%3A_The_Oxygen_Family_%28The_Chalcogens%29/8.11.05%3A_Chemistry_of_Tellurium_%28Z52%29.txt |
The halogens are located on the left of the noble gases on the periodic table. These five toxic, non-metallic elements make up Group 17 of the periodic table and consist of: fluorine (F), chlorine (Cl), bromine (Br), iodine (I), and astatine (At). Although astatine is radioactive and only has short-lived isotopes, it behaves similar to iodine and is often included in the halogen group. Because the halogen elements have seven valence electrons, they only require one additional electron to form a full octet. This characteristic makes them more reactive than other non-metal groups.
• 8.13.1: Physical Properties of the Halogens
It can be seen that there is a regular increase in many of the properties of the halogens proceeding down group 17 from fluorine to iodine. This includes their melting points, boiling points, the intensity of their color, the radius of the corresponding halide ion, and the density of the element. On the other hand, there is a regular decrease in the first ionization energy as we go down this group. As a result, there is a regular increase in the ability to form high oxidation states.
• 8.13.2: Chemical Properties of the Halogens
Covers the halogens in Group 17: fluorine (F), chlorine (Cl), bromine (Br) and iodine (I). Includes trends in atomic and physical properties, the redox properties of the halogens and their ions, the acidity of the hydrogen halides, and the tests for the halide ions.
• 8.13.3: Chemistry of Fluorine (Z=9)
Fluorine (F) is the first element in the Halogen group (group 17) in the periodic table. Its atomic number is 9 and its atomic weight is 19, and it's a gas at room temperature. It is the most electronegative element, given that it is the top element in the Halogen Group, and therefore is very reactive. It is a nonmetal, and is one of the few elements that can form diatomic molecules (F2).
• 8.13.4: Chemistry of Chlorine (Z=17)
Chlorine is a halogen in group 17 and period 3. It is very reactive and is widely used for many purposes, such as as a disinfectant. Due to its high reactivity, it is commonly found in nature bonded to many different elements.
• 8.13.5: Chemistry of Bromine (Z=35)
Bromine is a reddish-brown fuming liquid at room temperature with a very disagreeable chlorine-like smell. In fact its name is derived from the Greek bromos or "stench". It was first isolated in pure form by Balard in 1826. It is the only non-metal that is a liquid at normal room conditions. Bromine on the skin causes painful burns that heal very slowly. It is an element to be treated with the utmost respect in the laboratory.
• 8.13.6: Chemistry of Iodine (Z=53)
Elemental iodine is a dark grey solid with a faint metallic luster. When heated at ordinary air pressures it sublimes to a violet gas. The name iodine is taken from the Greek ioeides which means "violet colored". It was discovered in 1811 by Courtois.
• 8.13.7: Chemistry of Astatine (Z=85)
Astatine is the last of the known halogens and was synthesized in 1940 by Corson and others at the University of California. It is radioactive and its name, from the Greek astatos, means "unstable". The element can be produced by bombarding targets made of bismuth-209 with high energy alpha particles (helium nuclei). Astatine 211 is the product and has a half-life of 7.2 hours. The most stable isotope of astatine is 210 which has a half-life of 8.1 hours.
Thumbnail: Chlorine gas in an ampoule. (CC-BY-SA; W. Oelen (http://woelen.homescience.net/science/index.html)).
8.13: The Halogens
Some chemical and physical properties of the halogens are summarized in Table $1$. It can be seen that there is a regular increase in many of the properties of the halogens proceeding down group 17 from fluorine to iodine. This includes their melting points, boiling points, the intensity of their color, the radius of the corresponding halide ion, and the density of the element. On the other hand, there is a regular decrease in the first ionization energy as we go down this group. As a result, there is a regular increase in the ability to form high oxidation states and a decrease in the oxidizing strength of the halogens from fluorine to iodine.
Table $1$: Properties of Group 17 (The Halogens)
Property F Cl Br I
Atomic number, Z 9 17 35 53
Ground state electronic configuration [He]2s2 2p5 [Ne]3s2 3p5 [Ar]3d10 4s2 4p5 [Kr]4d10 5s2 5p5
color pale yellow gas yellow-green gas red-brown liquid blue-black solid
Density of liquids at various temperatures, /kg m-3 1.51 (85 °K) 1.66 (203 °K) 3.19 (273 °K) 3.96 (393 °K)
Melting point, /K 53.53 171.6 265.8 386.85
Boiling point, /K 85.01 239.18 331.93 457.5
Enthalpy of atomization, ΔaH° (298K) / kJ mol-1 79.08 121.8 111.7 106.7
Standard enthalpy of fusion of X2, ΔfusH°(mp) / kJ mol-1 0.51 6.4 10.57 15.52
Standard enthalpy of vaporization of X2, ΔvapH°(bp) / kJ mol-1 6.62 20.41 29.96 41.57
First ionization energy, IE1 / kJ mol-1 1681 1251.1 1139.9 1008.4
ΔEAH1°(298K) / kJ mol-1 -333 -348 -324 -295
ΔhydH°(X-,g) / kJ mol-1 -504 -361 -330 -285
ΔhydS°(X-,g) / JK-1 mol-1 -150 -90 -70 -50
ΔhydG°(X-,g) / kJ mol-1 -459 -334 -309 -270
Standard redox potential, E°(X2 /2X-) /V 2.87 1.36 1.09 0.54
Covalent radius, rcov = ½ X-X bond length /pm 72 100 114.2 133.3
Ionic radius, rion for X- /pm 133 181 196 220
van der Waals radius, rv /pm 135 180 195 215
X-X(g)bond energy /kJ mol-1 159 243 193 151
H-X(g)bond energy /kJ mol-1 562 431 366 299
C-X(g)bond energy /kJ mol-1 484 338 276 238
Pauling electronegativity, χP 3.98 3.16 2.96 2.66
Color
The origin of the color of the halogens stems from the excitation between the highest occupied π* molecular orbital and the lowest unoccupied σ* molecular orbital. The energy gap between the HOMO and LUMO decreases according to F2 > Cl2 > Br2 > I2. The amount of energy required for excitation depends upon the size of the atom. Fluorine is the smallest element in the group and the force of attraction between the nucleus and the outer electrons is very large. As a result, it requires a large excitation energy and absorbs violet light (high energy) and so appears pale yellow. On the other hand, iodine needs significantly less excitation energy and absorbs yellow light of low energy. Thus it appears dark violet. Using similar arguments, it is possible to explain the greenish yellow color of chlorine and the reddish brown color of bromine.
Figure $1$: Molecular orbital diagram for fluorine.
The halogens show a variety of colors when dissolved in different solvents. Solutions of iodine can be bright violet in CCl4, pink or reddish brown in aromatic hydrocarbons, and deep brown in alcohols, for example. This variety can be explained by weak donor-acceptor interaction and complex formation. The presence of charge-transfer bands further supports this since they are thought to be derived from interaction with the HOMO σu* orbital.
The X-ray structure of some of these complexes have been obtained, and often the intense color can be used for characterization and determination such as the bright blue color of iodine in the presence of starch. In the case of the solid formed between dibromine and benzene, the structure is shown below and a new charge transfer band occurs at 292 nm. The Br-Br bond length is essentially unchanged from that of dibromine (228 pm).
Figure $2$: Structure of dibromine and benzene complex
In a study of the reaction of dibromine with substituted phosphines in diethyl ether, all but one showed a tetrahedral arrangement where one bromine was linked to the phosphorus.[3]
$R_3P + Br_2 (Et_2O, N_2/r.t.) \rightarrow R_3PBr_2 \label{1}$
The X-ray study of the triethylphosphine was interpreted as [Et3PBr]Br, where the Br-Br separation was 330 pm. This is considerably longer than the 228 pm found above and was taken to mean that the compound was ionic. In the case of the tri(perfluorophenyl)phosphine, however, the structure showed both bromines linked to give a trigonal bipyramid arrangement with D3 symmetry. Why (C6F5)3PBr2 was the only R3PBr2 compound that adopted trigonal bipyramidal geometry was reasoned to be due to the very low basicity of the parent tertiary phosphine.
Melting and Boiling Points
Intermolecular forces are the attractive forces between molecules without which all substances would be gases. The various types of these interactions span large differences in energy and for the halogens and interhalogens are generally quite small. The dispersion forces involved in these cases are called London forces (after Fritz Wolfgang London, 1900-1954). They are derived from momentary oscillations of electron charge in atoms and hence are present between all particles (atoms, ions and molecules).
The ease with which the electron cloud of an atom can be distorted to become asymmetric is termed the molecule's polarizability. The greater the number of electrons an atom has, the farther the outer electrons will be from the nucleus, and the greater the chance for them to shift positions within the molecule. This means that larger nonpolar molecules tend to have stronger London dispersion forces. This is evident when considering the diatomic elements in group 17, the Halogens. All of these diatomic elements are nonpolar, covalently bonded molecules. Descending the group, fluorine and chlorine are gases, bromine is a liquid, and iodine is a solid. For nonpolar molecules, the farther you go down the group, the stronger the London dispersion forces.
To picture how this occurs, compare the situation 1) where the electrons are evenly distributed and then consider 2) an instantaneous dipole that would arise from an uneven distribution of electrons on one side of the nucleus. When two molecules are close together, the instantaneous dipole of one molecule can induce a dipole in the second molecule. This results in synchronized motion of the electrons and an attraction between them. 3) Multiply this effect over numerous molecules and the overall result is that the attraction keeps these molecules together, and for diiodine is sufficient to make this a solid.
Figure $3$: On average the electron cloud for molecules can be considered to be spherical in shape. When two non-polar molecules approach, attractions or repulsions between the electrons and nuclei can lead to distortions in their electron clouds (i.e., dipoles are induced). When more molecules interact, these induced dipoles lead to intermolecular attraction.
The changes seen in the variation of MP and BP for the dihalogens and binary interhalogens can be attributed to the increase in the London dispersion forces of attraction between the molecules. In general they increase with increasing atomic number.
Figure 4:
Redox Properties
The most characteristic chemical feature of the halogens is their oxidizing strength. Fluorine has the strongest oxidizing ability, so that a simple chemical preparation is almost impossible and it must be prepared by electrolysis. Note that since fluorine reacts explosively with water, oxidizing it to dioxygen, finding reaction conditions for any reaction can be difficult. When fluorine is combined with other elements they generally exhibit high oxidation states. Chlorine is the next strongest oxidizing agent, but it can be prepared by chemical oxidation. Most elements react directly with chlorine, bromine and iodine, with decreasing reactivity going down the group, but often the reaction must be activated by heat or UV light. [2] The energy changes in redox process are:
1. Enthalpy of atomization,
2. ΔEAH1,
3. ΔhydH°(X-,g)
The redox potential, E°, X2/2X-, measures a free-energy change, usually dominated by the ΔH term. The values in the Table below show that there is a decrease in oxidizing strength proceeding down the group (2.87, 1.36, 1.09, 0.54 V). This can be explained by comparing the steps shown above.
• 1) atomization of the dihalide is the energy required to break the molecule into atoms:
$½ X_{2(g)} \rightarrow X_{(g)} \label{2}$
Note that only F2 and Cl2 are gases in their natural states, so the energies associated with atomization of Br2 and I2 require converting the liquid or solid to gas first.
• 2) ΔEAH1 is the energy liberated when the atom is converted into a negative ion and is related to the Electron Affinity
$X_{(g)} + e^- \rightarrow X^-_{(g)} \label{3}$
Addition of an electron to the small F atom is accompanied by larger e-/e- repulsion than is found for the larger Cl, Br or I atoms. This would suggest that the process for F should be less exothermic than for Cl and not fit the trend that shows a general decrease going down the group.
• 3) ΔhydH°(X-,g) is the energy liberated upon the hydration of the ion, the Hydration energy.
$X^-_{(g)} + H_2O \rightarrow X^-_{(aq)} \label{4}$
The overall reaction is then:
$½ X_{2(g)} \rightarrow X^-_{(aq)} \label{5}$
Table $2$
Halogen atomization energy
(kJ mol-1)
ΔEAH1
(kJ mol-1)
hydration enthalpy
(kJ mol-1)
overall
(kJ mol-1)
F +79.08 -333 -504 -758
Cl +121.8 -348 -361 -587
Br +111.7 -324 -330 -542
I +106.7 -295 -285 -473
This shows a very negative energy change for the fluoride compared to the others in the group. This comes about because of two main factors: the high hydration energy and the low atomization energy. For F2 2) is less than for Cl2, but since the energy needed to break the F-F bond is also less and the hydration more, the total energy drop is much greater. In spite of their lower atomization energies, Br2 and I2 are weaker oxidizing agents than Cl2 and this is due to their smaller ΔEAH1 and smaller ΔhydH°.
It can be seen that the ΔEAH1 value for fluorine is in between those for chlorine and bromine and so this value alone does not provide a good explanation for the observed variation.
Each of the halogens is able to oxidize any of the heavier halogens situated below it in the group. They can oxidize hydrogen and nonmetals such as:
$X_2 + H_{2(g)} \rightarrow 2HX_{(g)} \label{6}$
In water, the halogens disproportionate according to:
$X_2 + H_2O_{(l)} \rightarrow HX_{(aq)} + HXO_{(aq)} \label{7}$
where $X=Cl, Br, I$. When base is added then the reaction goes to completion forming hypohalites, or at higher temperatures, halates; for example, heating dichlorine:
$3Cl_{2(g)} + 6OH^-_{(aq)} \rightarrow ClO^-_{3(aq)} + 5Cl^-_{(aq)} + 3H_2O(l) \label{8}$
First Ionization Energies
The trend seen for the complete removal of an electron from the gaseous halogen atoms is that fluorine has the highest IE1 and iodine the lowest. To overcome the attractive force of the nucleus means that energy is required, so the Ionization Energies are all positive. The variation with size can be explained, since as the size increases it take less energy to remove an electron. This inverse relationship is seen for all the groups, not just group 17. As the distance from the nucleus to the outermost electrons increases, the attraction decreases so that those electrons are easier to remove. The high value of IE1 for fluorine is such that it does not exhibit any positive oxidation states, whereas Cl, Br and I can exist in oxidation states as high as 7.
Oxidation states
Fluorine is the most electronegative element in the periodic table and exists in all its compounds in either the -1 or 0 oxidation state. Chlorine, bromine, and iodine, however, can be found in a range of oxidation states including: +1, +3, +5, and +7, as shown below.
Table $3$: Common Oxidation States for the Halogens
Oxidation States Examples
-1 CaF2, HCl, NaBr, AgI
0 F2, Cl2, Br2, I2
1 HClO, ClF
3 HClO2, ClF3
5 HClO3, BrF5, [BrF6]-, IF5
7 HClO4, BrF6+, IF7, [IF8]-
In general, odd-numbered groups (like group 17) form odd-numbered oxidation states, and this can be explained since all stable molecules contain paired electrons (free radicals are obviously much more reactive). When covalent bonds are formed or broken, two electrons are involved, so the oxidation state changes by 2.
When difluorine reacts with diiodine, initially iodine monofluoride is formed.
$I_2 + F_2 \rightarrow 2IF \nonumber$
Adding a second difluorine uses two more iodine valence electrons to form two more bonds:
$2IF + F_2 \rightarrow IF_3 \nonumber$
Contributors and Attributions
• The Department of Chemistry, University of the West Indies)
8.13.01: Physical Properties of the Halogens
This page discusses the trends in the atomic and physical properties of the Group 7 elements (the halogens): fluorine, chlorine, bromine and iodine. Sections below cover the trends in atomic radius, electronegativity, electron affinity, melting and boiling points, and solubility, including a discussion of the bond enthalpies of halogen-halogen and hydrogen-halogen bonds.
The figure above shows the increase in atomic radius down the group.
Explaining the increase in atomic radius
The radius of an atom is determined by:
• the number of layers of electrons around the nucleus
• the pull the outer electrons feel from the nucleus.
Compare the numbers of electrons in each layer of fluorine and chlorine:
F 2,7
Cl 2,8,7
In each case, the outer electrons feel a net +7 charge from the nucleus. The positive charge on the nucleus is partially neutralized by the negative inner electrons.
This is true for all the atoms in Group 7: the outer electrons experience a net charge of +7..
The only factor affecting the size of the atom is therefore the number of layers of inner electrons surrounding the atom. More layers take up more space due to electron repulsion, so atoms increase in size down the group.
Electronegativity is a measure of the tendency of an atom to attract a bonding pair of electrons. It is usually measured on the Pauling scale, on which the most electronegative element (fluorine) is assigned an electronegativity of 4.0. The figure below shows electronegativities for each halogen:
Notice that electronegativity decreases down the group. The atoms become less effective at attracting bonding pairs of electrons. This effect is illustrated below using simple dots-and-crosses diagrams for hydrogen fluoride and hydrogen chloride:
The bonding pair of electrons between the hydrogen and the halogen experiences the same net pull of +7 from both the fluorine and the chlorine. However, in the chlorine case, the nucleus is farther away from the bonding electrons, which are therefore not as strongly attracted as in the fluorine case.
The stronger attraction from the closer fluorine nucleus makes fluorine more electronegative than chlorine.
Summarizing the trend down the Group
As the halogen atoms increase in size, any bonding pair gets farther away from the halogen nucleus, and so is less strongly attracted toward it. Hence, down the group, the elements become less electronegative.
The first electron affinity is the energy released when 1 mole of gaseous atoms each acquire an electron to form 1 mole of gaseous 1- ions. In other words, it is the energy released in the following process:
$X(g) + e^- \rightarrow X^- (g) \nonumber$
First electron affinities have negative values by convention. For example, the first electron affinity of chlorine is -349 kJ mol-1. The negative sign indicates a release of energy.
The first electron affinities of the Group 7 elements
The electron affinity is a measure of the attraction between the incoming electron and the nucleus. There is a positive correlation between attraction and electron affinity. The trend down the group is illustrated below:
Notice that the trend down the group is inconsistent. The electron affinities generally decrease (meaning less heat is emitted), but the fluorine value deviates from this trend.
In the larger atom, the attraction from the more positive nucleus is offset by the additional screening electrons, so each incoming electron feels the effect of a net +7 charge from the center.
As the atom increases in size, the incoming electron is farther from the nucleus and so feels less attraction. The electron affinity therefore decreases down the group. However fluorine is a very small atom, with the incoming electron relatively close to the nucleus, and yet the electron affinity is smaller than expected.
Another effect must be considered in the case of fluorine. As the new electron comes approaches the atom, it enters a region of space already very negatively charged because of the existing electrons. The resulting repulsion from these electrons offsets some of the attraction from the nucleus.
Because the fluorine atom is very small, its existing electron density is very high. Therefore, the extra repulsion is particularly great and diminishes the attraction from the nucleus enough to lower the electron affinity below that of chlorine.
Melting and boiling points increase down the group. As indicated by the graph above, fluorine and chlorine are gases at room temperature, bromine is a liquid and iodine a solid.
All the halogens exist as diatomic molecules—F2, Cl2, and so on. van der Waals dispersion forces are the primary intermolecular attractions between one molecule and its neighbors. Larger molecules farther down the group have more electrons which can move around and form the temporary dipoles that create these forces.
The stronger intermolecular attractions down the group require more heat energy for melting or vaporizing, increasing their melting or boiling points.
Solubilities
Solubility in water
Fluorine reacts violently with water to produce aqueous or gaseous hydrogen fluoride and a mixture of oxygen and ozone; its solubility is meaningless. Chlorine, bromine, and iodine all dissolve in water to some extent, but there is again no discernible pattern. The following table shows the solubility of the three elements in water at 25°C:
solubility
(mol dm-3)
chlorine 0.091
bromine 0.21
iodine 0.0013
Chlorine dissolved in water produces a pale green solution. Bromine solution adopts a range of colors from yellow to dark orange-red depending on the concentration. Iodine solution in water is very pale brown. Chlorine reacts with water to some extent, producing a mixture of hydrochloric acid and chloric(I) acid (also known as hypochlorous acid). The reaction is reversible, and at any time only about a third of the chlorine molecules have reacted.
$Cl_2 + H_2O \rightleftharpoons HCl + HClO \nonumber$
Chloric(I) acid is sometimes symbolized as HOCl, indicating the actual bonding pattern. Bromine and iodine form similar compounds, but to a lesser extent. In both cases, about 99.5% of the halogen remains unreacted.
The solubility of iodine in potassium iodide solution
Although iodine is only slightly soluble in water, it dissolves freely in potassium iodide solution, forming a dark red-brown solution. A reversible reaction between iodine molecules and iodide ions gives I3- ions. These are responsible for the color. In the laboratory, iodine is often produced through oxidation of iodide ions. As long as there are any excess iodide ions present, the iodine reacts to form I3-. Once the iodide ions have all reacted, the iodine is precipitated as a dark gray solid.
Solubility in hexane
The halogens are much more soluble in organic solvents such as hexane than they are in water. Both hexane and the halogens are non-polar molecules, so the only intermolecular forces between them are van der Waals dispersion forces. Because of this, the attractions broken (between hexane molecules and between halogen molecules) are similar to the new attractions made when the two substances mix. Organic solutions of iodine are pink-purple in color.
Bond enthalpies (bond energies or bond strengths)
Bond enthalpy is the heat required to break one mole of covalent bonds to produce individual atoms, starting from the original substance in the gas state, and ending with gaseous atoms. For chlorine, Cl2(g), it is the heat energy required for the following reaction, per mole:
$Cl-Cl (g) \rightarrow 2Cl(g) \nonumber$
Although bromine is a liquid, the bond enthalpy is defined in terms of gaseous bromine molecules and atoms, as shown below:
Bond enthalpy in the halogens, X2(g)
Covalent bonding is effective because the bonding pair is attracted to both the nuclei at either side of it. It is that attraction which holds the molecule together. The extent of the attraction depends in part on the distances between the bonding pair and the two nuclei. The figure below illustrates such a covalent bond:
In all halogens, the bonding pair experiences a net +7 charge from either end of the bond, because the charge on the nucleus is offset by the inner electrons. As the atoms get larger down the group, the bonding pair is further from the nuclei and the strength of the bond should, in theory, decrease, as indicated in the figure below. The question is whether experimental data matches this prediction.
As is clear from the figure above, the bond enthalpies of the Cl-Cl, Br-Br and I-I bonds decreases as predicted, but the F-F bond enthalpy deviates.
Because fluorine atoms are so small, a strong bond is expected—in fact, it is remarkably weak. There must be another factor for consideration.
In addition to the bonding pair of electrons between the two atoms, each atom has 3 lone pairs of electrons in the outer shell. If the bond is very short,as in F-F, the lone pairs on the two atoms are close enough to cause significant repulsion, illustrated below:
In the case of fluorine, this repulsion is great enough to counteract much of the attraction between the bonding pair and the two nuclei. This weakens the bond.
Bond enthalpies in the hydrogen halides, HX(g)
If the halogen atom is attached to a hydrogen atom, this does not occur; there are no lone pairs on a hydrogen atom. Bond enthalpies for halogen-hydrogen bonds are given below:
As larger halogens are involved, the bonding pair is more distant from the nucleus. The attraction is lessened, and the bond should be weaker; this is supported by the data, without exception. This fact has significant implications for the thermal stability of the hydrogen halides— they are easily broken into hydrogen and the halogen on heating.
Hydrogen fluoride and hydrogen chloride are thermally very stable under typical laboratory conditions. Hydrogen bromide breaks down to some extent into hydrogen and bromine on heating, and hydrogen iodide is even less stable when heated. Weaker bonds are more easily broken.
Contributors and Attributions
Jim Clark (Chemguide.co.uk) | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/08%3A_Chemistry_of_the_Main_Group_Elements/8.13%3A_The_Halogens/8.13.01%3A_Physical_Properties_of_the_Halogens/8.13.1.01%3A_Atomic_and_Physical_Properties_of_Halogens.txt |
The halogens are located on the left of the noble gases on the periodic table. These five toxic, non-metallic elements make up Group 17 of the periodic table and consist of: fluorine (F), chlorine (Cl), bromine (Br), iodine (I), and astatine (At). Although astatine is radioactive and only has short-lived isotopes, it behaves similarly to iodine and is often included in the halogen group. Because the halogen elements have seven valence electrons, they only require one additional electron to form a full octet. This characteristic makes them more reactive than other non-metal groups.
Introduction
Halogens form diatomic molecules (of the form X2, where X denotes a halogen atom) in their elemental states. The bonds in these diatomic molecules are non-polar covalent single bonds. However, halogens readily combine with most elements and are never seen uncombined in nature. As a general rule, fluorine is the most reactive halogen and astatine is the least reactive. All halogens form Group 1 salts with similar properties. In these compounds, halogens are present as halide anions with charge of -1 (e.g., Cl-, Br-, etc.). Replacing the -ine ending with an -ide ending indicates the presence of halide anions; for example, Cl- is named "chloride." In addition, halogens act as oxidizing agents—they exhibit the property to oxidize metals. Therefore, most of the chemical reactions that involve halogens are oxidation-reduction reactions in aqueous solution. When in the -1 oxidation state, with carbon or nitrogen in organic compounds, the halogens often form single bonds. When a halogen atom is substituted for a covalently-bonded hydrogen atom in an organic compound, the prefix halo- can be used in a general sense, or the prefixes fluoro-, chloro-, bromo-, or iodo- can be used for specific halogen substitutions. Halogen elements can cross-link to form diatomic molecules with polar covalent single bonds.
Chlorine (Cl2) was the first halogen to be discovered in 1774, followed by iodine (I2), bromine (Br2), fluorine (F2), and astatine (At, discovered last in 1940). The name "halogen" is derived from the Greek roots hal- ("salt") and -gen ("to form"). Together these words combine to mean "salt former", referencing the fact that halogens form salts when they react with metals. Halite is the mineral name for rock salt, a natural mineral consisting essentially of sodium chloride (NaCl). Lastly, the halogens are also relevant in daily life, whether it be the fluoride that goes into toothpaste, the chlorine that disinfects drinking water, or the iodine that facilitates the production of thyroid hormones in one's body.
Elements
Fluorine - Fluorine has an atomic number of 9 and is denoted by the symbol F. Elemental fluorine was first discovered in 1886 by isolating it from hydrofluoric acid. Fluorine exists as a diatomic molecule in its free state (F2) and is the most abundant halogen found in the Earth's crust. Fluorine is the most electronegative element in the periodic table. It appears as a pale yellow gas at room temperature. Fluorine also has a relatively small atomic radius. Its oxidation state is always -1 except in its elemental, diatomic state (in which its oxidation state is zero). Fluorine is extremely reactive and reacts directly with all elements except helium (He), neon (Ne) and argon (Ar). In H2O solution, hydrofluoric acid (HF) is a weak acid. Although fluorine is highly electronegative, its electronegativity does not determine its acidity; HF is a weak acid due to the fact that the fluoride ion is basic (pH>7). In addition, fluorine produces very powerful oxidants. For example, fluorine can react with the noble gas xenon and form the strong oxidizing agent Xenon Difluoride (XeF2). There are many uses for fluorine, which will be discussed later.
Chlorine - Chlorine has the atomic number 17 and the chemical symbol Cl. Chlorine was discovered in 1774 by extracting it from hydrochloric acid. In its elemental state, it forms the diatomic molecule Cl2. Chlorine exhibits multiple oxidation states, such as -1, +1, +3, +5, and +7. At room temperature it appears as a light green gas. Since the bond that forms between the two chlorine atoms is weak, the Cl2 molecule is very reactive. Chlorine reacts with metals to produce salts called chlorides. Chloride ions are the most abundant ions that dissolve in the ocean. Chlorine also has two isotopes: 35Cl and 37Cl. Sodium chloride is the most prevalent compound of the chlorides.
Bromine - Bromine has an atomic number of 35 with a symbol of Br. It was first discovered in 1826. In its elemental form, it is the diatomic molecule Br2. At room temperature, bromine is a reddish- brown liquid. Its oxidation states vary from -1, +1, 3, 4 and 5. Bromine is more reactive than iodine, but not as reactive as chlorine. Also, bromine has two isotopes: 79Br and 81Br. Bromine consists of bromide salts, which have been found in the sea. The world production of bromide has increased significantly over the years, due to its accessibility and longer existence. Like all of the other halogens, bromine is an oxidizing agent, and is very toxic.
Iodine - Iodine has the atomic number 53 and symbol I. Iodine has oxidation states -1, +1, +5 and +7. Iodine exists as a diatomic molecule, I2, in its elemental state. At room temperature, it appears as a violet solid. Iodine has one stable isotope: 127I. It was first discovered in 1811 through the use of seaweed and sulfuric acid. Currently, iodide ions can be isolated in seawater. Although iodine is not very soluble in water, the solubility may increase if particular iodides are mixed into the solution. Iodine has many important roles in life, including thyroid hormone production. This will be discussed in Part VI of the text.
Astatine - Astatine is a radioactive element with an atomic number of 85 and symbol At. Its possible oxidation states include: -1, +1, +3, +5, and +7. It is the only halogen that is not a diatomic molecule, and it appears as a black, metallic solid at room temperature. Astatine is a very rare element, so there is not that much known about this element. In addition, astatine has a very short radioactive half-life, no longer than a couple of hours. It was discovered in 1940 by synthesis. Also, it is thought that astatine is similar to iodine. However, these two elements are assumed to differ in their metallic character.
Table 1.1: Electron configurations of the halogens.
Halogen Electronic Configuration
Fluorine 1s2 2s2 2p5
Chlorine [Ne]3s2 3p5
Bromine [Ar]3d10 4s2 4p5
Iodine [Kr]4d10 5s2 5p5
Astatine [Xe]4f14 5d10 6s2 6p5
The periodic trends observed in the halogen group:
Melting and Boiling Points (increase down the group)
The melting and boiling points increase down the group because of the van der Waals forces. The size of the molecules increases down the group. This increase in size means an increase in the strength of the van der Waals forces.
$F < Cl < Br < I < At \nonumber$
Table 1.2: Melting and Boiling Points of Halogens
Halogen Melting Point (˚C) Boiling Point (˚C)
Fluorine -220 -188
Chlorine -101 -35
Bromine -7.2 58.8
Iodine 114 184
Astatine 302 337
Atomic Radius (increases down the group)
The size of the nucleus increases down a group (F < Cl < Br < I < At) because the numbers of protons and neutrons increase. In addition, more energy levels are added with each period. This results in a larger orbital, and therefore a longer atomic radius.
Table 1.3: Atomic Radii of Halogens
Halogen Covalent Radius (pm) Ionic (X-) radius (pm)
Fluorine 71 133
Chlorine 99 181
Bromine 114 196
Iodine 133 220
Astatine 150
Ionization Energy (decreases down the group)
If the outer valence electrons are not near the nucleus, it does not take as much energy to remove them. Therefore, the energy required to pull off the outermost electron is not as high for the elements at the bottom of the group since there are more energy levels. Also, the high ionization energy makes the element appear non-metallic. Iodine and astatine display metallic properties, so ionization energy decreases down the group (At < I < Br < Cl < F).
Table 1.4 Ionization Energy of Halogens
Halogen First Ionization Energy (kJ/mol)
Fluorine 1681
Chlorine 1251
Bromine 1140
Iodine 1008
Astatine 890±40
Electronegativity (decreases down the group)
The number of valence electrons in an atom increases down the group due to the increase in energy levels at progressively lower levels. The electrons are progressively further from the nucleus; therefore, the nucleus and the electrons are not as attracted to each other. An increase in shielding is observed. Electronegativity therefore decreases down the group (At < I < Br < Cl < F).
Table 1.5: Electronegativity of Halogens
Halogen Electronegativity
Fluorine 4.0
Chlorine 3.0
Bromine 2.8
Iodine 2.5
Astatine 2.2
Electron Affinity (decreases down the group)
Since the atomic size increases down the group, electron affinity generally decreases (At < I < Br < F < Cl). An electron will not be as attracted to the nucleus, resulting in a low electron affinity. However, fluorine has a lower electron affinity than chlorine. This can be explained by the small size of fluorine, compared to chlorine.
Table 1.6: Electron Affinity of Halogens
Halogen Electron Affinity (kJ/mol)
Fluorine -328.0
Chlorine -349.0
Bromine -324.6
Iodine -295.2
Astatine -270.1
Reactivity of Elements (decreases down the group)
The reactivities of the halogens decrease down the group ( At < I < Br < Cl < F). This is due to the fact that the atomic radius increases in size with an increase of electronic energy levels. This lessens the attraction for valence electrons of other atoms, decreasing reactivity. This decrease also occurs because electronegativity decreases down a group; therefore, there is less electron "pulling." In addition, there is a decrease in oxidizing ability down the group.
Hydrogen Halides and Halogen Oxoacids
Hydrogen Halides
A halide is formed when a halogen reacts with another, less electronegative element to form a binary compound. Hydrogen, for example, reacts with halogens to form halides of the form HX:
• Hydrogen Fluoride: HF
• Hydrogen Chloride: HCl
• Hydrogen Bromide: HBr
• Hydrogen Iodide: HI
Hydrogen halides readily dissolve in water to form hydrohalic (hydrofluoric, hydrochloric, hydrobromic, hydroiodic) acids. The properties of these acids are given below:
• The acids are formed by the following reaction: HX (aq) + H2O (l) X- (aq) + H3O+ (aq)
• All hydrogen halides form strong acids, except HF
• The acidity of the hydrohalic acids increases as follows: HF < HCl < HBr < HI
Hydrofluoric acid can etch glass and certain inorganic fluorides over a long period of time.
It may seem counterintuitive to say that HF is the weakest hydrohalic acid because fluorine has the highest electronegativity. However, the H-F bond is very strong; if the H-X bond is strong, the resulting acid is weak. A strong bond is determined by a short bond length and a large bond dissociation energy. Of all the hydrogen halides, HF has the shortest bond length and largest bond dissociation energy.
Halogen Oxoacids
A halogen oxoacid is an acid with hydrogen, oxygen, and halogen atoms. The acidity of an oxoacid can be determined through analysis of the compound's structure. The halogen oxoacids are given below:
• Hypochlorous Acid: HOCl
• Chlorous Acid: HClO2
• Chloric Acid: HClO3
• Perchloric Acid: HClO4
• Hypobromous Acid: HOBr
• Bromic Acid: HBrO3
• Perbromic Acid: HBrO4
• Hypoiodous Acid: HOI
• Iodic Acid: HIO3
• Metaperiodic Acid: HIO4; H5IO6
In each of these acids, the proton is bonded to an oxygen atom; therefore, comparing proton bond lengths is not useful in this case. Instead, electronegativity is the dominant factor in the oxoacid's acidity. Acidic strength increases with more oxygen atoms bound to the central atom.
States of Matter at Room Temperature
Table 1.7: States of Matter and Appearance of Halogens
States of Matter (at Room Temperature) Halogen Appearance
Solid Iodine Violet
Astatine Black/Metallic [Assumed]
Liquid Bromine Reddish-Brown
Gas Fluorine Pale Yellow-Brown
Chlorine Pale Green
Explanation for Appearance
The halogens' colors are results of the absorption of visible light by the molecules, which causes electronic excitation. Fluorine absorbs violet light, and therefore appears light yellow. Iodine, on the other hand, absorbs yellow light and appears violet (yellow and violet are complementary colors, which can be determined using a color wheel). The colors of the halogens grow darker down the group:
• Fluorine pale yellow/brown
• Chlorine pale green
• Bromine red-brown
• www.crscientific.com/brominecell4.jpg
• Iodine violet
• genchem.chem.wisc.edu/lab/PTL...ments/I/I.jpeg
• Astatine* black/metallic
• www4.msu.ac.th/satit/studentP...t/astatine.jpg
In closed containers, liquid bromine and solid iodine are in equilibrium with their vapors, which can often be seen as colored gases. Although the color for astatine is unknown, it is assumed that astatine must be darker than iodine's violet (i.e., black) based on the preceding trend.
Oxidation States of Halogens in Compounds
As a general rule, halogens usually have an oxidation state of -1. However, if the halogen is bonded to oxygen or to another halogen, it can adopt different states: the -2 rule for oxygen takes precedence over this rule; in the case of two different halogens bonded together, the more electronegative atom takes precedence and adopts the -1 oxidation state.
Example 1.1: Iodine Chloride (ICl)
Chlorine has an oxidation state of -1, and iodine will have an oxidation of +1. Chlorine is more electronegative than iodine, therefore giving it the -1 oxidation state.
Example 1.2: Perbromic acid (HBrO4)
Oxygen has a total oxidation state of -8 (-2 charge x 4 atoms= -8 total charge). Hydrogen has a total oxidation state of +1. Adding both of these values together, the total oxidation state of the compound so far is -7. Since the final oxidation state of the compound must be 0, bromine's oxidation state is +7.
A third exception to the rule is this: if a halogen exists in its elemental form (X2), its oxidation state is zero.
Table 1.8: Oxidation States of Halogens
Halogen Oxidation States in Compounds
Fluorine (always) -1*
Chlorine -1, +1, +3, +5, +7
Bromine -1, +1, +3, +4, +5
Iodine -1, +1,+5, +7
Astatine -1, +1, +3, +5, +7
Example 1.3: Fluorine
Why does fluorine always have an oxidation state of -1 in its compounds?
Solution
Electronegativity increases across a period, and decreases down a group. Therefore, fluorine has the highest electronegativity of all of the elements, indicated by its position on the periodic table. Its electron configuration is 1s2 2s2 2p5. If fluorine gains one more electron, the outermost p orbitals are completely filled (resulting in a full octet). Because fluorine has a high electronegativity, it can easily remove the desired electron from a nearby atom. Fluorine is then isoelectronic with a noble gas (with eight valence electrons); all its outermost orbitals are filled. Fluorine is much more stable in this state.
Applications of Halogens
Fluorine: Although fluorine is very reactive, it serves many industrial purposes. For example, it is a key component of the plastic polytetrafluoroethylene (called Teflon-TFE by the DuPont company) and certain other polymers, often referred to as fluoropolymers. Chlorofluorocarbons (CFCs) are organic chemicals that were used as refrigerants and propellants in aerosols before growing concerns about their possible environmental impact led to their discontinued use. Hydrochlorofluorocarbons (HFCs) are now used instead. Fluoride is also added to toothpaste and drinking water to help reduce tooth decay. Fluorine also exists in the clay used in some ceramics. Fluorine is associated with generating nuclear power as well. In addition, it is used to produce fluoroquinolones, which are antibiotics. Below is a list of some of fluorine's important inorganic compounds.
Table 1.9: Important Inorganic Compounds of Fluorine
Compound Uses
Na3AlF6 Manufacture of aluminum
BF3 Catalyst
CaF2 Optical components, manufacture of HF, metallurgical flux
ClF3 Fluorinating agent, reprocessing nuclear fuels
HF Manufacture of F2, AlF3, Na3AlF6, and fluorocarbons
LiF Ceramics manufacture, welding, and soldering
NaF Fluoridating water, dental prophylaxis, insecticide
SF6 Insulating gas for high-voltage electrical equipment
SnF2 Manufacture of toothpaste
UF6 Manufacture of uranium fuel for nuclear reactors
Chlorine: Chlorine has many industrial uses. It is used to disinfect drinking water and swimming pools. Sodium hypochlorite (NaClO) is the main component of bleach. Hydrochloric acid, sometimes called muriatic acid, is a commonly used acid in industry and laboratories. Chlorine is also present in polyvinyl chloride (PVC) and several other polymers. PVC is used in wire insulation, pipes, and electronics. In addition, chlorine is very useful in the pharmaceutical industry. Medicinal products containing chlorine are used to treat infections, allergies, and diabetes. The neutralized form of hydrochloride is a component of many medications. Chlorine is also used to sterilize hospital machinery and limit infection growth. In agriculture, chlorine is a component of many commercial pesticides: DDT (dichlorodiphenyltrichloroethane) was used as an agricultural insecticide, but its use was discontinued.
Bromine: Bromine is used in flame retardants because of its fire-resistant properties. It also found in the pesticide methyl bromide, which facilitates the storage of crops and eliminates the spread of bacteria. However, the excessive use of methyl bromide has been discontinued due to its impact on the ozone layer. Bromine is involved in gasoline production as well. Other uses of bromine include the production of photography film, the content in fire extinguishers, and drugs treating pneumonia and Alzheimer's disease.
Iodine: Iodine is important in the proper functioning of the thyroid gland of the body. If the body does not receive adequate iodine, a goiter (enlarged thyroid gland) will form. Table salt now contains iodine to help promote proper functioning of the thyroid hormones. Iodine is also used as an antiseptic. Solutions used to clean open wounds likely contain iodine, and it is commonly found in disinfectant sprays. In addition, silver iodide is important for photography development.
Astatine: Because astatine is radioactive and rare, there are no proven uses for this halogen element. However, there is speculation that this element could aid iodine in regulating the thyroid hormones. Also, 211At has been used in mice to aid the study of cancer.
• Grube, Karl; Leffler, Amos J. "Synthesis of metal halides (ML)." J. Chem. Educ. 1993, 70, A204.
• This video provides information about some of the physical properties of chlorine, bromine, and iodine: http://www.youtube.com/watch?v=yP0U5rGWqdg
• The following video compares four halogens: fluorine, chlorine, bromine and iodine in terms of chemical reactions and physical properties. http://www.youtube.com/watch?v=u2ogMUDBaf4
• Color wheel referred to in the text: http://www.wou.edu/las/physci/ch462/c-wheel.gif
• Elson, Jesse. "A bonding parameter. III, Water solubilities and melting points of the alkali halogens." J. Chem. Educ.1969, 46, 86.
• Fessenden, Elizabeth. "Structural chemistry of the interhalogen compounds." J. Chem. Educ. 1951, 28, 619.
• Holbrook, Jack B.; Sabry-Grant, Ralph; Smith, Barry C.; Tandel, Thakor V. "Lattice enthalpies of ionic halides, hydrides, oxides, and sulfides: Second-electron affinities of atomic oxygen and sulfur." J. Chem. Educ. 1990, 67, 304.
• Kildahl, Nicholas K. "A procedure for determining formulas for the simple p-block oxoacids." J. Chem. Educ. 1991, 68, 1001.
• Liprandi, Domingo A.; Reinheimer, Orlando R.; Paredes, José F.; L'Argentière, Pablo C. "A Simple, Safe Way To Prepare Halogens and Study Their Visual Properties at a Technical Secondary School." J. Chem. Educ. 1999 76.
• Meek, Terry L. "Acidities of oxoacids: Correlation with charge distribution."J. Chem. Educ. 1992, 69, 270.
Practice Problems
1. Why does fluorine always have an oxidation state of -1 in its compounds?
2. Find the oxidation state of the halogen in each problem:
1. HOCl
2. KIO3
3. F2
3. What are three uses of chlorine?
4. Which element(s) exist(s) as a solid at room temperature?
5. Do the following increase or decrease down the group of halogens?
1. boiling point and melting point
2. electronegativity
3. ionization energy
Answers
1. Electronegativity increases across a period, and decreases down a group. Therefore, fluorine has the highest electronegativity out of all of the elements. Because fluorine has seven valence electrons, it only needs one more electron to acheive a noble gas configuration (eight valence electrons). Therefore, it will be more likely to pull off an electron from a nearby atom.
2. disinfecting water, pesticides, and medicinal products
1. +1 (Hydrogen has an oxidation state of +1, and oxygen has an oxidation state of -2. Therefore, chlorine must have an oxidation state of +1 so that the total charge can be zero)
2. +5 (Potassium's oxidation state is +1. Oxygen has an oxidation state of -2, so for this compound it is -6 (-2 charge x 3 atoms= -6). Since the total oxidation state has to be zero, iodine's oxidation state must be +5).
3. 0 (Elemental forms always have an oxidation state of 0.)
3. iodine and astatine
1. increases
2. decreases
3. decreases | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/08%3A_Chemistry_of_the_Main_Group_Elements/8.13%3A_The_Halogens/8.13.01%3A_Physical_Properties_of_the_Halogens/8.13.1.02%3A_General_Properties_of_Halogens.txt |
1. All elements are diatomic and molecular and the boiling and melting points increase as a result of the increasing van der Waals interactions between diatomic molecules for the heavier elements.
2. The elements are typical non-metals in their physical and chemical properties. They form anionic compounds based on X- (X = halogen), which is associated with a complete octet.
The ionic compounds MX become progressively less ionic as the relative atomic mass of X increases, because of the decreasing electronegativity of the halogens. Iodine has the greatest tendency to form cationic species, e.g., I2+, I5+, because it has the lowest ionization energy. The cation Br2+ is known in Br2+Sb3F16-, and Br5+ has been reported.
3. The atoms also form strong covalent bonds with other non-metals. The mean bond enthalpies for E-X bonds are particularly large for fluorine and therefore a wide range of molecular fluorides are known; fluorine is particularly effective at bringing out the highest valencies of the non-metals and highest oxidation states of the metals.
4. The oxidizing ability of the halogens decreases markedly down the group: F2 > Cl2 > Br2 > I2, and only iodine is oxidized by nitric acid.
5. The stabilities of the hydrogen halides decrease down the group, but their acid strengths increase.
6. Only H-F forms strong hydrogen bonds and this is reflected in the boiling and melting points of the hydrogen halides.
7. The halogens form many interhalogen compounds with the less electronegative halogen surrounded by the more electronegative halogens. Neutral, anionic, and cationic interhalogen compounds are known. ICl and IBr are widely used in organic synthesis and are commercially available.
The most extensive series of compounds exists for iodine, e.g., IF7, IF5, ICl4-, ICl2-.Fluorine does not form any interhalogen compounds where it occupies the central position within the molecule.
8. Oxygen fluorides are extremely strong and reactive oxidants and have been explored as potential rocket fuels; the oxides become less reactive down the column and more numerous. Iodine forms a particularly wide range of oxides.
9. The perhalates, EO4-, are only known for Cl, Br, and I. They exhibit an alternation in their oxidizing abilities, and the perbromates are particularly strong oxidizing agents.
10. In the highest oxidation state (+7) the relative oxidizing ability is:
Br> I > Cl
and results in the formation of the corresponding +5 oxoanions,
ClO4- + 2e- = ClO3- E° = 1.20 V
BrO4- + 2e- = BrO3- E° = 1.85 V
IO4- + 2e- = IO3- E° = 1.63 V
The hypohalite ions disproportionate according to the equation:
2XO- = 2X- + XO3-
The equilibrium constants are 1027 for ClO-/Cl- :(the reaction is slow at room temperature), 1015 for BrO-/Br-, and 1020for IO-/I-. HOF has been prepared from ice + F2 but is very reactive, decomposing to HF + O2.
8.13.1.04: Physical Properties of the Group 17 Elements
This page discusses the trends in some atomic and physical properties of the Group 17 elements (the halogens): fluorine, chlorine, bromine and iodine. Sections below describe the trends in atomic radius, electronegativity, electron affinity, melting and boiling points, and solubility. There is also a section on the bond enthalpies (and strengths) of halogen-halogen bonds (for example, the Cl-Cl bond) and of hydrogen-halogen bonds (e.g., the H-Cl bond).
You can see that the atomic radius increases as you go down the group.
The radius of an atom is governed by
• Electronegativity is a measure of the tendency of an atom to attract a bonding pair of electrons. It is usually measured on the Pauling scale, on which the most electronegative element (fluorine) is given an electronegativity of 4.0.
As shown in the figure above, electronegativity decreases from fluorine to iodine; the atoms become less effective at attracting bonding pairs of electrons as they grow larger. This can be visualized using dots-and-crosses diagrams for hydrogen fluoride and hydrogen chloride.
The bonding electrons between the hydrogen and the halogen experience the same net charge of +7 from either the fluorine or the chlorine. However, in the chlorine case, the nucleus is further away from the bonding pair. Therefore, electrons are not as strongly attracted to the chlorine nucleus as they are to the fluorine nucleus.
The stronger attraction to the closer fluorine nucleus makes fluorine more electronegative.
Summarizing the trend down the group
As the halogen atoms get larger, any bonding pair is farther and farther away from the halogen nucleus, and so is less strongly attracted towards it. Hence, the elements become less electronegative as you go down the group. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/08%3A_Chemistry_of_the_Main_Group_Elements/8.13%3A_The_Halogens/8.13.01%3A_Physical_Properties_of_the_Halogens/8.13.1.03%3A_Halogen_Group_%28Group_17%29_Trends.txt |
Covers the halogens in Group 17: fluorine (F), chlorine (Cl), bromine (Br) and iodine (I). Includes trends in atomic and physical properties, the redox properties of the halogens and their ions, the acidity of the hydrogen halides, and the tests for the halide ions.
8.13.02: Chemical Properties of the Halogens
This page examines the redox reactions involving halide ions and concentrated sulfuric acid, using these reactions to discuss the trend in reducing ability of the ions from fluoride to iodide. Two types of reactions might occur when concentrated sulfuric acid is added to a solid ionic halide like sodium fluoride, chloride, bromide or iodide. The concentrated sulfuric acid can act as both an acid and an oxidizing agent.
Concentrated sulfuric acid acting as an acid
Concentrated sulfuric acid transfers a proton to the halide ion to produce a gaseous hydrogen halide, which immediately escapes from the system. If the hydrogen halide is exposed to moist air, steam fumes are formed. For example, concentrated sulfuric acid reacts with solid sodium chloride at low temperatures to produce hydrogen chloride and sodium bisulfate, as in the following equation:
$NaCl + H_2SO_4 \rightarrow HCl + NaHSO_4 \nonumber$
All the halide ions behave similarly.
Concentrated sulfuric acid acting as an oxidizing agent
With fluoride or chloride
Concentrated sulfuric acid is not a strong enough oxidizing agent to oxidize fluoride or chloride. In those cases, only the steamy fumes of the hydrogen halide—hydrogen fluoride or hydrogen chloride—are produced. In terms of the halide ions, fluoride and chloride are not strong enough reducing agents to reduce the sulfuric acid. This is not the case for bromides and iodides.
With bromide
Bromide is a strong enough reducing agent to reduce sulfuric acid. Bromide is oxidized to bromine in the process, as in the half-equation below:
$2Br^- \rightarrow Br_2 + 2e^- \nonumber$
Bromide reduces sulfuric acid to sulfur dioxide gas, decreasing the oxidation state of sulfur from +6 to +4. The half-equation for this transition is as follows:
$H_2SO_4 + 2H^+ + 2e^- \rightarrow SO_2 + 2H_2O \nonumber$
These two half-equations can be combined into the overall ionic equation for the reaction:
$H_2SO_4 + 2H^+ + 2Br^- \rightarrow Br_2 +SO_2 +2H_2O \nonumber$
In practice, this reaction is confirmed by the steamy fumes of hydrogen bromide contaminated with the brown color of bromine vapor. The sulfur dioxide is a colorless gas, so its presence cannot be directly observed.
With Iodide
Iodide is a stronger reducing agent than bromide, and it is oxidized to iodine by the sulfuric acid:
$2I^- \rightarrow I_2 +2e^- \nonumber$
The reduction of the sulfuric acid is more complicated than with bromide. Iodide is powerful enough to reduce it in three steps:
• sulfuric acid to sulfur dioxide (sulfur oxidation state = +4)
• sulfur dioxide to elemental sulfur (oxidation state = 0)
• sulfur to hydrogen sulfide (sulfur oxidation state = -2).
The most abundant product is hydrogen sulfide. The half-equation for its formation is as follows:
$H_2SO_4 + 8H^+ + 8e^- \rightarrow H_2S + 4H_2O \nonumber$
Combining these two half-equations gives the following net ionic equation:
$H_2SO_4 + 8H^+ + 8I^- \rightarrow 4I_2 + H_2S + 4H_2O \nonumber$
This is confirmed by a trace of steamy fumes of hydrogen iodide and a large amount of iodine. The reaction is exothermic: purple iodine vapor is formed, with dark gray solid iodine condensing around the top of the reaction vessel. There is also a red color where the iodine comes into contact with solid iodide salts. The red color is due to the I3- ion formed by reaction between I2 molecules and I- ions. Hydrogen sulfide gas can be detected by its "rotten egg" smell, but this gas is intensely poisonous.
Summary of the trend in reducing ability
• Fluoride and chloride cannot reduce concentrated sulfuric acid.
• Bromide reduces sulfuric acid to sulfur dioxide. In the process, bromide ions are oxidized to bromine.
• Iodide reduces sulfuric acid to a mixture of products including hydrogen sulfide. Iodide ions are oxidized to iodine.
• The reducing ability of halide ions increases down the group.
Explaining the trend
An over-simplified explanation
The following explanation is only (partially) accurate if fluoride is neglected. When a halide ion acts as a reducing agent, it transfers electrons to something else. That means that the halide ion itself loses electrons. The larger the halide ion, the farther the outer electrons are from the nucleus, and the more they are shielded by inner electrons. It therefore gets easier for the halide ions to lose electrons down the group because there is less attraction between the outer electrons and the nucleus. This argument seems valid, but it is flawed. The energetics of the change must be examined.
A more detailed explanation
Enthalpy change variation between halogens
The amount of heat evolved or absorbed when a solid halide (like sodium chloride) is converted into an elemental halogen must be considered. Taking sodium chloride as an example, the following energetic quantities are important:
• The energy required to break the attractions between the ions in the sodium chloride (the lattice enthalpy).
• The energy required to remove the electron from the chloride ion. This is the reverse of the electron affinity of the chlorine (the electron affinity can be acquired from a data table and negated).
• The energy recovered when the chlorine atoms convert to diatomic chlorine. Energy is released when these bonds are formed. Chlorine is simple because it is a gas. In bromine and iodine, heat is also released during condensation to a liquid or a solid, respectively. To account for this, it is simpler to think in terms of atomization energy rather than bond energy. The value of interest is the reverse of atomization energy.
Atomization energy is the energy needed to produce 1 mole of isolated gaseous atoms starting from an element in its standard state (gas for chlorine, and liquid for bromine, for example - both of them as X2). The figure below shows how this information fits together:
The enthalpy change shown by the green arrow in the diagram for each of the halogens must be compared. The diagram shows that the overall change involving the halide ions is endothermic (the green arrow is pointing up toward a higher energy).
This is not the total enthalpy change for the whole reaction. Heat is emitted when the changes involving the sulfuric acid occur. That is the same irrespective of the halogen in question. The total enthalpy change is the sum of the enthalpy changes for the halide ion half-reaction and the sulfuric acid half-reaction.
The table below shows the energy changes that vary from halogen to halogen. The process is assumed to start from the solid sodium halide. The values for the lattice enthalpies for other solid halides would be different, but the pattern would be the same.
heat needed to break up NaX lattice
(kJ mol-1)
heat needed to remove electron from halide ion
(kJ mol-1)
heat released in forming halogen molecules
(kJ mol-1)
sum of these
(kJ mol-1)
F +902 +328 -79 +1151
Cl +771 +349 -121 +999
Br +733 +324 -112 +945
I +684 +295 -107 +872
The overall enthalpy change for the halide half-reaction:
The sum of the enthalpy changes, in the final column, is decreasingly endothermic down the group. The total change in enthalpy (including the sulfuric acid) is also less positive.
The amount of heat produced in the half-reaction involving the sulfuric acid must be great enough to make the reactions with the bromide or iodide feasible, but not enough to compensate for the more positive values produced by the fluoride and chloride half-reactions.
Exploring the changes in the various energy terms
In this section, the individual energy terms in the table that are most important in making the halogen half-reaction less endothermic down the group are determined.
Chlorine to iodine
From chlorine to iodine, the lattice enthalpy changes most, decreasing by 87 kJ mol-1. By contrast, the energy required to remove the electron decreases by only 54 kJ mol-1. Both of these terms matter, but the decrease in lattice enthalpy is the more significant. This quantity decreases because the ions are getting larger. That means that they are farther away from each other, and so the attractions between positive and negative ions in the solid lattice are lessened.
The simplified explanation mentioned earlier is misleading because it concentrates on the less-important decrease in the amount of energy needed to remove the electron from the ion.
Fluorine
Fluoride ions are very difficult to oxidize to fluorine. The table above shows that this has nothing to do with the amount of energy required to remove an electron from a fluoride ion. It actually takes less energy to remove an electron from a fluoride ion than from a chloride ion. The generalization that an electron becomes easier to remove as the ion becomes larger does not apply here.
Fluoride ions are so small that the electrons experience strong repulsion from each other. This outweighs the effect of their closeness to the nucleus and makes them easier to remove than the simplified argument predicts.
There are two important reasons why fluoride ions are so difficult to oxidize. The first is the comparatively high lattice enthalpy of the solid fluoride. This is due to the small size of the fluoride ion, which means that the positive and negative ions are very close together and therefore strongly attracted to each other. The other factor is the small amount of heat that is released when the fluorine atoms combine to make fluorine molecules (see the table above). This is due to the low bond enthalpy of the F-F bond. The reason for this low bond enthalpy is discussed on a separate page.
What if the halide ions were in solution rather than in a solid?
This discussion has focused on the energetics of the process starting from solid halide ions because that is the standard procedure when using concentrated sulfuric acid. Halides could also be oxidized in solution with another oxidizing agent.
The trend is exactly the same. Fluoride is difficult to oxidize and it becomes easier down the group toward iodide; in other words, fluoride ions are not good reducing agents, but iodide ions are.
The explanation starts from the hydrated ions in solution rather than solid ions. In a sense, this has already been done on another page. Fluorine is a very powerful oxidizing agent because it very readily forms its negative ion in solution. It is therefore energetically difficult to reverse the process. By contrast, for energetic reasons, iodine is relatively reluctant to form its negative ion in solution. Therefore, it is relatively easy for it to revert back to iodine. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/08%3A_Chemistry_of_the_Main_Group_Elements/8.13%3A_The_Halogens/8.13.02%3A_Chemical_Properties_of_the_Halogens/8.13.2.01%3A_Halide_Ions_as_Reducing_Agents.txt |
This page examines the trend in oxidizing ability of the Group 17 elements (the halogens): fluorine, chlorine, bromine and iodine. It considers the ability of one halogen to oxidize the ions of another, and how this changes down the group.
Basic facts
Consider a situation in which one halogen (chlorine, for example) is reacted with the ions of another (iodide, perhaps) from a salt solution. In the chlorine and iodide ion case, the reaction is as follows:
$\ce{Cl_2 + 2I^- \rightarrow 2Cl^- + I_2} \nonumber$
• The iodide ions lose electrons to form iodine molecules. In other words, they are oxidized.
• The chlorine molecules gain electrons to form chloride ions— they are reduced.
This is therefore a redox reaction in which chlorine acts as an oxidizing agent.
Fluorine
Fluorine must be excluded from this discussion because its oxidizing abilities are too strong. Fluorine oxidizes water to oxygen, as in the equation below, and so it is impossible to carry out reactions with it in aqueous solution.
$\ce{2F_2 + 2H_2O \rightarrow 4HF + O_2} \nonumber$
Chlorine, Bromine and Iodine
In each case, a halogen higher in the group can oxidize the ions of one lower down. For example, chlorine can oxidize bromide ions to bromine:
$\ce{Cl_2 + 2Br^- \rightarrow 2Cl^- + Br_2} \nonumber$
The bromine forms an orange solution. As shown below, chlorine can also oxidize iodide ions to iodine:
$\ce{Cl_2 +2I^- \rightarrow 2Cl^- + I_2} \nonumber$
The iodine appears either as a red solution if little chlorine is used, or as a dark gray precipitate if the chlorine is in excess.
Bromine can only oxidize iodide ions, and is not a strong enough oxidizing agent to convert chloride ions into chlorine. A red solution of iodine is formed (see the note above) until the bromine is in excess. Then a dark gray precipitate is formed.
$\ce{Br_2 + 2I^- \rightarrow 2Br^- + I_2} \nonumber$
Iodine won't oxidize any of the other halide ions, except possibly the extremely radioactive and rare astatide ions.
To summarize
• Oxidation is the loss of electrons. Each of the elements (for example, chlorine) could potentially take electrons from something else and are subsequently ionized (e.g., Cl-). This means that they are all potential oxidizing agents.
• Fluorine is such a powerful oxidizing agent that solution reactions are unfeasible.
• Chlorine has the ability to take electrons from both bromide ions and iodide ions. Bromine and iodine cannot reclaim those electrons from the chloride ions formed.
• This indicates that chlorine is a more powerful oxidizing agent than either bromine or iodine.
• Similarly, bromine is a more powerful oxidizing agent than iodine. Bromine can remove electrons from iodide ions, producing iodine; iodine cannot reclaim those electrons from the resulting bromide ions.
In short, oxidizing ability decreases down the group.
Explaining the trend
Whenever one of the halogens is involved in oxidizing a species in solution, the halogen end is reduced to a halide ion associated with water molecules. The following figure illustrates this process:
Down the group, the ease with which these hydrated ions are formed decreases; the halogens become less effective as oxidizing agents, taking electrons from something else less readily. The reason that the hydrated ions form less readily down the group is due to several complicated factors. Unfortunately, this explanation is often over-simplified, giving a faulty and misleading explanation. The wrong explanation is dealt with here before a proper explanation is given.
The Incorrect Explanation
The following explanation is normally given for the trend in oxidizing ability of chlorine, bromine and iodine. The ease of ionization depends on how strongly the new electrons are attracted. As the atoms get larger, the new electrons are further from the nucleus and increasingly shielded by the inner electrons (offsetting the effect of the greater nuclear charge). The larger atoms are therefore less effective at attracting new electrons and forming ions. This is equivalent to saying electron affinity decreases down the group. Electron affinity is described in detail on another page.
The problem with this argument is that it does not include fluorine. Fluorine's tendency to form a hydrated ion is much higher than that of chlorine. However, fluorine's electron affinity is less than that of chlorine. This contradicts the above argument. This problem stems from examining a single part of a very complicated process. The argument about atoms accepting electrons applies only to isolated atoms in the gas state picking up electrons to form isolated ions, also in the gas state. The argument must be generalized.
In reality:
• The halogen starts as a diatomic molecule, X2. This may be a gas, liquid or solid at room temperature, depending on the halogen.
• The diatomic molecule must split into individual atoms (atomization)
• Each atom gains an electron (electron affinity; this is the element of the process of interest in the faulty explanation.)
• The isolated ions are surrounded by water molecules; hydrated ions are formed (hydration).
The Correct Explanation
The table below shows the energy involved in each of these changes for atomization energy, electron affinity, and hydration enthalpy (hydration energy):
atomization energy
(kJ mol-1)
electron affinity
(kJ mol-1)
hydration enthalpy
(kJ mol-1)
overall
(kJ mol-1)
F +79 -328 -506 -755
Cl +121 -349 -364 -592
Br +112 -324 -335 -547
I +107 -295 -293 -481
Consider first the fifth column, which shows the overall heat evolved, the sum of the energies in the previous three columns.
The amount of heat evolved decreases quite dramatically from the top to the bottom of the group, with the biggest decrease between fluorine and chlorine. Fluorine generates a large amount of heat when it forms its hydrated ion, chlorine a lesser amount, and so on down the group.
The first electron affinity is defined as the energy released when 1 mole of gaseous atoms each acquire an electron to form 1 mole of gaseous 1- ions, as in the following equation: In symbol terms:
$X(g) + e^- \rightarrow X^-(g) \nonumber$
The fifth column measures the energy released when 1 mole of gaseous ions dissolves in water to produce hydrated ions, as in the following equation, which is not equivalent to that above:
$X^-(g) \rightarrow X^- (aq) \nonumber$
Why is fluorine a stronger oxidizing agent than chlorine?
There are two main factors. First, the atomization energy of fluorine is abnormally low. This reflects the low bond enthalpy of fluorine.
The main reason, however, is the very high hydration enthalpy of the fluoride ion. That is because fluoride is very small. There is a very strong attraction between fluoride ions and water molecules. The stronger the attraction, the more heat is evolved when the hydrated ions are formed.
Why does oxidizing ability decrease from chlorine to bromine to iodine?
The decrease in atomization energy between these three elements is relatively small, and would tend to make the overall change more negative down the group. It is helpful to look at the changes in electron affinity and hydration enthalpy down the group. Using the figures from the previous table:
change in electron affinity
(kJ mol-1)
change in hydration enthalpy
(kJ mol-1)
Cl to Br +25 +29
Br to I +29 +42
Both of these effects contribute, but the more important factor—the one that changes the most—is the change in the hydration enthalpy. Down the group, the ions become less attractive to water molecules as they get larger. Although the ease with which an atom attracts an electron matters, it is not as important as the hydration enthalpy of the negative ion formed.
The faulty explanation is incorrect even if restricted to chlorine, bromine and iodine:
• This is the energy needed to produce 1 mole of isolated gaseous atoms starting from an element in its standard state (gas for chlorine, and liquid for bromine, for example, both of the form X2).
• For a gas like chlorine, this is simply half of the bond enthalpy (because breaking a Cl-Cl bond produces 2 chlorine atoms, not 1). For a liquid like bromine or a solid like iodine, it also includes the energy that is needed to convert them into gases. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/08%3A_Chemistry_of_the_Main_Group_Elements/8.13%3A_The_Halogens/8.13.02%3A_Chemical_Properties_of_the_Halogens/8.13.2.02%3A_Halogens_as_Oxidizing_Agents.txt |
The halogens react with each other to form interhalogen compounds. The general formula of most interhalogen compounds is XYn, where n = 1, 3, 5 or 7, and X is the less electronegative of the two halogens. The compounds which are formed by the union of two different halogens are called interhalogen compounds. There are never more than two types of halogen atoms in an interhalogen molecule. These are of four general types:
1. AX- type : ClF, BrF, BrCl, ICl, IBr,
2. AX3-type: ClF3, BrF3, (ICl3)2,
3. AX5-type: ClF5, BrF5, IF5,
4. AX7-type: IF7.
The interhalogen compounds of type AX and AX3 are formed between halogens having a very low electronegativity difference (e.g., ClF, ClF3). The interhalogen compounds of type AX5 and AX7 are formed by larger atoms having a low electronegativity with the smaller atoms having a high electronegativity. This is because it is possible to fit a greater number of smaller atoms around a larger one (e.g., BrF5, IF7).
Interhalogens are all prone to hydrolysis and ionize to give rise to polyatomic ions. The interhalogens are generally more reactive than halogens except for F. This is because A-X bonds in interhalogens are weaker than the X-X bonds in dihalogen molecules. Reactions of interhalogens are similar to those of halogens. Hydrolysis of interhalogen compounds gives a halogen acid and an oxy-acid.
Nomenclature
To name an interhalogen compound, the less electronegative element is placed on to the left in formulas and naming is straightforward.
Properties
Some properties of interhalogen compounds are listed below. They are all prepared by direct combination of the elements, although since in some cases more than one product is possible, the conditions may be varied by altering the temperature and relative proportions. For example, under the same conditions, difluorine reacts with dichlorine to give ClF, with dibromine to give BrF3, but with diiodine to give IF5.
Compound ClF BrF BrCl ICl IBr ClF3 BrF3 IF3 I2Cl6 ClF5 BrF5 IF5 IF7
Appearance at 298K Colorless gas Pale brown gas impure Red solid Black solid Colorless gas Yellow liquid Yellow solid Orange solid Colorless gas Colorless liquid Colorless liquid Colorless gas
Stereochemistry linear linear linear linear linear T-shaped T-shaped T-shaped planar square-based pyramid square-based pyramid square-based pyramid pentagonal bipyramid
Melting point /K 117 ~240 dissoc. 300(a) 313 197 282 245 (dec) 337 (sub) 170 212.5 282.5 278 (sub)
Boiling point /K 173 ~293 ~278 ~373 389 285 399 - - 260 314 373 -
ΔfH°(298 K) /kJ mol-1 -50.3 -58.5 14.6 -23.8 -10.5 -163.2 -300.8 ~-500 -89.3 -255 -458.6 -864.8 -962
Dipole moment for gas-phase molecule /D 0.89 1.42 0.52 1.24 0.73 0.6 1.19 - 0 - 1.51 2.18 0
Bond distances in gas-phase molecules except for IF3 and I2Cl6 / pm 163 176 214 232 248.5 160 (eq), 170 (ax) 172 (eq), 181 (ax) 187 (eq), 198 (ax) 238 (terminal) 268 (bridge) 172 (basal), 162 (apical) 178 (basal), 168 (apical) 187 (basal), 185 (apical) 186 (eq), 179 (ax)
Structures
The structures found for the various interhalogens conform to what would be expected based on the VSEPR model. For XY3 the shape can be described as T-shaped with 2 lone pairs sitting in two of the equatorial positions of a trigonal bipyramid. For XY5 the shape is a square pyramid with the unpaired electrons sitting in an axial position of an octahedron, and XY7 is a pentagonal bipyramid.
XY diatomic interhalogens
The interhalogens with formula XY have physical properties intermediate between those of the two parent halogens. The covalent bond between the two atoms has some ionic character, with the larger element, X, becoming oxidized and having a partial positive charge. Most combinations of F, Cl, Br and I are known, but not all are stable.
• Chlorine monofluoride (ClF), the lightest interhalogen, is a colorless gas with a boiling point of 173 °K.
• Bromine monofluoride (BrF) has not been obtained in the pure form - it dissociates into the trifluoride and free bromine. Similarly, iodine monofluoride is unstable - iodine reacts with fluorine to form a pentafluoride.
• Iodine monofluoride (IF) is unstable and disproportionates rapidly and irreversibly at room temperature: $\ce{5IF → 2I2 + IF5} \nonumber$. However, its molecular properties have been determined by spectroscopy: the iodine-fluorine distance is 190.9 pm and the I-F bond dissociation energy is around 277 kJ mol-1. ΔHf° = -95.4 kJ mol-1 and ΔGf° = -117.6 kJ mol-1, both at 298 K.
$\ce{IF}$ can be generated by the following reactions:
I2 + F2 → 2IF at -45 °C in CCl3F;
I2 + IF3 → 3IF at -78 °C in CCl3F;
I2 + AgF → IF + AgI at 0 °C.
• Bromine monochloride (BrCl) is an unstable red-brown gas with a boiling point of 5 °C.
• Iodine monochloride (ICl) consists of red transparent crystals which melt at 27.2 °C to form a choking brownish liquid (similar in appearance and weight to bromine). It reacts with HCl to form the strong acid HICl2. The crystal structure of iodine monochloride consists of puckered zig-zag chains, with strong interactions between the chains.
• Iodine monobromide (IBr) is made by direct combination of the elements to form a dark red crystalline solid. It melts at 42 °C and boils at 116 °C to form a partially dissociated vapor.
XY3 interhalogens
• Chlorine trifluoride (ClF3) is a colorless gas that condenses to a green liquid and freezes to a white solid. It is made by reacting chlorine with an excess of fluorine at 250° C in a nickel tube. It reacts more violently than fluorine, often explosively. The molecule is planar and T-shaped.
• Bromine trifluoride (BrF3) is a yellow-green liquid that conducts electricity - it ionizes to form [BrF2+] + [BrF4-]. It reacts with many metals and metal oxides to form similar ionized entities; with some others it forms the metal fluoride plus free bromine and oxygen. It is used in organic chemistry as a fluorinating agent. It has the same molecular shape as chlorine trifluoride.
• Iodine trifluoride (IF3) is a yellow solid which decomposes above -28 °C. It can be synthesized from the elements, but care must be taken to avoid the formation of IF5. F2 attacks I2 to yield IF3 at -45 °C in CCl3F. Alternatively, at low temperatures, the fluorination reaction I2 + 3XeF2 → 2IF3 + 3Xe can be used. Not much is known about iodine trifluoride as it is so unstable.
• Iodine trichloride (ICl3) forms lemon yellow crystals which can be melted under pressure to a brown liquid. It can be made from the elements at low temperature, or from iodine pentoxide and hydrogen chloride. It reacts with many metal chlorides to form tetrachloriodides, and hydrolyses in water. The molecule is a planar dimer, with each iodine atom surrounded by four chlorine atoms. In the melt it is conductive, which may indicate dissociation: $\ce{I_2Cl_6 → ICl_2^{+} + ICl_4^{-}} \nonumber$
Chlorine trifluoride, ClF3, was first reported in 1931; it is primarily used for the manufacture of uranium hexafluoride, UF6, as part of nuclear fuel processing and reprocessing by the reaction:
$\ce{U + 3 ClF_3 → UF_6 + 3 ClF} \nonumber$
U isotope separation is difficult because the two isotopes have very nearly identical chemical properties, and can only be separated gradually using small mass differences. (235U is only 1.26% lighter than 238U.) A cascade of identical stages produces successively higher concentrations of 235U. Each stage passes a slightly more concentrated product to the next stage and returns a slightly less concentrated residue to the previous stage.
There are currently two generic commercial methods employed internationally for enrichment: gaseous diffusion (referred to as first generation) and gas centrifuge (second generation), which consumes only 6% as much energy as gaseous diffusion. These both make use of the volatility of UF6.
ClF3 has been investigated as a high-performance storable oxidizer in rocket propellant systems. Handling concerns, however, prevented its use.
$ClF_3$ is Hypergolic
Hypergolic means explodes on contact with no need for any activator. One observer made the following comment about $ClF_3$:
"It is, of course, extremely toxic, but that's the least of the problem. It is hypergolic* with every known fuel, and so rapidly hypergolic that no ignition delay has ever been measured. It is also hypergolic with such things as cloth, wood, and test engineers, not to mention asbestos, sand, and water with which it reacts explosively. It can be kept in some of the ordinary structural metals-steel, copper, aluminium, etc.-because of the formation of a thin film of insoluble metal fluoride which protects the bulk of the metal, just as the invisible coat of oxide on aluminium keeps it from burning up in the atmosphere. If, however, this coat is melted or scrubbed off, and has no chance to reform, the operator is confronted with the problem of coping with a metal-fluorine fire. For dealing with this situation, I have always recommended a good pair of running shoes."
It is believed that prior to and during World War II, ClF3 code named N-stoff ("substance N") was being stockpiled in Germany for use as a potential incendiary weapon and poison gas. The plant was captured by the Russians in 1944, but there is no evidence that the gas was actually ever used during the war.
XY5 interhalogens
• Chlorine pentafluoride (ClF5) is a colorless gas, made by reacting chlorine trifluoride with fluorine at high temperatures and high pressures. It reacts violently with water and most metals and nonmetals.
• Bromine pentafluoride (BrF5) is a colorless fuming liquid, made by reacting bromine trifluoride with fluorine at 200° C. It is physically stable, but reacts violently with water and most metals and nonmetals.
• Iodine pentafluoride (IF5) is a colorless liquid, made by reacting iodine pentoxide with fluorine, or iodine with silver fluoride. It is highly reactive, even slowly with glass. It reacts with elements, oxides and carbon halides. The molecule has the form of a tetragonal pyramid.
• Primary amines react with iodine pentafluoride to form nitriles after hydrolysis with water. $R-CH_2-NH_2 → R-CN \nonumber$
XY7 interhalogens
• Iodine heptafluoride (IF7) is a colorless gas. It is made by reacting the pentafluoride with fluorine. IF7 is chemically inert, having no lone pair of electrons in the valency shell; in this it resembles sulfur hexafluoride. The molecule is a pentagonal bipyramid. This compound is the only interhalogen compound possible where the larger atom is carrying seven of the smaller atoms.
• All attempts to form bromine heptafluoride have met with failure; instead, bromine pentafluoride and fluorine gas are produced.
Diatomic Interhalogens (AX)
The interhalogens of form XY have physical properties intermediate between those of the two parent halogens. The covalent bond between the two atoms has some ionic character, the less electronegative element, X, being oxidized and having a partial positive charge. Most combinations of F, Cl, Br and I are known, but not all are stable.
• Chlorine monofluoride (ClF): The lightest interhalogen compound, ClF is a colorless gas with a normal boiling point of -100 °C.
• Bromine monofluoride (BrF): BrF has not been obtained in a pure form; it dissociates into the trifluoride and free bromine.
• Iodine monofluoride (IF): IF is unstable and decomposes at 0 C, disproportionating into elemental iodine and iodine pentafluoride.
• Bromine monochloride (BrCl): A red-brown gas with a boiling point of 5 °C.
• Iodine monochloride (ICl): Red transparent crystals which melt at 27.2 °C to form a choking brownish liquid (similar in appearance and weight to bromine). It reacts with HCl to form the strong acid HICl2. The crystal structure of iodine monochloride consists of puckered zig-zag chains, with strong interactions between the chains.
• Iodine monobromide (IBr): Made by direct combination of the elements to form a dark red crystalline solid. It melts at 42 °C and boils at 116 °C to form a partially dissociated vapor.
Tetra-atomic Interhalogens (AX3)
• Chlorine trifluoride (ClF3) is a colorless gas which condenses to a green liquid and freezes to a white solid. It is made by reacting chlorine with an excess of fluorine at 250 °C in a nickel tube. It reacts more violently than fluorine, often explosively. The molecule is planar and T-shaped. It is used in the manufacture of uranium hexafluoride.
• Bromine trifluoride (BrF3) is a yellow-green liquid which conducts electricity and ionizes to form [BrF2+] + [BrF4-]. It reacts with many metals and metal oxides to form similar ionized entities; with some others it forms the metal fluoride plus free bromine and oxygen. It is used in organic chemistry as a fluorinating agent. It has the same molecular shape as chlorine trifluoride.
• Iodine trifluoride (IF3) is a yellow solid which decomposes above -28 °C. It can be synthesized from the elements, but care must be taken to avoid the formation of IF5. F2 attacks I2 to yield IF3 at -45 °C in CCl3F. Alternatively, at low temperatures, the fluorination reaction I2+ 3XeF2 --> 2IF3 + 3Xe can be used. Not much is known about iodine trifluoride as it is so unstable.
• Iodine trichloride (ICl3) forms lemon yellow crystals which can be melted under pressure to a brown liquid. It can be made from the elements at low temperature, or from iodine pentoxide and hydrogen chloride. It reacts with many metal chlorides to form tetrachloriodides, and hydrolyses in water. The molecule is a planar dimer, with each iodine atom surrounded by four chlorine atoms.
Hexa-atomic Interhalogens (AX5)
• Chlorine pentafluoride (ClF5) is a colorless gas, made by reacting chlorine trifluoride with fluorine at high temperatures and high pressures. It reacts violently with water and most metals and nonmetals.
• Bromine pentafluoride (BrF5) is a colorless fuming liquid, made by reacting bromine trifluoride with fluorine at 200Å C. It is physically stable, but reacts violently with water and most metals and nonmetals.
• Iodine pentafluoride (IF5) is a colorless liquid, made by reacting iodine pentoxide with fluorine, or iodine with silver fluoride. It is highly reactive, even slowly with glass. It reacts with elements, oxides and carbon halides.The molecule has the form of a tetragonal pyramid.
Octa-atomic interhalogens (AX7)
• Iodine heptafluoride (IF7) is a colourless gas. It is made by reacting the pentafluoride with fluorine. IF7 is chemically inert, having no lone pair of electrons in the valency shell; in this it resembles sulfur hexafluoride. The molecule is a pentagonal bipyramid. This compound is the only interhalogen compound possible where the larger atom is carrying seven of the smaller atoms
• All attempts to form bromine heptafluoride (BrF7) have failed and instead produce bromine pentafluoride (BrF5) gas.
Summary
All interhalogens are volatile at room temperature. All are polar due to differences in their electronegativity. These are usually covalent liquids or gases due to small electronegativity differences among them. Some compounds partially ionize in solution. For example: $\ce{2 ICl \rightarrow I^{+} + ICl_2^{-}} \nonumber$ Interhalogen compounds are more reactive than normal halogens, except for fluorine. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/08%3A_Chemistry_of_the_Main_Group_Elements/8.13%3A_The_Halogens/8.13.02%3A_Chemical_Properties_of_the_Halogens/8.13.2.03%3A_Interhalogens.txt |
This page describes reactions of the halogens that do not fall under the other categories in other pages in this section. All the reactions described here are redox reactions.
Reactions with hydrogen
The following examples illustrate the decrease in reactivity of the halogens down Group 7.
Fluorine combines explosively with hydrogen even under cold, dark conditions, evolving hydrogen fluoride gas.
A mixture of chlorine and hydrogen explodes if exposed to sunlight or a flame, producing hydrogen chloride gas. This reaction can be controlled by lighting a jet of hydrogen and then lowering it into a gas jar of chlorine. The hydrogen burns at a slower, constant rate, and hydrogen chloride gas is formed as before.
Bromine vapor and hydrogen combine with a mild explosion when ignited. Hydrogen bromide gas is formed.
Iodine and hydrogen combine only partially even on constant heating. An equilibrium exists between the hydrogen and the iodine and hydrogen iodide gas.
Each of these reactions has an equation of the form:
$H_2 + X_2 \rightarrow 2HX \nonumber$
A minor exception is made for iodine: the single arrow is replaced with a reversible sign.
Reactions with phosphorus
Care must be taken when analyzing the rates of these reactions; analogous reactions must be compared. For example, it is nonsensical to compare the rate at which phosphorus reacts with gaseous chlorine with the rate at which it reacts with liquid bromine. There is more contact between phosphorus and liquid bromine than between phosphorus and gaseous chlorine.
The formation of trihalides, PX3
All halogens react with phosphorus to form, in the first instance, phosphorus(III) halides of the form PX3.
There are two common forms of phosphorus: white phosphorus (sometimes called yellow phosphorus) and red phosphorus. White phosphorus is more reactive than red phosphorus. This video on YouTube shows the reaction between red phosphorus and bromine. This is a violent reaction under cold conditions, and white phosphorus behaves even more dramatically.
When writing the equations for these reactions, it is important to remember that white phosphorus is molecular, consisting of P4 molecules, whereas red phosphorus is polymeric, indicated by the symbol P. The reaction for white phosphorus and bromine is as follows:
$P_4 + 6Br_2 \rightarrow 4PBr_3 \nonumber$
The red phosphorus equation is shown below:
$2P + 3Br_2 \rightarrow 2PBr_3 \nonumber$
The formation of pentahalides, PX5
In excess chlorine or bromine, phosphorus reacts to form phosphorus(V) chloride or bromide. Most simply, using white phosphorus:
$P_4 + 10Cl_2 \rightarrow 4PCl_5 \nonumber$
The reaction between phosphorus(III) chloride and phosphorus(V) chloride is reversible:
$PCl_3 + Cl_2 \rightleftharpoons PCl_5 \nonumber$
An excess of chlorine pushes this equilibrium to the right. Phosphorus does not form a pentaiodide, in contrast; this is likely because five large iodine atoms cannot physically fit around the central phosphorus atom.
Reactions with sodium
All halogens react with sodium to produce sodium halides. A common reaction between hot sodium and chlorine gas produces a bright orange flame and white sodium chloride.
$2Na + Cl_2 \rightarrow 2NaCl \nonumber$
Hot sodium will also burn in bromine or iodine vapor to produce sodium bromide or sodium iodide. Each of these reactions produces an orange flame and a white solid.
Reactions with iron
With the exception of iodine, iron burns in halogen vapor, forming iron(III) halides. Iodine is less reactive, and produces iron(II) iodide.
Fluorine
Cold iron wool burns in cold fluorine to give iron(III) fluoride. Anhydrous iron(III) fluoride is described as either white or pale green. A standard inorganic chemistry textbook by Cotton and Wilkinson describes it as white. The reaction is given below:
$2Fe + 3F_2 \rightarrow 2FeF_3 \nonumber$
This is a rapid reaction, in which the iron burns and is oxidized to an iron(III) compound—in other words, from an oxidation state of zero in the elemental metal to an oxidation state of +3 in the iron(III) compound.
Chlorine
Chlorine gas in contact with hot iron forms iron(III) chloride. Anhydrous iron(III) chloride forms black crystals; any trace of water present in the apparatus or in the chlorine reacts with the crystals, turning them reddish-brown. The equation for this reaction is given below:
$2Fe + 3Cl_2 \rightarrow 2FeCl_3 \nonumber$
The iron is again oxidized from a state of zero to +3.
Bromine
Bromine vapor passed over hot iron triggers a similar, slightly less vigorous reaction, shown below; iron(III) bromide is produced. Anhydrous iron(III) bromide usually appears as a reddish-brown solid.
$2Fe + 3Br_2 \rightarrow 2FeBr_3 \nonumber$
In this reaction the iron is again oxidized to a +3 state.
Iodine
The reaction between hot iron and iodine vapor produces gray iron(II) iodide, and is much less vigorous. This reaction, the equation for which is given below, is difficult to carry out because the product is always contaminated with iodine.
$Fe + 2I_2 \rightarrow FeI_2 \nonumber$
Iodine is only capable of oxidizing iron to the +2 oxidation state.
Reactions with solutions containing iron(II) ions
Only the reactions of chlorine, bromine, and iodine can be considered. Aqueous fluorine is very reactive with water. Chlorine and bromine are strong enough oxidizing agents to oxidize iron(II) ions to iron(III) ions. In the process, chlorine is reduced to chloride ions, bromine to bromide ions.
This process is easiest to visualize with ionic equations:
For the bromine equation, Br is substituted for Cl.
The pale green solution containing the iron(II) ions turns into a yellow or orange solution containing iron(III) ions. Iodine is not a strong enough oxidizing agent to oxidize iron(II) ions, so there is no reaction. In fact, the reverse reaction proceeds. Iron(III) ions are strong enough oxidizing agents to oxidize iodide ions to iodine as shown:
$2Fe^{3+} + 2I^- \rightarrow 2Fe^{2+} + I_2 \nonumber$
Reactions with sodium hydroxide solution
Once again, only chlorine, bromine, and iodine are considered.
The reaction of chlorine with cold sodium hydroxide solution
Chlorine and cold, dilute sodium hydroxide react as follows:
$2NaOH + Cl_2 \rightarrow NaCl + NaClO + H_2O \nonumber$
NaClO (sometimes written as NaOCl) symbolizes sodium chlorate(I). The traditional name for this compound is sodium hypochlorite; the solution on the product side of the equation is commonly sold as bleach.
Consider this reaction in terms of oxidation states. Chlorine displays an obvious state change from its elemental form to ionic compounds. The oxidation numbers for each element are shown below:
Chlorine is the only element that changes oxidation state—it is both oxidized and reduced. One atom is reduced because its oxidation state has decreased; the other is oxidized. This is a good example of a disproportionation reaction, a reaction in which a single substance is both oxidized and reduced.
The reaction of chlorine with hot sodium hydroxide solution
Chlorine reacts with hot, concentrated sodium hydroxide as follows:
$6NaOH + 3Cl_2 \rightarrow 5NaCl + NaClO_3 + 3H_2O \nonumber$
The product formed is sodium chlorate(V) - NaClO3. As before, the oxidation states of each element are calculated. Once again, the only change is in chlorine, from 0 in the chlorine molecules on the reactant side to -1 (in the NaCl) and +5 (in the NaClO3). This is another example of a disproportionation reaction.
Balancing equations for these reactions
The first equation is simple to balance. The second one is more difficult; oxidation states are used to derive it.
The two main products of the reaction are NaCl and NaCIO3, so the reaction can be tentatively written as follows:
$NaOH + Cl_2 \rightarrow NaCl + NaClO_3 + ? \nonumber$
In its conversion to NaCl, the oxidation state of the chlorine decreases from 0 to -1. When converted to NaClO3, it increases from 0 to +5. The positive and negative oxidation state changes must cancel out, so for every NaClO3 formed, there must be 5 NaCl:
$NaOH + Cl_2 \rightarrow 5NaCl + NaClO_3 + ? \nonumber$
Now it is a simple task to balance the sodium and the chlorine atoms, after which there are enough hydrogen and oxygen atoms to make 3H2O.
The reactions involving bromine and iodine
These are essentially similar to that of chlorine; the difference lies in the reaction temperatures. The tendency to form the ion with the halogen in the +5 oxidation state increases rapidly down the group.
Bromine and sodium hydroxide solution
For bromine, the formation of the sodium bromate(V) happens at around room temperature. Sodium bromate(I) must be formed at about 0°C.
Iodine and sodium hydroxide solution
In this case, sodium iodate(V) is formed at any temperature. Cotton and Wilkinson (Advanced Inorganic Chemistry 3rd edition page 477) say that the iodate(I) ion is unknown in solution.
Contributors and Attributions
Jim Clark (Chemguide.co.uk) | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/08%3A_Chemistry_of_the_Main_Group_Elements/8.13%3A_The_Halogens/8.13.02%3A_Chemical_Properties_of_the_Halogens/8.13.2.04%3A_More_Reactions_of_Halogens.txt |
Consider a reaction between one halogen—chlorine, for example—and the ions of another—iodide, in this case. The iodide ions are dissolved from a salt such as sodium iodide or potassium iodide. The sodium or potassium ions are spectator ions and therefore irrelevant to the reaction, which proceeds as follows:
$Cl_2 + 2I^- \rightarrow 2Cl^- + I_2 \nonumber$
• The iodide ions lose electrons to form iodine molecules; they are oxidized.
• The chlorine molecules gain electrons to form chloride ions. They are reduced.
This is a redox reaction in which chlorine is acting as an oxidizing agent. The driving force force this reaction is straightforward to identify from the table of Standard Reduction Potentials (Table P2).
Contributors and Attributions
• Jim Clark (ChemGuide)
8.13.2.06: Testing for Halide Ions
This page discusses the tests for halide ions (fluoride, chloride, bromide and iodide) using silver nitrate and ammonia.
Using silver nitrate solution
This test is carried out in a solution of halide ions. The solution is acidified by adding dilute nitric acid. The nitric acid reacts with, and removes, other ions that might also form precipitates with silver nitrate. Silver nitrate solution is then added, and the halide can be identified from the following products:
ion present observation
F- no precipitate
Cl- white precipitate
Br- very pale cream precipitate
I- very pale yellow precipitate
The chloride, bromide and iodide precipitates are shown in the photograph:
The chloride precipitate is easily identified, but the other two are quite similar to each other. They can only be differentiated in a side-by-side comparison. All the precipitates change color if they are exposed to light, taking on gray or purple tints. The absence of a precipitate with fluoride ions is unhelpful unless it is known that a halogen is present; otherwise, it indicates that there is no chloride, bromide, or iodide.
The chemistry of the test
The precipitates are insoluble silver halides: silver chloride, silver bromide or silver iodide. The formation of these is illustrated in the following equations:
$Ag^+_{aq} + Cl^-_{(aq)} \rightarrow AgCl_{(s)} \nonumber$
$Ag^+_{aq} + Br^-_{(aq)} \rightarrow AgBr_{(s)} \nonumber$
$Ag^+_{aq} + I^-_{(aq)} \rightarrow AgI_{(s)} \nonumber$
Silver fluoride is soluble, so no precipitate is formed.
$Ag^+_{aq} + F^-_{(aq)} \rightarrow Ag^+_{aq} + F^-_{(aq)} \nonumber$
Confirming the precipitate using ammonia solution
Ammonia solution is added to the precipitates.
original precipitate Observation
AgCl precipitate dissolves to give a colorless solution
AgBr precipitate is almost unchanged using dilute ammonia solution, but dissolves in concentrated ammonia solution to give a colorless solution
AgI precipitate is insoluble in ammonia solution of any concentration
There are no absolutely insoluble ionic compounds. A precipitate forms if the concentrations of the ions in solution in water exceed a certain value, unique to every compound. This value is known as the solubility product. For the silver halides, the solubility product is given by the expression:
$K_{sp} = [Ag^+][X^-] \nonumber$
The square brackets indicate molar concentrations, with units of mol L-1.
• If the product of the concentrations of ions is less than the solubility product, no precipitate is formed.
• If the product of the concentrations exceeds this value, a precipitate is formed.
Essentially, the product of the ionic concentrations is never greater than the solubility product value. Enough solid is always precipitated to lower the ionic product to the solubility product. The table below lists solubility products from silver chloride to silver iodide (a solubility product for silver fluoride cannot be reported because it is too soluble).
Ksp (mol2dm-6)
AgCl 1.8 x 10-10
AgBr 7.7 x 10-13
AgI 8.3 x 10-17
The compounds are all quite insoluble, but become even less so down the group.
The purpose of ammonia
The ammonia combines with silver ions to produce a complex ion called the diamminesilver(I) ion, [Ag(NH3)2]+. This is a reversible reaction, but the complex is very stable, and the position of equilibrium lies well to the right. The equation for this reaction is given below:
A solution in contact with one of the silver halide precipitates contains a very small concentration of dissolved silver ions. The effect of adding the ammonia is to lower this concentration still further. If the adjusted silver ion concentration multiplied by the halide ion concentration is less than the solubility product, some precipitate dissolves to restore equilibrium.
This occurs with silver chloride, and with silver bromide if the ammonia is concentrated. The more concentrated ammonia pushes the equilibrium even further to the right, lowering the silver ion concentration even more.
The silver iodide is so insoluble that ammonia cannot lower the silver ion concentration enough for the precipitate to dissolve.
An alternative test using concentrated sulfuric acid
Adding concentrated sulfuric acid to a solid sample of one of the halides gives the following results:
ion present observation
F- steamy acidic fumes (of HF)
Cl- steamy acidic fumes (of HCl)
Br- steamy acidic fumes (of HBr) contaminated with brown bromine vapor
I- some HI fumes with large amounts of purple iodine vapor and a red compound in the reaction vessel
The only possible confusion is between a fluoride and a chloride—they behave identically under these conditions. They can be distinguished by dissolving the original solid in water and then testing with silver nitrate solution. The chloride gives a white precipitate; the fluoride produces none. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/08%3A_Chemistry_of_the_Main_Group_Elements/8.13%3A_The_Halogens/8.13.02%3A_Chemical_Properties_of_the_Halogens/8.13.2.05%3A_Oxidizing_Ability_of_the_Group_17_Elements.txt |
This page discusses the acidity of the hydrogen halides: hydrogen fluoride, hydrogen chloride, hydrogen bromide and hydrogen iodide. It begins by describing their physical properties and synthesis and then explains what happens when they react with water to make acids such as hydrofluoric acid and hydrochloric acid.
Physical properties
The hydrogen halides are colorless gases at room temperature, producing steamy fumes in moist air. The boiling points of these compounds are shown in the figure below:
Hydrogen fluoride has an abnormally high boiling point for a molecule of its size (293 K or 20°C), and can condense under cool conditions. This is due to the fact that hydrogen fluoride can form hydrogen bonds. Because fluorine is the most electronegative of all the elements, the fluorine-hydrogen bond is highly polarized. The hydrogen atom carries a high partial positive charge (δ+); the fluorine is fairly negatively charged (δ-).
In addition, each fluorine atom has 3 lone pairs of electrons. Fluorine's outer electrons are at the n=2 level, and the lone pairs represent small, highly charged regions of space. Hydrogen bonds form between the δ+ hydrogen on one HF molecule and a lone pair on the fluorine of another one.The figure below illustrates this association:
The other hydrogen halides do not form hydrogen bonds because the larger halogens are not as electronegative as fluorine; therefore, the bonds are less polar. In addition, their lone pairs are at higher energy levels, so the halogen does not carry such an intensely concentrated negative charge; therefore, other hydrogen atoms are not attracted as strongly.
Making hydrogen halides
There are several ways of synthesizing hydrogen halides; the method considered here is the reaction between an ionic halide, like sodium chloride, and an acid like concentrated phosphoric(V) acid, H3PO4, or concentrated sulfuric acid.
Making hydrogen chloride
When concentrated sulfuric acid is added to sodium chloride under cold conditions, the acid donates a proton to a chloride ion, forming hydrogen chloride. In the gas phase, it immediately escapes from the system.
$Cl^- + H_2SO_4 \rightarrow HCl + HSO_4^- \nonumber$
The full equation for the reaction is:
$NaCl + H_2SO_4 \rightarrow HCl + NaHSO_4 \nonumber$
Sodium bisulfate is also formed in the reaction. Concentrated phosphoric(V) acid reacts similarly, according to the following equation:
$Cl^- + H_3PO_4 \rightarrow HCl + H_2PO_4^- \nonumber$
The full ionic equation showing the formation of the salt, sodium biphosphate(V), is given below:
$NaCl + H_3PO_4 \rightarrow HCl + NaH_2PO_4 \nonumber$
Making other hydrogen halides
All hydrogen halides can be formed by the same method, using concentrated phosphoric(V) acid.
Concentrated sulfuric acid, however, behaves differently. Hydrogen fluoride can be made with sulfuric acid, but hydrogen bromide and hydrogen iodide cannot.
The problem is that concentrated sulfuric acid is an oxidizing agent, and as well as producing hydrogen bromide or hydrogen iodide, some of the halide ions are oxidized to bromine or iodine. Phosphoric acid does not have this ability because it is not an oxidant.
The acidity of the hydrogen halides
Hydrogen chloride as an acid
By the Bronsted-Lowry definition of an acid as a proton donor, hydrogen chloride is an acid because it transfers protons to other species. Consider its reaction with water.
Hydrogen chloride gas is soluble in water; its solvated form is hydrochloric acid. Hydrogen chloride fumes in moist air are caused by hydrogen chloride reacting with water vapor in the air to produce a cloud of concentrated hydrochloric acid.
A proton is donated from the hydrogen chloride to one of the lone pairs on a water molecule.
A coordinate (dative covalent) bond is formed between the oxygen and the transferred proton.
The equation for the reaction is the following:
$H_2O + HCl \rightarrow H_3O^+ + Cl^- \nonumber$
The H3O+ ion is the hydroxonium ion (also known as the hydronium ion or the oxonium ion). This is the normal form of protons in water; sometimes it is shortened to the proton form, H+(aq), for brevity.
When hydrogen chloride dissolves in water (to produce hydrochloric acid), almost all the hydrogen chloride molecules react in this way. Hydrochloric acid is therefore a strong acid. An acid is strong if it is fully ionized in solution.
Hydrobromic acid and hydriodic acid as strong acids
Hydrogen bromide and hydrogen iodide dissolve in (and react with) water in exactly the same way as hydrogen chloride does. Hydrogen bromide forms hydrobromic acid; hydrogen iodide gives hydriodic acid. Both of these are also strong acids.
Hydrofluoric acid as an exception
By contrast, although hydrogen fluoride dissolves freely in water, hydrofluoric acid is only a weak acid; it is similar in strength to organic acids like methanoic acid. The complicated reason for this is discussed below.
The bond enthalpy of the H-F bond
Because the fluorine atom is so small, the bond enthalpy (bond energy) of the hydrogen-fluorine bond is very high. The chart below gives values for all the hydrogen-halogen bond enthalpies:
bond enthalpy
(kJ mol-1)
H-F +562
H-Cl +431
H-Br +366
H-I +299
In order for ions to form when the hydrogen fluoride reacts with water, the H-F bond must be broken. It would seem reasonable to say that the relative reluctance of hydrogen fluoride to react with water is due to the large amount of energy needed to break that bond, but this explanation does not hold.
The energetics of the process from HX(g) to X-(aq)
The energetics of this sequence are of interest:
All of these terms are involved in the overall enthalpy change as you convert HX(g) into its ions in water.
However, the terms involving the hydrogen are the same for every hydrogen halide. Only the values for the red terms in the diagram need be considered. The values are tabulated below:
bond enthalpy of HX
(kJ mol-1)
electron affinity of X
(kJ mol-1)
hydration enthalpy of X-
(kJ mol-1)
sum of these
(kJ mol-1)
HF +562 -328 -506 -272
HCl +431 -349 -364 -282
HBr +366 -324 -335 -293
HI +299 -295 -293 -289
There is virtually no difference in the total HF and HCl values.
The large bond enthalpy of the H-F bond is offset by the large hydration enthalpy of the fluoride ion. There is a very strong attraction between the very small fluoride ion and the water molecules. This releases a lot of heat (the hydration enthalpy) when the fluoride ion becomes wrapped in water molecules.
Other attractions in the system
The energy terms considered previously have concerned HX molecules in the gas phase. To reach a more correct explanation, the molecules must first be considered as unreacted aqueous HX molecules. The equation for this is given below:
The equation is incorporated into an improved energy cycle as follows:
Unfortunately, values for the first step in the reaction are not readily available. However, in each case, the initial separation of the HX from water molecules is endothermic. Energy is required to break the intermolecular attractions between the HX molecules and water.
That energy is much greater for hydrogen fluoride because it forms hydrogen bonds with water. The other hydrogen halides experience only the weaker van der Waals dispersion forces or dipole-dipole attractions.
The overall enthalpy changes (including all the stages in the energy cycle) for the reactions are given in the table below:
$HX(aq) + H_2O (l) \rightarrow H_3O^+ (aq) + X^- (aq) \nonumber$
enthalpy change
(kJ mol-1)
HF -13
HCl -59
HBr -63
H-I -57
The enthalpy change for HF is much smaller in magnitude than that for the other three hydrogen halides, but it is still negative exothermic change. Therefore, more information is needed to explain why HF is a weak acid.
Entropy and free energy considerations
The free energy change, not the enthalpy change, determines the extent and direction of a reaction.
Free energy change is calculated from the enthalpy change, the temperature of the reaction and the entropy change during the reaction.
For simplicity, entropy can be thought of as a measure of the amount of disorder in a system. Entropy is given the symbol S. If a system becomes more disordered, then its entropy increases. If it becomes more ordered, its entropy decreases.
The key equation is given below:
In simple terms, for a reaction to happen, the free energy change must be negative. But more accurately, the free energy change can be used to calculate a value for the equilibrium constant for a reaction using the following expression:
The term Ka is the equilibrium constant for the reaction below:
$HX(aq) + H_2O(l) \rightleftharpoons H_3O^+(aq) + X^-(aq) \nonumber$
The values for TΔS (needed to calculate ΔG) for the four reactions at a temperature of 298 K are tabulated below:
TS
(kJ mol-1)
HF -29
HCl -13
HBr -4
H-I +4
Notice that at the top of the group, the systems become more ordered when the HX reacts with the water. The entropy of the system (the amount of disorder) decreases, particularly for the hydrogen fluoride.
The reason for this is that the very strong attraction between H3O+ and F-(aq) imposes a lot of order on the system, as does the attraction between the water molecules and the various ions present. These attractions are each greatest for the small fluoride ions.
The total effect on the free energy change, and therefore the value of the equilibrium constant, can now be considered. These values are calculated in the following table:
H
(kJ mol-1)
TS
(kJ mol-1)
G
(kJ mol-1)
Ka
(mol dm-3)
HF -13 -29 +16 1.6 x 10-3
HCl -59 -13 -46 1.2 x 108
HBr -63 -4 -59 2.2 x 1010
HI -57 +4 -61 5.0 x 1010
The values for these estimated equilibrium constants for HCl, HBr and HI are so high that the reaction can be considered "one-way". The ionization is virtually 100% complete. These are all strong acids, increasing in strength down the group.
By contrast, the estimated Ka for hydrofluoric acid is small. Hydrofluoric acid only ionizes to a limited extent in water. Therefore, it is a weak acid.
The estimated value for HF in the table can be compared to the experimental value:
• Experimental value: 5.6 x 10-4 mol dm-3
• Estimated value: 1.6 x 10-3 mol dm-3
These values differ by an order of magnitude, but because of the logarithmic relationship between the free energy and the equilibrium constant, a very small change in ΔG has a very large effect on Ka.
To have the values in close agreement, ΔG would have to increase from +16 to +18.5 kJ mol-1. Given the uncertainty in the values used to calculate ΔG, the difference between the calculated value and the experimental value could easily fall within this range.
Summary: Why is hydrofluoric acid a weak acid?
The two main factors are:
• Entropy decreases dramatically when the hydrogen fluoride reacts with water. This is particularly noticeable with hydrogen fluoride because the attraction of the small fluoride ions produced imposes significant order on the surrounding water molecules and nearby hydronium ions. The effect decreases with larger halide ions.
• Very strong hydrogen bonding exists between the hydrogen fluoride molecules and water molecules. This costs a large amount of energy to break. This effect does not occur in the other hydrogen halides.
Contributors and Attributions
Jim Clark (Chemguide.co.uk) | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/08%3A_Chemistry_of_the_Main_Group_Elements/8.13%3A_The_Halogens/8.13.02%3A_Chemical_Properties_of_the_Halogens/8.13.2.07%3A_The_Acidity_of_the_Hydrogen_Halides.txt |
Fluorine (F) is the first element in the Halogen group (group 17) in the periodic table. Its atomic number is 9 and its atomic weight is 19, and it's a gas at room temperature. It is the most electronegative element, given that it is the top element in the Halogen Group, and therefore is very reactive. It is a nonmetal, and is one of the few elements that can form diatomic molecules (F2). It has 5 valence electrons in the 2p level. Its electron configuration is 1s22s22p5. It will usually form the anion F- since it is extremely electronegative and a strong oxidizing agent. Fluorine is a Lewis acid in weak acid, which means that it accepts electrons when reacting. Fluorine has many isotopes, but the only stable one found in nature is F-19.
Quick Reference Table
Symbol F
Atomic Number 9
Group 17 (Halogens)
Electron Configuration 1s22s22p5
Atomic Weight 18.998 g
Density 1.7 g/L
Melting Point -219.62oC
Boiling Point -188.12oC
Critical Point 144.13K, 5.172 MPa
Oxidation States -1
Electronegativity 3.98
Stable Isotopes F-19
Brief History
In the late 1600's minerals which we now know contain fluorine were used in etching glass. The discovery of the element was prompted by the search for the chemical substance which was able to attack glass (it is HF, a weak acid). The early history of the isolation and work with fluorine and hydrogen fluoride is filled with accidents since both are extremely dangerous. Eventually, electrolysis of a mixture of KF and HF (carefully ensuring that the resulting hydrogen and fluorine would not come in contact) in a platinum apparatus yielded the element.
Fluorine was discovered in 1530 by Georgius Agricola. He originally found it in the compound Fluorspar, which was used to promote the fusion of metals. It was used in this application until 1670, when Schwanhard discovered its usefulness in etching glass. Pure fluorine (from the Latin fluere, for "flow") was not isolated until 1886 by Henri Moissan, burning and even killing many scientists along the way. It has many uses today; a particular one was being used in the Manhattan project to help create the first nuclear bomb.
Electronegativity of Fluorine
Fluorine is the most electronegative element on the periodic table, which means that it is a very strong oxidizing agent and accepts other elements' electrons. Fluorine's atomic electron configuration is 1s22s22p5 (see Figure 2).
Fluorine is the most electronegative element because it has 5 electrons in its 2p shell. The optimal electron configuration of the 2p orbital contains 6 electrons, so since fluorine is so close to ideal electron configuration, the electrons are held very tightly to the nucleus. The high electronegativity of fluorine explains its small radius because the positive protons have a very strong attraction to the negative electrons, holding them closer to the nucleus than the bigger and less electronegative elements do.
Reactions of Fluorine
Because of its reactivity, elemental fluorine is never found in nature, and no other chemical element can displace fluorine from its compounds. Fluorine bonds with almost any element, both metals and nonmetals, because it is a very strong oxidizing agent. It is very unstable and reactive since it is so close to its ideal electron configuration. It forms covalent bonds with nonmetals, and since it is the most electronegative element, is always going to be the element that is reduced. It can also form a diatomic element with itself ($F_2$), or covalent bonds where it oxidizes other halogens ($ClF$, $ClF_3$, $ClF_5$). It will react explosively with many elements and compounds such as hydrogen and water. Elemental fluorine is slightly basic, which means that when it reacts with water it forms $OH^-$.
$2F_2+2H_2O \rightarrow O_2+4HF \tag{1}$
When combined with hydrogen, fluorine forms hydrofluoric acid ($HF$), which is a weak acid. This acid is very dangerous and when dissociated can cause severe damage to the body because while it may not be painful initially, it passes through tissues quickly and can cause deep burns that interfere with nerve function.
$HF+H_2O \rightarrow H_3O^++F^- \tag{2}$
There are also some organic compounds made of fluorine, ranging from nontoxic to highly toxic. Fluorine forms covalent bonds with carbon, which sometimes form into stable aromatic rings. When carbon reacts with fluorine, the reaction is complex and forms a mixture of $CF_4$, $C_2F_6$, and $C_5F_{12}$.
$C_{(s)} + F_{2(g)} \rightarrow CF_{4(g)} + C_2F_6 + C_5F_{12} \tag{3}$
Fluorine reacts with oxygen to form $OF_2$ because fluorine is more electronegative than oxygen. The reaction goes:
$2F_2 + O_2 \rightarrow 2OF_2 \tag{4}$
Fluorine is so electronegative that sometimes it will even form molecules with noble gases like Xenon, such as the molecule xenon difluoride, $XeF_2$.
$Xe + F_2 \rightarrow XeF_2 \tag{5}$
Fluorine also forms strong ionic compounds with metals. Some common ionic reactions of fluorine are:
$F_2 + 2NaOH \rightarrow O_2 + 2NaF +H_2 \tag{6}$
$4F_2 + HCl + H_2O \rightarrow 3HF + OF_2 + ClF_3 \tag{7}$
$F_2 + 2HNO_3 \rightarrow 2NO_3F + H_2 \tag{8}$
Applications of Fluorine
Compounds of fluorine are present in fluoridated toothpaste and in many municipal water systems, where they help to prevent tooth decay. And, of course, fluorocarbons such as Teflon have made a major impact on life in the 20th century. There are many applications of fluorine:
• Rocket fuels
• Polymer and plastics production
• Teflon and tefzel production
• When combined with oxygen, used as a refrigerator cooler
• Hydrofluoric acid used for glass etching
• Purify public water supplies
• Uranium production
• Air conditioning
Sources
Fluorine can either be found in nature or produced in a lab. To make it in a lab, compounds like potassium fluoride are put through electrolysis with hydrofluoric acid to create pure Fluorine and other compounds. It can be carried out with a variety of compounds, usually ionic ones involving fluorine and a metal. Fluorine can also be found in nature in various minerals and compounds. The two main compounds it can be found in are Fluorspar ($CaF_2$) and Cryolite ($Na_3AlF_6$).
Problems
(Highlight to view answers)
1. Q. What is the electron configuration of fluorine? of F-?
A. 1s22s22p5
1s22s22p6
2. Q. Is fluorine usually oxidized or reduced? explain.
A. Fluorine is usually reduced because it accepts an electron from other elements since it is so electronegative.
3. Q. What are some common uses of fluorine?
A. Toothpaste, plastics, rocket fuels, glass etching, etc.
4. Q. Does fluorine form compounds with nonmetals? if so, give two examples, one of them being of an oxide.
A. OF2, ClF
5. Q. What group is fluorine in? (include name of group and number)
A. 17, Halogens
Contributors and Attributions
• Rachel Feldman (University of California, Davis)
• Stephen R. Marsden | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/08%3A_Chemistry_of_the_Main_Group_Elements/8.13%3A_The_Halogens/8.13.03%3A_Chemistry_of_Fluorine_%28Z9%29.txt |
Chlorine is a halogen in group 17 and period 3. It is very reactive and is widely used for many purposes, such as as a disinfectant. Due to its high reactivity, it is commonly found in nature bonded to many different elements.
Chlorine, which is similar to fluorine but not as reactive, was prepared by Sheele in the late 1700's and shown to be an element by Davy in 1810. It is a greenish-yellow gas with a disagreeable odor (you can detect it near poorly balanced swimming pools). Its name comes from the Greek word chloros, meaning greenish-yellow. In high concentration it is quite toxic and was used in World War I as a poison gas.
Properties
Atomic Number 17
Atomic Weight 35.457
Electron Configuration [Na]3s23p5
1st Ionization Energy 1251 kJ/mol
Ionic Radius 181 pm
Density (Dry Gas) 3.2 g/L
Melting Point -101°C
Boiling Point -34.05°C
Specific Heat 0.23 g cal/g/°C
Heat of Vaporization 68 g cal/g
Heat of Fusion 22 g cal/g
Critical Temperature 114°C
Standard Electron Potential
$Cl_2 + 2e^- \rightarrow 2Cl^-$
1.358V
At room temperature, pure chlorine is a yellow-green gas. Chlorine is easily reduced, making it a good oxidation agent. By itself, it is not combustible, but many of its reactions with different compounds are exothermic and produce heat. Because chlorine is so highly reactive, it is found in nature in a combined state with other elements, such as NaCl (common salt) or KCl (sylvite). It forms strong ionic bonds with metal ions.
Like fluorine and the other members of the halogen family, chlorine is diatomic in nature, occurring as $Cl_2$ rather than Cl. It forms -1 ions in ionic compounds with most metals. Perhaps the best known compound of that type is sodium chloride, common table salt (NaCl).
Small amounts of chlorine can be produced in the lab by oxidizing $HCl$ with $MnO_2$. On an industrial scale, chlorine is produced by electrolysis of brines or even sea water. Sodium hydroxide (also in high demand) is a by-product of the process.
In addition to the ionic compounds that chlorine forms with metals, it also forms molecular compounds with non-metals such as sulfur and oxygen. There are four different oxides of the element. Hydrogen chloride gas (from which we get hydrochloric acid) is an important industrial product.
Reactions with Water
Usually, reactions of chlorine with water are for disinfection purposes. Chlorine is only slightly soluble in water, with its maximum solubility occurring at 49° F. After that, its solubility decreases until 212° F. At temperatures below that range, it forms crystalline hydrates (usually $Cl_2$) and becomes insoluble. Between that range, it usually forms hypochlorous acid ($HOCl$). This is the primary reaction used for water/wastewater disinfection and bleaching.
$Cl_2+H_2O \rightarrow HOCl + HCl \nonumber$
At the boiling temperature of water, chlorine decomposes water:
$2Cl_2+2H_2O \rightarrow 4HCl + O_2 \nonumber$
Reactions with Oxygen
Although chlorine usually has a -1 oxidation state, it can have oxidation states of +1, +3, +4, or +7 in certain compounds, such as when it forms oxoacids with the alkali metals.
Oxidation State Compound
+1 NaClO
+3 NaClO2
+5 NaClO3
+7 NaClO4
Reactions with Hydrogen
When H2 and Cl2 are exposed to sunlight or high temperatures, they react quickly and violently in a spontaneous reaction. Otherwise, the reaction proceeds slowly.
$H_2+Cl_2 \rightarrow 2HCl \nonumber$
HCl can also be produced by reacting chlorine with compounds containing hydrogen, such as hydrogen sulfide
Reactions with Halogens
Chlorine, like many of the other halogens, can form interhalogen compounds (examples include BrCl, ICl, ICl2). The heavier element in one of these compounds acts as the central atom. For chlorine, this occurs when it is bonded to fluorine in ClF, ClF3, and ClF5.
Reactions with Metals
Chlorine reacts with most metals and forms metal chlorides, with most of these compounds being soluble in water. Examples of insoluble compounds include $AgCl$ and $PbCl_2$. Gaseous or liquid chlorine usually does not have an effect on metals such as iron, copper, platinum, silver, and steel at temperatures below 230°F. At high temperatures, however, it reacts rapidly with many of the metals, especially if the metal is in a form that has a high surface area (such as when powdered or made into wires).
Example: Oxidizing Iron
Chlorine can oxidize iron:
$Cl_2+Fe \rightarrow FeCl_2 \nonumber$
Half Reactions:
$Fe \rightarrow Fe^{+2} +2e^- \nonumber$
$Cl_2+2e^- \rightarrow 2Cl^- \nonumber$
Isotopes
$\ce{^35}Cl$ and $\ce{^37}Cl$ are the two natural, stable isotopes of chlorine. $\ce{^36}Cl$, a radioactive isotope, occurs only in trace amounts as a result of cosmic rays in the atmosphere. Chlorine is usually a mixture of 75% $\ce{^35}Cl$ and 25% $\ce{^37}Cl$. Besides these isotopes, the other isotopes must be artificially produced. A table containing some common isotopes is found below:
Isotope Atomic Mass Half-Life
$\ce{^33}Cl$ 32.986 2.8 seconds
$\ce{^34}Cl$ 33.983 33 minutes
$\ce{^35}Cl$ 34.979 Stable ($\infty$)
$\ce{^36}Cl$ 35.978 400,000 years
$\ce{^37}Cl$ 35.976 Stable ($\infty$)
$\ce{^38}Cl$ 37.981 39 Minutes
Production and Uses
Chlorine is a widely used chemical with many applications.
Water Treatment
Chlorine is used in the disinfection (removal of harmful microorganisms) of water and wastewater. In the United States, it is almost exclusively used. Chlorine was first used to disinfect drinking water in 1908, using sodium hypochlorite (NaOCl):
$NaOCl+ H_2O \rightarrow HOCl+NaOH \nonumber$
Following widespread use of sodium hypochlorite to disinfect water, diseases caused by unclean water decreased greatly. Compared to other methods, it is effective at lower concentrations and is inexpensive.
Polyvinyl Chloride (PVC)
Polyvinyl chloride is a plastic which is widely manufactured throughout the globe, and is responsible for nearly a third of the world’s use of chlorine. It is usually manufactured by first taking EDC (ethylene dichloride) and then making it into a vinyl chloride, the basic unit for PVC. From then on, vinyl chloride monomers are linked together to form a polymer. PVC becomes malleable at high temperatures, making it flexible and ideal for many purposes from pipes to clothing. However, PVC is toxic. When in gaseous form and inhaled, it can cause damage to the lungs, the body’s blood circulation, and the nervous system. The production of PVC has many regulations surrounding it due to the many harmful effects that the plastic itself and the intermediates involved have on the environment and on human health.
Paper Bleaching
Paper is one of the most widely consumed products in the world. Before wood is made into a paper product, however, it must be turned into pulp (separated fibrous material). This pulp has a color that ranges from light to dark brown. Chlorine is used to bleach the pulp to turn it into a bright, white color, which makes it desirable for consumers. The process usually involves a number of steps, depending on the nature of the pulp.
Problems
1) Solve and balance the following equations
1. $H_2S + Cl_2 + H_2O \rightarrow$
2. $Sb + Cl_2 +H_2O \rightarrow$
2) Write the electron configuration for chlorine.
3) What is the molecular geometry of the following? (See Valence Bond Theory)
1. $ClO_2$
2. $ClF_5$
4) What are the naturally occurring chlorine isotopes?
5) When does chlorine have an oxidation state of +5?
Answers
1) Solve and balance the following equations:
1. H2S + 4Cl2 + 4H20 --> H2S04 + 8HCl
2. 2Sb + 3Cl2 --> 2SbCl3
2) The electron configuration of chlorine is: 1s22s22p63s23p5
3) What is the molecular geometry of the following?
1. $ClO_2$ -Bent or angular; the Cl is bonded to two ligands, has one lone pair and one unpaired electron.
2. $ClF_5$ -Square pyramid; the Cl is bonded to five ligands and has one lone pair.
4) The naturally occurring chlorine isotopes are chlorine-35 and chlorine-37. While chlorine-36 does occur naturally, it is radioactive and unstable.
5) Chlorine has an oxidation state of +5 when it reacts with oxoacids with the Alkali Metals.
References
1. Sconce, J.S. Chlorine: Its Manufacture, Properties, and Uses. Reinhold Corporation, 1962.
2. Stringer, Ruth, and Paul Johnston. Chlorine and the Enviroment. Norwell: Kluwer Academic, 2001.
3. Reynolds, Tom D. Unit Operations and Processes in Environmental Engineering. Brooks/Cole Engineering Division, a Division of Wadsworth Inc, 1982. 523-532
4. Davis, Stanley N., DeWayne Cecil, Marek Zreda, and Pankaj Sharma. "Chlorine-36 and the Initial Value Problem." Hydrogeology Journal 6.1 (1998): 104-14. SpringerLink. Web. 23 May 2010. <www.springerlink.com/content/3205uburlwx2x48g/>
5. Pettrucci, Ralph H. General Chemistry: Principles and Modern Applications. 9th. Upper Saddle River: Pearson Prentice Hall, 2007
Contributors and Attributions
• Judy Hsia (University of California, Davis)
8.13.04: Chemistry of Chlorine (Z17)
This page describes the manufacture of chlorine by the electrolysis of sodium chloride solution using a diaphragm cell and a membrane cell. Both cells rely on the same underlying chemistry, but differ in detail.
Background chemistry
Chlorine is manufactured by electrolyzing sodium chloride solution. This process generates three useful substances: chlorine, sodium hydroxide, and hydrogen.
The chemistry of the electrolysis process
Sodium chloride solution contains the following:
• sodium ions
• chloride ions
• protons (from the water)
• hydroxide ions (from the water)
The protons and hydroxide ions come from the equilibrium:
$H_2O (l) \rightleftharpoons H^+(aq) + OH^- (aq) \label{1}$
At any time, the concentration of protons or hydroxide ions is very small; the position of equilibrium lies well to the left.
At the anode
The negative ions, chloride and hydroxide, are attracted to the positively charged anode. It is easier to oxidize hydroxide ions to oxygen than to oxidize chloride ions to chlorine, but there are far more chloride ions arriving at the anode than hydroxide ions.
The major reaction at the anode is therefore:
$2Cl^-_{(aq)}\rightarrow Cl_{2(g)}+2e^- \label{2}$
Two chloride ions each give up an electron to the anode, and the atoms produced combine into chlorine gas. The chlorine is, however, contaminated with small amounts of oxygen because of a reaction involving hydroxide ions, which also give up electrons:
$4OH^-_{(aq)} \rightarrow 2H_2O_{(l)} + O_{2(g)} + 4e^- \label{3}$
The chlorine must be purified by removing this oxygen.
At the cathode
Sodium ions and protons (from the water) are attracted to the negative cathode. It is much easier for a proton to pick up an electron than for a sodium ion to do so. Therefore, the following reaction occurs:
$2H^+_{(aq)} + 2e^- \rightarrow H_{2(g)} \label{4}$
As protons convert into hydrogen gas, the equilibrium below shifts to the right to replace them:
The net effect of this process is a buildup of sodium ions and newly-produced hydroxide ions at the cathode. In other words, sodium hydroxide solution is formed.
The necessity of keeping all products separate
If chlorine comes into contact with hydrogen, it produces a mixture that explodes violently on exposure to sunlight or heat, producing hydrogen chloride gas. Clearly these gases must remain separated. However, chlorine also reacts with sodium hydroxide solution to produce a mixture of sodium chloride and sodium chlorate(I), also known as sodium hypochlorite; this mixture is commonly sold as bleach. In addition, when the desired products are chlorine and sodium hydroxide rather than bleach, chlorine and sodium hydroxide must also be kept apart. The diaphragm and membrane cells are designed to keep all the products separate.
The diaphragm
The diaphragm is made of a porous mixture of asbestos and polymers. The solution can seep through it from the anode compartment into the cathode compartment. Notice that there is a higher level of liquid on the anode side. This ensures that the liquid always flows from left to right, preventing any of the sodium hydroxide solution from coming into contact with chlorine products.
Production of chlorine at the anode
Chlorine is produced at the titanium anode according to the following equation:
$2Cl^- (aq) - 2e^- \rightarrow Cl_2(g) \nonumber$
The product is contaminated with some oxygen because of the reaction below:
$4OH^-(aq) - 4e^- \rightarrow 2H_2O(l) + O_2(g) \nonumber$
The chlorine is purified by liquefaction under pressure. The oxygen remains a gas when compressed at ordinary temperatures.
Production of hydrogen at the cathode
Hydrogen is produced at the steel cathode by the following process:
$2H^+ (aq) + 2e^- \rightarrow H_2 (g) \nonumber$
Production of the sodium hydroxide
A dilute solution of sodium hydroxide solution is also produced at the cathode (see above for the explanation of what happens at the cathode). It is highly contaminated with unreacted sodium chloride.
The sodium hydroxide solution leaving the cell is concentrated by evaporation. During this process, most of the sodium chloride crystallizes out as solid salt. The salt can be separated, dissolved in water, and passed through the cell again. Even after concentration, sodium hydroxide still contains a small percentage of sodium chloride.
The membrane
The membrane is made from a polymer that only allows the diffusion of positive ions. That means that only the sodium ions can pass through the membrane; the chloride ions are blocked. The advantage of this is that the sodium hydroxide formed in the right-hand compartment is never contaminated with sodium chloride. The sodium chloride solution must be pure. If it contains any other metal ions, these can also pass through the membrane and so contaminate the sodium hydroxide solution.
Production of chlorine
Chlorine is produced at the titanium anode according to the following equation:
$2Cl^- (aq) + 2e^- \rightarrow Cl_2(g) \nonumber$
It is contaminated with some oxygen because of the parallel reaction below:
$4OH^-(aq) + 4e^- \rightarrow 2H_2O(l) + O_2 \nonumber$
The chlorine is purified by liquefaction under pressure. The oxygen remains in the gas phase when compressed at normal temperatures.
Production of hydrogen
Hydrogen is produced at the nickel cathode as follows:
$2H^+ (aq) +2e^- \rightarrow H_2 (g) \nonumber$
Production of sodium hydroxide
An approximately 30% solution of sodium hydroxide solution is also produced at the cathode (see the background chemistry section for an explanation of what happens at the cathode).
8.13.05: Chemistry of Bromine (Z35)
Bromine is a reddish-brown fuming liquid at room temperature with a very disagreeable chlorine-like smell. In fact its name is derived from the Greek bromos or "stench". It was first isolated in pure form by Balard in 1826. It is the only non-metal that is a liquid at normal room conditions. Bromine on the skin causes painful burns that heal very slowly. It is an element to be treated with the utmost respect in the laboratory.
Most bromine is produced by displacement from ordinary sea water. Chlorine (which is more active) is generally used to dislodge the bromine from various compounds in the water. Before leaded gasolines were removed from the market, bromine was used in an additive to help prevent engine "knocking". Production now is chiefly devoted to dyes, disinfectants and photographic chemicals.
Contributors and Attributions
Stephen R. Marsden
8.13.06: Chemistry of Iodine (Z53)
Elemental iodine is a dark grey solid with a faint metallic luster. When heated at ordinary air pressures it sublimes to a violet gas. The name iodine is taken from the Greek ioeides which means "violet colored". It was discovered in 1811 by Courtois.
Commercially iodine is recovered from seaweed and brines. It is an important trace element in the human diet, required for proper function of the thyroid gland. Thus iodine is added to table salt ("iodized") to insure against iodine deficiencies. Radioactive isotopes of iodine are used in medical tracer work involving the thyroid and also to treat diseases of that gland.
Contributors and Attributions
Stephen R. Marsden
8.13.07: Chemistry of Astatine (Z85)
Astatine was formerly known as alabamine. It has no stable isotopes and was first synthetically produced (1940) at the University of California.
Name: Astatine
Symbol: At
Atomic Number: 85
Atomic Mass: (210.0) amu
Melting Point: 302.0 °C (575.15 K, 575.6 °F)
Boiling Point: 337.0 °C (610.15 K, 638.6 °F)
Number of Protons/Electrons: 85
Number of Neutrons: 125
Classification: Halogen
Crystal Structure: Unknown
Density @ 293 K: Unknown
Color: Unknown
Date of Discovery: 1940
Discoverer: D.R. Corson
Name Origin: From the Greek word astatos (unstable)
Uses: No uses known
Obtained From: Man-made
Oxidation Number: -1, +5
Astatine is the last of the known halogens and was synthesized in 1940 by Corson and others at the University of California. It is radioactive and its name, from the Greek astatos, means "unstable". The element can be produced by bombarding targets made of bismuth-209 with high energy alpha particles (helium nuclei). Astatine 211 is the product and has a half-life of 7.2 hours. The most stable isotope of astatine is 210, which has a half-life of 8.1 hours.
Not much is known about the chemical properties of astatine, but it is expected to react like the other halogens, although much less vigorously, and it should be more metallic than iodine. There should be tiny quantities of astatine in the earth's crust as products of other radioactive decays, but their existence would be short-lived.
Contributors and Attributions
Stephen R. Marsden | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/08%3A_Chemistry_of_the_Main_Group_Elements/8.13%3A_The_Halogens/8.13.04%3A_Chemistry_of_Chlorine_%28Z17%29/8.13.4.01%3A_The_Manufacture_of_Chlorine.txt |
The noble gases (Group 18) are located in the far right of the periodic table and were previously referred to as the "inert gases" due to the fact that their filled valence shells (octets) make them extremely nonreactive. The noble gases were characterized relatively late compared to other element groups.
Thumbnail: Vial of glowing ultrapure neon. (CC SA; Jurii via http://images-of-elements.com/neon.php).
8.14: The Noble Gases
The noble gases (Group 18) are located in the far right of the periodic table and were previously referred to as the "inert gases" due to the fact that their filled valence shells (octets) make them extremely nonreactive. The noble gases were characterized relatively late compared to other element groups.
The History
The first person to discover the noble gases was Henry Cavendish in the late 180th century. Cavendish distinguished these elements by chemically removing all oxygen and nitrogen from a container of air. The nitrogen was oxidized to $NO_2$ by electric discharges and absorbed by a sodium hydroxide solution. The remaining oxygen was then removed from the mixture with an absorber. The experiment revealed that 1/120 of the gas volume remained un-reacted in the receptacle. The second person to isolate, but not typify, them was William Francis (1855-1925). Francis noted the formation of gas while dissolving uranium minerals in acid.
Argon
In 1894, John William Strutt discovered that chemically-obtained pure nitrogen was less dense than the nitrogen isolated from air samples. From this breakthrough, he concluded that another, unknown gas was present in the air. With the aid of William Ramsay, Strutt managed to replicate and modify Cavendish's experiment to better understand the inert component of air in his original experiment. The researchers' procedure differed from the Cavendish procedure: they removed the oxygen by reacting it with copper, and removed the nitrogen in a reaction with magnesium. The remaining gas was properly characterized and the new element was named "argon," which originates from the Greek word for "inert."
Helium
Helium was first discovered in 1868, manifesting itself in the solar spectrum as a bright yellow line with a wavelength of 587.49 nanometers. This discovery was made by Pierre Jansen. Jansen initially assumed it was a sodium line. However, later studies by Sir William Ramsay (who isolated helium on Earth by treating a variety of rare elements with acids) confirmed that the bright yellow line from his experiment matched up with that in the spectrum of the sun. From this, British physicist William Crookes identified the element as helium.
Neon, Krypton, Xenon
These three noble gases were discovered by Morris W. Travers and Sir William Ramsay in 1898. Ramsay discovered neon by chilling a sample of the air to a liquid phase, warming the liquid, and capturing the gases as they boiled off. Krypton and xenon were also discovered through this process.
Radon
In 1900, while studying the decay chain of radium, Friedrich Earns Dorn discovered the last gas in Group 18: radon. In his experiments, Dorn noticed that radium compounds emanated radioactive gas. This gas was originally named niton after the Latin word for shining, "nitens". In 1923, the International Committee for Chemical Elements and International Union of Pure Applied Chemistry (IUPAC) decided to name the element radon. All isotopes of radon are radioactive. Radon-222 has the longest half-life at less than 4 days, and is an alpha-decay product of Radium-226 (part of the U-238 to Pb-206 radioactive decay chain).
The Electron Configurations for Noble Gases
• Helium 1s2
• Neon [He] 2s2 2p6
• Argon [Ne] 3s2 3p6
• Krypton [Ar] 3d10 4s2 4p6
• Xenon [Kr] 4d10 5s2 5p6
• Radon [Xe] 4f14 5d10 6s2 6p6
Table 1: Trends within Group 18
Atomic # Atomic mass Boiling point (K) Melting point (K) 1st Ionization (E/kJ mol-1) Density (g/dm3) Atomic radius (pm)
He 2 4.003 4.216 0.95 2372.3 0.1786 31
Ne 10 20.18 27.1 24.7 2080.6 0.9002 38
Ar 18 39.948 87.29 83.6 1520.4 1.7818 71
Kr 36 83.3 120.85 115.8 1350.7 3.708 88
Xe 54 131.29 166.1 161.7 1170.4 5.851 108
Rn 86 222.1 211.5 202.2 1037.1 9.97 120
The Atomic and Physical Properties
• Atomic mass, boiling point, and atomic radii INCREASE down a group in the periodic table.
• The first ionization energy DECREASES down a group in the periodic table.
• The noble gases have the largest ionization energies, reflecting their chemical inertness.
• Down Group 18, atomic radius and interatomic forces INCREASE resulting in an INCREASED melting point, boiling point, enthalpy of vaporization, and solubility.
• The INCREASE in density down the group is correlated with the INCREASE in atomic mass.
• Because the atoms INCREASE in atomic size down the group, the electron clouds of these non polar atoms become increasingly polarized, which leads to weak van Der Waals forces among the atoms. Thus, the formation of liquids and solids is more easily attainable for these heavier elements because of their melting and boiling points.
• Because noble gases’ outer shells are full, they are extremely stable, tending not to form chemical bonds and having a small tendency to gain or lose electrons.
• Under standard conditions all members of the noble gas group behave similarly.
• All are monotomic gases under standard conditions.
• Noble gas atoms, like the atoms in other groups, INCREASE steadily in atomic radius from one period to the next due to the INCREASING number of electrons.
• The size of the atom is positively correlated to several properties of noble gases. The ionization potential DECREASES with an INCREASING radius, because the valence electrons in the larger noble gases are further away from the nucleus; they are therefore held less tightly by the atom.
• The attractive force INCREASES with the size of the atom as a result of an INCREASE in polarizability and thus a DECREASE in ionization potential.
• Overall, noble gases have weak interatomic forces, and therefore very low boiling and melting points compared with elements of other groups.
For covalently-bonded diatomic and polyatomic gases, heat capacity arises from possible translational, rotational, and vibrational motions. Because monatomic gases have no bonds, they cannot absorb heat as bond vibrations. Because the center of mass of monatomic gases is at the nucleus of the atom, and the mass of the electrons is negligible compared to the nucleus, the kinetic energy due to rotation is negligible compared to the kinetic energy of translation (unlike in di- or polyatomic molecules where rotation of nuclei around the center of mass of the molecule contributes significantly to the heat capacity). Therefore, the internal energy per mole of a monatomic noble gas equals its translational contribution, $\frac{3}{2}RT$, where $R$ is the universal gas constant and $T$ is the absolute temperature.
For monatomic gases at a given temperature, the average kinetic energy due to translation is practically equal regardless of the element. Therefore at a given temperature, the heavier the atom, the more slowly its gaseous atoms move. The mean velocity of a monatomic gas decreases with increasing molecular mass, and given the simplified heat capacity situation, noble gaseous thermal conductivity decreases with increasing molecular mass.
Applications of Noble Gases
Helium
Helium is used as a component of breathing gases due to its low solubility in fluids or lipids. This is important because other gases are absorbed by the blood and body tissues when under pressure during scuba diving. Because of its reduced solubility, little helium is taken into cell membranes; when it replaces part of the breathing mixture, helium causes a decrease in the narcotic effect of the gas at far depths. The reduced amount of dissolved gas in the body means fewer gas bubbles form, decreasing the pressure of the ascent. Helium and Argon are used to shield welding arcs and the surrounding base metal from the atmosphere.
Helium is used in very low temperature cryogenics, particularly for maintaining superconductors (useful for creating strong magnetic fields) at a very low temperatures. Helium is also the most common carrier gas in gas chromatography.
Neon
Neon has many common and familiar applications: neon lights, fog lights, TV cine-scopes, lasers, voltage detectors, luminous warnings, and advertising signs. The most popular application of neon is the neon tubing used in advertising and elaborate decorations. These tubes are filled with neon and helium or argon under low pressure and submitted to electrical discharges. The color of emitted light is depends on the composition of the gaseous mixture and on the color of the glass of the tube. Pure Neon within a colorless tube absorbs red light and reflects blue light, as shown in the figure below. This reflected light is known as fluorescent light.
One of the many colors of neon lights.
Argon
Argon has a large number of applications in electronics, lighting, glass, and metal fabrications. Argon is used in electronics to provide a protective heat transfer medium for ultra-pure silicon crystal semiconductors and for growing germanium. Argon can also fill fluorescent and incandescent light bulbs, creating the blue light found in "neon lamps." By utilizing argon's low thermal conductivity, window manufacturers provide a gas barrier needed to produce double-pane insulated windows. This insulation barrier improves the windows' energy efficiencies. Argon also creates an inert gas shield during welding, flushes out melted metals to eliminate porosity in casting, and provides an oxygen- and nitrogen-free environment for annealing and rolling metals and alloys.
Argon plasma light bulb.
Krypton
Similarly to argon, krypton can be found in energy efficient windows. Because of its superior thermal efficiency, krypton is sometimes chosen over argon for insulation. It is estimated that 30% of energy efficient windows sold in Germany and England are filled with krypton; approximately 1.8 liters of krypton are used in these countries. Krypton is also found in fuel sources, lasers and headlights. In lasers, krypton functions as a control for a desired optic wavelength. It is usually mixed with a halogen (most likely fluorine) to produce excimer lasers. Halogen sealed beam headlights containing krypton produce up to double the light output of standard headlights. In addition, Krypton is used for high performance light bulbs, which have higher color temperatures and efficiency because the krypton reduces the rate of evaporation of the filament.
Krypton laser.
Xenon
Xenon has various applications in incandescent lighting, x-ray development, plasma display panels (PDPs), and more. Incandescent lighting uses xenon because less energy can be used to obtain the same light output as a normal incandescent lamp. Xenon has also made it possible to obtain better x-rays with reduced amounts of radiation. When mixed with oxygen, it can enhance the contrast in CT imaging. These applications have had great impact on the health care industries. Plasma display panels (PDPs) using xenon as one of the fill gases may one day replace the large picture tubes in television and computer screens.
Nuclear fission products may include several radioactive isotopes of xenon, which absorb neutrons in nuclear reactor cores. The formation and elimination of radioactive xenon decay products are factors in nuclear reactor control.
Radon
Radon is reported as the second most frequent cause of lung cancer, after cigarette smoking. However, it also has beneficial applications in radiotherapy, arthritis treatment, and bathing. In radiotherapy, radon has been used in implantable seeds, made of glass or gold, primarily used to treat cancers. It has been said that exposure to radon mitigates auto-immune diseases such as arthritis. Some arthritis sufferers have sought limited exposure to radioactive mine water and radon to relieve their pain. "Radon Spas" such as Bad Gastern in Austria and Onsen in Japan offer a therapy in which people sit for minutes to hours in a high-radon atmosphere, believing that low doses of radiation will boost up their energy. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/08%3A_Chemistry_of_the_Main_Group_Elements/8.14%3A_The_Noble_Gases/8.14.02%3A_Properties_of_Nobel_Gases.txt |
The noble gases (Group 18) are located in the far right of the periodic table and were previously referred to as the "inert gases" due to the fact that their filled valence shells (octets) make them extremely nonreactive.
The Chemical Properties
Noble gases are odorless, colorless, nonflammable, and monotonic gases that have low chemical reactivity.
Atomic Number Element Number of Electrons/Shell
2 Helium 2
10 Neon 2,8
18 Argon 2,8,8
36 Krypton 2,8,18,8
54 Xenon 2,8,18,18,8
86 Radon 2,8,18,32,18,8
The full valence electron shells of these atoms make noble gases extremely stable and unlikely to form chemical bonds because they have little tendency to gain or lose electrons. Although noble gases do not normally react with other elements to form compounds, there are some exceptions. Xe may form compounds with fluoride and oxide.
Example 1: Xenon Fluorides
Xenon Difluoride (\(XeF_2\))
• Dense white crystallized solid
• Powerful fluorinating agent
• Covalent inorganic fluorides
• Stable xenon compound
• Decomposes on contact with light or water vapor
• Linear geometry
• Moisture sensitive
• Low vapor pressure
Xenon Tetrafluoride (\(XeF_4\))
• Colorless Crystals
• Square planar geometry
• Discovered in 1963
Xenon Hexafluoride (\(XeF_6\))
• Strongest fluorinating agent
• Colorless solid
• Highest coordination of the three binary fluorides of xenon (\(XeF_2\) and \(XeF_4\))
• Formation is exergonic, and the compound is stable at normal temperatures
• Readily sublimes into intense yellow vapors
• Structure lacks perfect octahedral symmetry
Example 2: Xenon Oxide
Xenon Tetroxide (XeO4)
• Yellow crystalline solid
• Relatively stable
• Oxygen is the only element that can bring xenon up to its highest oxidation state of +8
Two other short-lived xenon compounds with an oxidation state of +8, XeO3F2 and XeO2F4, are produced in the reaction of xenon tetroxide with xenon hexafluoride.
Example 3: Radon Compounds
Radon difluoride (RnF2) is one of the few reported compounds of radon. Radon reacts readily with fluorine to form a solid compound, but this decomposes on attempted vaporization and its exact composition is uncertain. The usefulness of radon compounds is limited because of the noble gas's radioactivity. The longest-lived isotope, 222Ra, has a half-life of only 3.82 days.
8.14.05: Chemistry of Helium (Z2)
Helium is at the top of the noble gas group (which also contains neon, argon, krypton, xenon, and radon) and is the least reactive element. Helium has many interesting characteristics, such as making balloons float and raising the pitch of one's voice; these applications are discussed below.
Introduction
Helium is the second most abundant element in the universe, next to hydrogen. Helium is colorless, odorless, and tasteless. It has a very low boiling point, and is monatomic. Helium is small and extremely light, and is the least reactive of all elements; it does not react with any other elements or ions, so there are no helium-bearing minerals in nature. Helium was first observed by studying the sun, and was named after the Greek word for the sun, Helios.
Physical Properties
Color Colorless
Phase at Room Temperature Gas
Density 0.0002 g/cm3
Boiling Point 4.2 K
Heat of Vaporization 0.1 kJ/mol
Thermal Conductivity 0.15 J/m sec K
Source Natural gas
Atomic Properties
Electron Configuration 1s2
Number of Isotopes 7 (2 liquid)
Electron Affinity 0 kJ/mol
First Ionization Energy 2372.3 kJ/mol
Second Ionization Energy 5250.3 kJ/mol
Polarizability 0.198 Å3
Atomic Weight 4.003
Atomic Volume 27.2 cm3/mol
Atomic Radius 31 pm
Abundance
In Earth's Crust 8x10-3
In Earth's Ocean 7×10-6
In Human Body 0%
Occurrence and production
Helium is one of the most abundant elements in the universe. Large quantities are produced in the energy-producing fusion reactions in stars. Previously, helium was rarely used, because only .0004% of Earth's atmosphere is helium—that equates to one helium molecule for every 200,000 air molecules, including oxygen, hydrogen, and nitrogen. However, the discovery of helium-rich wells in Texas, Russia, Poland, Algeria, China, and Canada has made helium more accessible.
Helium is produced in minerals through radioactive decay. Helium is extracted from natural gas deposits, which often contain as much as 10% helium. These natural gas reserves are the only industrially-available source of helium. The total world helium resources theoretically add up to 25.2 billion cubic meters; the United States contains 11.1 billion cubic meters. The extracted gas is subjected to chemical pre-purification, using an alkaline wash to remove carbon dioxide and hydrogen sulfide. The remaining gas is cooled to -200°C, where all materials, except helium gas, are liquefied.
History
Helium was first discovered in 1868 by the French astronomer P. J. C. Jenssen, who was studying the chromosphere of the Sun during a solar eclipse. He used a spectrometer to resolve the light into its spectrum, in which each color represents a different gaseous element. He observed a new yellow light, concluding that it indicated the presence of an element not previously known. In 1895, the existence of helium on Earth was proved by Sir William Ramsay. Heating cleveite (a radioactive mineral) released an inert gas, which was found to be helium; this helium is a by-product of the natural decay of radioactive elements. The chemists Norman Lockyer and Edward Frankland confirmed helium as an element and named it after helios, the Greek word for the Sun.
Applications and hazards
Helium has a number of applications due to its inert nature. Liquefied helium has cryogenic properties, and is used to freeze biological materials for long term storage and later use. Twenty percent of industrial helium use is in wielding and industrial applications. Helium protects the heated parts of metals such as aluminum and titanium from air. Mixtures of helium and oxygen are used in tanks for underwater breathing devices: due to its low density, helium gas allows oxygen to stream easily through the lungs. Because helium remains a gas, even at temperatures low enough to liquefy hydrogen, it is used as pressure gas to move liquid hydrogen into rocket engines. Its inert nature also makes helium useful for cooling nuclear power plants.
The most commonly known characteristic of helium is that it is lighter than air. It can levitate balloons during parties and fly blimps over sports stadiums. Helium has 92% of the lifting power of hydrogen; however, it is safer to use because it is noncombustible and has lower rate of diffusion than that of hydrogen gas. The famous Hindenburg disaster is an example of the hazards of using combustible gas like hydrogen. Because helium was previously very expensive only available from natural gas reserves in U.S., Nazi Germany had only hydrogen gas at its disposal. The consequences were devastating, as shown below:
Currently, helium is found in other natural gas reserves around the world. The cost of helium has decreased from \$2500/ft3 in 1915 to \$0.15/ft3 in 1989. Helium is what keeps the Goodyear blimps afloat over stadiums.
Helium is often inhaled from balloons to produce a high, squeaky voice. This practice can be very harmful. Inhaling helium can lead to loss of consciousness and cerebral arterial gas embolism, which can temporarily lead to complete blindness. This occurs when blood vessels in the lungs rupture, allowing the gas to gain access to the pulmonary vasculature and subsequently the brain.
Characteristics
Gas and plasma phases
Helium is naturally found in the gas state. Helium is the second least reactive element and noble gas (after neon). Its low atomic mass, thermal conductivity, specific heat, and sound speed are greatest after hydrogen. Due to the small size of helium atoms, the diffusion rate through solids is three times greater than that of air and 65% greater than that of hydrogen. The element is inert, monatomic in standard conditions, and the least water soluble gas.
At normal ambient temperatures, helium has a negative Joule Thomson coefficient. Thus, upon free expansion, helium naturally heats up. However, below its Joule Thomson inversion temperature (32-50 K at 1 atm), it cools when allowed to freely expand. Once cooled, helium can be liquefied through expansion cooling. Helium is commonly found throughout the universe as plasma, a state in which electrons are not bound to nuclei. Plasmas have high electrical conductivities and are highly influenced by magnetic and electric fields.
Solid and liquid phases
Helium is the only element that cannot be solidified by lowering the temperature at ordinary pressures; this must be accompanied by a pressure increase. The volume of solid helium, 3He and 4He, can be decreased by more than 30% by applying pressure. Solid helium has a projected density of 0.187 ± 0.009 g/mL at 0 K and 25 bar. Solid helium also has a sharp melting point and a crystalline structure. There are two forms of liquid helium: He4I and He4II.
Helium I
Helium I is formed when temperature falls below 4.22 K and above the lambda point of 2.1768 K. It is a clear liquid that boils when heat is applied and contracts when temperature is lowered. Below the lambda point, helium does not boil, but expands. Helium I has a gas-like index of refraction of 1.026 which makes its surface difficult to see. It has a very low viscosity and a density 1/8th that of water. This property can be explained with quantum mechanics. Both helium I and II are quantum fluids, displaying atomic properties on a macroscopic scale due to the fact that the boiling point of helium is so close to absolute zero.
Helium II
At 2.174 K, helium I forms into helium II. Its properties are very unusual, and the substance is described as superfluid. Superfluid is a quantum-mechanical state of matter; the two-fluid model for helium II explains why one portion of helium atoms exists in a ground state, flowing with zero viscosity, and another portion is in an excited state, behaving like an ordinary fluid. The viscosity of He4II is so low that there is no internal friction.
He4II can conduct heat 300 times more effectively than silver, making it the best heat conductor known. Its thermal conductivity is a million times that of helium I and several hundred times that of copper. The conductivity and viscosity of helium II do not obey classical rules, but are consistent with the rules of quantum mechanics. When temperature is lowered, helium II expands in volume. It cannot be boiled, but evaporates directly to gas when heated.
In this superfluid state, liquid helium can flow through thin capillaries or cracks much faster than helium gas. It also exhibits a creeping effect, moving along the surface seemingly against gravity. Helium II creeps along the sides of a open vessel until it reaches a warmer region where it evaporates. As a result of the creeping behavior and the ability to leak rapidly through tiny openings, helium II is very difficult to confine. Helium II also exhibits a fountain effect. Suppose a chamber allows a reservoir of helium II to filter superfluid and non-superfluid helium. When the interior of the container is heated, superfluid helium converts to non-superfluid helium to maintain equilibrium. This creates intense pressure on the superfluid helium, causing the liquid to fountain out of the container.
Isotopes
Helium has eight known isotopes but only two are stable: 3He and 4He. 3He is found in only very small quantities compared to 4He. It is produced in trace amounts by the beta decay of tritium. This form is found in abundance in stars, as a product of nuclear fusion. Extraplanetary materials have trace amounts of 3He from solar winds. 4He is produced by the alpha decay of heavier radioactive elements on Earth. It is an unusually stable isotope because its nucleons are arranged in complete shells.
Problems
1. What happens when a lit cigarette is thrown at a leaking, high-pressured helium cylinder?
a). nothing
b). the cigarette is incinerated before touching the cylinder
c). the cylinder explodes
d). the cylinder becomes a flame thrower
1. How many isotopes of helium are known? ______
2. (Helium gas, or helium II liquid) leaks faster than the other when stored in a opened cylinder (at STP).
3. What happens when a section of divided petri dish is filled with helium II at 2.173K?
a). It starts boiling.
b). It starts "creeping" over the divider, soon filling up the other sections of the dish.
c). It evaporates and soon leaves the dish.
d.) It solidifies and expands, breaking the dividers of the petri dish, and filling up the whole dish.
1. Helium was first discovered through ________.
Contributors and Attributions
• Jun-Hyun Hwang - University of California, Davis | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/08%3A_Chemistry_of_the_Main_Group_Elements/8.14%3A_The_Noble_Gases/8.14.04%3A_Reactions_of_Nobel_Gases.txt |
Contributors and Attributions
• Jonathan Molina - (UCD)
8.14.07: Chemistry of Argon (Z18)
Argon is it is colorless, tasteless and odorless noble gas that is located in Group 18 on the Periodic Table. It was discovered by Henry Cavendish in 1785 and was named Argon, which is derived from the Greek word "argos" meaning inactive. Cavendish formed oxides of nitrogen by passing electric currents through air, then dissolved them in water to get nitric acid, but was unable to get all of the air to react. He suspected that there was a then unidentified gas component of air; Ramsay and Rayleigh went on to isolate this component in 1894, and the new found element was thus named Argon.
Contributors and Attributions
• Katherine Cubbon (UCD)
Stephen R. Marsden
8.14.08: Chemistry of Krypton (Z36)
Krypton is one of the six Noble Gas elements (Group 18), which are widely known for their relative "inertness" and difficulty in forming chemical compounds with any other elements, due to these elements having full valence shells. Contrary to original thinking, however, Krypton has been made to react with the highly electronegative elements and is used in lighting and other commercial purposes.
Facts
• Element number: 36
• Electron configuration: [Ar]3d104s24p6
• Atomic weight: 83.798g/mol
• Color: colorless, odorless, tasteless
• Light: large number of spectral lines, strongest being green and yellow/ whitish emission
• Solidified: white and crystalline/ face-centered cubic crystal structure
• Melting point: 115.79K
• Boiling point: 119.92K
• Critical point: 209.41K
• Specific heat capacity: 20.786 J/mol K
• 0.000108-0.000114% of atmosphere
• 6 Stable isotopes
• Produced by breakdown of uranium and plutonium in the earth's crust at a very small %
The Origin and History
Krypton is found in the Group 18 elements, otherwise known as the Noble Gases. In 1785, Henry Cavendish suggested that air contained nonreactive gases after he was unsuccessful in getting a sample of air to react. A century later, British chemists John Rayleigh and William Ramsey began to isolate these inert gases (beginning with Argon) and seperated them in their own group on the periodic table since each of these elements had full electron valence shells. One of these gases, Krypton, was discovered along with Neon and Xenon by Rayleigh and fellow chemist Morris Travers in 1898 in a residue left from evaporating almost all components of liquid air. The name Krypton is derived from the Greek word "kryptos", meaning "hidden". However, the inert quality of these gases was disproved when Xenon compounds were created in 1962 and a Krypton compound (KrF2) was synthesized successfully a year later. This proved that this group of gases is not necessarily inert. Although both Kr and Xe have full valence shells, they are both the most easily ionized of the group. It simply took an element of high electronegativity, in this case Fluorine, to force Xe and Kr to react under high temperatures.
Isolation
It ranks sixth in abundance in the atmosphere. As with the other noble gases, krypton is isolated from the air by liquefaction.
Compounds & Isotopes
Although Krypton is naturally chemically nonreactive, krypton difluoride was synthesized in 1963.
$Kr_{(g)}+F_{2(g)} \rightarrow KrF_{2(g)} \nonumber$
It has also been discovered that Krypton can bond with other atoms besides Fluorine, however such compounds are much more unstable than krypton difluoride. For example, KrF2 can bond with nitrogen when it reacts with [HC≡NH]+[AsF6] under -50°C to form [HC≡N–Kr–F]+. There have been other reports of successfully synthesizing additional Krypton compounds, but none have been verified. Krypton has 6 stable isotopes: 78Kr, 80Kr, 82Kr, 83Kr, 84Kr, and 86Kr. There are a total of 31 isotopes of Krypton, and the only isotope besides the six given that occur naturally is 81Kr which is a product of atmospheric reactions between the other natural isotopes.
Applications
Krypton gas is used in various kinds of lights, from small bright flashlight bulbs to special strobe lights for airport runways. Due to Krypton's large number of spectral lines, it's ionized gas is white, which is why light bulbs that are krypton based are used in photography and studio lighting in the film industry. In neon lights, Krypton reacts with other gases to produce a bright yellow light as well. The isotope 85Kr can also be used in combination with phosphors to produce materials that shine in the dark due to the fact that this particular isotope of Krypton reflects off of phosphors. Krypton is also used in lasers as a control for a desired wavelength, especially in red lasers because Krypton has a much higher light density in the red spectral region than other gases such as Neon, which is why krypton-based lasers are used to produce red light in laser-light shows.
Perhaps one of the most significant uses of Krypton is in the krypton-fluoride laser which is used in nuclear fusion energy research.
other.nrl.navy.mil/LaserFusio...ercreation.htm
Isotopes of Krypton
Yet another important application of Krypton, specifically 83Kr, is in Magnetic Resonance Imaging (MRI), which is used instead of other gases because of its high spin and smaller/less polar electron cloud compared to other noble gases such as Xenon. It is used to distinguish hydrophobic and hydrophillic regions containing an airway.
An international agreement was made in 1960 to base the length of the meter on the wavelength of light emitted by 86Kr (605.78 nm). However, this was changed in 1983 when the International Bureau of Weights and Measures determined the meter to be the distance that light travels in a vacuum in 1/299,792,458 s.
Problems
1. How is the element Krypton isolated?
2. Why can fluorine react with Krypton to form a compound?
3. What is the electron configuration of Krypton?
4. What colors are most pronounced in the Krypton spectral emission?
5. How many stable isotopes of Krypton are there?
Solutions
1. By evaporating the componnts of liquid air.
2. Because fluorine is highly electronegative and Krypton is readily ionizable.
3. [Ar]3d104s24p6
4. Green and yellow
5. 6
Contributors and Attributions
• Megan Meadows (UCD)
8.14.09: Chemistry of Radon (Z86)
Contributors and Attributions
• Boundless
• Stephen R. Marsden
8.14.10: Chemistry of Xenon (Z54)
Xenon is an element under the Noble gases group and is on period 7 of the periodic table. This element is most notable for its bright luminescence in light bulbs. Xenon is unique for being the first noble gas element to be synthesized into a compound.
Contributors and Attributions
• Albert Young (UCD) | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/08%3A_Chemistry_of_the_Main_Group_Elements/8.14%3A_The_Noble_Gases/8.14.06%3A_Chemistry_of_Neon_%28Z10%29.txt |
Coordination compounds are important to all areas of chemistry, engineering, the life and environmental sciences, and beyond. In the synthetic laboratory catalytic amounts of coordination compounds enable organic chemists to synthesize new compounds selectively and in high yield under mild conditions. Applied industrially, coordination compound catalysts serve as vital catalysts that facilitate the conversion of raw petrochemical or bio-derived feedstocks into useful industrial and consumer products. Without them life as we know it would be impossible, as many biochemical systems are coordination complexes. Examples include the hemoglobin that transports oxygen around our bodies and the myoglobin that stores it, the photosystems that harvest light and use light energy in photosynthesis, the constituents of the respiratory chain, and many of the enzymes involved in the expression and transmission of genetic information. In studying coordination chemistry you are about to take your first steps into a vast and exciting world.
09: Coordination Chemistry I - Structure and Isomers
What is a coordination compound?
Coordination compounds consist of one or more metals bound to one or more Lewis base ligands. For example, hexamineruthenium (3+) ion is a coordination complex in which six ammonia ligands coordinate a Co3+ ion, as shown in Scheme $\sf{1}$.
Such complexes are called coordination complexes because the ligand-metal bond may be thought of as a coordinate covalent bond in which both of the bonding electrons come from the ligand, which is then said to coordinate the metal. This coordinate covalent model is a very useful formalism for understanding the basic features of coordination chemistry, although it does not always accurately reflect the actual details of the bonding in every coordination complex. Nevertheless, even in those cases where the simple coordinate covalent bond model breaks down, the coordinate covalent bond concept supplies the language sophisticated models employ to describe the more complex bonding involved.
Additional important terms
Some of the common widely used terms that follow from the coordinate covalent model of bonding in coordination complexes include:
• Coordination compounds are also called coordination complexes, metal complexes, or just complexes. The term complex refers to coordination compounds' composite nature, in that they may be thought of as comprising multiple ligand and metal ion parts that can be restored by breaking the coordinate covalent bonds holding the complex together. This is in contrast to inorganic or organic molecules which are more commonly thought about as whole molecules held together by the sharing of electrons contributed by all the atoms.
• Coordination complexes that are ions are called complex ions.
• Ligands bound to the coordination complex are said to reside in the primary or inner coordination sphere. These bound ligands are not readily exchangeable, in contrast to nearby counterions and solvent molecules, which are said to reside in the secondary or outer coordination sphere.
• The portion of the complex contributing the electron pairs is said to be the donor and the portion which receives them the acceptor. In conventional coordination compounds the ligand is the donor and the metal the acceptor. In these cases it would be equally convenient to refer to the ligand donor as the Lewis base and the metal acceptor as the Lewis acid. However, in more complex bonding scenarios there may be multiple electron pair donation and acceptance interactions taking place between each pair of atoms and donor-acceptor language will be more convenient.
• The number of ligand sites donating lone pairs to the central atom is referred to as the coordination number. For most complexes this will just be equal to the number of ligand atoms bound to the metal. In simple complexes it is just equal to the number of ligands. For instance, the cobalt in Scheme $\sf{1}$ has a coordination number of six.
• Although technically compounds with metal-carbon bonds are coordination complexes, the term coordination complex is sometimes used to refer to complexes which do not possess metal-carbon bonds in their primary coordination sphere. Complexes which possess metal-carbon bonds are called organometallic compounds instead. The use of the terms organometallic and coordination to distinguish organometallic compounds from other types of coordination compounds is often convenient since many organometallic ligands engage in more than simple $\sigma$ donor-acceptor coordinate covalent bond formation with the metal center. However, this is true of some wholly inorganic ligands too, so it should always be kept in mind that organometallic compounds are just a type of coordination compound and that inorganic ligands can in principle be tuned to interact with a metal center in much the same way an organic ligand does.
A summary of some of the concepts and terms used to describe coordination compounds is given in Scheme $\sf{2}$.
The formulae of coordination complexes
The way the formulae of coordination complexes are written reflects the fact that it is often convenient to think of coordination compounds as a composite of metals and ligands. When writing the formula of a complex
• the atoms of a ligand are not added to those of the rest of the compound. Instead, the ligand atoms are kept together, if necessary by enclosing the ligand formula in parentheses or giving an abbreviation for the ligand.
• For complex ions the metal and ligands are enclosed in square brackets. Sometimes this is also done for neutral coordination compounds as well. In either case the brackets enclose those parts of the compound which comprise the primary coordination sphere; anything else is in the secondary coordination sphere outside.
A careful perusal of the examples in Scheme $\sf{3}$ should make the important features of this system clear. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/09%3A_Coordination_Chemistry_I_-_Structure_and_Isomers/9.01%3A_Prelude_to_Coordination_Chemistry_I_-_Structure_and_Isomers.txt |
History of the Coordination Compounds
Coordination compounds have been known and used since antiquity; one of the oldest synthetic pigments is the blue pigment Egyptian blue, a copper complex of formula CaCuSi4O10 used by the Egyptians since the third Millenium B.C. (in ancient China the Ba analogue, Han blue, was discovered independently). The blue color of Egyptian blue is due to interlocked Cu(Si2O7)4 units in which each copper is coordinated by four O atoms in a square planar arrangement. Later, in 1706, the Berlin painter Diesbach would discover another deep blue pigment, Prussian blue: $\ce{KFe2(CN)6}$.
Despite their long use, the chemical nature of coordination compounds 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:$\ce{AlF3}$ 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. Of these the most influential was the Blomstrand-Jørgensen chain theory of bonding in coordination compounds, which predicted the “Chattanooga choo-choo” model for CoCl3·4NH3 shown in Scheme $\sf{\PageIndex{I}}$.
Nevertheless, this theory was not wholly illogical and, in fact, explained much of the analytical data on coordination compounds available to chemists of the time. This data included the electrical conductivity of aqueous solutions of these compounds, which was roughly proportional to the number of ions formed per mole, and the number of free chloride ions present, which could be determined by precipitating them gravimetrically as AgCl. In the case of CoCl3·4NH3, two ions and one chloride were produced when the compound was dissolved in water, which Jørgensen was able to explain using the chain structure shown above by postulating that chlorides attached to NH3 could dissociate while those attached to Co could not. The modern theory of coordination chemistry, which overthrew the chain theory, 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 $1$. 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 did not give free chloride ions when dissolved in water.
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 ligands bound to the metal ion. If Pt had a primary valence of 4 and a secondary valence of 6, Werner could explain the properties of the 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} \nonumber
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+.
Exercise $1$. The CoCl3xNH3 series.
The series CoCl3·xNH3 was particularly important in establishing the correctness of Werner's coordination theory over the rival chain theory. By ~1900 conductivity measurements suggested that the members of the series gave the number of ions shown in Table $\sf{1}$.
Table $\sf{1}$. The CoCl3·xNH3 series according to coordination theory, chain theory, and experiment.
Compound Color Werner formulation Blomstrand-Jørgensen chain theory formulation Number of ions in solution
CoCl3·6NH3 yellow [Co(NH3)6]Cl3 4
CoCl3·5NH3 violet [Co(NH3)5Cl]Cl2 3
CoCl3·4NH3 green [Co(NH3)4Cl2]Cl 2
CoCl3·3NH3 orange [Co(NH3)4Cl3] 0
What does this data suggest about the relative explanatory power of Werner's coordination theory and chain theory? Explain.
Answer
Remember that
• chain theory predicts that the number of ions is the number formed when the Cl atoms bound in a chain with NH3 dissociate.
• coordination theory predicts the number of ions based on the number of complex ions and their counterions.
Based on this the predictions of coordination theory and chain theory can be compared with the experimental data, as is done in Table $\sf{2}$.
Table $\sf{2}$: Comparison of ions predicted for the CoCl3·xNH3 series by coordination theory and chain theory with the number observed experimentally.
Compound Color Ions predicted by Werner's coordination theory Number of ions predicted by coordination theory Ions predicted by Blomstrand-Jørgensen chain theory Number of ions predicted by chain theory Observed Number of ions in solution
CoCl3·6NH3 yellow
[Co(NH3)6]3+
Cl-
Cl-
Cl-
4 4 4
CoCl3·5NH3 violet
[Co(NH3)5Cl]2+
Cl-
Cl-
3 3 3
CoCl3·4NH3 green
[Co(NH3)4Cl2]+
Cl-
2 2 2
CoCl3·3NH3 orange None 0 2 0
As can be seen by comparing the number of ions predicted by coordination and chain theory in Table $\sf{2}$, coordination theory successfully explains all the observed ion counts, while chain theory fails to explain the lack of ions observed for CoCl3·3NH3.
Nevertheless, as is often the case when developing theoretical models using data from real experimental investigations, these observations did not convince Jørgensen, who could point to the experimental difficulty of determining the number of ions present from solution conductivity data.
What ultimately convinced Jørgensen of the correctness of Werner's coordination model over his own chain theory was how Werner's explanation of the structure of cobalt coordination complexes using an octahedral coordination geometry explained the existence of isomers in Co complexes containing Cl and NH3 ligands. In the case of [Co(NH3)4Cl2]Cl two isomers were known: one red and the other green. 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, stereoisomers (Figure $1$). His conclusion was also corroborated by the existence of two and only two stereoisomers of the next compound in the series: Co(NH3)3Cl3.
Example $1$: Why did Werner propose an octahedral geometry for 6-coordinate complexes?
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 that this suggests these complexes possess an octahedral geometry, calculate the maximum number of different structures possible for six-coordinate MA4B2 complexes with each of the three most symmetrical possible structures the ligands will form about the central metal - a hexagon, a trigonal prism, and an octahedron.
Assuming that the absence of evidence for additional compounds in this case serves as reasonable circumstantial evidence for their absence, what does the fact that no more than two forms of any MA4B2 complex were known suggest about the three-dimensional structures of these complexes?
Solution
In this problem you are given
• the stochiometry of the complexes, MA4B2
• three possible coordination geometries - hexagonal, trigonal prismatic, and octahedral.
In order to calculate the number of isomers that could be present for each geometry it is best to follow a systematic approach. Since there are fewer B type ligands than A type ligands, the easiest way to do this for each geometry is to start by placing a B ligand at one vertex and then to determine how many different positions are available for the second B ligand.
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:
• position 2 (or 6)
• position 3 (or 5)
• position 4
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. Those of you who remember your ogranic chemistry might recognize that this situation is formally analogous to the ortho-, meta-, and para- isomerism in disubstituted benzenes.
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)
• 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:
• position 2 (or 3 or 4 or 5)
• 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
Even Werner's explanation of isomerism in coordination complexes in terms of octahedral and other recognized coordination geometries did not convince all chemists until he was able to resolve a racemic mixture of d- and l-[Co{Co(NH3)4(OH)2}3] into its enantiomers, which are shown in Scheme $\sf{\PageIndex{II}}$. By doing so Werner demonstrated to chemists of his time (virtually none of whom knew group theory) that tetrahedral carbon atoms were not required for chirality; D3 octahedral complexes were also chiral. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/09%3A_Coordination_Chemistry_I_-_Structure_and_Isomers/9.02%3A_History.txt |
Systems of nomenclature and formulae are intended as tools, to be employed insofar as they are useful
There are well-established rules for both naming and writing the formulae of coordination compounds. The purpose of these rules is to facilitate clear and precise communication among chemists. As with all such rules, some are more burdensome than others to employ, and some serve more crucial roles in the communication process while others are more peripheral - and all are poorly used in the service of pedantic tyranny, especially when used against those who are otherwise doing good work. For this reason, you are urged to approach the rules in a spirit of generosity towards others in
• naming and writing formulas as reasonably accurately as you can so that ligands and metals are where readers expect them and thus can understand what you mean more easily.
• being gracious towards the many professional inorganic chemists who adhere loosely to some of the rules you are about to learn.
• recognizing that in cases when a structure is particularly complex and a picture may make be particularly useful, you should supply one (See the note below).
Note $\sf{1}$: Sometimes the most helpful name to give a compound is 42.
Even though the IUPAC nomenclature rules permit specification of even the most complex structures, it is often much easier and more effective to supply a numbered structure that can be referred to instead of the IUPAC name. Consider bis{[($\mu$-2-mercaptoethyl)(2-mercaptoethyl)-methylthioethylaminato (2-)]Nickel(II)}. Which is easier, to expect readers and hearers to work out the structure from that name or to just refer them to compound 42 in Scheme $\sf{\PageIndex{I}}$?
Coordination Complexes are named as the ligand derivatives of a metal
A variety of systems have been used for naming coordination compounds since the development of the discipline in the time of Alfred Werner. In this section the most common approaches as they are currently used by practicing chemists will be described. Those who need a more thorough and accurate acquaintance with the full IUPAC nomenclature rules are encouraged to consult the IUPAC brief guide to inorganic nomenclature followed by complete guidelines, commonly known as the IUPAC red book. If those are still not enough a careful read of note $\sf{1}$ is suggested.
The systems for naming coordination compounds used at present are additive, meaning that they consider coordination compounds as comprising a central metal to which are added ligands. To specify the structure and bonding in this metal-ligand complex then involves:
1. when there are several different ways of attaching the metal and ligands, specifying the structural or stereoisomer
2. systematically listing the ligands in a way that, as necessary, conveys information about how they are linked to the metal and their stereochemistry
3. providing the identity of the metal and its oxidation state, or if the oxidation state is unclear, at least the overall charge on the complex
4. specifying any counterions present
Since the stereochemistry of coordination compounds forms the subject of the next section, in this section it will be addressed by simply giving the prefixes that designate stereochemistry as if they were self-evident. Do not worry about these for now. They will make sense after you have learned more about stereochemistry in the next section. At that time, you can go back over the examples in that section to solidify your understanding of how to name coordination compounds.
However, in order to name coordination compounds accurately you will need to learn how to think about and name ligands first.
Ligands
Ligands are classified based on whether they bind to the metal center through a single site on the ligand or whether they bind at multiple sites. Ligands that bind through only a single site are called monodentate from the Latin word for tooth; in contrast, those which bind through multiple sites are called chelating after the Greek $\chi \alpha \lambda \epsilon$ for “claw”. These relationships are summarized in Figure $\sf{1}$.
Following naturally from the classification of non-chelating ligands as monodentate, chelating ligands are further classified according to the number of sites which they can use to bind a metal center. This number of binding sites is called the denticity and ligands are referred to as monodentate (non- chelating), bidentate, tridentate, etc., based on the number of sites available. Ligands with two binding sites have a denticity of two and are said to be bidentate; those with three are tridentate, four tetradentate, and so on. To illustrate this classification system examples of chelating ligands classified according to denticity are given in Scheme $\sf{\PageIndex{I}}$.
Scheme $\sf{\PageIndex{I}}$. A selection of chelating ligands classified according to denticity.
Although only one carboxylate oxygen usually binds to a metal, it is still possible to bind a metal using both oxygens. As shown in Scheme $\sf{\PageIndex{IIB}}$, complexes in which both carboxylates bind to a metal are known, and in fact are common in the active sites of some enzymes. It is just that the binding of both oxygens gives a strained four-membered ring that is usually unstable.
Scheme $\sf{\PageIndex{II}}$. (A) Only one oxygen per carboxylate counts towards the denticity of EDTA since on binding the other oxygen generally points away from the metal center, as in the structure of Fe(EDTA)-. This does not mean that both oxygens of a carboxylate can never both bind to metal centers in a complex. (B) Structures in which both oxygens of a carboxylate side chain bind to a metal are sometimes found in the active sites of some of the nonheme iron enzymes your body uses to break down amino acids.
Scheme $\sf{\PageIndex{III}}$. As in this complex, dithiocarbamates commonly bind metals through both sulfur atoms. Consequently, dithiocarbamates are classified as bidentate. This work by Stephen Contakes is licensed under a Creative Commons Attribution 4.0 International License.
Because of these factors it is technically more correct to say that carboxylates usually act as monodentate ligands and dithiocarbamates bidentate ones than it is to say that carboxylates are monodentate ligands and dithiocarbamates bidentate ones. So in other words the ligand classifications presented here just represent common binding modes.
Exercise $1$
Determine the denticity of each ligand in the list below and classify them as monodentate, tridentate, etc.
Answer
(a) bidentate
(b) tridentate
(c) bidentate
(d) tridentate (only the lower N on each ring has a lone pair that can be used to bind the metal)
(e) hexadentate (remember that each carboxylate only counts as one point of attachment)
(f) bidentate
(g) monodentate (through the lone pair on the isocyanide C)
(h) bidentate
This experimentally-based classification of dithiocarbamates as bidentate and carboxylates as monodentate can be confusing to a beginner. Fortunately, such experimentally based classifications are embedded in the lists of common monodentate ligands as given in Table $\sf{1}$ and common chelating ligands in Table $\sf{2}$.
A perusal of the ligands in Table $\sf{1}$ reveals that several can bind to a metal in multiple ways. For example, thiocyanate, SCN- can bind metals through its S or N atoms. Such ligands are called ambidentate ligands. In naming an ambidentate ligand, the atom through which it attaches to the metal is commonly specified after the ligand name using the italicized element symbol or, more formally, a $\kappa$ followed by the italicized element symbol. An example is given in Scheme $\sf{\PageIndex{I}}$.2
Scheme $\sf{\PageIndex{I}}$. Two possible binding modes of nitrite acting as a ligand.3
Table $\sf{1}$. Common monodentate ligands. Most chemists still prefer common names over the IUPAC ones.
Ligand Common name IUPAC name
H- (H ligands are always considered anions for naming purposes) hydrido hydrido
F- fluoro fluorido
Cl- chloro chlorido
Br- bromo bromido
I- iodo iodido
CN-, as M-CN cyano cyanido or cyanido-$\kappa$C or cyanido-C
CN-, as M-NC isocyano isocyanido or cyanido-$\kappa$N or cyanido-N
CH3NC methylisocyanide methylisocyanide
N3- azido azido
SCN-, e.g. thiocyanate as M-SCN thiocyanato thiocyanato-$\kappa$S or thiocyanato-S
NCS-, e.g. thiocyanate as M-NCS isothiocyanato thiocyanato-$\kappa$N or thiocyanato-N
CH3CO2- acetato ethanoato
N3- nitrido nitrido
NH2- imido azanediido
NH2- amido azanido
NH3 ammine ammine
RNH2, R2NH, R3N
alkylamine, dialkylamine, trialkalyamine
(e.g. methylamine for CH3NH2)
alkylamine, dialkylamine, trialkalyamine
(e.g. methylamine for CH3NH2)
, piperidine, abbreviated pip piperidine piperidine
, pyridine, abbreviated py pyridine pyridine
CH3CN, acetonitrile, abbreviated MeCN acetonitrile acetonitrile
P3- phosphido phosphido
PH3 phosphine phosphane
PR3 trialkylphosphine (e.g. trimethylphosphine for Me3P) trialkylphosphane (e.g. trimethylphosphane for Me3P)
PAr3 triarylphosphine (e.g. triphenylphosphine for Ph3P) triarylphosphine (e.g. triphenylphosphane for Ph3P)
, dimethylsulfoxide or DMSO or dmso
dimethylsulfoxide
(sometimes called dimethylsulfoxo but this usage is rare and violates the nomenclature rules for neutral ligands)
(methanesulfinyl)methane or dimethyl(oxido)sulfur
, thiourea or tu thiourea thiourea
O2- oxo oxido
OH- hydroxo hydroxido
H2O aqua aqua
S2- sulfo sulfo
HS- hydrosulfido hydrosulfido
RS- alkanethiolate (e.g. ethanthiolate for EtS-) thioalkanoate
H2S hydrogen sulfide hydrogen sulfide
R2S alkylsulfanylalkane (e.g. ethylsulfanylethane for Et2S) dialkyl sulfide
O2 dioxygen dioxygen
O2-, superoxide superoxido dioxido(1-) or superoxido
O22-, peroxide peroxido dioxido(2-) or peroxido
N2 dinitrogen dinitrogen
NO (are always considered neutral for naming purposes) nitrosyl nitrosyl
CO carbonyl carbonyl
CS thiocarbonyl thiocarbonyl
SO, as M-SO sulfino sulfur monoxide-$\kappa$S or sulfur monoxide-S
NO2, as M-NO2 nitryl nitrogen dioxide-$\kappa$N or nitrogen dioxide-N
CO32- carbonato carbonato
NO2-, as M-NO2 nitro or nitrito-N nitrito-$\kappa$N or nitrito-N
NO2-, as M-ONO nitrito or nitrito-O nitrito-$\kappa$O or nitrito-O
NO3- nitrato nitrato
SO32- sulfito sulfito
SO42- sulfato sulfato
S2O32-, as M-S-SO2-O- thiosulfato-S thiosulfato-$\kappa$S or thiosulfato-S
S2O32-, as M-O-SO2-S- thiosulfato-O thiosulfato-$\kappa$O or thiosulfato-O
Table $\sf{2}$. Common chelating ligands organized by denticity. Most chemists use the common names and abbreviations to describe these ligands.
Common Ligand name IUPAC ligand name abbreviation (if applicable) structure or representative/parent structure (shown in the ionization state in which they bind to a metal)
bidentate ligands
acetylacetonato 2,4-pentanediono acac
R-BINAP and S-BINAP R- or S-2,2'-bis(diphenylphosphino)-1,1'-binapthyl BINAP
2,2'-bipyridine 2,2'-bipyridine bpy or bipy
cyclooctadiene 1,5-cyclooctadiene COD
(binding to the metal occurs through the alkene $\pi$ cloud)
dialkyldithiocarbamato dialkylcarbamodithiolato R2NCS2- or dtc
dimethylgloximato butanedienedioxime
Hdmg
or DMG
diphenylphosphinoethane
or 1,2-(diphenylphosphino)ethane
Ethane-1,2-diylbis(diphenylphosphane) dppe
ethylenediamine Ethane-1,2-diamine en
ethylenedithiolato Ethane-1,2-dithiolato C2H2S22-
nacnac N,N'-diphenyl-2,4-pentanediiminato nacnac
oxalato oxalato ox
1,10-phenanthroline or o-phenanthroline 1,10-phenanthroline phen or o-phen
phenylpyridinato
2-phenylpyridinato-C2,N
or 2-phenylpyridinato-$\kappa$C2,N
ppy
tridentate ligands
triazacyclononane 1,3,7-triazacyclononane tacn
diethylenetriamine 1,4,7-triazaheptane dien
pyrazoylborato
(scorpionate)
hydrotris(pyrazo-1-yl)borato Tp
terpyridine
or 2,2';6',2"-terpyridine
12,22:26,32-terpyridine
or 2,6-bis(2-pyridyl)pyridine, tripyridyl, 2,2′:6′,2″-terpyridine
tpy or terpy
tetradentate ligands
$\beta$, $\beta$', $\beta$''-triaminotriethylamine $\beta$, $\beta$', $\beta$''-tris(2-aminoethyl)amine tren
triethylenetetramine 1,4,7,10-tetraazadecane trien
corroles variable and generally not used cor or Cor
12-crown-4 1,4,7,10-tetraoxacyclododecane 12-crown-4
tetramethylcyclam 1,4,8,11-tetramethyl-1,4,8,11-tetraazacyclotetradecane TMC or cyclam
cyclam 1,4,8,11-tetraazacyclotetradecane cyclam
cyclen 1,4,7,10-tetraazacyclododecane cyclen
tris(2-pyridylmethyl)amine 1-pyridin-2-yl-N,N-bis(pyridin-2-ylmethyl)methanamine tpa or TPA
phthalocyanines variable and generally not used variable, usually a modified Pc
porphyrins variable and generally not used
variable, usually a modified por, Por, or P
(e.g. TPP = tetraphenylporphyrin)
salen 2,2'-ethylenebis(nitrilomethylidene)diphenoxido salen
pentadentate ligands
15-crown-5 1,4,7,10,13-Pentaoxacyclopentadecane 15-crown-5
tetraethylenepentamine 1,4,7,10,13-pentaazatridecane tepa or TEPA
hexadentate ligands
18-crown-6 1,4,7,10,13,16-hexaoxacyclooctadecane 18-crown-6
2,1,1-cryptand 4,7,13,18-Tetraoxa-1,10-diazabicyclo[8.5.5]icosane
2,1,1-crypt
or [2.1.1]-cryptand
kryptofix 211
and variations thereof
ethylenediaminetetraaceto 2,2′,2″,2‴-(Ethane-1,2-diyldinitrilo)tetraaceto EDTA, edta, Y4-
heptadentate ligands
2,2,1-cryptand 4,7,13,16,21-pentaoxa-1,10-diazabicyclo[8.8.5]icosane
2,2,1-crypt
or [2.2.1]-cryptand
kryptofix 221
and variations thereof
octadentate ligands
2,2,2-cryptand 4,7,13,16,21,24-Hexaoxa-1,10-diazabicyclo[8.8.8]hexacosan
2,2,2-crypt
or [2.2.2]-cryptand
kryptofix 222
and variations thereof
pentetato acid or diethylenetriaminepentaacetato or DTPA 2-[bis[2-[bis(carboxylatomethyl)amino]ethyl]amino]acetato DTPA
DOTA or tetraxetan 1,4,7,10-Tetraazacyclododecane-1,4,7,10-tetraacetic acid Dota, DOTA
Rules for Naming Coordination Compounds
As explained above, the name of a coordination compound communicates
• as appropriate, information about isomerism
• systematically listing the ligands in a way that as necessary conveys information about their oxidation state and how they are linked to the metal
• the identity of the metal and its oxidation state
• any counterions present
Before going into these rules it is worth pointing out a few things.
1. It is easiest to learn these rules by starting with one or two of the rules, learning how to apply them, and then adding additional rules one at a time. To that end, instructors who wish to use a more programmed approach may find it convenient to first direct their students to this page which focuses on getting the names of the ligands and metal right, without worrying about isomerism or stereochemistry.
2. The rules also assume some familiarity with common coordination geometries and patterns of isomerism in metal complexes. Thus it might be easiest to learn about common coordination geometries first, followed by common patterns of isomerism in metal complexes before beginning this section. If you decide to dive right in to this section you might find it helpful to know that when applying nomenclature and formula rules most textbooks assume
• complexes in which the metal has a coordination number of six are octahedral
• complexes in which the metal has a coordination number of five are trigonal bipyramidal
• complexes in which PtII , PdII , or RhI, or IrI have a coordination number of four are square planar
• other complexes in which the metal has a coordination number of four are tetrahedral
Like all assumptions these don't always work in real life but they should be good enough to get you through your first inorganic chemistry course.
For the pedants among you, note that the complexes given as examples and in exercises on this page have been selected for pedagogical utility. Although many are well-known compounds, others are hypothetical.
Now on to the rules.
Rule 1: If ions are present, name the cation first, followed by the anion.
Examples:
K2[PtIICl4] potassium tetrachloroplatinate(2-)
[CoIII(NH3)6](NO3)3 hexaamminecobalt(3+) nitrate
[CoIII(NH3)6][CrIII(C2O4)3] hexaamminecobalt(3+) tris(oxalato)chromium(3-)
Rule 2: When multiple isomers are possible, designate the particular isomer in italics at the front of the name of each complex
If you have not yet learned about isomerism in coordination compounds skip this rule for now and return to it after you have.
When a complex might exist as one of two stereoisomers, prefixes are commonly used to designate which isomer is present. The most common cases are listed in Table $\sf{3}$.
Table $\sf{3}$. Prefixes used to specify isomerism about a metal center when naming and writing coordination compounds' formulae.
Type of isomerism Graphical reminder Prefixes
Geometric, cis- , trans- cis- or trans-
Geometric, fac- /mer- fac- or mer-
Enantiomers, $\Lambda$-, $\Delta$- $\Lambda$- or $\Delta$-
Examples of how isomerism about a metal center is designated are given in Scheme $\sf{\PageIndex{II}}$.
Scheme $\sf{\PageIndex{II}}$. Application of nomenclature rules for stereosimerism about a metal.3
There are a number of other cases where it might be advisable to specify the stereochemistry of a complex. These cases involve specifying
• the coordination geometry about a metal center (octahedral, trigonal prismatic, tetrahedral, square planar, etc. )
• the geometry cannot be unambigouously described by a single cis/trans or fac/mer relaionship of ligands
These cases may also be handled by using a designator to specify the coordination geometry and, as necessary, giving the position of ligated atoms in terms of designated numbered positions for that geometry. See the IUPAC red book for details as such cases fall outside the scope of what is normally advisable for an undergraduate course.
Rule 3: Specify the identity, number, and as appropriate, isomerism of the ligands present in alphabetical order by ligand name.
Before specifying the metal, the ligands are written as prefixes of the metal.
In specifying the ligands several rules are followed.
1. The ligands are written in alphabetical order by the ligand name only; symbols are not considered and prefixes do not count in determining alphabetical order.
Example: In the name of the complex ion [Co(NH3)5Cl]2+, pentamminechlorocobalt(II), the ammine ligand is named before the chloro ligand because the order is alphabetical by the ligand name by virtue of which ammine comes before chloro.
1. Prefixes are used to indicate the number of each ligand present. Specifically, di-, tri-, tetra- , penta-, hexa-, etc. prefixes are used to indicate multiple ligands of the same type EXCEPT when the ligand is polydentate or its name already has a di-, tri-, tetra- etc. In that case bis-,tris-, tetrakis-, etc. are used instead. These prefix rules are summarized in Table $\sf{3}$.
Table $\sf{3}$. Prefixes used to specify the number of a given ligand present.
Number of identical ligands prefix used when the ligand name is simple prefix used when the ligand is polydentate or its name already has a di-, tri-, tetra- etc.
2 di- bis-
3 tri- tris-
4 tetra- tetrakis-
5 penta- pentakis-
6 hexa- hexakis-
7 hepta- heptakis-
8 octa- octakis-
9 nona- nonakis-
10 deca- decakis-
An example of the application of the prefix rule is given in Scheme $\sf{\PageIndex{III}}$.
Scheme $\sf{\PageIndex{III}}$. Example of the use of prefixes to specify the number of ligands of each type in a complex.3
1. Ligand names are based on their charge.
• Neutral and cationic ligand names are the same as the names of their neutral compounds with two caveats:
1. names that involve spaces should either be put in parentheses or the spaces should be eliminated (preferred)
Example: cis-dichlorobis(dimethyl sulfoxide)platinum(II) or cis-dichlorobis(dimethylsulfoxide)platinum(II)
1. A few ligands are given common names.
• H2O = aqua
• NH3 = ammine (notice that there are two n's)
• CO = carbonyl
• CS = thiocarbonyl
• NO = nitrosyl
• For anionic ligands, the vowel at the end of their anion names is changed to an -”o”
Examples: Cl- = chloro, NH2- = amido, N3- = azido
Caveat: some anionic ligands have common names that may also be used
Examples:
I- = iodo or iodino
CN- = cyano or cyanido
O2- = oxo or oxido
The IUPAC and common names of many ligands are given in Tables $\sf{1}$. and $\sf{2}$.
1. When an ambidentate ligand is present, the atom through which it is bound to the metal is indicated by giving either its element symbol or a $\kappa$ and its element symbol in italics after the ligand name.
Example:
M-SCN is thiocyanato-S or thiocyanato-$\kappa$S
M-NCS = thiocyanato-N or thiocyanato-$\kappa$N
The use of $\kappa$ and an element symbol to indicate how a ligand and metal are linked is called a k-term. More complex k-terms might also involve specifying the atoms by number, though their use is outside the scope of this text.
1. As appropriate, additional information about the way a ligand is bound to the metal center and/or its stereochemistry is specified using a prefix. The prefixes to provide linkage and stereochemistry for ligands are given in Table $\sf{4}$.
Table $\sf{4}$. Prefixes used to specify ligands' isomerism when naming and writing coordination compounds' formulae. Some of these types of isomerism will be discussed in later pages.
Type of isomerism Graphical reminder Prefixes
when a multidentate ligand binds through less than the full number of atoms $\kappa$n where n is the number of attached atoms; used when the attached atoms are not directly connected by a chemical bond. The metal-ligand bonding usually involves $\sigma$-type coordination.
hapticity
$\eta$n where n is the number of attached atoms; used when the coordinated atoms are all connected by bonds. Usually the metal-ligand bonding involves $\pi$-type coordination.
In speech, $\eta^1$=monohapto; $\eta^2$=dihapto; $\eta^3$=trihapto, etc.
bridging ligands $\mu$n where n is the number of atoms bridged. The number n is usually omitted when n =2.
chelating ligand ring twist $\lambda$- or $\delta$-
A example showing how the nomenclature rule is applied to a ligand that can have two coordination modes is given in Scheme $\sf{\PageIndex{IV}}$.
Scheme $\sf{\PageIndex{IV}}$. Use of the $\kappa$ notation to specify the number of attached groups in a multidentate ligand.
Scheme $\sf{\PageIndex{V}}$. Use of the $\mu$ notation to specify bridging ligands in metal complexes.3
1. If desired, parentheses may be used to delineate a ligand name to make it easier to identify in the name of the complex. This can be particularly helpful when the entire name contains a lot of information to keep track of. An example is given in Scheme $\sf{\PageIndex{VI}}$.
Scheme $\sf{\PageIndex{VI}}$. When naming the complex shown, cis-diaquabis(ethylenediamine)chromium(III) nitrate is easier to read than cis-diaquabisethylenediaminechromium(III) nitrate.
Rule 4: Specify the identity of the metal
• In neutral and cationic complexes the metal's name is used directly
- e.g. as in hexammineruthenium(III) for [Ru(NH3)6]3+
• In anionic complexes, -ate replaces -ium, -en, or –ese or adds to the metal name,
e.g., as in hexachloromanganate(IV) for MnCl62-
• In anionic complexes of some metals a Latin-derived name is used instead of the element's English name. These names are given in Table $\sf{5}$.
Table $\sf{5}$. Latin terms for select metal Ions. Redrawn from this page describing the nomenclature of coordination complexes.
Transition Metal Latin
Copper Cuprate
Gold Aurate
Iron Ferrate
Lead Plumbate
Silver Argentate
Tin Stannate
An example of the application of the metal naming rules is given in Scheme \sf{\PageIndex{VII}}\).
Scheme $\sf{\PageIndex{VII}}$. Example of the application of the metal specification rules to a cationic and an anionic platinum complex.3
Rule 5: Specify the oxidation state of the metal.
Two different systems are used to specify the oxidation state of the metal.
1. In the Stock system the metal's oxidation state is indicated in Roman numerals after the metal name.
Examples:
[CoCl(NH3)5]Cl2 = pentamminechlorocobalt(III) chloride
[PtBr2(bpy)] = dibromobipyridineplatinium(II)
K[Ag(SCN)2] = potassium di-S-thiocyanatoargentate(I)
1. In the Ewing-Bassett system the charge on the complex is specified in Arabic numerals after the complex name. This provides a way of specifying a complex even when the oxidation state of the metal isn't known and, in cases where it is known, the value of the metal's oxidation state may be inferred from the complex ion's charge.
[CoCl(NH3)5]Cl2 = pentamminechlorocobalt(2+) chloride
[PtBr2(bpy)] = dibromobipyridineplatinium(0)
K[Ag(SCN)2] = potassium di-S-thiocyanatoargentate(1-)
Summary of rules for naming coordination complexes.
A graphical summary of the rules for naming complexes along with a few examples that you can use to review the nomenclature rules is given in Figure $\sf{1}$. This work by Stephen Contakes is licensed under a Creative Commons Attribution 4.0 International License.
When learning chemical nomenclature, practice makes perfect. The following examples and exercises are provided to give you this practice. Additional examples and exercises on the https://chem.libretexts.org/ site include a set of simple examples with explained solutions, a set of simple exercises with solutions, and a set of more challenging exercises without solutions.
Exercise $2$. Assigning metal oxidation states in a complex
In order to name a complex in the Stock system it is necessary to assign a formal oxidation state to the metal. For this reason it is important to be able to assign the oxidation state of a metal in a complex. Fortunately, this is easy to do if you remember
1. The sum of all atoms' oxidation states will equal the overall charge on the complex
2. When determining the metal's oxidation state the ligands can be treated as having an oxidation state equal to their charge - i.e., the charge they possess in the form in which they coordinate the metal, so if they need to lose a proton to bind, don't forget to account for that.
Given the above, assign the oxidation state of the metal in the following real and hypothetical complexes.
1. K3[Fe(CN)6]
2. [CoCl(NH3)5](NO3)2
3. K2[PtCl4]
4. [MnCl(por)]
5. [Ru(bpy)3]Cl2
6. [PdCl2(dppe)]
7. [Mn(en)2(SCN)2]
Answer for K3[Fe(CN)6]3-.
This contains [Fe(CN)6]3-; so O.S.Fe + 6 x (-1) (for CN-) = -3 (the complex's charge) so O.S.Fe = +3 or Fe3+.
Answer for [CoCl(NH3)5](NO3)2.
This contains [CoCl(NH3)5]2+; so O.S.Co + 1 x (-1) (for Cl-) + 0 x 5 (for NH3) = +2 (the complex's charge) so O.S.Co = +3 or Co3+.
Answer for K2[PtCl4].
This contains 2K+ and [PtCl4]2-; so O.S.Pt + 4 x (-1) (for Cl-) = -2 (the complex's charge) so O.S.Pt = +2 or Pt2+.
Answer for [MnCl(por)].
O.S.Pt + 1 x (-2) (for por; see table 9.2.2) + 1 x (-1) (for Cl-) = +0 (the complex's charge) so O.S.Mn = +3 or Mn3+.
Answer [Ru(bpy)3]Cl2.
The complex is [Ru(bpy)3]2+ so O.S.Ru + 0 x 3 (for bpy) = +2 (the complex's charge) so O.S.Ru = +2 or Ru2+.
Answer [PdCl2(dppe)].
O.S.Pd + 2 x (-1) (for Cl-) + 0 x 3 (for dppe) = +0 (the complex's charge) so O.S.Pd = +2 or Pd2+.
Answer [Mn(en)2(SCN)2].
O.S.Mn + (2 x 0) (for en) + 2 x (-1) (for SCN-) = +0 (the complex's charge) so O.S.Mn = +2 or Mn2+.
Exercise $3$: Simple Nomenclature Problems.
Name the following compounds in both the Stock and Ewing-Bassett systems:
1. [Ru(NH3)6](NO3)3
2. K2[PtCl4]
3. K[Ag(CN)2]
4. Cs[CuBrCl2F]
5. [Cu(acac)2]
Answer
Complex Stock system name Ewing-Bassett System name
a [Ru(NH3)6](NO3)3 hexammineruthenium(III) nitrate hexammineruthenium(3+) nitrate
b K2[PtCl4] potassium tetrachloroplatinate(II) potassium tetrachloroplatinate(2-)
c K[Ag(CN)2] potassium dicyanoargentate(I) potassium dicyanoargentate(1-)
d Cs2[CuBrCl2F] cesium bromodichlorofluorocuprate(II) cesium bromodichlorofluorocuprate(2-)
e [Cu(acac)2] bis(acetylacetonato)copper(II) bis(acetylacetonato)copper(0)
Exercise $4$: More simple nomenclature problems.
Name the following compounds and ions in both the Stock and Ewing-Bassett systems.
1. Cu(OH)4-
2. [AuXe4]2+
3. AuCl4-
4. Fe(CN)63-
5. K4[Fe(CN)6]
6. trans-[Cu(en)2(NO2)2] (the N is bound to Cu)
7. cis-IrCl2(CO)(PPh3) (ignore stereochemistry)
8. IrCl(PPh3)3
Answer
Compound Stock System Name Ewing-Bassett System Name
a Cu(OH)4- tetrahydroxidocuprate(III) or tetrahydroxidocuprate(III) tetrahydroxidocuprate(1-) or tetrahydroxidocuprate(1-)
b [AuXe4]2+ tetraxenongold(II) tetraxenongold(2+)
c AuCl4- tetrachloroaurate(III) tetrachloroaurate(1-)
d Fe(CN)63- hexacyanoferrate(III) or hexacyanidoferrrate(III) hexacyanoferrate(3-) or hexacyanidoferrate(3-)
e K4[Fe(CN)6] potassium hexacyanoferrate(II) or potassium hexacyanidoferrrate(II) potassium hexacyanoferrate(4-) or potassium hexacyanidoferrrate(4-)
f trans-[Cu(en)2(NO2)2] (the N is bound to Cu) bis(ethylenediamine)bisnitrocopper(II) or bis(ethylenediamine)bis(nitrito-$\kappa$N)copper(II) bis(ethylenediamine)bisnitrocopper(0) or bis(ethylenediamine)bis(nitrito-$\kappa$N)copper(0)
g cis-IrCl2(CO)(PPh3)
cis-dichlorocarbonyltriphenylphosphineiridium(I)
or cis-dichloro(carbonyl)(triphenylphosphine)iridium(I)
cis-dichlorocarbonyltriphenylphosphineiridium(0)
or cis-dichloro(carbonyl)(triphenylphosphine)iridium(0)
h IrCl(PPh3)3 chlorotris(triphenylphosphine)iridium(I) chlorotris(triphenylphosphine)iridium(0)
Exercise $5$: Even more simple nomenclature problems.
Name the following compounds and ions in both the Stock and Ewing-Bassett systems. Ignore prefixes for designating isomers if you haven't learned about those.
1. Fe(acac)3
2. K2[CuBr4]
3. ReH9
4. [Ag(NH3)2]BF4
5. [Ag(NH3)2][Ag(CN)2]
6. [Ni(CN)4]2-
7. [Co(N3)(NH3)5]SO4
8. [CoBrCl(H2O)(NH3)]I (ignore stereochemistry)
Sample Answers
Complex Stock system name Ewing-Bassett System name
a Fe(acac)3 tris(acetoacetato)iron(III) tris(acetoacetato)iron(0)
b Na2[CuBr4] sodium tetrabromocuprate(II) sodium tetrabromocuprate(2-)
c [Co(NH3)6][Co(ox)3] hexamminecobalt(III) tris(oxalato)cobalt(III) hexamminecobalt(3+) tris(oxalato)cobalt(3-)
d [Ag(NH3)2]BF4 diamminesilver(I) tetrafluoroborate diamminesilver(1+) tetrafluoroborate
e [Ag(NH3)2][Ag(CN)2]
diamminesilver(I) dicyanoargentate(I)
or diamminesilver(I) dicyanidoargentate(I)
diamminesilver(1+) dicyanoargentate(1-)
or diamminesilver(1+) dicyanidoargentate(1-)
f [Ni(CN)4]2- tetracyanonickelate(II) or tetracyanonickelate(II) ion tetracyanonickelate(2-) or tetracyanonickelate(2-) ion
g [Co(N3)(NH3)5]SO4 pentammineazidocobalt(III) sulfate pentammineazidocobalt(2+) sulfate
h
[CoBrCl(H2O)(NH3)]I
(ignore stereochemistry)
ammineaquabromochlorocobalt(III) iodide ammineaquabromochlorocobalt(1+) iodide
Exercise $6$: Nomenclature problems, some of which involve consideration of isomerism.
Name the following compounds and ions in both the Stock and Ewing-Bassett systems. Ignore prefixes for designating isomers if you haven't learned about those.
1. trans-[Cu(dppe)2(NO2)2] (the N is bound to Cu)
2. [Pd(en)2(SCN)2], with the thiocyanates bound Pd-SCN
3. [Mn(CO)6]BPh4 (BPh4 = tetraphenylborate)
4. Rb[AgF4]
5. K2ReH9
6. K3CrCl6
7. [Ru(H2O)6]Cl2
8. [cis-Fe(CO)4I2]
9. K2[trans-Fe(CN)4(CO)2]
10. [cis-MnCl(H2O)4(NH3)](NO3)
11. K3[fac-RuCl3(PMe3)3]
Answer
Complex Stock system name Ewing-Bassett System name
a trans-[Cu(dppe)2(NO2)2] trans-bis(diphenylphosphinoethane)bisnitrocopper(II) or trans-bis(diphenylphosphinoethane)bis(nitrito-$\kappa$N)copper(II) trans-bis(diphenylphosphinoethane)bisnitrocopper(0) or trans-bis(diphenylphosphino)ethanebis(nitrito-$\kappa$N)copper(0)
b [Pd(en)2(SCN)2], with the thiocyanates bound Pd-SCN
bis(ethylenediamine)bisthiocyanatopalladium(II)
or bis(ethylenediamine)bis(thiocyanato-S)palladium(II)
or bis(ethylenediamine)bis(thiocyanato-$\kappa$S)palladium(II)
bis(ethylenediamine)bisthiocyanatopalladium(0)
or bis(ethylenediamine)bis(thiocyanato-S)palladium(0)
or bis(ethylenediamine)bis(thiocyanato-$\kappa$S)palladium(0)
c [Mn(CO)6]BPh4 hexacarbonylmanganese(I) tetraphenylborate hexacarbonylmanganese(1+) tetraphenylborate
d Rb[AgF4] rubidium tetrafluoroargentate(III) rubidium tetrafluoroargentate(1-)
e K2ReH9 potassium nonahydridorhenium(VII) potassium nonahydridorhenium(2-)
f K3CrCl6 or K3[CrCl6]
potassium hexachlorochromium(III) or
potassium hexachloridochromium(III)
potassium hexachlorochromium(3-) or
potassium hexachloridochromium(3-)
g [Ru(H2O)6]Cl2 hexaaquaruthenium(II) chloride hexaaquaruthenium(2+) chloride
h [cis-Fe(CO)4I2] cis-tetracarbonyldiiodoiron(II) cis-tetracarbonyldiiodoiron(0)
i K2[trans-Fe(CN)4(CO)2] potassium trans-dicarbonyltetracyanoferrate(II) potassium trans-dicarbonyltetracyanoferrate(2-)
j [cis-MnCl(H2O)4(NH3)](NO3) cis-amminetetraaquachloromanganese(0) nitrate cis-amminetetraaquachloromanganese(1+) nitrate
k K[fac-RuCl3(PMe3)3] potassium fac-trichlorotris(triphenylphosphine)ruthenium(II) potassium fac-trichlorotris(triphenylphosphine)ruthenium(1-)
Exercise $7$
Name the compounds and ions below using both the Stock and Ewing-Bassett systems. Ignore prefixes for designating isomers if you haven't learned about those.
Answer
# Structure Stock system name Ewing-Bassett system name
1
cis-tetraacetonitriledicyanoiron(II)
or cis-tetraacetonitriledicyanidoiron(II)
cis-tetraacetonitriledicyanoiron(0)
or cis-tetraacetonitriledicyanidoiron(0)
2
trans-tetraacetonitriledicyanoiron(II)
or trans-tetraacetonitriledicyanidoiron(II)
trans-tetraacetonitriledicyanoiron(0)
or trans-tetraacetonitriledicyanidoiron(0)
3
trans-bromochlorobis(ethylenediamine)iron(III)
or trans-bromidochloridobis(ethylenediamine)iron(III)
trans-bromochlorobis(ethylenediamine)iron(1+)
or trans-bromidochloridobis(ethylenediamine)iron(1+)
4
fac-tricarbonyltricyanomolybdate(0)
or fac-tricarbonyltricyanidomolybdate(0)
fac-tricarbonyltricyanomolybdate(3-)
or fac-tricarbonyltricyanidomolybdate(3-)
5
mer-tricarbonyltricyanomolybdate(0)
or mer-tricarbonyltricyanidomolybdate(0)
mer-tricarbonyltricyanomolybdate(3-)
or mer-tricarbonyltricyanidomolybdate(3-)
6
pentamminenitrito-N-cobalt(III),
pentamminenitrito-$\kappa$N-cobalt(III),
or pentamminenitrocobalt(III)
pentamminenitrito-N-cobalt(2+),
pentamminenitrito-$\kappa$N-cobalt(2+),
or pentamminenitrocobalt(2+)
7
pentamminenitrito-O-cobalt(2+),
pentamminenitrito-$\kappa$O-cobalt(2+),
or pentamminenitritocobalt(2+)
pentamminenitrito-O-cobalt(2+),
pentamminenitrito-$\kappa$O-cobalt(2+),
or pentamminenitritocobalt(2+)
Exercise $8$
Draw structural formulae for the following compounds and ions. You may assume that
• complexes in which the metal has a coordination number of six are octahedral
• complexes in which the metal has a coordination number of five are trigonal bipyramidal
• complexes in which PtII , PdII , or RhI, or IrI have a coordination number of four are square planar
• other complexes in which the metal has a coordination number of four will be tetrahedral
1. (2,2'-bipyridine)tetracyanoruthenium(2-)
2. sodium tetrachloroaluminate (note that since Al is a main group metal with a generally fixed oxidation state no oxidation state is given)
3. pentaamminechlorocobalt(2+) sulfate
4. carbonylhydridotris(triphenylphosphine)rhodium(I) (the ligands in this complex occupy sterically preferred positions)
5. bromotrichlorocobaltate(III)
6. hexaaquacopper(2+) sulfate
7. sodium tris(oxalato)cobalt(III)
8. fac-(1,10-phenanthroline)tricarbonylchlororhenium(I)
9. mer-triaquatrichlorochromium(III)
10. trans-dichlorobis(ethylenediamine)platinum(IV)
Answers
a
b
c
d
e
f
g
h
i
j
Exercise $9$.
The name of the structure named tris(tetraammine-$\mu$-dihydroxocobalt)cobalt(6+) in Scheme $\sf{\PageIndex{III}}$ (reproduced below) is incomplete. Give the complete name of the structure in both the Stock and Ewing-Basset systems.
Answer
$\Delta$-tris(tetraammine-$\mu$-dihydroxocobalt)cobalt(6+) or $\Delta$-tris(tetraammine-$\mu$-dihydroxocobalt(III)cobalt(III)
Rules for writing the formulae of coordination compounds
The rules for writing the formulae of coordination compounds follow the same convention used to specify their names. However, according to a myth that has been widely propagated on educational websites, the charge need not be listed. As organized predictable formulae can be easier to read and understand than haphazardly written ones, the rules for writing the formulae of coordination compounds have value. To support those who wish to employ them they are given in this section. The rules as given here are adapted from a summary by Robert Lancashire.
1. If there are multiple ions present list the cations before anions.
2. Enclose all the constituents of each complex ion in square brackets.
3. For each complex ion,
• Give the central metal atom first.
• Then ligands next, listed in alphabetical order, ignoring prefixes according to the first letters in the ligand's symbol as written. This is true regardless of whether the symbol is an element symbol (like C, N, O, etc.) or a symbol for the ligand name (bpy, en, MeCN, etc.). Contrary to widely-circulated myths, the ligand's charge does not matter.4
• The formulae or abbreviations (e.g. en) for all polyatomic ligands should be enclosed in ordinary parentheses.
• As appropriate, use italicized atom symbols to indicate linkage isomerism and prefixes such as cis-, trans-, fac-, mer-, $\Lambda$-, $\Delta$-, κn, ηn , μn, $\lambda$-, or $\delta$- to indicate stereochemistry.
• When a ligand is bound to a metal through a particular atom, preferably place that atom closest to the metal - e.g., [Fe(CN)6]3- not [Fe(NC)6]3- (Note: this rule is used primarily for ambidentate ligands; although IUPAC recommends that aqua ligands be written as OH2 when the O would be closest to its coordinated metal, they are still usually written as H2O).
These rules are graphically summarized in Figure $\sf{2}$.
Exercise $10$
Give the formulae of the following complexes.
.
Answer
Exercise $11$
Write formulae for each of the following compounds and ions. When multidentate ligands are present use suitable abbreviations.
1. pentaamminechlorocobalt(2+) sulfate
2. $\Delta$-diamminebis(oxalato)manganate(III)
3. trans-tetraacetonitriledicyanoiron(II)
4. tricarbonyldichlorobis(triphenylphosphine)molybdenum
Answers
a.
[CoCl(NH3)5](SO4)
b.
$\Delta$-[Mn(NH3)2(ox)2]-
c.
trans-[Fe(CN)2(NCMe)4]
d.
[MoCl2(CO)3(PPh3)] or MoCl2(CO)3(PPh3)
e.
mer-[CrBrCl(H2O)3I]
f.
[Co(O2)py(salen)]
g.
[Fe(NO)2(SEt)2]-
h.
[MnCl(por)]- or [MnCl(porphyrin)]-
i.
[Ni(DMG)2] or [Ni(Hdmg)2]
j.
potassium trans-[Fe(CN)2(CO)4]
k.
trans-[CuCl(H2O)4(NH3)]SO4
l.
trans-[PtCl2(en)2]2+
m.
cis-[CoBrCl(NH3)4]SO4
n.
K2[Fe(CN)5NO]
Multinuclear coordination complexes
Multinuclear coordination complexes contain multiple metals connected by one or more bridging ligands. The structures of bridging complexes can usually be inferred from their $\mu$-tagged ligands in their names and formulae. For the benefit of instructors who wish to have their students name multinuclear complexes, the rules are presented below.
Multinuclear complexes are named differently depending on whether the groups on either side of the bridging ligands are identical or different, as shown in Scheme $\sf{\PageIndex{VIII}}$.
Scheme $\sf{\PageIndex{VIII}}$. Multinuclear coordination complexes may be named differently depending on whether the groups on either side of the bridging ligands are the same or different. The groups are the same if the metal, ligands, and ligand arrangement are identical. This is true in A but in B the metals differ, in C the ligands differ, and in D both the metal and ligands differ.
Naming multinuclear complexes
Let's look at the rules for naming symmetric and asymmetric multinuclear complexes.
Application of the IUPAC system and variants thereof to symmetric complexes
The IUPAC naming system helpfully avoids the sort of ambiguities and ad hoc choices involved in most textbook-level nomenclature systems for multimetallic complexes. Unfortunately, it is correspondingly difficult for beginners to employ. Thus it will only be applied in depth to the case of symmetric complexes. Symmetric complexes are particularly easy to name in the IUPAC system and several of its variants that find common use. In these
1. The ligands are given in alphabetical order. When there are bridging and non-bridging ligands of the same type the bridging ligands are given first. When there are multiple bridging ligands of the same type but which use different bridging modes (e.g., $\mu$4-, $\mu$3-, $\mu$2-), the bridging ligands are specified in decreasing order of bridging multiplicity, e.g., $\mu$3 -sulfido-di-$\mu$-sulfido.
2. The groups bridged are given afterwards using names that follow the ordinary rules. Generally this involves either
1. naming all the ligands followed by all the metals, in both cases using prefixes to indicate the number of each, or
2. naming each group of atoms individually as in the less formal naming system, using prefixes to indicate the number of ligands (this is an unofficial variant of the IUPAC system that some textbook authors seem to prefer).
These two rules are sufficient to describe simple symmetric bridging complexes.
Example: The compound in Scheme $\sf{\PageIndex{VIIIA}}$
may be named
[$\mu$-amido-$\mu$-hydroxo-octaamminedichromium(4+)] ion
or
[$\mu$-amido-$\mu$-hydroxo-bis(tetraamminechromium(III))] ion
or
[$\mu$-amido-$\mu$-hydroxo-bis(tetraamminechromium)(4+)] ion
Example: The complex5
may be named [tri-$\mu$-carbonyl-bis(tricarbonyliron)(0)], [tri-$\mu$-carbonyl-bis(tricarbonyliron(0))], or [tri-$\mu$-carbonyl-hexacarbonyldiiron)(0)]
Example: The complex
may be named di-$\mu$-chlorido-tetrachloridodicopper(II), di-$\mu$-chlorido-bis(dichlorocopper(II)), or di-$\mu$-chlorido-bis(dichlorocopper)(0)
A note on the application of the IUPAC system and variants thereof to asymmetric complexes
For a symmetrical complex like the [Cu2Cl4($\mu$-Cl)2] considered above it is enough to specify the existence of the two bridging and four terminal chloro ligands; there is no need to number the chloro ligands or to specify exactly which ones are involved in bridging or to clarify that the bridging involves $\kappa$2 coordination of the chloro ligands. In cases where the two metal centers or the chloro ligands differ it is necessary to specify the exact ligands and metals involved and how they are connected.
For example, a more extensive IUPAC name for [Cu2Cl4($\mu$-Cl)2] in which the chloro ligands are individually and more completely specified would read di-$\mu$2-chlorido-tetrachlorido-1$\kappa$2Cl,2$\kappa$2Cl-dicopper(II).
Even more extensive systems would involve numbering the metals and specifying how they are connected together too. The details of how this is done are typically beyond the level of most undergraduate and graduate courses in inorganic chemistry. Those who want to know the details should consult the red book. In many cases it is sufficient to reserve the use of formal IUPAC names for use in publications and to employ a convenient shorthand naming system for everyday use. An example of one type of system that is sometimes employed is given next.
Unofficial but commonly used methods applicable to complexes in which the groups bridged are identical or different.
It can be quite complicated to use the IUPAC system to name asymmetric multinuclear coordination complexes. Fortunately it is rarely necessary and there are a variety of simpler somewhat ad hoc methods that work well for such cases. Since these methods are commonly used by textbooks and inorganic instructors it is likely worth your while to learn about their general features. In these methods
1. A multinuclear complex is named as a derivative of one of the metal centers and the other metal centers and their ligands are treated as ligands to the prioritized metal center.
2. Unfortunately, most of the time the choice of which metals are part of the ligands and which one is central is made haphazardly. In other words, there is no agreed upon system for assigning which metal center is the central metal and which should be regarded as part of the ligands around it. One way to be more systematic about the selection of the central and ligand-embedded metals is to assign the central metal as the metal of highest priority in the IUPAC priority rules. In the IUPAC priority rules, the central metal is the highest priority metal alphabetically by name; then if there is a tie the ligands on each tied metal center are ranked alphabetically and the tied metal center with the highest priority (or most highest priority) ligands wins.
3. The metal centers not chosen as the primary metal center and all ligands around them, including those bridging to the winning metal center, are then treated as a single ligand that coordinates the winning metal center.
4. When naming the ligands and the overall complex, again there is much that is haphazard. However, in the best systems each coordination sphere is named using the same rules as for mononuclear complexes - i.e., the ligands are given in alphabetical order, etc.
• This is true of any metal center-containing ligands.
• When there is a bridging and non-bridging ligand of the same type, the bridging ligands are given first. When there are multiple bridging ligands of the same type but which use different bridging modes (e.g. $\mu$4-, $\mu$3-, $\mu$2-), the ligands are specified in decreasing order of bridging multiplicity, e.g. $\mu$3 -sulfido-di-$\mu$-sulfido.
Let's apply these rules to the examples in Scheme $\sf{\PageIndex{VIII}}$.
Compound A:
Using hydroxo for the OH-, it may be named
[($\mu$-amido-tetraammine-$\mu$-hydroxochromium(III))tetraamminechromium(III)] ion
or
[($\mu$-amido-tetraammine-$\mu$-hydroxochromium)chromium(4+)] ion
Compound B:
Using hydroxo for the OH-, it may be named
[(pentaammine-$\mu$-hydroxocobalt(III))tetraamminechromium(III)] ion
or
[(pentaammine-$\mu$-hydroxocobalt)tetraamminechromium(5+)] ion
Compound C:
Using hydroxo for OH-, it may be named
[pentaammine(pentachloro-$\mu$-hydroxocobalt(III))cobalt(III)]
or
[pentaammine(pentachloro-$\mu$-hydroxocobalt)cobalt](0)]
Compound D:
Using hydroxo for OH- and chloro for Cl-, it may be named
[pentaammine(pentachloro-$\mu$-hydroxocobalt(III))chromate(III)] ion
or
[(tetrammine-$\mu$-ammine-$\mu$-hydroxocobalt)tetrachlorochromate](1-)] ion
As can be seen from the examples above this system gives serviceable names for multimetallic complexes but those names are not the IUPAC names and so should not be used to describe complexes outside of pedagogical and informal settings.
Writing and interpreting formulae for multinuclear complexes
The rules for writing formulae for multinuclear complexes are the same as for mononuclear ones with two added details:
1. Write bridging ligands after nonbridging ligands of the same type. For example, Cu2Cl6 should be written as [Cu2Cl4-($\mu$-Cl)2]
2. However, you may ignore that and other rules if you find it helpful to keep groups of metals and ligands together in a way that better conveys how the atoms are connected. Just as the structure of ethane may be more clearly conveyed as H3CCH3 instead of CH3CH3, the complex [Cu2Cl4($\mu$-Cl)2] may be written as [(Cl2Cu)($\mu$-Cl)2(CuCl2)].
Other Examples:
For dichromate write, [Cr2O6($\mu$-O)]2- or [O5Cr-$\mu$-O-CrO5]2-
For compound C of Scheme $\sf{\PageIndex{VIII}}$.
write [Co2Cl5(NH3)5($\mu$-OH)] or, even better, [(H3N)5Co($\mu$-OH)CoCl5].
Exercise $10$
Write reasonable formulae for complexes A, B, and D in Scheme $\sf{\PageIndex{VIII}}$, which for convenience is reproduced below.
Sample answers for A
[Cr2(NH3)8-$\mu$(NH2)-$\mu$(OH)]4+ or [(H3N)4CrIII-$\mu$(NH2)-$\mu$(OH)-CrII(NH3)4] and variants thereof involving writing the H3N as NH3, $\mu$ as $\mu_2$, etc.
Sample answers for B
[CoCr(NH3)10-$\mu$(OH)]5+ or, better, [(H3N)5CoIII-$\mu$(OH)-CrIII(NH3)5]5+
Sample answers for C
This was already done as an example above, [Co2Cl5(NH3)5-$\mu$(OH)] or, better, [(NH3)5CoIII-$\mu$(OH)-CoIIICl5] and variants thereof
Sample answers for D
[CoCrCl4(NH3)4-$\mu$(NH2)-$\mu$(OH)] or, better, [(H3N)4CoIII-$\mu$(NH2)-$\mu$(OH)-CrIICl4]
Contributors and Attributions
Stephen Contakes, Westmont College, to whom comments, corrections, and criticisms should be addressed.
with some examples taken from Naming Transition Metal Complexes by Kathryn Haas.
Consistent with the policy for original artwork made as part of this project, all unlabeled drawings of chemical structures are by Stephen Contakes and licensed under a Creative Commons Attribution 4.0 International License. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/09%3A_Coordination_Chemistry_I_-_Structure_and_Isomers/9.03%3A_Nomenclature_and_Ligands.txt |
Metal complexes present a rich, interesting, and diverse structural chemistry. Major points of variation in the structural chemistry of metal complexes include
1. coordination number and coordination geometry, which involve differences in how many ligands surround a central metal and their overall geometric arrangement. Examples of the latter include the tetrahedral and square planar geometries commonly observed for four-coordinate metal centers. Tetrahedral, square planar, and octahedral coordination serve as a backdrop to the discussion of isomerism on this page. Readers who are unfamiliar with these structures might consider reading the section on coordination geometries before this one.
2. structural iosmerism involves differences in how potential ligand atoms are bound to metals in a complex. Possibilities for structural isomerism include
• linkage or ambidentate isomerism
• hydrate/solvate isomerism
• ionization isomerism
• coordination isomerism.
Of these, linkage isomerism should always be considered when working with ambidentate ligands. As classifications, the last three forms of structural isomerism are mainly of academic interest since they represent permutations of ordinary structure patterns for solvates, salts, and coordination complexes, respectively.
1. stereochemistry, which includes
• which coordination geometry is adopted for a metal with a given set of ligands in a particular oxidation state. For instance, [NiIICl4]2- is tetrahedral while [PtIICl4]2- is square planar. Since this geometry is usually fixed by the metal and ligands it is typically not a source of stereoisomeric varierty.
• metal-centric stereoisomerism involving possible variations in where ligands are located relative to one another around the metal. The possibilities depend on the coordination number, geometry, and number of different types of ligands. For instance, square planar complexes can exhibit cis-/trans- isomerism while tetrahedral ones cannot. Tetrahedral complexes with four different ligands exhibit R and S chirality but complexes of that type are relatively rare. The more common cases include cis and trans isomerism in square planar complexes and cis & trans-, mer & fac-, and $\Lambda$ & $\Delta$ isomerism in octahedral ones, although some multidentate ligands present additional possibilities for stereochemical variation
• Stereoisomerism centered on the ligands bound to a metal center. Possibilities include
• stereoisomerism inherent to a ligand. This would include cases where the ligand itself is chiral or, as in the case of amines, exists as a rapidly interconverting mixture of isomers. Thus this sort of stereoisomerism is an extension of the isomerism encountered in ordinary organic and other main group compounds. The main new issues introduced by binding of such ligands to a metal center involve the creation of new possibilities for diastereoisomerism based on the stereochemistry of the metal center and the freezing out of stereocenter inversion on binding to a metal. The latter is particularly important for amines, which as free amines racemize rapidly by nitrogen inversion.
• or conformational isomerism involving five-membered chelate rings created when a multidentate ligand binds to a metal. This isomerism is often called chelate ring twist since individual rings can exhibit one of two conformations depending on how they twist on binding.
A summary of these forms of isomerism is given in Figure $\sf{1}$.
Structural isomers
Structural isomerism involves different topological linkages of atoms. Differences in atomic linkages distinctive to coordination chemistry involve the linkages between metals and ligands. The main variations are:
1. Hydrate/solvate isomerism
Solvate isomers differ in terms of whether a molecule acts as a ligand or whether it acts as a solvate by occupying a lattice site in the crystal. Among solvate isomers, the case in which water is the ligand or solvate is the best known. The resulting isomers are called hydrate isomers after the term for water acting as a solvate, hydrate. A well-known example that illustrates how solvate isomerism works is the series [CrClx(H2O)6-x]Cl3-x·xH2O, for which x = 0-2. The structures of the complex ions involved in this series are shown in Figure $\sf{2}$.1
As can be seen from the compounds in Figure $\sf{2}$, hydrate isomers differ both in terms of whether water acts as a ligand or hydrate and in terms of whether a potential counterion acts as a counterion or ligand. Thus, in trans-[CrCl2(H2O)4]Cl·2H2O two chlorides act as chloro ligands and four waters as aqua ligands while in [CrCl(H2O)5]Cl2·H2O five water molecules act as aqua ligands while only one chloride acts as a chloro ligand. In this way, solvate isomers simply represent cases in which two or more of the possible permutations for metal ligand binding among a set of solvent and counterion molecules are stable.
2. Ionization isomerism
In ionization isomerism there are two or more potential ions that can act as ligands. The ionization isomers differ in terms of which of these ions act as counterions and which act as ligands. Consider, for instance, the complexes shown in Figure $\sf{3}$. These complexes differ in terms of whether the chloride or sulfate acts as a ligand, with the other acting as a counterion.
3. Coordination isomerism
Coordination isomers exist in compounds containing two or more complexes, each of which possesses a different set of ligands that can in principle be swapped with a ligand of the other complex. An example involves [CoIII(NH3)6][CrIII(ox)3], depicted in Figure $\sf{\PageIndex{4A}}$.1 Coordination isomers of this complex involve swapping the ammine ligands around the Co3+ center for oxalato ligands surrounding the Cr3+ center. For instance, swapping all the ammine and oxalato ligands between the metal centers gives the coordination isomer [CrIII(NH3)6][CoIII(ox)3] shown in Figure $\sf{\PageIndex{4B}}$.
As classifications, hydrate/solvate, ionization, and coordination isomerism represent permutations of ordinary structure patterns for solvates, salts, and coordination complexes, respectively. As a result these forms of isomerism are rarely used as independent conceptual frameworks when thinking and talking about the structure of coordination compounds. Not so the final type of structural isomerism, linkage isomerism. That is because linkage isomerism has to do with the special capacity of ambidentate ligands to bind metals in multiple ways.
4. Linkage or ambidentate isomerism
Linkage or ambidentate isomers differ in how one or more ambidentate ligands bind to metal centers in a complex. The classic example dating from the work of Jørgensen and Werner is given in Scheme $\sf{\PageIndex{I}}$.
Just as alkenes exist as E and Z isomers, compounds possessing ambidentate ligands exist as one among the possible linkage isomers. As such, when working with ambidentate ligands the particular linkage isomer formed should be determined experimentally and considered when interpreting the complex's chemical and physical behavior.
The main ambidentate ligands which give rise to linkage isomerism are
• cyanide, CN-
• thiocyanate, SCN-, and the O and Se analogues, OCN- and SeCN-
• nitrite, NO2-
• sulfite, SO3-
• nitrosyl, NO
Although ambidentate ligands can bind metals in multiple ways, most exhibit a preferred binding mode (i.e., prefer to bind metal centers in one of the possible ways). For instance, cyanide almost always binds through its carbon atom and thiocyanate almost always binds through its nitrogen ($\mu$-N). However, the other binding mode can sometimes be formed kinetically or by using conditions that particularly favor its formation. Thus thiocyanate forms M-SCN- linkages in the presence of exceptionally soft metal centers or in hard polar solvents (in which Lewis acid groups can stablize the terminal nitrogen of a bound thiocyanate ligand).
Ambidentate ligands are of significant research interest because many don't just bind to metal centers in two ways; the linkage isomers sometimes have significantly different physical properties. In addition, the possibility of two binding modes introduces the possibility of exploiting linkage isomerism to take advantage of some of these ligands:
1. different structural chemistry in different coordination modes. In mononuclear complexes the coordination mode influences which side of the ligand faces away from the complex and might be susceptible to stabilization by interaction with solvent. Additionally, the orientation of the ligand relative to the metal center might differ between one coordination mode and another. For instance, as predicted from the minor contributor to its resonance structure, thiocyanate binds nonlinearly via its S atom and linearly via its N (Figure $\sf{5}$). Because of this the apparent steric bulk of the ligand around the metal center might differ between forms.
1. bind two different metal centers at the same time, linking them together. Perhaps the best-known examples involve thiocyanate and cyanide. The latter forms linkages of the type M-C≡N-M'. In the dye Prussian blue these take the form FeII-C≡N-FeIII. An example from the author's graduate research involved tetrahedral clusters containing four metal atoms linked by six cyano ligands, as shown in Figure $\sf{6}$.
1. can be induced to change from one binding mode to another in response to a stimulus. The classic example involves the light-driven transformation of the thermodynamically more stable yellow nitro complex of pentamminecobalt(III), [Co(NH3)5NO2]2+ to the less stable red O-nitrito complex, [Co(NH3)5ONO]2+. Less stable cyanometallate coordination networks like those in the red K{FeII[CrIII(CN)6]} transform into the more stable green K{FeII[CrIII(NC)6]} form on heating, as shown in Figure $\sf{7}$.3
Exercise $1$: Bridging ambidentate ligands and the Hard-Soft Acid- Base Principle.
The hard-soft acid-base principle helps explain the preference of bridging ambidentate ligands for particular binding modes. How might the greater stability of the green K{FeII[CrIII(NC)6]} over red K{FeII[CrIII(CN)6]} be explained in terms of the hard and soft acid-base concept?
Answer
The greater stability of K{FeII[CrIII(NC)6]} over K{FeII[CrIII(CN)6]} reflects the greater stability of the FeII-CN-CrIII linkages in the former over the FeII-NC-CrIII linkages in the latter. This is consistent with the preference for hard-hard and soft-soft Lewis acid-base interactions of the Hard-Soft Acid-Base Principle.
The FeII-CN-CrIII linkages are more stable because they possess bonds between
• the softer Lewis acid (FeII) and Lewis base (the C end of CN-)
• the harder Lewis acid (CrIII) and Lewis base (the N end of CN-)
In contrast, the FeII-NC-CrIII linkages in the less stable linkage isomer involve bonds between
• the softer base (the C end of CN-) and harder acid (CrIII)
• the harder base (the N end of CN-) and softer acid (FeII)
Stereoisomerism
Optical Isomerism/Chirality
Molecules of Dn, Cn, or C1 symmetry with only proper rotation axes (including E = C1) are chiral and exhibit optical isomerism. As described in Figure $\sf{1}$, the main sources of such optical isomerism in coordination chemistry are:
1. Chiralityinherent to an organic or main group ligand.
This type of isomerism is just an extension of the sort described in undergraduate organic texts and consequently does not merit separate discussion here, other than to note that some forms of optical isomerism which are of little importance in organic chemistry lead to optical activity in coordination compounds. In particular, the nitrogen inversion process which serves to rapidly racemize chiral amines is frozen out on formation of a metal-ligand bond. Because of this chiral amine, ligands bound to a metal form non-interconvertible R and S enantiomers, as shown in Figure $\sf{8}$.
1. Chirality arising from the symmetry of ligands about a metal center.
The two common situations through which such chirality arises involve:
1. Chirality at a tetrahedral metal center surrounded by four different ligand attachment points.
Such cases are often referred to as MABCD or Mabcd where M stands for the metal and a, b, c, and d the four different ligand attachment points. An example of such a complex is given in Figure $\sf{\PageIndex{9A}}$. The chirality of such complexes is analogous to the chirality arising from a tetrahedral carbon stereocenter. There are two caveats, though. First, it is not enough to have a four-coordinate complex; the metal must possess tetrahedral symmetry since, as shown in Figure $\sf{\PageIndex{9B}}$, square planar M complexes with four different ligands are identical to their mirror images. Second, such chirality is most easily realized using macrocyclic ligands like proteins. Metal ligand bonds involving sterically unhindered monodentate complexes like that in the example of Figure $\sf{\PageIndex{9A}}$ are in general weaker and more flexible than analogous carbon-carbon bonds. Consequently, in such cases it is likely that the enantiomers would interconvert through a fluxional process, making their resolution difficult if not impossible.
1. $\Lambda$ and $\Delta$ isomerism at octahedral metal centers surrounded by two or three chelating ligands.
Specifically,
1. Octahedral tris-chelates of formula M(L~L)3 exhibit D3 symmetry.
2. cis-octahedral bis-chelates of formula M(L~L)2XY exhibit C2 or C1 symmetry: C2 if X = Y and C1 if X $\neq$ Y.
As shown in Figure $\sf{\PageIndex{10A}}$, octahedral tris-chelates like [V(ox)3]3- are chiral and can exist as nonsuperimposable $\Lambda$ and $\Delta$ enantiomers. The chirality of these tris-chelates is analogous to that of pinwheels. As shown in Figure $\sf{\PageIndex{10B}}$, when looking directly at the head of any individual pinwheel from above, the blades will be angled from back to front going around the pinwheel in either the clockwise or counterclockwise direction. As shown in Figure $\sf{\PageIndex{10C}}$, the case for which the blades are angled in the clockwise direction is analogous to the $\Lambda$ enantiomer and the case when they are angled in the counterclockwise direction the $\Delta$.
The $\Lambda$ and $\Delta$ chirality in octahedral bis-chelates is analogous to that in the tris-chelates except this time the third chelating ligand is replaced by an arbitrary pair of ligands. This lowers the symmetry of the complex to either C2 or C1, as shown for the C2 complex [CoCl2(en)2]+ in Figure $\sf{\PageIndex{11A}}$. In either case the result is that such complexes exist as $\Lambda$ and $\Delta$ enantiomers analogous to those in the metal tris-chelates, as shown in Figure $\sf{\PageIndex{11B}}$.
Octahedral complexes containing ligands with denticities of four or more also exhibit $\Lambda$ and $\Delta$ chirality; it is just that in such cases the bound ligand defines multiple rings within the complex that can be defined as existing in either $\Lambda$ or $\Delta$ orientations relative to each other. The procedure for making these assignments is beyond the scope of most introductory inorganic courses but the result is that such complexes are designated as $\Lambda\Lambda$, $\Delta\Delta$, $\Delta\Lambda$, $\Delta\Delta\Delta$, etc., depending on the number of relationships involving the rings defined by the bound ligand. Interested readers should consult reference 4 for more details.
Circular dichroism may be used to characterize optically active metal complexes.
At the present the most definitive way to determine the absolute configuration of an optically active coordination compound is to determine its 3D molecular structure using single crystal X-ray crystallography.5 However, as that is not always possible or convenient, the optical activity of coordination complexes is also commonly studied using circular dichroism (CD) spectroscopy. CD spectroscopy is used because, unlike most chiral organics, optically active coordination complexes possess low-energy electronic transitions that occur in the far UV and visible range. This means that it is often convenient to use the ability of a complex to refract and absorb different circular polarizations of light to derive information about both the configuration of an optically active complex and its electronic structure. The focus of this section will be to explain how the determination of absolute configuration by circular dichroism and optical rotatory dispersion work, as well as the relationship of these techniques to the more ordinary sort of polarimetry used to measure optical rotation in organic systems. Only brief notes will be made about the instrumental and electronic structure applications of these techniques since the electronic spectra of metal complexes will be discussed in Chapter 11: Coordination Chemistry III - Electronic Spectra and the instrumental aspects of CD are well-treated elsewhere.
Both polarimetry and circular dichroism spectroscopy are grounded in the recognition that plane polarized light is equivalent to a superposition of left and right circularly polarized light, as shown in Figure $\sf{12}$.
When plane polarized light passes through a chiral medium, its left and right circularly polarized components move at different rates (i.e., have different indices of refraction). This causes the plane of polarization to rotate in the direction of the faster component according to the relationship
$\sf{\alpha ~=~\dfrac{n_l~-~n_r}{\lambda}} \nonumber$
where $\alpha$ is the angle of rotation, $\lambda$ the light's wavelength, and nl and nr the refractive indices experienced by left and right circularly polarized light.
An example of such a rotation is shown in Figure $\sf{13}$.
Chemists typically call this rotation of light by a chiral medium optical rotation. In organic chemistry it is common to measure the optical rotation at the sodium D wavelength of 589.29 nm, a much longer wavelength than most organics are able to absorb. Because of this the optical rotations commonly measured by organic chemists are largely independent of absorption and mainly serve as a characteristic physical property similar to melting points or are used to establish the purity of a mixture of enantiomers. The sort of spectroscopic data used to characterize chiral coordination compounds more commonly makes use of the relationship between absorption and optical rotation.
In coordination chemistry the wavelength-dependence of optical rotation is used to discern information about compounds' electronic spectra. This measurement of the wavelength dependence of optical rotation is called optical rotatory dispersion (ORD) and the resulting plots are called optical rotatory dispersion or ORD spectra or curves. Such ORD curves are useful for characterizing the chiral environment in a coordination complex because the ORD curves exhibit a characteristic shape. This characteristic shape is because according to the Cotton Effect, the optical rotation changes sign at the absorption maximum of a chromophore in the compound. For one configuration the rotation will go from negative to positive with increasing wavelength and is called a positive cotton effect; for the other configuration the rotation will go from positive to negative or exhibit a negative cotton effect. This behavior is summarized in the top two curves of Figure $\sf{14}$.
A technique used even more often than ORD is circular dichroism (CD), represented by the lowest spectrum in Figure $\sf{14}$. Circular dichroism arises from the differential absorption of left and right circularly polarized light by a chiral compound. This gives elliptically polarized light as shown in Figure $\sf{15}$.
In circular dichroism (CD) spectroscopy a sample is irradiated with polarized light resulting in the transmission of elliptically polarized light due to differential refraction and absorption of the polarized light's left and right circularly polarized components. For achiral molecules the handedness doesn't affect absorbance, but for chiral molecules, which chiral ground and excited states the handedness of an EM wave affects can determine how readily that wave can distort the chiral ground state into the chiral excited one. For this reason chiral molecules absorb left and right circularly polarized light differently.
In CD spectroscopy the magnitudes of the left and right circularly polarized components of light that is transmitted by the sample are measured and used to determine how much light of each was absorbed (just as in regular absorbance spectroscopy). The difference in absorption between left and right circularly polarized light constitutes the primary signal in CD spectroscopy. More specifically, circular dichroism spectra (CD spectra) show the difference in extinction coefficients for left and right circularly polarized light, $\Delta \epsilon$, as a function of wavelength, where
$\sf{\Delta \epsilon\ ~=~ \epsilon_l ~-~\epsilon_r} \nonumber$
where $\epsilon_\sf{l}$ and $\epsilon_\sf{r}$ are the extinction coefficients observed with left and right circularly polarized light, respectively.
As with ORD spectra, the signs of the features in CD spectra like that of Δ–[Co(en)3]Cl3 given in Figure $\sf{16}$ are enantiomer-dependent. As illustrated in the lower spectrum in Figure $\sf{14}$ , the CD spectra of enantiomers mirror one another. Because of this difference in behavior it is possible in principle to determine the absolute configuration of a complex from its CD spectrum.
There are several ways that ORD and CD spectra can be useful for determining the absolute configuration of a chiral complex.
1. First, the sign of the Cotton effect and peaks in CD spectra exhibited by a given absolute configuration is often consistent across a series of compounds. Thus if a new member of the series is isolated, its configuration can be determined based on which enantiomer has the same Cotton effect. For instance, if both the $\Lambda$ isomer of a metal diimine complex like [Ru(bpy)3]2+ exhibit a positive Cotton effect and a newly resolved metal diimine exhibits a negative Cotton effect it might reasonably be inferred that the newly-resolved complex is in the opposite or $\Delta$ configuration.
2. Second, for some systems the expected sign of the Cotton effect and CD peaks for each enantiomer may be predicted semi-empirically based on the spatial locations of substituents relative to the chromophore (the part of the molecule that changes electronic structure during the transition), although the details are beyond the scope of the present discussion.
3. Consideration of interactions between multiple chromophores in a compound can be used to infer absolute configuration. Again, the details are beyond the scope of the present discussion. Interested readers are referred to reference 7 for details.
Common patterns of Diastereomerism
Geometric isomerism
Geometric isomerism involves differences in the geometric placement of atoms in a compound. In coordination chemistry, geometric isomerism involves differences in the relative placement of a set of ligands about a metal center. There are two main types:
cis and trans isomerism
This type of isomerism has to do with how two ligands are oriented relative to one another in a square planar or octahedral complex. As shown in Figure $\sf{17}$, ligands that are next to one another with a L-M-L bond angle of 90$^{\circ}$ are said to be cis; those on opposite sides of the metal with a L-M-L bond angle of 180$^{\circ}$ are in the trans arrangement.
Simple cis and trans geometric isomers can be identified when there are two identical ligands (A) or chelating ligands with distinguishable attachment points (A~B) oriented about either a square planar or octahedral center. Examples are given in Figure $\sf{18}$. Notice from the example given in Figure $\sf{\PageIndex{18C}}$ that multidentate ligands have the potential to constrain complexes to adopt a particular geometry.
Slightly more involved examples involve perturbations of the simple cases above in which there are multiple distinguishable cis and/or trans relationships. Two square planar examples are given in Figure $\sf{19}$.
More complex examples of cis and trans relationships between ligands in octahedral complexes are given in the exercises that conclude this page, which also demonstrates how a systematic approach may be used to identify isomers.
mer and fac isomerism
This type of isomerism has to do with how three ligands are oriented relative to one another in an octahedral complex. The arrangements are represented in Figure $\sf{20}$.
Since it can help to visualize the mer and fac geometry from multiple points of view, several representations are given in Figure $\sf{21}$. As shown in Figure $\sf{21}$, in fac or facial arrangements, the ligands occupy the same "face" of the octahedral coordination sphere, while in the mer or meridional geometry the ligands form a T shape in the same plane or its "meridian."8
Octahedral complexes with three identical ligands oriented about an octahedral center can only exist in mer and fac arrangements. Consequently, such complexes can be designated as mer and fac isomers. Examples are given in Figure $\sf{22}$.
Diastereomerism arising from multiple ligand-associated chiral centers
Diastereomerism can also arise when two or more chiral ligands bind to a metal center. Such cases typically give rise to a complex mixture of isomers, as shown by the example in Figure $\sf{23}$. As may be seen from Figure $\sf{23}$, the differences between these isomers can arise from both changes in the stereochemical configuration of the ligand and the relationship of particular ligand centers relative to one another. For instance, the leftmost two structures in Figure $\sf{23}$ possess one R and S nitrogen center each but differ in whether the centers are oriented so that the R and S centers on the two ligands are trans or cis to one another.
Diastereomerism arising from $\lambda$ and $\delta$ ring conformation isomerism.
Chelate rings are formed when a chelating ligand binds a metal
As may also be seen from the structures in Figure $\sf{23}$, when a multidentate ligand coordinates a metal, the ligand and metal center comprise one or more chelate rings. Examples of such chelate rings are shown in Figure $\sf{24}$.
As may be inferred from the examples in Figure $\sf{24}$, chelate rings may be of different sizes, although four, six, and especially five-membered rings are particularly common. Exactly which ring sizes will be more stable for a given system depends on the coordination geometry and, due to differences in M-L bond lengths, to a lesser degree on the metal. As a result, for octahedral and square planar complexes with 90$^{\circ}$ L-M-L bond angles between cis ligands, five-membered rings tend to be especially stable (Figure $\sf{25}$).
Not all metal chelates prefer to form five-membered rings. The steric requirements of chelate rings depend on both the preferred coordination geometry of the metal center and the stereochemistry of the ligand. Both of these effects are typically considered in terms of preferred L-M-L bond angles. From the perspective of the metal center, the preferred bond angle is determined by the coordination geometry. Octahedral and square planar metals prefer 90$^{\circ}$ L-M-L bond amgles, trigonal bipyramidal systems 90 and 120$^{\circ}$ L-M-L bond angles, and tetrahedral complexes 109.5$^{\circ}$ L-M-L angles. The larger preferred bond angles in tetrahedral and trigonal bipyramidal systems often require the formation of larger six or seven-membered chelate rings for maximum stability. The size of the chelate ring actually formed between a metal and ligand is determined by the ligand's structure. As the contributor of all the atoms in the chelate ring but one, the ligand directly determines the chelate ring size. More subtly, ligands naturally prefer to coordinate metals at a particular L-M-L angle, called the bite angle, as shown in Figure $\sf{\PageIndex{26A}}$. Ligands with bite angles corresponding to the ideal L-M-L angle for a metal's preferred geometry tend to form more stable complexes, although in turn ligand bite angles can cause metals with a weak coordination geometry preference to adopt the ligand's preferred geometry instead.10 As may be seen from the values in Figure $\sf{\PageIndex{26B}}$, bite angles roughly increase with the size of the chelate ring formed but are also influenced by the types of structures used to connect the ligand's Lewis base sites. This facilitates the use of diphosphine ligands to tailor the structure and reactivity of phosphine-containing organometallic catalysts.
$\lambda$ and $\delta$ isomerism involves differences in the conformation of nonplanar five-membered chelate rings
As shown in Figure $\sf{27}$, some chelate ring systems are planar while others are not.
Among the nonplanar chelate ring systems are ligands like en and dppe, which contain tetrahedral C, N, O, P, and S atoms. Because these atoms create kinks in the chelate ring they introduce additional opportunities for diastereomerism due to differences in the chelate ring conformation, called ring twist. Ring twist in nonplanar systems has been most extensively explored for five-membered chelates like those formed by dppe and by en and other chelating amines. Unlike the more rigid four-membered chelate rings, five-membered chelates tend to be conformationally flexible, but not so conformationally flexible that their conformers are rapidly interconverting at room temperature (at least not when the rings are substituted to introduce additional steric strain). This balance between flexibility and rigidity enables some five-membered chelate rings to exist as mixtures of distinguishable conformational isomers at room temperature.
To understand the two most stable conformers that five-membered chelates rings tend to form, it is helpful to think of five-membered chelate rings as involving two components (Figure $\sf{28}$):
• a planar L-M-L group, where L are the atoms directly attached to the metal, M, and
• the remaining two ring atoms, E, which form a rigid E-E bar across the back of the chelate ring.
As can be seen from Figure $\sf{28}$ , rotation about the M-L and L-E bonds causes twisting of the E-E bar relative to the L-M-L plane. The two most stable conformers produced by these motions are the $\lambda$ or $\delta$ conformations as shown in Figures $\sf{29}$ and $\sf{29}$.
The relative stability of the $\lambda$ and $\delta$ conformers will depend on the configuration of any stereocenters present and steric factors. As with organic ring systems, steric effects tend to favor conformers in which bulky groups are placed in the less sterically strained equatorial positions while the configuration of the stereocenters present can serve to restrict the allowable ring twist conformations, as illustrated in Figure $\sf{30}$. A detailed treatment of such systems is beyond the scope of this page. Interested readers are referred to reference 12 for more details.
In closing, it should be noted that it might seem to be making much of small effects in pointing out that one source of diastereomerism in metal complexes involves the freezing out of chelate ring conformations. However, that would be to mistake the impacts that ring conformations can have on the steric accessibility of a coordination site and shaping the course of processes that might occur there. As illustrated in Figure $\sf{31}$, a simple shift in the conformation of one ring in a square planar, pyramidal, or octahedral complex containing coplanar ethylene diamine ligands can exert a significant effect on the steric profile of the complex perpendicular to the MN4 square plane. Because of effects like these, ring conformation isomerism plays a role in the design of stereoselective transition metal catalysts.
Exercises
Exercise $1$. Identify the type of isomerism
What type of isomers are
1. [CoCl(NH3)5](NO3) and [Co(NH3)5(N-NO2)]Cl?
2. [Co(NH3)5-CN-Ru(NH3)5]4+ and [Co(NH3)5-NC-Ru(NH3)5]4+
3. [Cr(CN)5-CN-Co(NH3)5] and [Co(CN)5-CN-Cr(NH3)5]
4. cis-[Mn(en)2(CN)2] and trans-[Mn(en)2(CN)2]
Answer a.
Ionization isomers.
Answer b.
Linkage isomers.
Answer c.
Coordination isomers.
Answer d.
Geometric isomers, specifically cis/trans isomers.
Exercise $2$. Assigning metal oxidation states in a complex
Many properties of transition metal complexes depend on the metal's oxidation state. For instance,
• octahedral complexes of CoII lose and gain ligands rapidly
• octahedral complexes of CoIII lose and gain ligands very slowly
• four-coordinate complexes are generally tetrahedral
• EXCEPT four-coordinate complexes of metals like PtII , PdII , RhI, and IrI, among others, are square planar
For this reason it is important to be able to estimate the formal oxidation state of a metal in a complex. Fortunately, this is easy to do if you remember
1. The sum of all atoms' oxidation states will equal the overall charge on the complex
2. When determining the metal's oxidation state, the ligands can be treated as having an oxidation state equal to their charge - i.e., the charge they possess in the form in which they coordinate the metal - so if they need to lose a proton to bind, don't forget to account for that.
Given the above, estimate the oxidation state of the metal in the following real and hypothetical complexes.
1. Na4[Fe(CN)6]
2. [Cu(phen)2]BF4
3. [PtF4(NH3)2]
4. [Ni(en)2]SO4
5. Co(acac)3
6. [MnCl(O)(salen)]
Answer for Na4[Fe(CN)6].
This contains [Fe(CN)6]4-; so O.S.Fe + 6 x (-1) (for CN-) = -4 (the complex's charge) so O.S.Fe = +2 or Fe2+.
Answer for [Cu(phen)2]BF4.
Since tetrafluoroborate is a monoanion, the complex is [Cu(phen)2]+ so O.S.Cu + 0 x 2 (for phen) = +1 (the complex's charge) so O.S.Cu = +1 or Cu+.
Answer for [PtF4(NH3)2].
O.S.Pt + 4 x (-1) (for F-) + 0 x 2 (for NH3) = +0 (the complex's charge) so O.S.Pt = +4 or Pt4+.
Answer [Ni(en)2]SO4.
The complex is [Ni(en)2]+ so O.S.Ni + 0 x 2 (for en) = +2 (the complex's charge) so O.S.Ni = +2 or Ni2+.
Answer Co(acac)3.
O.S.Co + 3 x (-1) (for acac; see table 9.2.2) = +0 (the complex's charge) so O.S.Co = +3 or Co3+.
Answer [MnCl(O)(salen)].
O.S.Mn + 1 x (-1) (for Cl-) + 1 x (-2)(for oxo) + 1 x (-2) (for salen; see table 9.2.2) = +0 (the complex's charge) so O.S.Mn = +5 or Mn5+.
Exercise $3$. Drawing isomers from descriptions
Draw structures that match the descriptions given, assuming that
• complexes in which the metal has a coordination number of six are octahedral
• complexes in which the metal has a coordination number of five are trigonal bipyramidal
• complexes in which PtII , PdII , or RhI, or IrI have a coordination number of four are square planar
• other complexes in which the metal has a coordination number of four will be tetrahedral
1. mer-triammineaqua-trans-dichlorocobalt(III) ion
2. $\Delta$-diaminebis(oxalato)manganate(III)
3. [CoCl4]2-
4. trans-diamminebis(ethylenediamine)Nickel(2+) tetracyanopalladate(2-)
5. $\Lambda$-bis(ethylenediamine)cobalt(III)- μ-amido μ-hydroxo-$\Delta$-bis(ethylenediamine)cobalt(III)
Answer a. mer-triammineaqua-trans-dichlorocobalt(III) ion.
Answer b. $\Delta$-diaminebis(oxalato)manganate(III).
Answer c. [CoCl4]2- .
Answer d. trans-diamminebis(ethylenediamine)Nickel(2+) tetracyanopalladate(2-).
Note that since the Pd in [Pd(CN)4]2- is Pd2+ it will be square planar.
Answer e. $\Lambda$-bis(ethylenediamine)cobalt(III)- μ-amido μ-hydroxo-$\Delta$-bis(ethylenediamine)cobalt(III).
.
Exercise $4$. Stereoisomers of complexes containing only monodentate ligands.
Draw all the stereoisomers of the following real and hypothetical complexes. You may assume that
• complexes in which the metal has a coordination number of six are octahedral
• complexes in which the metal has a coordination number of five are trigonal bipyramidal
• complexes in which PtII , PdII , or RhI, or IrI have a coordination number of four are square planar
• other complexes in which the metal has a coordination number of four will be tetrahedral
1. [IrCl(CO)(PPh3)2]
2. [CoCl3(NH3)3]
3. [CoCl2(H2O)2(NH3)2]+
4. [CoBrCl(H2O)2(NH3)2]+
5. [CoBrClI(NH3)3]
Answer a. [IrCl(CO)(PPh3)2]
This is a 4-coordinate IrI complex and, as such will be square planar. Since two of the ligands are identical, it will have cis and trans isomers. The trans isomer is famous for its ability to form adducts and is called Vaska's complex.
Answer b. [CoCl3(NH3)3]
This complex contains three ammine and three chloro ligands and so will have fac and mer isomers.
Answer c. [CoCl2(H2O)2(NH3)2]+
This is a case of an MA2B2C2 system (A = Cl-, B = H2O, C = NH3) involving multiple cis and trans relationships. The six stereoisomers are shown below.
Since it may not be obvious how to arrive at a set of isomers like the ones above, it is worth considering how one might work through the possibilities for a system of ligands like this one. Several systems may be used to systematically identify isomers. Once approach is the Macbdef system described in Note $\sf{1}$ at the end of these exercises. The solutions presented here and in subsequent problems employ a variant of that approach.
• Start by fixing one set of ligands. In this case the chloro ligands were first fixed as cis to one another. Then the remaining ligands might be cis or trans to one another, although since the chloro ligands are already cis, the ammine and aqua ligands cannot both be trans at the same time. Thus for cis chloro ligands the possibilities for the others are:
• cis-H2O ,cis-NH3,
• cis--H2O, trans-NH3
• trans-H2O, cis-NH3
• Swap the configuration of the ligand set you started with. In this case it means placing the chloro ligands trans to one another. From that configuration, the possibilities for the ammine and aqua ligands are
• cis-H2O ,cis-NH3,
• trans-H2O, trans-NH3
• Check for enantiomers and create mirror images of any chiral complexes you drew. The easiest way to do this is to assign point groups. However, it turns out that the octahedral all-cis case is D3 and formally equivalent to the symmetry of a tris chelate, as shown below.
• finally, check to make sure you didn't include the same complex twice. Humans do make mistakes after all.
Answer d. [CoBrCl(H2O)2(NH3)2]+
This problem is analogous to the one above except that this complex contains a bromo and chloro ligand in place of the two chloro ligands in [CoCl2(H2O)2(NH3)2]+. Consequently, when the bromo and chloro ligands are cis to one another, additional possibilities for isomers arise based on whether ammine or aqua ligands are trans to the chloro and bromo. The result is eight isomers.
Answer e. [CoBrClI(NH3)3]
There are two possibilities for this complex - a fac-(NH3)3 and a mer-(NH3)3 arrangement. These may be taken as starting points for examining the possible permutations for the Cl-, Br-, and I- ligands. The top row presents the three possibilities within the mer-(NH3)3 arrangement; each corresponds to a different ligand trans to an ammine ligand. The bottom row presents the two possible isomers with a fac-(NH3)3 arrangement. In each, the Cl-, Br-, and I- ligands are trans to ammine ligands. However, the complexes possess C1 symmetry and so are chiral and exist as enantiomers.
Exercise $5$. Stereoisomers of complexes with bidentate ligands (ignoring ring twist).
Ignoring ring twist, draw all the stereoisomers of the following real and hypothetical octahedral complexes.
1. [CoBr2Cl2(en)]-
2. [CoBrCl(en)2]+
3. [CoBrCl(gly)2]-
4. [Co(en)3]3+
5. [Co(gly)3]
Answer a. [CoBr2Cl2(en)]-
This complex is analogous to [CoCl2(H2O)2(NH3)2]+ in possessing pairs of identical ligating groups. The main difference is that in [CoBr2Cl2(en)]- the two amine ligating group of the en ligand are restricted to a cis arrangement. Under the MABCDEF system explained in Note $1$, [CoCl2(H2O)2(NH3)2]+ may be MA2B2C2, but this complex, [CoBr2Cl2(en)]-, is M(AA)B2C2. The isomers are:
Answer b. [CoBrCl(en)2]+
The key to problems like this is to focus on the unique ligands, in this case Br- and Cl-. These can exist in either a cis or trans arrangement. The trans form is achiral, but in the case where Br- and Cl- are cis, $\Lambda$ and $\Delta$ enantiomers are possible. The result is three isomers.
Answer c. [CoBrCl(gly)2]-
This case is analogous to the one above, but this time the bidentate ligand, gly, possesses distinguishable binding sites. Thus there are additional permutations involving how the gly ligands are bound relative to one another and/or whether the O or N end of the gly is bound trans to Br or Cl. The result is 11 different isomers, which are given below.
Answer d. [Co(en)3]3+
As a tris-chelate of symmetric bidentate ligands, ignoring ring twist, [Co(en)3]3+ will only exhibit $\Lambda$ and $\Delta$ isomerism.
Answer e. [Co(gly)3]
Since there are three gly ligands, each of which have carboxy and amine ends, the allowable arrangements involve whether the resulting three carboxy and three amine ends are oriented in a mer or fac arrangement. This gives two possibilities, each of which can exist as either the $\Lambda$ or $\Delta$ enantiomer, for a total of four isomers, which are given below.
Exercise $6$. More stereoisomers: Now featuring linkage isomerism
Ignoring ring twist, draw all chemically reasonable stereoisomers for the following real and hypothetical complexes. You may assume that
• complexes in which the metal has a coordination number of six are octahedral
• complexes in which the metal has a coordination number of five are trigonal bipyramidal
• complexes in which PtII , PdII , or RhI, or IrI have a coordination number of four are square planar
• other complexes in which the metal has a coordination number of four will be tetrahedral
1. [Pt(en)(SCN)2]
2. [Pt(NH3)2(SCN)2]
3. [Co(en)2(SCN)2]+
Answer a. [Pt(en)(SCN)2]
The Pt in this neutral complex must be Pt2+ to balance the negative charges of the two SCN- ligands. Complexes of Pt2+ are square planar and four coordinate, consistent with the bidentate en and two thiocyanato ligands. Of the ligands, the en ligand must bind in a cis arrangement. Consequently, the two SCN- ligands will be in a cis arrangement as well. The isomers will therefore only differ in whether the two SCN- bind $\kappa$N or $\kappa$S. The possibilities are:
Answer b. [Pt(NH3)2(SCN)2]
As with the preceding example, this will be a square planar Pt2+ complex. The main difference is that this time the coordinated nitrogens are not constrained to adopt a cis arrangement so there will be both cis and trans isomers.
Answer c. [Co(en)2(SCN)2]+
Exercise $7$. Stereoisomer free for all.
Ignoring ring twist, draw all chemically reasonable stereoisomers for the following real and hypothetical complexes. You may assume that
• complexes in which the metal has a coordination number of six are octahedral
• complexes in which the metal has a coordination number of five are trigonal bipyramidal
• complexes in which PtII , PdII , or RhI, or IrI have a coordination number of four are square planar
• other complexes in which the metal has a coordination number of four will be tetrahedral
1. [Ru(bpy)(phen)(dppe)]2+
2. [CoClF(PPh3)(py)]-
3. [Ni(en)2(NO2)2]
4. [Fe(H2O)3(SCN)3]
5. [PtClF(PPh3)(py)]
6. [CoBr2Cl(NH3)3]
Answer a. [Ru(bpy)(phen)(dppe)]2+
This complex is an octahedral tris-chelate containing symmetric ligands. As such it will exhibit $\Lambda$ and $\Delta$ isomers:
Answer b. [CoClF(PPh3)(py)]-
This complex has a coordination number of 4 and contains a Co2+ ion with a d7 electron configuration, so a tetrahedral geometry is expected. Tetrahedral complexes like this one with four different ligands are chiral and can form R and S enantiomers.
Answer c. [Ni(en)2(NO2)2]
This is an octahedral bis-chelate and will exist in $\Lambda$ and $\Delta$ configurations. Within each configuration the NO2- ligands can exist as $\kappa$N and $\kappa$O, leading to the following possibilities:
Answer d. [Fe(H2O)3(SCN)3]
As an octahedral complex with three identical ligands it can exist in mer and fac configurations, with the ambidentate SCN ligand providing additional possibilities for isomerism.
Answer e. [PtClF(PPh3)(py)]
As a square planar complex with four nonidentical ligands, this complex exists as a single isomer.
Answer f. [CoBr2Cl(NH3)3]
As an octahedral complex with three identical ligands it will exhibit mer and fac isomerism. In addition, it will have two mer configurations that differ in terms of the cis and trans relationships between the bromo and chloro ligands.
Exercise $8$
Complexes of formula Ru(TPP)py2 have been prepared, in which TPP is tetraphenylporphyrin, which binds metals in the form given below. Draw all isomers of Ru(TPP)py2.
Answer
There is only one isomer. While normally complexes containing two identical ligands (in this case py) can exhibit cis and trans isomerism, in this case the planar tetraphenylporphyrin ring ligates the Ru2+ ion in a square planar arrangement, as shown at left in the image below. This leaves only a pair of trans coordination sites for the chloro ligands to occupy, giving the isomer shown at right below.
Exercise $9$
Draw the diastereomers formed due to the chirality of the amine nitrogen atoms in [PtCl2(N,N'-dimethylethane-1,2-diamine)] which has the atomic connectivity represented below
Answer
The diastereomers will differ in whether the N atoms adopt an R or S configuration at the nitrogen atoms. The possible permutations are (R,R), (S,S), (R,S), and (S,R). However, as may be seen from the image below, the (R,S) and (S,R) configurations are identical (the two projections shown can be interconverted through a C2 rotation along an axis bisecting the Cl-Pt-Cl unit). As a result there are only three unique isomers. Of these, the (R,R) and (S,S) configurations are enantiomers (they are mirror images by reflection in the PtCl2N2 plane). Note that in solving problems like this one it can be helpful to keep track of possible isomers by assigning the stereochemistry at each chiral center as R or S using the Cahn-Ingold-Prelog-convention, though it is not strictly necessary to do so.
Exercise $10$
Label the conformations of all chelate rings in the structure below.
Answer
.
Appendix: The MABCDEF bookkeeping system for identifying isomers
The Mabcdef or MABCDEF system is one method for identifying the number of isomers in an octahedral complex, although since it is really just a bookkeeping and organizational system it can easily be extended to other geometries as well. The M in Mabcdef stands for metal and the other letters are used to stand in for ligands. The basic approach involves
• classifying the ligands as A, B, C, D, E, or F in order of multiplicity. Thus for
[Cr(NH3)6]3+ A = NH3; there are no B, C, D,E, or F ligands; and the complex is classified as MA6
[CoCl(NH3)5]2+ A = NH3, B = Cl-, and the complex is MA5B
[CrCl2(H2O)2(NH3)2]+ A = Cl-, B = H2O, C = NH3, and the complex is MA2B2C2
• in the classification above, multidentate ligands are typically classified before monodentate ones and are designated AA, AB, ABC, ABA, etc. based on the symmetry of the attachment points. Thus
en has identical attachment amine points and is AA
gly has a carboxy and amine attachment points and is AB
trien is ABA
CoCl2(en)2+ is M(AA)2B2
CoCl2(gly)2- is M(AB)2C2
CoCl2(en)(gly) is M(AA)(AB)C2
• Systematically list out the possible trans arrangements of ligands by
1. assigning one pair of ligands to be trans to one another.
2. Then systematically list out the other possible trans pairs by permuting the remaining trans arrangements. It can help to organize the permutations in a table, such as that shown below for an MABCDEF complex where A and B are assigned trans. In looking at the table notice how the second set of trans permutations is systematically varied. This helps ensure that no possibility is skipped.
3. Go through the list of isomers and remove any duplicates you generated so far. For instance, notice in the table below that the last three stereoisomers are identical with the first three (e.g., stereoisomer 4 is identical to stereoisomer 3, 5 with 2, and 6 with 1).
Stereoisomer 1 Stereoisomer 2 Stereoisomer 3
Stereoisomer 4
(same as isomer 3)
Stereoisomer 5
(same as isomer 2)
Stereoisomer 6
(same as isomer 1)
trans AB (fixed) trans AB (fixed) trans AB (fixed) trans AB (fixed) trans AB (fixed) trans AB (fixed)
trans CD trans CE trans CF trans DE trans DF trans EF
trans EF trans DF trans DE trans CF trans CE trans CD
1. If the complex possesses multidentate ligands that demand that certain groups exist and cis pairs, then also remove any configurations which do not agree with the known binding capability of the ligand (e.g., the two amine groups of ethylenediamine cannot be trans to one another, so if you have an en ligand, remove configurations like that from the list).
2. Next, swap or permute the original trans pair and repeat the process you just followed. In the case of Mabcdef this gives the following results:
Isomer "fixed" trans pair Additional trans pairs
1 AB CD EF
2 AB CE DF
3 AB CF DE
4 AC BD EF
5 AC BE BF
6 AC BF DE
7 AD BC EF
8 AD BE CF
9 AD BF CE
10 AE BC DF
11 AE BD CF
12 AE BF CD
13 AF BC DE
14 AF BD CE
15 AF BE CD
• Since the procedure explained above only identifies isomers based on unique trans pairings, it does not identify when a configuration is chiral and corresponds to a pair of enantiomers. Any enantiomers can be identified by drawing out the complexes and either classifying their point groups or by drawing their mirror images and checking if they are superimposable. In the MABCDEF case - i.e., where all the ligands are different - all of the isomers identified have C1 symmetry so are chiral. This means there will be a pair of enantiomers for each, giving 15 x 2 = 30 different stereoisomers. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/09%3A_Coordination_Chemistry_I_-_Structure_and_Isomers/9.04%3A_Isomerism.txt |
Why do coordination complexes form the structures they do?
As with all chemical structure, coordination complexes form the structures they do so as to best stabilize the metal center and ligands through the formation of metal-ligand bonds while avoiding destabilizing interactions like steric repulsions. The issue then is how many metal-ligand bonds should be formed and how those bonds should be arranged spatially to give the largest net stabilization possible. This question will eventually be considered in detail in connection with the nature of bonding in coordination compounds. For now, it will be helpful to think about it in terms of seven factors:
The stabilizing effect of metal-ligand bond formation.
The driving force for complex formation is the stabilization of electrons in covalent chemical bonds. In the vast majority of cases, this largely involves stabilization of the ligand lone pair as it experiences the effective nuclear charge of the metal, although a few instances involve stabilization of metal electrons by ligand nuclei (inverse ligand fields). Regardless, metal-ligand bond formation is stabilizing and classified by the way it preferences the addition of ligands to the complex.
Steric effects, specifically steric repulsions between ligands.
One reason coordination numbers do not increase indefinitely is that only so many ligands can fit around a metal. Exactly how many depends on
• the size of the metal center. This is one of the more important factors. Many metals tend to exhibit preferred coordination numbers, which depend on their oxidation state and size as shown in Figure $\sf{1}$. Larger inner transition metals like the Lanthanides and Actinides can accommodate 9-12 sterically undemanding ligands, while the smaller transition metals tend to accommodate up to six, although larger coordination numbers are more common for low valent metals and as size increases on moving from right to left across the transition metal block of the periodic table. Thus, the early transition metal molybdenum forms seven- and eight-coordinate [MoIII(CN)7]4- and [MoIV(CN)8]4- with the sterically undemanding cyano ligand.
• the size of the ligands. As long as ligands are not excessively rigid and bulky, their size is less important than the size of the metal in determining the number of ligands that coordinate. Ligands' ability to donate electrons to the metal center also tends to influence coordination number more than ligand size. However, all other things being equal for a given metal and ligand donor ability, small ligands allow higher coordination numbers while fewer bulky ligands will fit around the metal center.
• how the ligand bonds to the metal. Again, all other things being equal, ligands that are more sterically demanding in the vicinity of the metal center tend to limit the ability of other ligands to bind more than those which bind through a small extended group. For example, a bulky isocyanide like t-BuCN will sterically crowd the metal less than a bulky phosphine like t-BuH2P would. For this reason the effective size of a ligand is sometimes rated in terms of either a cone angle of space they are estimated to occupy around the metal (called the Tolman cone angle, it is commonly used to evaluate phosphines' steric bulk) or in terms of the percentage of the metal's coordination sphere the ligand occupies (called the percent buried volume, it is used to estimate the steric impact of N-heterocyclic carbenes).
Repulsion between M-L bonding electrons on different ligands.
For many complexes, steric effects are neither the only effects nor the most important. Among the additional factors that should be considered are the repulsions that occur between the electrons that different ligands donate to the metal-ligand bonding. These electron-electron repulsions affect the
• Coordination number. When a ligand donates its electrons to a metal center to form a new metal-ligand bond, the electron density around the metal increases, raising the overall energy of the other M-L bonding electrons. This increased repulsion often limits the number of coordinated ligands. As more ligands are added, the electron-electron repulsions keep increasing until the lowering of energy of the ligand electrons in the new bond is insufficient to compensate for the raising of energy of the existing M-L bonding electrons. Based on this effect alone,
• larger metals tend to achieve higher coordination numbers than smaller ones because the electron-electron repulsions are spread across a larger coordination sphere.
• With a given metal, ligands that are more electron donating have a greater tendency to form complexes with lower coordination numbers with a given metal than similar neutral ones do. This is why anionic ligands (which tend to be better electron donors) tend to give lower coordination numbers than comparable neutral ligands (which tend to be weaker donors). Thus Co2+ forms CoCl42- with chloro ligands but [Co(H2O)6]2+ with aqua ligands.
• Coordination geometry. In the Kepert model for the shapes of coordination complexes, this intraligand repulsion determines the most stable coordination geometry by causing the ligands to move as far apart from one another on the metal's coordination sphere as possible.
Formally, according to the Kepert model
• any of the metal's valence electrons not involved in metal-ligand bonds occupy (n-1)d orbitals and function as core electrons. As core electrons they do not influence the molecular shape.
• electrons involved in bonding to a given ligand constitute an electron group that repels all other electron groups around the metal.
• all other things being equal, the complex will form the geometry that maximizes the intra-electron group repulsions.
Notice the similarities of these postulates to those of VSEPR theory. In predicting coordination geometries in terms of electron-electron repulsions, the Kepert model is just an extension of VSEPR theory to coordination compounds. The difference between VSEPR theory and the Kepert model is that in the Kepert model, only electrons involved in metal-ligand bonds count.
Coordination geometries predicted by the Kepert model for coordination numbers two through nine are given in Figure $\sf{2}$. As may be seen from the geometries listed in Figure $\sf{2}$, these are just equivalent to VSEPR geometries for cases in which the number of electron groups is equal to the coordination number.
The difference between optimal and suboptimal coordination geometries is greater with few ligands, and becomes smaller as ligands become increasingly dispersed across the metal's coordination sphere. In complexes containing five, seven, eight, or higher coordinate metals, there are a number of geometries that are similar in energy to the preferred geometry. These geometries, which should be regarded as accessible, are also listed in Figure $\sf{2}$.
d-electron effects
A few coordination geometries are noticeably absent from the Kepert-preferred and Kepert-accessible geometries in Figure $\sf{1}$. These include the trigonal prismatic geometries formed by compounds like W(CH3)6 and the very common square planar geometry illustrated by complexes like [PtCl4]2- and [IrClH(PPh3)2]. One of the reasons the Kepert model fails to predict the existence of such structures is its neglect of directional interactions involving d electrons on the metal center. Metal d electrons exert a profound influence on almost all properties of transition metal complexes, including their structures. The way in which this occurs will be explored at length in the next chapter. For now, it is enough to note that both the ligand-donated electrons surrounding a metal center and the electrons occupying particular d orbitals on that metal are oriented in specific directions relative to one another. Because of this, the strength of the interactions between the ligand and metal d electrons depends on the number of d electrons present, how strongly metal-ligand binding affects their energy, and how the ligands are arranged about the metal center. The impact of these effects, here termed ligand field effects, differs from case to case and can include
• distortions of the complex's geometry. For instance, an ideal octahedral coordination geoemtry might be tetragonally distorted by flattening or elongating it.
• imparting a strong preference for non-Kepert coordination geometries. This is why, for example, 2nd and 3rd row complexes in which the metal has a d8 electron configuration are almost always square planar.
• stabilizing non-Kepert geometries enough to permit complexes to adopt them in the presence of a rigid or semirigid ligand that prefers to coordinate the metal in that geometry.3
Because of these effects, square planar and trigonal prismatic geometries are also observed, and the list of coordination geometries given in Figure $\sf{2}$ may be extended to that shown in Figure $\sf{3}$.
Ligand constraints imposed by rigid or semirigid ligands
Rigid or semirigid ligands influence the coordination geometry of metal complexes in two main ways:
1. Bulky rigid ligands that crowd the metal center prevent other ligands from binding. Thus such ligands are useful for preparing low-coordinate complexes.
2. Rigid and semirigid ligands can impose their preferred coordination geometry on a metal center. This is because these ligands energetically prefer to adopt a particular conformation when they bind a metal center. In doing so they shift the coordination geometry energy landscape toward that preferred geometry. If the shift is large enough relative to the native preference due to ligand repulsion and ligand field effects, the complex will either adopt the ligand-preferred geometry or be distorted in the direction of the ligand-preferred geometry. Examples are given in Figure $\sf{4}$.
The influence of ligands on coordination geometry is important in living systems, in which proteins and nucleic acids can act as rigid or semirigid ligands. The ability of these ligands to distort the coordination geometries of metal atoms in ways that enable them to perform specific functions is so common that the resulting distorted geometries are termed entactic states. A particularly spectacular case of an entactic state involves the blue copper proteins azurin and plastocyanin, the structure of which is given in Figure $\sf{5}$.
As may be seen from the structure in Figure $\sf{5}$, the copper in plastocyanin exibits a distorted tetrahedral coordination geometry. The protein is said to act like a medieval torture device called a rack in stretching the metal into its distorted geometry. This distortion makes it easier for the copper center to undergo facile redox reactions, enabling it to better function as an electron carrier.
Crystal packing effects, in which the energy-lowering packing of molecules and ions in a crystal drives the distortion of a complex's structure away from what it would adopt in the gas phase or solution
This effect is similar to that of ligand constraints except that in this case it arises not from the structure internal to a ligand but out of the forces involves in maximizing the stabilization energy of a crystal. With lower coordination number complexes, packing effects can shift the conformations of flexible ligands but only give rise to very small distortions of the overall coordination geometry. Packing effects can drive a shift in the overall coordination geometry of higher coordination number complexes, for which packing effects are significant relative to the small difference in energy between geometries. Thus while [Mo(CN)8]4- has a square antiprismatic coordination geometry, in solution it exhibits a dodecahedral coordination geometry in the crystals of many of its salts.
Relativistic effects on orbital energies
The proximity of fast moving electrons to massive nuclei in the heavier transition elements results in relativistic expansion and contraction of orbitals. The net results are that
• heavier elements tend to be smaller than expected. This effect preferences lower coordination numbers.
• the relative energies of orbitals shift. Orbitals which become contracted are lowered in energy while those which are expanded increase in energy, as shown for the case of gold in Figure $\sf{\PageIndex{6A}}$.
The combination of smaller sizes and altered orbital energies affects coordination preferences. Relativistic effects contribute to the greater tendency of AuI relative to other group 11 metals to form linear two-coordinate complexes. As shown in Figure $\sf{\PageIndex{6B}}$, the relative closeness in energy of the 6s and 5d orbitals of gold makes mixing of these orbitals more favorable, facilitating the ability of gold to form two-coordinate complexes with strong sigma bonds oriented 180/(^{\circ}\) from one another.
What structures do coordination complexes form?
Metal complexes with coordination numbers ranging from one to 16 are known, although values greater than seven are rare for the transition metals. In this section, examples of common coordination geometries will be presented in order of coordination number.
Coordination Number 1.
Condensed phase monocoordinate complexes are unknown for the transition metals, although the post-transition metals Tl and In form monocoordinate complexes with the bulky ligands triazapentadienyl and 2,6-tris(2,4,6-triisopropylphenyl)benzene as shown in Figure $\sf{7}$.
Coordination Number 2.
A coordination number of 2 is rare outside of d10 complexes of the group 11 metals and mercury, specifically, Cu+, Ag+, Au+, and Hg2+. In accordance with the predictions of the Kepert model these give linear complexes.
Among these,
• Cu+ more commonly gives tetrahedral complexes but can be coaxed to give linear ones. The most prominent example is [CuCl2]-, which forms when CuCl is treated with concentrated HCl under anerobic conditions.
• Ag+ also commonly forms tetrahedral or trigonal planar complexes but can give linear ones. The most prominent example is [Ag(NH3)2]+, which can be formed by treating silver slats with concentrated aqueous or liquid ammonia.
• Au+ almost always forms linear complexes, but many of these formally two-coordinate complexes associate as depicted in Figure $\sf{8}$. The ability of Au+ to form linear complexes with cyanide is even used to selectively extract metallic gold from low grade ores. The stability of [Au(CN)2]- means that the dissolution of metallic gold in aqueous cyanide is thermodynamically favorable under aerobic conditions.
$\sf{4~Au~~+~~8~CN^-~~+~~O_2~~+~~2~H_2O~~\longrightarrow~~4~[Au(CN)_2]^-~~+~~4~OH^-} \nonumber$
• Hg2+, like Au+, benefits from relativistic effects and more commonly forms two-coordinate complexes with a linear geometry. Among these is [Hg(CN)2]. However, its preference for linearity is not as rigid as for Au+, and so complexes with a variety of coordination geometries are known.
And by means of honorary mention, the mercury(I) ion, Hg22+, forms linear complexes of the type L-Hg-Hg-L, although since Hg22+ is often considered as a single unit, these aren't always considered to be two-coordinate complexes.
Two-coordinate complexes may also be formed through the use of bulky ligands that only allow for the binding of two to the metal center. The classic examples are given in Figure $\sf{9}$.
Coordination Number 3
Three-coordinate complexes are similar to two-coordinate ones in that they are rare and, aside from the constraining influence of ligands, usually limited to d10 metal ions such as Cu+ , Ag+, Au+, Hg2+, and Pt(0). As expected from the Kepert model, in the absence of constraining ligands, three-coordinate complexes are trigonal planar.
Coordination Number 4
The two common four-coordinate geometries are tetrahedral and square planar.
Tetrahedral complexes are commonly formed by metals possessing either a d0 or d10 electron configuration. Monometallic examples of d0 configurations include TiCl4, VO43-, WS42-, MnO4-, CrO42-, and OsO4, while d10 examples are [Ni(CO)4], [HgBr4]2-, [ZnCl4]2-, and [CdI4]2-. For other electron configurations, tetrahedral complexes are known but much less common. Examples usually involve good donor ligands and include [FeCl4]- (d5), [CoCl42-] (d6), and [NiCl4]2- (d7).
Second and third row transition metal centers with d8 electron configurations like Rh+, Ir+, Pd2+, Pt2+, and Au3+ almost exclusively exhibit square planar geometries. Beyond this, square planar geometries are often formed by Ni2+ (d8), Ni3+ (d7), and Cu2+ (d9). Examples of square planar complexes include [Cu(acac)2]; [PtCl4]2- ;Wilkinson's catalyst, [RhCl(PPh3)3]; and Vaska's complex, trans-[Ir(CO)Cl(PPh3)2].
Coordination Number 5
The two common coordination geometries for five-coordinate complexes are trigonal bipyramidal and square pyramidal.
Homoleptic [Ni(CN)5]3- possesses a square pyramidal structure, although the geometry is more common for macrocyclic complexes like the iron protoporphyrin of deoxymyoglobin shown in figure $\sf{4}$ and for complexes containing oxo and nitrido ligands, examples of which are shown in Figure $\sf{12}$.
In the absence of rigid constraining ligands, the relatively low energy difference between the trigonal bipyramidal and square pyramidal coordination geometries provides a mechanism for interconversion of the axial and equatorial ligands in a trigonal planar complex. For example, pentacarbonyliron(0) exhibits fluxionality involving a square pyramidal intermediate via a Berry pseudorotation mechanism, as shown in Figure $\sf{13}$.
Coordination Number 6
The two common coordination geometries for coordination number 8 are octahedral and trigonal prismatic.
Trigonal prismatic coordination is related to octahedral coordination as shown in Figure $\sf{15}$. As may be seen in Figure $\sf{15}$, an octahedral coordination sphere is just a trigonal antiprism in which all edge lengths are identical. Rotation of one triangular face relative to its opposite until the two are eclipsed gives a trigonal prismatic geometry. In fact, since continuation of this rotation gives another octahedral complex, the trigonal prismatic geometry is an intermediate in isomerization reactions involving octahedral complexes.
In contrast to octahedral coordination geometries, trigonal prismatic coordination (and distorted versions thereof) are rare and occur mostly for d0, d1, and d2 configurations. Examples of trigonal prismatic metal centers include the d2 Mo4+ centers in MoS2, d1 [Re(S2C2Ph2)3]-, and d0 [Ta(CH3)6]-, of which the latter two structures are given in Figure $\sf{16}$. Semirigid ligands like that shown in Figure $\sf{\PageIndex{4C}}$ may be used to encourage the adoption of a trigonal prismatic geometry, although once the number of d electrons present exceeds two, the preference for octahedral coordination is too great for a trigonal prismatic geometry to occur.
Coordination Number 7
Seven-coordinate complexes are rare outside of the relatively large early transition metals, lanthanides, and actinides. The three common seven-coordinate geometries are pentagonal bipyramidal, monocapped octahedral, and monocapped trigonal prismatic. The latter two are often called capped octahedral and capped trigonal prismatic, with the mono- prefix being understood.
Although intraligand repulsions are smaller in the pentagonal bipyramidal coordination geometry than the capped octahedral and capped trigonal prismatic geometries, the difference is small, and the three structures are often close in energy. As a result the structure observed is often dependent on ligand-based constraints, crystal packing, and solvent effects that preference one geometry over the others.
Heptacyano complexes are often pentagonal bipyramidal. Examples include [Mo(CN)7]3-, [W(CN)7]3-, and [Os(CN)7]3-. Seven-coordinate complexes containing oxo ligands commonly are pentagonal bipyramidal with the oxo ligand(s) in the less sterically hindered axial position. Examples include [NbOF6]3- and, for the inner transition metals, [UO2F5]3-. Ligands that have been used to promote formation of seven-coordinate species include 15-crown-5 and 2,2':6',2'':6'',2'''-quaterpyridine. Representative complexes are given in Figure $\sf{17}$.
Capped trigonal prismatic geometries are common for complexes of the early transition metals. Examples include [NbF7]2- ,[TaF7]2-, and [ZrF7]3- in (NH4)3[ZrF7].
Capped octahedral geometries are found in [MoMe7]-, [WMe7]-, and [WBr3(CO)4], which contains three pairs of trans-Br and CO with the final CO capping the octahedron's (CO)3 face, as shown in Figure $\sf{18}$.
In seven- and higher-coordinate complexes, ligand and crystal packing effects frequently give distorted coordination geometries. These geometries are intermediate between two or more of the idealized seven coordinate geometries, making it difficult to tell exactly which structure they are a distortion of (Figure $\sf{19}$).
Coordination Number 8
Eight-coordinate complexes are rare and occurs in discrete molecules and ions only for the relatively large early transition metals, lanthanides, and actinides. The three common eight-coordinate geometries are square antiprismatic, dodecahedral, and bicapped trigonal prismatic. In contrast, the cubic coordination geometry is only found in ionic lattices like that of CsCl and in complexes of the inner transition metals such as Na3[UF8].
As with other high-coordinate structures, the energy difference between these eightfold coordination geometries is small enough that packing effects can significantly influence the observed structure. For example, octacyanomolybdates commonly adopt a square antiprismatic coordination geometry but depending on the counterions present can give dodecahedral or bicapped trigonal prismatic complexes. Examples are given in Figure $\sf{21}$.
Coordination Number 9
Again, nine-coordinate complexes typically require larger transition metals, lanthanides, and actinides. Coordination geometries are typically either tricapped trigonal prismatic or idiosyncratically determined by the ligands. Simple examples include the aqua complexes [Sc(H2O)9]3+, [Y(H2O)9]3+, and [La(H2O)9]3+, as well as [TcH9]2- and [ReH9]2-.
Coordination Numbers 10-16
Coordination numbers higher than nine are extremely rare for compounds that bind in $\kappa$ fashion (form conventional metal-ligand bonds)14 and usually involve some combination of large metals, sterically undemanding ligands, and special ligand structures that promote higher coordination. Noteworthy examples include
1. Twelve-coordinate [Hf(BH4)4], which illustrates how small multidentate ligands promote higher coordination numbers. As shown in Figure $\sf{23}$, [Hf(BH4)4] has a cubooctahedral structure in which BH4- acts as a tridentate ligand, with BH3 units occupying triangular faces of the cubooctahedron to give a tetrahedron of BH4- ligands around the Hf.
1. Twelve-coordinate [Ce(NO3)6]2-, in which the nitrate oxygens define an icosahedral coordination geometry as shown in Figure $\sf{24}$. The nitrates in the structure bind the Ce center in bidentate fashion in an octahedral array.
1. Fifteen-coordinate [Th(H3BNMe2BH3)4], which also uses bridging H-B-H units that occupy little of the coordination sphere. In [Th(H3BNMe2BH3)4], three of the four H3BNMe2BH3 ligands bind in $\kappa$4 fashion and one binds $\kappa$3, giving the fifteen fold coordination.16
1. Sixteen-coordinate [CoB16], which possesses the highest coordination number yet observed. Its structure is given in Figure $\sf{25}$. The coordination geometry is an octahedral antiprism, and the complex should be considered to involve a Co center in the midst of a B16- "molecular drum" held together by cluster bonds. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/09%3A_Coordination_Chemistry_I_-_Structure_and_Isomers/9.05%3A_Coordination_Numbers_and_Structures.txt |
Bridging ligands link metals in multicenter coordination complexes
Some ambidentate and multidentate ligands have the capacity to bridge two metal centers. For instance, the cyano ligands can bridge two metal centers by forming linear links of the type
$\ce{M-CN-M'} \nonumber$
Linkages like these can be used to bridge multiple metal centers in a cluster or network solid, the latter of which is sometimes called a coordination framework. When the cyano ligand is combined with octahedral metal centers possessing 90$^{\circ}$ L-M-L bond angles, a cubic coordination network results. An example is the Prussian Blue coordination framework, which possesses the unit cell depicted in Figure $\sf{\PageIndex{1A}}$.
The cubic network shown in Figure $\sf{\PageIndex{1A}}$ reflects the 3D structure of the octahedron, in which the M-L bonds point along the x, y, and z axes. If a square planar metal center is used, a 2D coordination network comprising interlinked squares is produced instead. An example is the Hofmann clathrate shown in Figure $\sf{\PageIndex{1B}}$, which depicts a Hofmann clathrate in which square planar [Ni(CN)4]2- units are linked by trans-Ni(NH3)22+ units (which are octahedral but have a square plane of open coordination sites perpendicular to the NH3-Ni-NH3 axis).
The structures of coordination polyhedra and frameworks reflect ligand and metal coordination geometries
As illustrated by the Prussian Blue and Hofmann clathrate coordination networks shown in Figure $\sf{1}$, the structures formed when a bridging ligand links multiple metal centers depends on the geometry of the centers, the ligand, and the metal-ligand bonds formed. Because the directionality of metal-ligand bonding in many types of complexes is well-characterized and predictable, it is possible to design metal complexes and linkers that can serve as building blocks for clusters and networks. In particular
• metal complexes with combinations of labile (substitutable) ligands in particular topologies. These insure that the linking ligand will be arranged around the metal center in those defined directions.
• Ligands that possess binding sites oriented in defined directions so that they link the metal centers together in particular topologies.
Examples of the sort of metal centers that can be used and examples of linking ligands are given in Figure $\sf{2}$.
Examples of how the building blocks shown in Figure $\sf{2}$ might be used to prepare molecular polyhedra and coordination networks are depicted schematically as shown in Figure $\sf{3}$. Among the examples shown in Figure $\sf{3}$, those depicted on the right schematize the Hofmann clathrate (top) and Prussian Blue (bottom) coordination networks of Figure $\sf{1}$.
A wider variety of structures are possible than those hinted at by the building blocks and structures shown in Figures $\sf{2}$ and $\sf{3}$. More may be obtained using the principles of molecular structure outlined in the sections on coordination geometry, ligands, and isomerism. In fact, one of the first coordination polyhedra prepared involved Jean-Marie Lehn's recognition that planar bidentate ligands in a tetrahedral bis-chelate are oriented perpendicular to one another, as shown in Figures $\sf{\PageIndex{5A}}$. This enabled him to prepare a molecular trigonal prism by combining a source of Cu2+ with the trigonal planar and linear-capable organic linkers that bind the Cu2+ in bidentate fashion, as shown in Figure $\sf{\PageIndex{5B}}$.
Metal Organic Frameworks (MOFs) are porous coordination polymers in which metal "building units" are connected by rigid organic ligand "struts"
One class of coordination frameworks that has received much attention over the past 25 years is the metal organic frameworks (MOFs). These consist of metal centers linked by rigid organic ligands to give a porous structure similar to those of the aluminosilicate-based zeolites. The structure of one MOF, called MIL-53, is given in Figure $\sf{\PageIndex{6A}}$.
As can be seen from the overall structure of MIL-53 depicted in Figure $\sf{\PageIndex{6A}}$, the structure possesses large rhombic prismatic voids, represented by the yellow spheres. These voids can in principle be occupied by small molecule substrates. Because of MOF's potential to bind and store substrates, there is considerable interest in developing MOFs that are useful for storing and separating particular gases.
A closer look at the structure of MIL-53 reveals that the carboxylate oxygens in the benzenedicarboxylato (BDU) linkers span two different metal centers rather than binding to only one. As explained in Figure $\sf{7}$, this binding mode enables the formation of rigid stable networks, although it requires the use of metal building blocks in which two or more metal centers are held in close proximity.
In MIL-53, the metal centers that the DBU ligands coordinate are chains of octahedral metal centers bridged by $\mu_2$ hydroxo ligands. These chains form spontaneously as the network forms under the high temperature conditions of its synthesis. In other cases, metal secondary building units (SBUs) containing the necessary bridging dicarboxylate ligands are used. One common class of metal SBUs used is the paddlewheel carboxylates shown in Figure $\sf{\PageIndex{8A}}$. These consist of two metals spanned by a square plane of four carboxylates in a paddlewheel arrangment.
The use of paddlewheel carboxylate metal SBUs and bridging organic carboxylate ligands enables the construction of coordination frameworks like that of HKUST-1 shown in Figure $\sf{9}$. The structure of these networks depends on the geometry of the linking ligand. For instance, in HKUST-1 the 1,3,5-benzenetricarboxylato bridging ligands each link three paddlewheel SBUs to give the cubic network shown in Figure $\sf{9}$. Other organic carboxylate ligands give different network topologies. In addition, the ability of some paddlewheel carboxylates to coordinate ligands perpendicular to the carboxylates, represented by the ligands marked L? in Figure $\sf{8}$, affords additional opportunities to use these networks to bind substrates that can coordinate those sites.
Another common metal SBU is the basic zinc acetate structure shown in Figure $\sf{10}$. As shown in Figure $\sf{\PageIndex{10A}}$, the structure is analogous to those of the basic beryllium acetates and involves an OZn4 tetrahedron in which the tetrahedrally arranged Zn2+ are connected by carboxylate ligands spanning the tetrahedron's six edges. When these carboxylate ligands are replaced by rigid dicarboxylates (Figure $\sf{\PageIndex{10B}}$), the result is that the Zn4O core is surrounded by an octahedral coordination sphere of carboxylate ligands (Figure $\sf{\PageIndex{10C}}$).
The structure of MOF-5 also illustrates how the size of the pores in an MOF structure may be tailored by lengthening or shortening the organic linker. In particular, from Figure $\sf{11}$ it can be seen that the size of the MOF-5's cubic unit cell depends on the length of the organic portion of the carboxylate linker. In MOF-5 this linker consists of a single benzene unit and gives 9-Angstrom pores, while the use of two and three benzene rings increases the pore size to 13 and 16 Angstroms, respectively, as shown in $\sf{12}$. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/09%3A_Coordination_Chemistry_I_-_Structure_and_Isomers/9.06%3A_Coordination_Frameworks.txt |
19.1: Occurrence, Preparation, and Properties of Transition Metals and Their Compounds
Q19.1.1
Write the electron configurations for each of the following elements:
1. Sc
2. Ti
3. Cr
4. Fe
5. Ru
S19.1.1
The electron configuration of an atom is the representation of the arrangement of electrons distributed among the orbital shells and sub-shells. The electron configuration of each element is unique to its position on the periodic table where the energy level is determined by the period and the number of electrons is given by the atomic number of the element. There are four different types of orbitals (s, p, d, and f) which have different shapes and each orbital can hold a maximum of 2 electrons, but the p, d and f orbitals have different sub-levels, meaning that they are able to hold more electrons.
The periodic table is broken up into groups which we can use to determine orbitals and thus, write electron configurations:
Group 1 & 2: S orbital
Group 13 - 18: P orbital
Group 3 - 12: D orbital
Lanthanide & Actinides: F orbital
Each orbital (s, p, d, f) has a maximum number of electrons it can hold. An easy way to remember the electron maximum of each is to look at the periodic table and count the number of periods in each collection of groups.
Group 1 & 2: 2 (2 electrons total = 1 orbital x max of 2 electrons = 2 electrons)
Group 13 - 18: 6 (6 electrons total = 3 orbitals x 2 electrons max = 6 electrons)
Group 3 - 12: 10 (10 electrons total = 5 orbitals x 2 electrons max = 10 electrons)
Lanthanide & Actinides: 14 (14 electrons total = 7 orbitals x 2 electrons max = 14 electrons)
Electron fills the orbitals in a specific pattern that affects the order in which the long-hand versions are written:
Electron filling pattern: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f
An easier and faster way to write electron configurations is to use noble gas configurations as short-cuts. We are able to do this because the electron configurations of the noble gases always have all filled orbitals.
He: 1s22s2
Ne: 1s22s22p6
Ar: 1s22s22p63s23p6
Kr: 1s22s22p63s23p64s23d104p6
Xe: 1s22s22p63s23p64s23d104p65s24d105p6
Rn: 1s22s22p63s23p64s23d104p65s24d105p66s24f145d106p6
The most common noble gas configuration used is Ar. When you want to use the noble gas configuration short-cut, you place the noble gas's symbol inside of brackets:
[Ar]
and then write it preceding the rest of the configuration, which is solely the orbitals the proceed after that of the noble gas.
a. Sc
Let's start off by identifying where Scandium sits on the periodic table: row 4, group 3. This identification is the critical basis we need to write its electron configuration.
By looking at Scandium's atomic number, 21, it gives us both the number of protons and the number of electrons. At the end of writing its electron configuration, the electrons should add up to 21.
At row 4, group 3 Sc, is a transition metal; meaning that its electron configuration will include the D orbital.
Now, we can begin to assign the 21 electrons of Sc to orbitals. As you assign electrons to their orbitals, you move right across the periodic table.
Its first 2 electrons are in the 1s orbital which is denoted as
1s2
where the "1" preceding the s denotes the fact that it is of row one, and it has an exponent of 2 because it fulfills the s orbital's maximum electron number. Now we have 21-2=19 more electrons to assign.
Its next 2 electrons are in the 2s orbital which is denoted as
2s2
where the "2" preceding the s indicates that it is of row two, and it has an exponent of 2 because it fulfills the s orbital's maximum electron number. Now we have 19-2=17 more electrons to assign.
Its next 6 electrons are in the 2p orbital which is denoted as
2p6
where the "2" preceding the p indicates that it is of row two, and it has an exponent of 6 because it fulfills the p orbital's maximum electron number. Now we have 17-6=11 more electrons to assign.
Its next 2 electrons are in the 3s orbital which is denoted as
3s2
where the "3" preceding the s indicates that it is of row three, and it has an exponent of 2 because it fulfills the s orbital's maximum electron number. Now we have 11-2=9 more electrons to assign.
Its next 6 electrons are in the 3p orbital which is denoted as
3p6
where the "3" preceding the p indicates that it is of row three, and it has an exponent of 6 because it fulfills the p orbital's maximum electron number. Now we have 9-6=3 more electrons to assign.
Its next 2 electrons are in the 4s orbital which is denoted as
4s2
where the "4" preceding the s indicates that it is of row four, and it has an exponent of 2 because it fulfills the s orbital's maximum electron number. Now we have 3-2=1 more electron to assign.
Its last electron would be alone in the 3 d orbital which is denoted as
3d1
where the "3" preceding the d indicates that, even though it is technically of row 4, by disregarding the first row of H and He, this is the third row and it has an exponent of 1 because there is only 1 electron to be placed in the d orbital. Now we have assigned all of the electrons to the appropriate orbitals and sub-orbitals, so that the final, entire electron configuration is written as:
1s22s22p63s23p64s23d1
This is the long-hand version of its electron configuration.
So for Sc, its short-hand version of its electron configuration would therefore be:
[Ar] 4s23d1
b. Ti
Start off by identifying where Titanium sits on the periodic table: row 4, group 4, meaning it has 22 electrons total. Titanium is one element to the right of the previous problem's Sc, so we will basically use the same method except, in the end, there will be 2 electrons remaining, so therefore the final orbital will be denoted as:
3d2
If needed, look above to the exact steps for how to do it in detail again; the long-hand electron configuration for Titanium will be:
1s22s22p63s23p64s23d2
So for Ti, its short-hand version of its electron configuration would therefore be:
[Ar] 4s23d2
c. Cr
Start off by identifying where Chromium sits on the periodic table: row 4, group 6, that means it has a total of 24 electrons. But first, Cr, along with Mo, Nb, Ru, Rh, Pd, Cu, Sg, Pt and Au, is a special case. You would think that since it has 24 electrons that its configuration would look like:
1s22s22p63s23p64s23d4
which is how we learned it earlier. However, this electron configuration is very unstable because of the fact that there are 4 electrons in its 3 d orbital. The most stable configurations are half-filled (d5) and full orbitals (d10), so the elements with electrons resulting in ending with the d4 or d9 are so unstable that we write its stable form instead, where an electron from the preceding s orbital will be moved to fill the d orbital, resulting in a stable orbital.
If needed, look above to the exact steps for how to do the beginning of the configuration in detail again. However we have to apple the new rule to attain stability so that the long-hand electron configuration for Chromium will be:
1s22s22p63s23p64s13d5
So for Cr, its short-hand version of its electron configuration would therefore be:
[Ar] 4s13d5
d. Fe
Start off by identifying where Iron sits on the periodic table: row 4, group 8, meaning it has 26 electrons total. This is 5 elements to the right of the previous problem's Sc, so we will basically use the same method except, in the end, there will be 6 electrons remaining, so therefore the final orbital will be denoted as:
3d6
If needed, look above to the exact steps for how to do it in detail again; the long-hand electron configuration for Iron will be:
1s22s22p63s23p64s23d6
So for Fe, its short-hand version of its electron configuration would therefore be:
[Ar] 4s23d6
e. Ru
Start off by identifying where Ruthenium sits on the periodic table: row 5, group 8, that means it has a total of 44 electrons. But first, as stated earlier, Ru, along with Cr, Mo, Nb, Rh, Pd, Cu, Sg, Pt and Au, is a special case. You would think that since it has 44 electrons that its configuration would look like:
1s22s22p63s23p64s23d104p65s2 4d6
which is how we learned it earlier. However, this electron configuration is very unstable because of the fact that, even though there are 4 paired electrons, there are also 4 electrons unpaired. This results in a very unstable configuration, so to restore stability, we have to use a configuration that has the most paired electrons, which would be to take an electron from the s orbital and place it in the d orbital to create:
5s14d7
If needed, look above to the exact steps for how to do the beginning of the configuration in detail again. However we have to apple the new rule to attain stability so that the long-hand electron configuration for Ru will be:
1s22s22p63s23p64s23d104p65s14d7
So for Cr, its short-hand version of its electron configuration would therefore be:
[Kr] 5s14d7
A19.1.1
1. Sc: [Ar]4s23d1
2. Ti: [Ar]4s23d2
3. Cr: [Ar]4s13d5
4. Fe: [Ar]4s23d6
5. Ru: [Kr]5s14d7 (anomalous configuration)
Q19.1.2
Write the electron configurations for each of the following elements and its ions:
1. Ti
2. Ti2+
3. Ti3+
4. Ti4+
S19.1.2
Electrons are distributed into molecular orbitals, the $s, p, d, and f$ blocks. An orbital will have a number in front of it and a letter that corresponds to the block. The s block holds two electrons, the p block holds six, the d block holds ten, and the f block holds fourteen. So, based on the number of electrons an atom has, the molecular orbitals are filled up in a certain way. The order of the orbitals is $1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p$. An exponent will be put after the letter for each orbital to signify how many electrons are in that orbital. Noble gas notation can also be used by putting the noble gas prior to the element you are writing the configuration for, and then proceed by writing the orbitals filled after the noble gas. Metal ions of the d-block will have the two electrons removed from the s block prior to any electrons being removed from the proceeding d-block.
Solutions:
1. $Ti$
Titanium has an atomic number of 22, meaning it has 22 electrons. The noble gas prior to Titanium is Argon. Looking at row 4 of the periodic table, Titanium still has 4 electrons to be placed in orbitals since Argon has 18 electrons that are already placed. The remaining electrons will fill the $4s$ orbital and the remaining two electrons will go into the $3d$ orbital. [Ar]4s23d2
2. $Ti^{+2}$
This is an ion with a plus 2 charge, meaning 2 electrons have been removed. The electrons will be removed from the $4s$ orbital and the 2 remaining electrons will be placed in the $3d$ orbital. Like number 1, the prior noble gas is Argon. [Ar]3d2
3. $Ti^{+3}$
This is an ion with a plus 3 charge, meaning 3 electrons have been removed. The first 2 electrons will be removed from the $4s$ orbital, and the third will be taken from the $3d$ orbital, and the 1 remaining electron will be placed in the 3d orbital. Like number 1, the prior noble gas is Argon. [Ar]3d1
4. $Ti^{+4}$
This is an ion with a plus 4 charge, meaning 4 electrons have been removed. The first 2 electrons will be removed from the $4s$ orbital and the second 2 will be removed from the $3d$ orbital. This results in the ion having the same electron configuration as Argon. [Ar]
Answers:
1. $[Ar]4s^23d^2$
2. $[Ar]3d^2$
3. $[Ar]3d^1$
4. $[Ar]$
A19.1.2
1. $[Ar]4s^2 3d^2$
2. $[Ar]3d^2$
3. $[Ar]3d^1$
4. $[Ar]$
Q19.1.3
Write the electron configurations for each of the following elements and its 3+ ions:
1. La
2. Sm
3. Lu
S19.1.3
In order to write the electron configuration, we begin by finding the element on the periodic table. Since La, Sm, and Lu are all a period below the noble gas Xenon, we can abbreviate ${1s^2}{2s^2}{2p^6}{3s^2}{3p^6}{3d^{10}}{4s^2}{4p^6}{4d^{10}}{5s^2}{5p^6}$ as [Xe] when writing the orbital configurations. We then find the remaining of the orbital configurations using the Aufbau Principle. For other elements not just those in period 6, the shorthand notation using noble gases would be the noble gas in the period above the given element.
1. La has three additional electrons. Two of them fill the 6s shell and the other single electron is placed on the 5d shell.
$La:$ [Xe] ${6{s}^2} {5{d}^1}$
2. Sm has eight more electrons. The 6s orbital is filled as previously and the 4f orbital receives 6 electrons because pairing electrons requires lower energy on the 4f shell than on the 5d shell.
$Sm:$ [Xe] ${6{s}^2} {4{f}^6}$
3. Lu has seventeen more electrons. Two electrons fill the 6s orbital, 14 electrons fill the 4f orbital, and extra single electron goes to the 5d orbital .
$Lu:$ [Xe] ${6s^2}{4f^{14}}{5d^1}$
To find the 3+ ion electron configuration, we remove 3 electrons from the neutral configuration, starting with the 6s orbital.
1. The ionization of La3+ removes the three extra electrons. So it reverts back to the stable Xenon configuration.
${La^{3+}:}$ [Xe]
2. The ionization of Sm3+ removes two electrons from the 6s shell and one from the outermost (4f) shell
${Sm^{3+}}:$ [Xe] ${4f^5}$
3. The ionization of Lu3+ removes its two 6s shell and one from the outermost (5d) shell, leaving only a full 4f shell
$Lu^{3+}:$ [Xe] $4f^{14}$
A19.1.3
La: [Xe]6s 25d 1, La3+: [Xe]; Sm: [Xe]6s 24f 6, Sm3+: [Xe]4f 5; Lu: [Xe]6s 24f 145d 1, Lu3+: [Xe]4f 14
Q19.1.4
Why are the lanthanoid elements not found in nature in their elemental forms?
A19.1.4
Lanthanides are rarely found in their elemental forms because they readily give their electrons to other more electronegative elements, forming compounds instead of staying in a pure elemental form. They have very similar chemical properties with one another, are often found deep within the earth, and difficult to extract. They are the inner transition elements and have partially filled d orbitals that can donate electrons. Because of this, they are very reactive and electropositive.
Q19.1.5
Which of the following elements is most likely to be used to prepare La by the reduction of La2O3: Al, C, or Fe? Why?
S19.1.5
An activity series is a list of elements in decreasing order of their reactivity. Elements on the top of the list are good reducing agents because they easily give up an electron, and elements on the bottom of the series are good oxidizing agents because they are highly electronegative would really want to accept an electron.
Step 1: Compare Aluminum, Carbon, and Iron on an activity series. Many activity series include carbon and hydrogen as references. An activity series can be found here
The activity series goes in the order (from top to bottom): Aluminum, Carbon, and Iron.
Step 2: Identify which element is the best reducing agent.
Elements on the top of the list are the best reducing agents, because they give up electrons the best.
Aluminum is the best reducing agent of the options available.
Therefore aluminum will be the best reducing agent to prepare La by the reduction of La2O3 because it is the most reactive in the series amongst the three elements.
A19.1.5
Al is used because it is the strongest reducing agent and the only option listed that can provide sufficient driving force to convert La(III) into La.
Q19.1.6
Which of the following is the strongest oxidizing agent: $\ce{VO4^{3-}}$, $\ce{CrO4^2-}$, or $\ce{MnO4-}$?
S19.1.6
Oxidizing agents oxidize other substances. In other words, they gain electrons or become reduced. These agents should be in their highest oxidation state. In order to determine, the strength of the compounds above as oxidizing agents, determine the oxidation numbers of each constituent elements.
$\\mathrm{VO_4^{3-}}$
We know that $\mathrm{O}$ has a -2 oxidation state and the overall charge of the ion is -3. We just need to determine Vanadate's oxidation number in this compound.
$\\mathrm{V} + \mathrm{-2(4)} = \mathrm{-3}$
$\\mathrm{V} = \mathrm{+5}$
Vanadate has an oxidation number of +5, which is its highest possible oxidation state.
$\\mathrm{CrO_4^{2-}}$
Like in the previous calculation, $\mathrm{O}$ has a -2 oxidation state. The overall charge is -2. So calculate for chromium.
$\\mathrm{Cr} + \mathrm{-2(4)} = \mathrm{-2}$
$\\mathrm{Cr} = \mathrm{+6}$
Chromium is in its highest possible oxidation state of +6 in this compound.
$\\mathrm{MnO_4^-}$
$\mathrm{O}$ has a -2 oxidation state and the overall charge is -1.
$\\mathrm{Mn} + \mathrm{-2(4)} = \mathrm{-1}$
$\\mathrm{Mn} = \mathrm{+7}$
Manganese is also in its highest oxidation state, +7.
An oxidizing agent has to be able to gain electrons which, in turn, reduces its oxidation state. Here manganese has the greatest oxidation state which allows it to experience a greater decrease in its oxidation state if needed, meaning it can gain the most electrons. So among the three compounds, $\mathrm{MnO_4^-}$ is the strongest oxidizing agent. This method assumes the metals have similar electronegativities.
Alternatively, check a redox table.
A19.1.6
$MnO_4^-$
Q19.1.7
Which of the following elements is most likely to form an oxide with the formula MO3: Zr, Nb, or Mo?
S19.1.7
Mo because Zr has an oxidation state of +4 and Nb has a oxidation state of +5 and those would not balance out the charge of 3 oxygens in the state of -2 which creates a charge of -6. Mo however has multiple oxidation states, the most common being +6 which balances out the -6 charge created by 3 oxygen ions. This is why its most likely to form an oxide with the formula MO3 or $\ce{MoO3}$.
Mo
Q19.1.8
The following reactions all occur in a blast furnace. Which of these are redox reactions?
1. $\ce{3Fe2O3}(s)+\ce{CO}(g)⟶\ce{2Fe3O4}(s)+\ce{CO2}(g)$
2. $\ce{Fe3O4}(s)+\ce{CO}(g)⟶\ce{3FeO}(s)+\ce{CO2}(g)$
3. $\ce{FeO}(s)+\ce{CO}(g)⟶\ce{Fe}(l)+\ce{CO2}(g)$
4. $\ce{C}(s)+\ce{O2}(g)⟶\ce{CO2}(g)$
5. $\ce{C}(s)+\ce{CO2}(g)⟶\ce{2CO}(g)$
6. $\ce{CaCO3}(s)⟶\ce{CaO}(s)+\ce{CO2}(g)$
7. $\ce{CaO}(s)+\ce{SiO2}(s)⟶\ce{CaSiO3}(l)$
S19.1.8
o identify redox reaction, we have to determine if have to see if the equation is an oxidation-reduction reaction-meaning that the species are changing oxidation states during the reaction, which involves the transfer of electrons between two species. If a species is losing electrons, then that species is being oxidized. If a species is gaining electrons, then that species is being reduced. A way to remember this is using the acronyms OIL RIG. Oxidation Is Loss, and Reduction Is Gain, referring to electrons. Both of these must occur for an equation to be a redox reaction. Let's see if these equations are redox reactions or not:
a. In the reactants side $\ce{Fe2O3}$, Fe is has an oxidation number of +3. In the product $\ce{Fe3O4}$, Fe has an oxidation number of +2.67. Since Fe changed from +3 to +2.67, we can say that Fe had gained electrons and therefore reduced. In the reactant, CO, carbon has an oxidation number of +2, and in $\ce{CO2}$ (product) carbon has an oxidation number of +4. Therefore, carbon has lost electrons and it has been oxidized. Since there is oxidation and reduction of species- we can conclude that this is a redox reaction.
b. In the reactant, $\ce{Fe3O4}$, Fe has an oxidation number of +2.67. In the product, FeO, Fe has an oxidation number of +2. Since the oxidation of Fe has changed from +2.67 to +2, electrons have been added therefore Fe has been reduced. In the reactant, CO, carbon has an oxidation number of +2, and in $\ce{CO2}$ (product) carbon has an oxidation number of +4. Therefore, carbon has lost electrons and it has been oxidized. Since there is oxidation and reduction of species- we can conclude that this is a redox reaction.
c. In the reactant side, in FeO, Fe has an oxidation number of +2 and in the products side Fe has an oxidation number of 0. Since the oxidation number of Fe changed from +2 to 0, electrons have been gained and therefore Fe has been reduced. In the reactant, CO, carbon has an oxidation number of +2, and in $\ce{CO2}$ (product) carbon has an oxidation number of +4. Therefore, carbon has lost electrons and it has been oxidized. Since there is oxidation and reduction of species- we can conclude that this is a redox reaction.
d. In the reactants C has an oxidation number of 0, and in the products side in $\ce{CO2}$, C has an oxidation number of +4. Since the oxidation number of C has changed from 0 to +4, we can say that C has been oxidized. In the reactants, in $\ce{O2}$ oxygen has an oxidation number of 0, and in the products CO2, oxygen has an oxidation number of -2. Since the oxidation number of oxygen has changed from 0 to -2, oxygen has been reduced. Since there is oxidation and reduction of species- we can conclude that this is a redox reaction.
e. In the reactants $\ce{CO2 }$ has an oxidation number of +4, and in the products side in CO, C has an oxidation number of +2. Since carbon went from +4 to +2, carbon has been reduced. In the reactants, in $\ce{CO2 }$ oxygen has an oxidation number of -4 and in the products CO carbon has an oxidation number of -2. Since oxygen went from -4 to -2, it has been oxidized. Since there is oxidation and reduction of species- we can conclude that this is a redox reaction.
f. In the reactants, $\ce{CaCO3}$ Ca has an oxidation number of +2, and in products CaO Ca has an oxidation number of +2. Since the oxidization number doesn't change- we can conclude that this equation is not a redox reaction.
g. In the products CaO Ca has an oxidation number of +2, and in the products $\ce{CaSiO3}$ Ca has an oxidation number of +2. Since the oxidization number doesn't change- we can conclude that this equation is not a redox reaction.
a, b, c, d, e
Q19.1.9
Why is the formation of slag useful during the smelting of iron?
S19.1.9
Slag is a substance formed as a byproduct of iron ore or iron pellets melting together in a blast furnace. Slag is also the byproduct that is formed when a desired metal has been separated from its raw ore. It is important to note that slag from steel mills is created in a manner that reduces the loss of the desired iron ore. The $\ce{CaSiO3}$slag is less dense than the molten iron, so it can easily be separated. Also, the floating slag layer creates a barrier that prevents the molten iron from exposure to $\ce{O2}$, which would oxidize the $\ce{Fe}$ back to $\ce{Fe2O3}$. Since Fe has a low reduction potential of -0.440 this means it has a high oxidation potential so it would easily oxidize in the presence of O2. Creating a barrier between iron and oxygen allows the maximum product of iron to be obtained in the end of smelting.
A19.1.9
The CaSiO3 slag is less dense than the molten iron, so it can easily be separated. Also, the floating slag layer creates a barrier that prevents the molten iron from exposure to O2, which would oxidize the Fe back to Fe2O3.
Q19.1.10
Would you expect an aqueous manganese(VII) oxide solution to have a pH greater or less than 7.0? Justify your answer.
S19.1.10
Manganese(VII) oxide, can be written as Mn2O7.
In relation to the Lewis acid-base theory, a Lewis acid accepts lone pair electrons, and is also known as the electron pair acceptor. Based on this theory, acidity can be measured by the element's ability to accept electron pairs. By doing the math, we find that Manganese has an oxidation state of +7 (Oxygen has an oxidation state of -2, and 2x-7=-14 or this can be shown as $-7(2)+2(x)=0$ and $x=7$ since the whole compound has a charge of zero, in order to balance the ion's charge, Mn must be +7). Therefore Mn has high capability of accepting electrons due to its high positive charge. For most metals, as the oxidation number increases, so does its acidity, because of its increased ability to accept electrons.
A19.1.10
In relation to the Lewis acid-base theory, the Lewis acid accepts lone pair electrons; thus, it is also known as the electron pair acceptor. This may be any chemical species. Acids are substances that must be lower than 7. Therefore, oxides of manganese is most likely going to become more acidic in (aq) solutions if the oxidation number increases.
Q19.1.11
Iron(II) can be oxidized to iron(III) by dichromate ion, which is reduced to chromium(III) in acid solution. A 2.5000-g sample of iron ore is dissolved and the iron converted into iron(II). Exactly 19.17 mL of 0.0100 M Na2Cr2O7 is required in the titration. What percentage of the ore sample was iron?
S19.1.11
To answer this question, we must first identify the net ionic equation from the given half-reactions. We can write the oxidation and reduction half-reactions:
$\text{ oxidation:} \text{ Fe}^\text{ 2+} \rightarrow \text{ Fe}^\text{ 3+}$
$\text{ reduction: }\ce{Cr2O7^2-} \rightarrow \text{ Cr}^\text{ 3+}$
We can quickly balance the oxidation half-reaction by adding the appropriate number of electrons to get
$\text{ Fe}^\text{ 2+} \rightarrow \text{ Fe}^\text{ 3+}+\text{ e}^-$
The first step in balancing the reduction half-reaction is to balance elements in the equation other than O and H. In doing so, we get
$\ce{Cr2O7^2-} \rightarrow 2 \text{ Cr}^\text{ 3+}$
The second step would be to add enough water molecules to balance the oxygen.
$\ce{Cr2O7^2-} \rightarrow 2 \text{ Cr}^\text{ 3+}+ \ce{7H2O}$
Next, we add the correct amount of H+ to balance the hydrogen atoms.
$\ce{Cr2O7^2-}+14 \text{ H}^+ \rightarrow 2 \text{ Cr}^\text{ 3+}+ \ce{7H2O}$
Finally, we add enough electrons to balance charge.
$\ce{Cr2O7^2-}+14 \text{ H}^++6 \text{ e}^- \rightarrow 2 \text{ Cr}^\text{ 3+}+ \ce{7H2O}$
The electrons involved in both half-reactions must be equal in order for us to combine the two to get the net ionic equation. This can be done by multiplying each equation by the appropriate coefficient. Scaling the oxidation half-reaction by 6, we get
$6 \text{ Fe}^\text{ 2+} \rightarrow 6 \text{ Fe}^\text{ 3+}+6 \text{ e}^-$
Now we can combine both half-reactions to get
$6 \text{ Fe}^\text{ 2+}+\ce{Cr2O7^2-}+14 \text{ H}^++6 \text{ e}^- \rightarrow 6 \text{ Fe}^\text{ 3+}+6 \text{ e}^-+2 \text{ Cr}^\text{ 3+}+ \ce{7H2O}$
The electrons cancel out, so you get:
$6 \text{ Fe}^\text{ 2+}+\ce{Cr2O7^2-}+14 \text{ H}^+ \rightarrow 6 \text{ Fe}^\text{ 3+}+2 \text{ Cr}^\text{ 3+}+ \ce{7H2O}$
From this we can see that the mole ratio of Cr2O72- to Fe2+ is 1:6. Given that 19.17 mL (or 0.01917 L) of 0.01 M Na2Cr2O7 was needed for titration we know that
$0.01917\text{ L} \times 0.01 \text{ M} = 1.917 \times 10^{-4} \text{ mol}$
of Na2Cr2O7 reacted. Also, since any number of moles of Na2Cr2O7 produces the same number of moles of Cr2O72- in solution
$1.917 \times 10^{-4} \text{ mol} \text{ of } \ce{Na2Cr2O7} =1.917 \times 10^{-4} \text{ mol} \text{ of } \ce{Cr2O7^2-}$
We can use the mole ratio of Cr2O72- to Fe2+ to determine how many moles of iron (ii) was in the solution. The number of moles of iron (ii) is the same as the number of moles of pure iron in the sample since all of the iron was converted into iron (ii).
$1.917 \times 10^{-4} \text{ mol} \text{ of } \ce{Cr2O7^2-}\times \frac{6\text{ mol}\text{ of } \text{ Fe}^\text{ 2+}}{1\text{ mol}\text{ of }\ce{Cr2O7^2-}} = 0.0011502\text{ mol}\text{ of } \text{ Fe}^\text{ 2+}$
$0.0011502\text{ mol}\text{ of } \text{ Fe}^\text{ 2+} = 0.0011502\text{ mol}\text{ of } \text{ Fe}$
Now we can find the number of grams of iron that were present in the 2.5 g iron ore sample.
$0.0011502\text{ mol}\text{ of } \text{ Fe}\times\frac{55.847\text { g}}{1\text{ mol}} = 0.0642352194\text{ g}\text{ of }\text{ Fe}$
Finally, we can answer the question and find what percentage of the ore sample was iron.
$\frac{0.0642352194\text{ g}}{2.5\text{ g}} \times 100 \approx 2.57\text{%}$
So 2.57% of the ore sample was iron.
2.57%
Q19.1.12
How many cubic feet of air at a pressure of 760 torr and 0 °C is required per ton of Fe2O3 to convert that Fe2O3 into iron in a blast furnace? For this exercise, assume air is 19% oxygen by volume.
S19.1.12
This question uses a series of unit conversions and the $PV=nRT$ equation.
The first step is to write out the balanced chemical equation for the conversion of Fe2O3 to pure iron.
$2\;Fe_2O_3(s)\rightarrow 4\;Fe(s)+3\;O_2(g)$
Next, we need to analyze the original question to determine the value that we need to solve for. Because the question asks for a value of cubic feet, we know we need to solve for volume. We can manipulate $PV=nRT$ to solve for volume.
$V={nRT}/P$
Now determine the known variables and convert into units that will be easy to deal with.
$n = 2000\:lbs\; Fe_2O_3\frac{453.592\: grams\: Fe_2O_3}{1\: lb \:Fe_2O_3}\frac{1\: mole \:Fe_2O_3}{159.69\: grams \:Fe_2O_3}\frac{3 \;moles\: O_2}{2\; moles \:Fe_2O_3}$
$n=8521\: moles\: of \:O_2$
Convert to atm for easier calculations
$R=\frac{.0821\:L\:atm}{mol\:K}$
$T=0^{\circ}C=273\:K$
$P=760 \:torr= 1 \:atm$
Now plug the numbers into the manipulated gas law to get to an answer for V.
$V=190991.8\: liters \:of\: O_2$
From here we convert liters to cubic feet.
use the conversion
$1\;L=.0353 ft^3$
thus we have 6744.811 ft3 of O2
We then refer back to the initial question and remember that this value is only 19% of the volume of the total air. So use a simple equation to determine the total volume of air in cubic feet.
$6744.811ft^3=.19x$
x=35499 ft3 of air
35499 ft3 of air
Q19.1.13
Find the potentials of the following electrochemical cell:
Cd | Cd2+ (M = 0.10) ‖ Ni2+ (M = 0.50) | Ni
S19.1.13
Step 1 Write out your two half reactions and identify which is oxidation and which is reduction using the acronym OIL RIG to remember that oxidation is loss of electrons and reduction is gain of electrons
Cd(s)⟶Cd2+(aq)+2e- (oxidation)
Ni2+(aq)+2e-⟶Ni(s) (reduction)
Step 2 Write out complete balanced equation
Cd(s)+ Ni2+(aq)⟶Cd2+(aq)+Ni(s)
Step 3 Find Eocell
Eocell = Ecathode-Eanode
oxidation: Cd(s)⟶Cd2+(aq)+2e- Eo=-0.40V
reduction: Ni2+(aq)+2e-⟶Ni(s) Eo=-0.26V
* E values come from standard reduction potentials table given above. Also, remember anode is where oxidation happens, and cathode is where reduction happens.
Eocell=-0.26-(-.40)
Eocell=0.14V
Step 4 Find Q
Q=[products]/[reactants] (look at complete balanced equation) (remember that [x] means the concentration of x typically given in molarity and that we ignore solids or liquids)
Q=[Cd2+]/[Ni2+]
Q=0.10/0.50
Q=0.2
Step 5 Calculate E using E= Eocell-(.0592/n)logQ (n is number of moles of electrons transferred and in our case the balanced reaction transfers 2 electrons)
E= 0.14-(.0592/2)log(0.2)
E= 0.14-(-.207)
E=0.16 V
0.16 V
Q19.1.14
A 2.5624-g sample of a pure solid alkali metal chloride is dissolved in water and treated with excess silver nitrate. The resulting precipitate, filtered and dried, weighs 3.03707 g. What was the percent by mass of chloride ion in the original compound? What is the identity of the salt?
S19.1.14
A 2.5624-g sample of a pure solid alkali metal chloride is dissolved in water and treated with excess silver nitrate. The resulting precipitate, filtered and dried, weighs 3.03707 g. What was the percent by mass of chloride ion in the original compound? What is the identity of the salt?
Assuming that metal chloride is XCl
The balance equation for the reaction would be:
$XCl(aq)+AgNO_{3}(aq)\rightarrow XNO_{3}(aq) +AgCl(s)$
The mass of AgCl = 3.03707g
To find the moles of AgCl present:
Next, we can determine the moles of AgCl present in the reaction since 1) the mass of the precipitate is given to us and 2) this value can help us determine the moles of alkali metal chloride compound present. Given the mass of AgCl is 3.03707g in the problem and the molecular mass of AgCl per mole is 143.32g, we can solve for how many moles of AgCl is in the reaction:
$moles of Agcl=\tfrac{3.03707g}{143.32g/mol}=0.0211 mol$
Since the molar ratio of the compounds are 1:1 so the number of moles of XCl used = 0.0211 mol
We can calculate the weight of Cl- with the equation:
$0.0211 mol \times 35.5g/mol = 0.7490g$
the amount of metal present in the original compound is the weight of the compound subtracted by weight of the Cl ion:
$(2.5624- 0.7490)g= 1.8134g$
And the percentage can be calculate by
$\frac{0.7490}{2.5624}\times 100= 29.23 \%$
the molar ratio of XCl is 1:1 so then
Atomic mass of metal = $=\frac{1.8134\;g\; metal}{0.0211\; mol\; RbCl}=85.943g/mol$
So the atomic mass is 85.943 g/mol which is of Rb hence the identity of the salt is RbCl
Q19.1.15
The standard reduction potential for the reaction $\ce{[Co(H2O)6]^3+}(aq)+\ce{e-}⟶\ce{[Co(H2O)6]^2+}(aq)$ is about 1.8 V. The reduction potential for the reaction $\ce{[Co(NH3)6]^3+}(aq)+\ce{e-}⟶\ce{[Co(NH3)6]^2+}(aq)$ is +0.1 V. Calculate the cell potentials to show whether the complex ions, [Co(H2O)6]2+ and/or [Co(NH3)6]2+, can be oxidized to the corresponding cobalt(III) complex by oxygen.
S19.1.15
To calculate the cell potential, we need to know the potentials for each half reaction. After doing so, we need to determine which one is being oxidized and which one is being reduced. The one that is oxidized is the anode and the one that is reduced is the cathode. To find the cell potential, you use this formula and the reduction potential values found in a reduction potential table. If E°cell is positive, $\Delta$G is negative and the reaction is spontaneous.
cell= E°cathode - E°anode
Because it states that $[Co(H_{2}O)_{6}]^{3+}$ will be oxidized, this means it is the anode.
$O_{2}(g) + 4 H^{+}(aq) + 4 e^{-} \rightarrow 2 H_{2}O$ +1.229 V
$O_{2}$ is being reduced, so it is the cathode.
1.229V - 1.8V= -.571 V, or -0.6 V using significant figures. This cannot happen spontaneously because E°cell is negative.
For $[Co(NH_{3})_{6}]^{3+}$, it is again being oxidized, meaning it’s the anode.
1.229-.1= 1.129 V or 1.1 V using significant figures. This reaction is spontaneous because E°cell is positive.
A19.1.15
a) E ° = −0.6 V, E ° is negative so this reduction is not spontaneous.
b) E ° = +1.1 V, E ° is positive so this reduction is spontaneous.
Q19.1.16
Predict the products of each of the following reactions. (Note: In addition to using the information in this chapter, also use the knowledge you have accumulated at this stage of your study, including information on the prediction of reaction products.)
1. $\ce{MnCO3}(s)+\ce{HI}(aq)⟶$
2. $\ce{CoO}(s)+\ce{O2}(g)⟶$
3. $\ce{La}(s)+\ce{O2}(g)⟶$
4. $\ce{V}(s)+\ce{VCl4}(s)⟶$
5. $\ce{Co}(s)+xs\ce{F2}(g)⟶$
6. $\ce{CrO3}(s)+\ce{CsOH}(aq)⟶$
S19.1.16
There is a myriad of reactions that can occur, which include: single replacement, double replacement, combustion, acid-base/neutralization, decomposition or synthesis. The first step to determine the products of a reaction is to identify the type of reaction. From then on, the next steps you take to predict the products will vary based on the reaction type.
1. This reaction is a double displacement reaction, in which the cations and anions of the reactants switch places to form new compounds. Writing out the equation in terms of it's aqueous ions will help you visualize what exactly is getting moved around:
$2H^{+}(aq)+2I^{-}(aq)\rightarrow 2H^{+}(aq)+CO_{3}^{2-}(aq)+Mn^{2+}(aq)+2I^{-}(aq)$
In this case, the hydrogen cations will recombine with carbonate anions whilst manganese cations will recombine with iodide anions giving us the following equation:
$\ce{MnCO3}(s)+\ce{2HI}(aq)⟶\ce{MnI2}(aq)+\ce{H2CO3}(aq)$
This is still not the final answer however, as carbonic acid is unstable and decomposes to carbon dioxide and water under standard conditions. Taking this into account, our final equation is:
$\ce{MnCO3}(s)+\ce{2HI}(aq)⟶\ce{MnI2}(aq)+\ce{CO2}(g)+\ce{H2O}(l)$
2. This reaction is a synthesis reaction, in which two or more reactants combine to form a more complex compound. In this case we are also reacting a metal oxide with oxygen which would result in another metal oxide as the product. The resulting product would be the mixed valence oxide Co3O4 in which one cobalt atom has a +2 oxidation state whilst the other two have a +3 oxidation state. Now all is left is to balance the equation:
$\ce{6CoO}(s)+\ce{O2}(g)⟶\ce{2Co3O4}(s)$
3. Like equation 2, this reaction is also a synthesis reaction involving a metal and oxygen which should result in the formation of a metal oxide. It is a matter now of balancing the oxidation states to attain a neutral compound. Oxygen will always hold a -2 oxidation state in compounds whilst lanthanum will always exhibit a +3 oxidation state. As such, a combination of 2 lanthanum atoms with a +3 oxidation state and 3 oxygen atoms with a -2 oxidation state will give us a molecule with an overall charge of 0 (3(-2)+2(+3)=0). We know our product now, La2O3, and now just need to balance the overall equation, giving us:
$\ce{4La}(s)+\ce{3O2}(g)⟶\ce{2La2O3}(s)$
4. This reaction is slightly harder to define as it encapsulates both the properties of synthesis and decomposition reactions, wherein vanadium reacts with vanadium tetrachloride to produce vanadium trichloride. This reaction however is primarily a synthesis reaction since we are combining two reactants to produce one complex compound. With vanadium trichloride as our product, we can balance the equation:
$\ce{V}(s)+\ce{3VCl4}(s)⟶\ce{4VCl3}(s)$
5. This is another synthesis reaction combining cobalt and fluorine. This equation includes the "xs" notation in front of fluorine which is short for 'excess', meaning more fluorine than actually required is present in the reactants, ensuring the reaction goes to completion. Finding the products if a simple matter of combining the cobalt and fluorine into one molecule, which already leaves us with a balanced equation:
$\ce{Co}(s)+xs\ce{F2}(g)⟶\ce{CoF2}(s)$
6. It may not be obvious here, but the reaction we've been given here is actually an acid-base/neutralization reaction, with chromium trioxide acting as the acid and cesium hydroxide as the base. Chromium trioxide is referred to as an acidic oxide which means that it will react with water to form an acid. Note that this reaction can still proceed even if the reactants aren't in the same phases. The basic rule for acid-base/neutralization reactions is they form a salt (salt being the general term for any ionic compound formed from acid-base reactions) and water. Since we know water is one of our products, our other product must be a salt composed of cesium, chromium and oxygen. Thus, our other product should be cesium chromate, and you can now balance the equation accordingly:
$\ce{CrO3}(s)+\ce{2CsOH}(aq)⟶\ce{Cs2CrO4}(aq)+\ce{H2O}$
A19.1.16
1. $\ce{MnCO3}(s)+\ce{2HI}(aq)⟶\ce{MnI2}(aq)+\ce{CO2}(g)+\ce{H2O}(l)$
2. $\ce{6CoO}(s)+\ce{O2}(g)⟶\ce{2Co3O4}(s)$
3. $\ce{4La}(s)+\ce{3O2}(g)⟶\ce{2La2O3}(s)$
4. $\ce{V}(s)+\ce{3VCl4}(s)⟶\ce{4VCl3}(s)$
5. $\ce{Co}(s)+xs\ce{F2}(g)⟶\ce{CoF2}(s)$
6. $\ce{CrO3}(s)+\ce{2CsOH}(aq)⟶\ce{Cs2CrO4}(aq)+\ce{H2O}$
Q19.1.17
Predict the products of each of the following reactions. (Note: In addition to using the information in this chapter, also use the knowledge you have accumulated at this stage of your study, including information on the prediction of reaction products.)
1. $\ce{Fe}(s)+\ce{H2SO4}(aq)⟶$
2. $\ce{FeCl3}(aq)+\ce{NaOH}(aq)⟶$
3. $\ce{Mn(OH)2}(s)+\ce{HBr}(aq)⟶$
4. $\ce{Cr}(s)+\ce{O2}(g)⟶$
5. $\ce{Mn2O3}(s)+\ce{HCl}(aq)⟶$
6. $\ce{Ti}(s)+xs\ce{F2}(g)⟶$
S19.1.17
Predict the products of each of the following reactions.
• Fe(s)+H2SO4(aq)⟶ ?
Whenever a metal reacts with an acid, the products are salt and hydrogen. Because Fe is lower on the activity series, we know that when it reacts with an acid it will result in the formation of Hydrogen gas. To simplify the equation is:
$Metal + Acid ⟶ Salt + Hydrogen$
The salt produced will depend on the metal and in this case, the metal is iron (Fe) so the resulting equation would be:
$\ce{Fe}(s)+\ce{H2SO4}(aq)⟶ \ce{FeSO4}(aq) + \ce{H2}(g)$
This equation works out as the H2 is removed from H2SO2, resulting in a SO42- ion where Fe will take on an oxidation state of Fe+2 to form FeSO4 which will be the salt in this example.
But since FeSO4 and H2SO4 are aqueous, the reactants and products can also be written as its ions where the overall equation can be:
$\ce{Fe}(s)+\ce{2H3O+}(aq)+\ce{SO2^{−4}}(aq)⟶\ce{Fe2+}(aq)+\ce{SO2^{−4}}(aq)+\ce{H2}(g)+\ce{2H2O}(l)$
• FeCl3(aq)+NaOH(aq)⟶ ?
In this case, adding a metal hydroxide (NaOH) to a solution with a transition metal ion (Fe) will form a transition metal hydroxide (XOH). As iron is bonded to three chlorine atoms in the reactants side, it has the oxidation state of +3 where three hydroxide ions (OH-) are needed to balance out the charges when they are bonded in the products. The remaining ions are Na+ and Cl- where they bond together in a 1:1 ratio where there are 3 molecules of NaCl once the reaction is balanced.
The overall reaction will be:
$\ce{FeCl3}(aq)+\ce{NaOH}(aq)⟶ \ce{Fe(OH)3}(s) + \ce{3NaCl}(aq)$
NOTE: Fe(OH)3(s) is a solid as it is a rule that all all transition metal hydroxides are insoluble and a precipitate is formed.
Since NaOH(aq) and NaCl(aq) are aqueous, we can write them out in their ion forms:
$\ce{FeCl3}(aq)+\ce{3Na+}(aq)+\ce{3OH^{−}}(aq)+\ce{Fe(OH)3}(s)+\ce{3Na+}(aq)+\ce{3Cl+}(aq)$
• Mn(OH)2(s)+HBr(aq)⟶ ?
This is an example of a metal hydroxide reacting with an acid where a metal salt and water will always be formed:
$Metal Hydroxide + Acid ⟶ Metal Salt + Water$
When this rule is applied to this equation, we will get the following:
$\ce{Mn(OH)2}(s)+\ce{HBr}(aq)⟶ \ce{MnBr2}(aq)+\ce{2H2O}(l)$
But to follow through with this question, the aqueous solutions such as HBr(aq) and MnBr2(aq) can be re-written as:
$\ce{Mn(OH)2}(s)+\ce{2H3O+}(aq)+\ce{2Br-}(aq)⟶\ce{Mn2+}(aq)+\ce{2Br-}(aq)+\ce{4H2O}(l)$
• Cr(s)+O2(g)⟶ ?
This is the general reaction of a metal reacting with oxygen which will always result in a metal oxide. However, the metal oxide is determined by the oxidation state of the metal so there may be several outcomes of this reaction such as:
$\ce{4Cr}(s)+\ce{3O2}(g)⟶\ce{2Cr2O3}(s)$
$\ce{Cr}(s)+\ce{O2}(g)⟶\ce{2CrO}(s)$
$\ce{Cr}(s)+\ce{O2}(g)⟶\ce{CrO2}(s)$
$\ce{2Cr}(s)+\ce{3O2}(g)⟶\ce{CrO3}(s)$
However, Cr2O3 is the main oxide of chromium so it can be assumed that this is the general product of this reaction.
• Mn2O3(s)+HCl(aq)⟶?
This follows the general reaction of a metal oxide and an acid will always result in a salt and water
$Metal Oxide + Acid ⟶ Salt + Water$
Using this general reaction, similar to the general reactions above, the reaction will result in:
$\ce{Mn2O3}(s)+\ce{HCl}(aq)⟶\ce{2MnCl3}(s)+\ce{9H2O}(l)$
However, since HCl is an aqueous solution, the overall equation can also be re-written as:
$\ce{Mn2O3}(s)+\ce{6H3O+}(aq)+\ce{6Cl-}(aq)⟶\ce{2MnCl3}(s)+\ce{9H2O}(l)$
• Ti(s)+xsF2(g)⟶?
Titanium is able to react with the halogens where there are two oxidation state that titanium can be: +3 and +4. The following reactions follow each oxidation state accordingly:
$\ce{2Ti}(s)+\ce{3F2}(g)⟶ \ce{2TiF3}(s)$
$\ce{Ti}(s)+\ce{2F2}(g)⟶\ce{TiF4}(s)$
However, since there is the symbol "xs", this indicates that F2 is added in excess so the second reaction is favored more as it drives the reaction to completion.
OVERALL:
$\ce{Ti}(s)+\ce{xsF2}(g)⟶\ce{TiF4}(g)$
A19.1.17
1. $\ce{Fe}(s)+\ce{2H3O+}(aq)+\ce{SO4^2-}(aq)⟶\ce{Fe^2+}(aq)+\ce{SO4^2-}(aq)+\ce{H2}(g)+\ce{2H2O}(l)$;
2. $\ce{FeCl3}(aq)+\ce{3Na+}(aq)+\ce{3OH-}(aq)+\ce{Fe(OH)3}(s)+\ce{3Na+}(aq)+\ce{3Cl+}(aq)$;
3. $\ce{Mn(OH)2}(s)+\ce{2H3O+}(aq)+\ce{2Br-}(aq)⟶\ce{Mn^2+}(aq)+\ce{2Br-}(aq)+\ce{4H2O}(l)$;
4. $\ce{4Cr}(s)+\ce{3O2}(g)⟶\ce{2Cr2O3}(s)$;
5. $\ce{Mn2O3}(s)+\ce{6H3O+}(aq)+\ce{6Cl-}(aq)⟶\ce{2MnCl3}(s)+\ce{9H2O}(l)$;
6. $\ce{Ti}(s)+xs\ce{F2}(g)⟶\ce{TiF4}(g)$
Q19.1.18
Describe the electrolytic process for refining copper.
S19.1.18
By electrolysis, copper can be refined and purely made. The reason why copper needs to remove the impurities is because it helps increase the electrical conductivity in electrical wire. You can refine copper and remove the impurities through electrolysis. Pure copper is important in making electrical wire, because it creates better electrical conductivity when transferring electricity. In order for better electrical conductivity, the impurities needs to be removed and this can be done by firing the impure copper to remove the impurities, such as sulfur, oxygen, etc. and shaping them into electrical anodes that can be used in electrolysis. Then the copper electrodes are placed into an electrical cell (into separate beakers) where electrical current can pass through the beakers and onto the electrodes. Through this process, the copper is stripped off of the anode and deposited onto the cathode. This process helps remove the impurities and refine copper because all the copper has been deposited onto the cathode all in one electrode. This process increases the weight of the cathode due to copper being deposited onto the cathode. This is a prime example of how to tell if an electrode is a cathode or an anode, as stated in Q17.2.9 above.
Q19.1.19
Predict the products of the following reactions and balance the equations.
1. Zn is added to a solution of Cr2(SO4)3 in acid.
2. FeCl2 is added to a solution containing an excess of $\ce{Cr2O7^2-}$ in hydrochloric acid.
3. Cr2+ is added to $\ce{Cr2O7^2-}$ in acid solution.
4. Mn is heated with CrO3.
5. CrO is added to 2HNO3 in water.
6. FeCl3 is added to an aqueous solution of NaOH.
S19.1.19
1. $\bg_black Zn$ is added to a solution of $\bg_black Cr_2(SO_4)_3$ in acid.
• Oxidized half-reaction:$3Zn(s) \rightarrow 3Zn^{2+}(aq)+6e^-$
• Reduction half reaction:
• Overall reaction:
• Chromium will precipitate out of the solution because it has a higher reduction potential than Zinc; the reaction is a single replacement.
2. $\bg_black FeCl_2$ is added to a solution containing an excess of $\bg_black Cr_2O_7^{2-}$in hydrochloric acid.
• Dissociation reaction:$6FeCl_2(s) \rightarrow 6Fe^{2+}(aq)+12Cl^-(aq)$
• Oxidation half-reaction:$6Fe^{2+}(aq) \rightarrow 6Fe^{3+}(aq)+6e^-$
• Reduction half-reaction:$Cr_2O_7^{2-}(aq) + 14 H^+(aq) + 6 e^- \rightarrow 2Cr^{3+}(aq) + 7H_2O(l)$
• Overall reaction:
• The reduction potential for permanganate is larger so the reaction is still favorable even when the oxidation of $Fe^{2+}$ is negative.
3. $\bg_black Cr^{2+}$ is added to $\bg_black Cr_2O_7^{2-}$ in acid solution.
• Reduction half-reaction:
• Oxidation half-reaction:
• Overall reaction:$6Cr^{2+}(aq)+Cr_2O_7^{2-}(aq)+14H^+(aq) \rightarrow 8Cr^{3+}(aq)+7H_2O(l)$
• The reaction is favorable with a high positive $E^\circ$
4. $\bg_black Mn$ is heated with $\bg_black CrO_3$.
• Reduction half-reaction:$8CrO_3(aq)+24H^++24e^- \rightarrow 4Cr_2O_3(aq)+12H_2O(l)$
• Oxidation half-reaction:$9Mn(s)+12H_2O(l) \rightarrow 3Mn_3O_4(aq)+24H^+(aq)+24e^-$
• Overall reaction:$8CrO_3(aq)+9Mn(s) \rightarrow^{\Delta} 4Cr_2O_3(aq)+3Mn_3O_4(aq)$
• Heat creates a product with higher energy than both previous reactants.
5. $\bg_black CrO$ is added to $\bg_black HNO_3$ in water.
• Strong acid dissociation:$2HNO_3(aq)+2H_2O(l) \rightarrow 2H_3O^+(aq)+2NO_3^-(aq)$
• Overall reaction:$2HNO_3(aq)+CrO(aq) \rightarrow Cr^{2+}(aq)+2NO_3^-(aq)+H_2O(l)$
• This reaction works by exchange of electrons to yield Chromium ions.
6. $\bg_black FeCl_3$ is added to an aqueous solution of $\bg_black NaOH$.
• Overall reaction:$3NaOH(aq)+FeCl_3(s) \rightarrow 3Na^+(aq)+FeOH_3^-(s)+3Cl^-(aq)$
• Iron hydroxide will precipitate because the two metals will exchange anions.
A19.1.19
1. $\ce{Cr2(SO4)3}(aq)+\ce{2Zn}(s)+\ce{2H3O+}(aq)⟶\ce{2Zn^2+}(aq)+\ce{H2}(g)+\ce{2H2O}(l)+\ce{2Cr^2+}(aq)+\ce{3SO4^2-}(aq)$;
2. $\ce{4TiCl3}(s)+\ce{CrO4^2-}(aq)+\ce{8H+}(aq)⟶\ce{4Ti^4+}(aq)+\ce{Cr}(s)+\ce{4H2O}(l)+\ce{12Cl-}(aq)$;
3. In acid solution between pH 2 and pH 6, $\ce{CrO4^2-}$ forms $\ce{HrCO4-}$, which is in equilibrium with dichromate ion. The reaction is $\ce{2HCrO4-}(aq)⟶\ce{Cr2O7^2-}(aq)+\ce{H2O}(l)$. At other acidic pHs, the reaction is $\ce{3Cr^2+}(aq)+\ce{CrO4^2-}(aq)+\ce{8H3O+}(aq)⟶\ce{4Cr^3+}(aq)+\ce{12H2O}(l)$;
4. $\ce{8CrO3}(s)+\ce{9Mn}(s)\overset{Δ}{⟶}\ce{4Cr2O3}(s)+\ce{3Mn3O4}(s)$;
5. $\ce{CrO}(s)+\ce{2H3O+}(aq)+\ce{2NO3-}(aq)⟶\ce{Cr^2+}(aq)+\ce{2NO3-}(aq)+\ce{3H2O}(l)$;
6. $\ce{CrCl3}(s)+\ce{3NaOH}(aq)⟶\ce{Cr(OH)3}(s)+\ce{3Na+}(aq)+\ce{3Cl-}(aq)$
Q19.1.20
What is the gas produced when iron(II) sulfide is treated with a nonoxidizing acid?
S19.1.20
Formula for iron(II) sulfide: $FeS$
Definition of non-oxidizing acid: A non-oxidizing acid is an acid that doesn't act the oxidizing agent. Its anion is a weaker oxidizing agent than H+, thus it can't be reduced. Examples of non-oxidizing acids: $HCl, HI, HBr, H_3PO_4, H_2SO_4$
Step 2: Choose one of the non-oxidizing acid, in this case HCl, and write the chemical reaction:
$FeS(s)+2HCl(aq)\rightarrow FeCl_2(s)+H_2S(g)$
The gas produced when iron (II) sulfide treated with a non-oxidizing acid, HCl, is H2S (dihydrogen sulfide) gas.
Q19.1.21
Predict the products of each of the following reactions and then balance the chemical equations.
1. Fe is heated in an atmosphere of steam.
2. NaOH is added to a solution of Fe(NO3)3.
3. FeSO4 is added to an acidic solution of KMnO4.
4. Fe is added to a dilute solution of H2SO4.
5. A solution of Fe(NO3)2 and HNO3 is allowed to stand in air.
6. FeCO3 is added to a solution of HClO4.
7. Fe is heated in air.
S19.1.21
a. Steam is water ($\ce{H_{2}O}$)
We can write out the reaction as:
$\ce{Fe}$ + $\ce{H_{2}O}$ → ?
This is a single replacement reaction, so $\ce{Fe}$ replaces $\ce{H_{2}}$. So, one of the products is $\ce{Fe_{3}O_{4}}$ since it is a combination of iron(II) oxide, $\ce{FeO}$, and iron(III) oxide, $\ce{Fe_{2}O_{3}}$.
The $\ce{Fe}$ is heated in an atmosphere of steam. $\ce{H_{2}}$ becomes neutrally charged and becomes another product.
After balancing the coefficients, the final reaction is:
$\ce{3Fe}(s)$ + $\ce{4H_{2}O}(g)$ → $\ce{Fe_{3}O_{4}}(s)$ + $\ce{4H_{2}}(g)$
b. $\ce{NaOH}$ added to a solution of $\ce{Fe(NO_{3})_{3}}$ is a double replacement and precipitation reaction.
We can write out the reaction as:
$\ce{NaOH}$ + $\ce{Fe(NO_{3})_{3}}$ → ?
The $\ce{Na}$ and $\ce{Fe}$ switch to form $\ce{Fe(OH)_{3}}(s)$ and $\ce{NaNO_{3}}(aq)$.
$\ce{Fe(OH)_{3}}$ is solid because it is insoluble according to solubility rules.
After balancing the coefficients in the reaction, the final reaction is:
$\ce{Fe(NO_{3})_{3}}(aq)$ + $\ce{3NaOH}(aq)$ → $\ce{Fe(OH)_{3}}(s)$ + $\ce{NaNO_{3}}(aq)$
c. For instance, the acid used to make the acidic solution is $\ce{H_{2}SO_{4}}$, then the reaction is:
$\ce{FeSO_{4}}$ + $\ce{KMnO_{4}}$ + $\ce{H_{2}SO_{4}}$ → $\ce{Fe_{2}(SO_{4})_{3}}$ + $\ce{MnSO_{4}}$ + $\ce{H_{2}O}$ + $\ce{K_{2}SO_{4}}$
Next, the net ionic reaction has to be written to get rid of the spectator ions in the reaction, this is written as:
$\ce{Fe^{2+}}$ + $\ce{MnO_{4}^{-}}$ + $\ce{H^{+}}$ → $\ce{Fe^{3+}}$ + $\ce{Mn^{2+}}$ + $\ce{H_{2}O}$
As seen in the net ionic equation above, $\ce{Fe^{2+}}$ is oxidized to $\ce{Fe^{3+}}$ and $\ce{MnO_{4}^{-}}$ is reduced to $\ce{Mn^{2+}}$. These can be written as two half reactions:
$\ce{Fe^{2+}}$ → $\ce{Fe^{3+}}$
$\ce{MnO_{4}^{-}}$ → $\ce{Mn^{2+}}$
To balance the oxidation half reaction, one electron as to be added to the $\ce{Fe^{3+}}$, this is shown as:
$\ce{Fe^{2+}}$ → $\ce{Fe^{3+}}$ + $\ce{e^{-}}$
The reduction half reaction also has to be balanced, but with $\ce{H^{+}}$ ions and $\ce{H_{2}O}$, this is shown as:
$\ce{MnO_{4}^{-}}$ + $\ce{8H^{+}}$ → $\ce{Mn^{2+}}$ + $\ce{4H_{2}O}$
After the charge of the $\ce{Mn}$ atoms are balanced, the overall charge has to be balanced on both sides because on the reactants side, the charge is $\ce{7+}$, and the charge on the products side is $\ce{2+}$. The overall charge can be balanced by adding electrons, this is shown as:
$\ce{MnO_{4}^{-}}$ + $\ce{8H^{+}}$ + $\ce{5e^{-}}$ → $\ce{Mn^{2+}}$ + $\ce{4H_{2}O}$
Now since both half reactions are balanced, the electrons in both half reactions have to be equal, and then the half reactions are added together. After this is done, the reaction looks like this:
$\ce{MnO_{4}^{-}}$ + $\ce{8H^{+}}$ + $\ce{5Fe^{2+}}$ + $\ce{5e^{-}}$ → $\ce{Mn^{2+}}$ + $\ce{4H_{2}O}$ + $\ce{5Fe^{3+}}$ + $\ce{5e^{-}}$
The $\ce{5e^{-}}$ on both sides cancel out and the final balanced reaction is:
$\ce{MnO_{4}^{-}}$ + $\ce{8H^{+}}$ + $\ce{5Fe^{2+}}$ →$\ce{Mn^{2+}}$ + $\ce{4H_{2}O}$ + $\ce{5Fe^{3+}}$
d. $\ce{Fe}$ added to a dilute solution of $\ce{H_{2}SO_{4}}$ is a single replacement reaction.
The $\ce{Fe}$ is added to a dilute solution so the $\ce{H_{2}SO_{4}}$ is written as separate ions.
We can write out the reaction as:
$\ce{Fe}(s)$ + $\ce{2H^+}(aq)$ + $\ce{(SO_{4})^{2-}}(aq)$ → ?
The Fe replaces the $\ce{H^+}$ ion, and becomes an $\ce{Fe^{2+}}$ ion.
$\ce{H_{2}O}$ is also a product because the solution is dilute.
Furthermore, the $\ce{FeSO_{4}}$ also has to be separated into ions as a result of the $\ce{Fe}$ being added to a dilute solution.
After balancing all of the coefficients, the final reaction is:
$\ce{Fe}(s)$ + $\ce{(2H_{3}O)^+}(aq)$ + $\ce{(SO_{4})^{2-}}(aq)$ → $\ce{Fe^{2+}}(aq)$ + $\ce{SO_{4}^{2-}}(aq)$ + $\ce{H_{2}}(g)$ + $\ce{2H_{2}O}(l)$
Note: $\ce{H^+}$ can also be written as the the hydronium ion, $\ce{(H_{3}O)^{+}}$.
e. We initially can initially write out:
$\ce{4Fe(NO_{3})_{2}}$ + $\ce{4HNO_{3}}$ + $\ce{O_{2}}$ → ?
We write the oxygen term in the reactants because it is stated that the solution is allowed to stand in air.
We just have to analyze the possible products that can be formed and we can see that the hydrogen from nitric acid can combine with oxygen gas to form water and then combining everything together, we get the final reaction to be:
$\ce{4Fe(NO_{3})_{2}}(aq)$ + $\ce{4HNO_{3}}(aq)$ + $\ce{O_{2}}(g)$ → $\ce{2H_{2}O}(l)$ + $\ce{4Fe(NO_{3})_{3}}(aq)$
f. When $\ce{FeCO_{3}}$ is added to $\ce{HClO_{4}}$, a double replacement reaction occurs.
The $\ce{Fe^{2+}}$ ion switches spots with the $\ce{H^+}$ ion to form $\ce{Fe(ClO_{4})_{2}}$ as a product.
When the $\ce{H^+}$ ion is added to the $\ce{(CO_{3})^{2-}}$ ion, $\ce{H_{2}CO_{3}}$ is formed.
After balancing the coefficients, the final reaction is:
$\ce{FeCO_{3}}(s)$ + $\ce{HClO_{4}}(aq)$ → $\ce{Fe(ClO_{4})_{2}}(aq)$ + $\ce{H_{2}O}(l)$ + $\ce{CO_{2}}(g)$
g. Air is composed of oxygen gas, which is a diatomic molecule, so it is $\ce{O_{2}}$.
Adding $\ce{Fe}$ to $\ce{O_{2}}$ will cause a synthesis reaction to occur forming $\ce{Fe_{2}O_{3}}$.
After balancing coefficients, the final reaction is:
$\ce{3Fe}(s)$ + $\ce{2O_{2}}(g)$ → $\ce{Fe_{2}O_{3}}(s)$
A19.1.21
1. $\ce{3Fe}(s)+\ce{4H2O}(g)⟶\ce{Fe3O4}(s)+\ce{4H2}(g)$;
2. $\ce{3NaOH}(aq)+\ce{Fe(NO3)3}(aq)\xrightarrow{\ce{H2O}}\ce{Fe(OH)3}(s)+\ce{3Na+}(aq)+\ce{3NO3-}(aq)$;
3. $\ce{MnO_{4}^{-}}$ + $\ce{8H^{+}}$ + $\ce{5Fe^{2+}}$ →$\ce{Mn^{2+}}$ + $\ce{4H_{2}O}$ + $\ce{5Fe^{3+}}$
4. $\ce{Fe}(s)$ + $\ce{(2H_{3}O)^+}(aq)$ + $\ce{(SO_{4})^{2-}}(aq)$ → $\ce{Fe^{2+}}(aq)$ + $\ce{SO_{4}^{2-}}(aq)$ + $\ce{H_{2}}(g)$ + $\ce{2H_{2}O}(l)$
5. $\ce{4Fe(NO_{3})_{2}}(aq)$ + $\ce{4HNO_{3}}(aq)$ + $\ce{O_{2}}(g)$ → $\ce{2H_{2}O}(l)$ + $\ce{4Fe(NO_{3})_{3}}(aq)$
6. $\ce{FeCO_{3}}(s)$ + $\ce{HClO_{4}}(aq)$ → $\ce{Fe(ClO_{4})_{2}}(aq)$ + $\ce{H_{2}O}(l)$ + $\ce{CO_{2}}(g)$
7. $\ce{3Fe}(s)$ + $\ce{2O_{2}}(g)$ → $\ce{Fe_{2}O_{3}}(s)$
Q19.1.22
Balance the following equations by oxidation-reduction methods; note that three elements change oxidation state.
$\ce{Co(NO3)2}(s)⟶\ce{Co2O3}(s)+\ce{NO2}(g)+\ce{O2}(g)$
S19.1.22
Balance the following equations by oxidation-reduction methods; note that three elements change oxidation state.
$Co(NO_3){_2(s)}⟶Co_2O{_3(s)}+NO{_2(g)}+O{_2(g)}$
In this reaction, N changes oxidation states from +5 to +4 (reduced), Co changes oxidation states from +2 to +3 (oxidized), and O changes oxidation states from -2 to 0 (also oxidized).
First, split this reaction into an oxidation and reduction half reaction set, and balance all of the elements that are not hydrogen or oxygen (we will deal with these later):
$Reduction: 2Co(NO_3){_2}\rightarrow Co_2O_{3}+4NO_2$
Now, for the oxidation reaction, we are only dealing with O2 on the products side. In order to balance this, we will need to add water and hydrogen to both sides:
$Oxidation: 2H_2O\rightarrow O_2+4H^+$
Balance the amount of oxygens on each side by adding the correct number of water molecules (H2O), and balance the amount of hydrogen by adding the correct number of H+ atoms:
$Reduction: 2H^++2Co(NO_3){_2}\rightarrow Co_2O_{3}+4NO_2+H_2O$
$Oxidation: 2H_2O\rightarrow O_2+4H^+$
Finally, balance the charges by adding electrons to each side of the equation. For the reduction reaction, we will add 2 electrons to balance out the 2H+, and to the oxidation reaction, we will add 4 electrons to balance out the 4H+. Remember, the goal of this step is to make sure that the charges are balanced, so we can cancel them out in the end.
$Reduction: 2e^- + 2H^++2Co(NO_3){_2}\rightarrow Co_2O_{3}+4NO_2+H_2O$
$Oxidation: 2H_2O\rightarrow O_2+4H^+ +4e^-$
Multiply the reduction reaction by two, in order to balance the charges so there are 4 electrons on each side of the reaction.
$Reduction: 2(2e^- + 2H^++2Co(NO_3){_2}\rightarrow Co_2O_{3}+4NO_2+H_2O)$
and combine both reactions which comes out to:
$2H_2O + 4Co(NO_3)_2 + 4H^+ \rightarrow 2CO_2O_3 + 8NO_2 + 2H_2O + O_2 + 4H^+$
Cancel out like terms:
$4Co(NO_3){_2(s)} \rightarrow 2CO_2O{_3(s)} + 8NO{_2(g)} + O{_2(g)}$
Both sides have overall charges of 0 and can be checked to see if they are balanced.
A19.1.22
$4Co(NO_3){_2(s)} \rightarrow 2CO_2O{_3(s)} + 8NO{_2(g)} + O{_2(g)}$
Q19.1.23
Dilute sodium cyanide solution is slowly dripped into a slowly stirred silver nitrate solution. A white precipitate forms temporarily but dissolves as the addition of sodium cyanide continues. Use chemical equations to explain this observation. Silver cyanide is similar to silver chloride in its solubility.
S19.1.23
Dilute sodium cyanide solution is slowly dripped into a slowly stirred silver nitrate solution. A white precipitate forms temporarily but dissolves as the addition of sodium cyanide continues. Use chemical equations to explain this observation. Silver cyanide is similar to silver chloride in its solubility.
A: Step 1: look at the question and begin to write out a general product to reactant formula for this reaction.
Step 2: try to reason out why a precipitate will form but only for a finite period of time before reforming in an aqueous substance.
Step 3: With step 2 you should have noticed that the reaction is a multiple step reaction and using the rough formula that you derived in step 1, you should try and see what the series of steps are that lead to the overall product of liquid AgCN2
In this reaction we see how NaCN is added to AgNO3 .A precipitate forms but then disappears with the addition of even more NaCN, this must mean that its an intermediate reaction which will not appear as the final product. The silver and the cyanide temporarily bond, but the bond is too weak to hold them together so they are pulled apart again when NaCN is added because a new, more stronger and stable compound is formed: [Ag(CN)2]- (aq).
The actual reaction equation when it is first taking place is $AgCl(aq)+NaCN(aq)\rightarrow AgCN(s)+NaCl(aq)$
This can be written out in the following way: as CN is added, the silver and the cyanide combine : Ag+(aq)+CN(aq)→AgCN (s)
As more CN- is added the silver and two cyanide combine to create a more stable compound: Ag+(aq)+2CN(aq)→[Ag(CN)2]- (aq)
AgCN(s) + CN- (aq) → [Ag(CN)2]- (aq)
A19.1.23
As CN is added,
$\ce{Ag+}(aq)+\ce{CN-}(aq)⟶\ce{AgCN}(s)$
As more CN is added,
$\ce{Ag+}(aq)+\ce{2CN-}(aq)⟶\ce{[Ag(CN)2]-}(aq)$
$\ce{AgCN}(s)+\ce{CN-}(aq)⟶\ce{[Ag(CN)2]-}(aq)$
Q19.1.24
Predict which will be more stable, [CrO4]2− or [WO4]2−, and explain.
S19.1.24
According to the rules associated with Crystal Field Stabilizing Energies, stable molecules contain more electrons in the lower-energy molecular orbitals than in the high-energy molecular orbitals. In this case, both complexes have O4 as ligands, and both have a -2 charge. Therefore, you determine stability by comparing the metals. Chromium is in the 3d orbital, according to the periodic table. Tungsten (W) is in the 5d orbital. 3d is a lower energy level than 5d.Higher-level orbitals are more easily ionized, and make their base elemental form more stable. If the elemental form is more stable the oxidized form is less stable. Therefore, [CrO4]2− is more stable than [WO4]2−.
A19.1.24
[CrO4]2- is more stable because Chromium is in the 3d orbital while Tungsten is in the 4d orbital, which has a higher energy level and makes it less stable.
Q19.1.25
Give the oxidation state of the metal for each of the following oxides of the first transition series. (Hint: Oxides of formula M3O4 are examples of mixed valence compounds in which the metal ion is present in more than one oxidation state. It is possible to write these compound formulas in the equivalent format MO·M2O3, to permit estimation of the metal’s two oxidation states.)
1. Sc2O3
2. TiO2
3. V2O5
4. CrO3
5. MnO2
6. Fe3O4
7. Co3O4
8. NiO
9. Cu2O
S19.1.25
The first step to solving this problem is looking at the rules of Oxidizing states for various elements:
chem.libretexts.org/Core/Analytical_Chemistry/Electrochemistry/Redox_Chemistry/Oxidation_State
The main rules that will be used in these problems will be the oxidation state rule 6 which states that oxidation state for Oxygen is (-2) and rule 2 which is that the total sum of the oxidation state of all atoms in any given species is equal to the net charge on that species. Solving these problems requires simple algebra. The oxidation states of both elements in the compound is equal to zero, so set the unknown oxidation of the element that is not oxygen to a variable ${x}$, and the oxidation state of Oxygen equal to ${-2}$. Then multiply both oxygen states by the number of atoms of the element present. Add the values together, set the equation equal to zero and solve for ${x}$.
1. $\ce{Sc2O3}={3{(-2)}}+{2{x}}={0}⟶{-6}+{2{x}}={0}⟶{x}={Sc}={+3}$ $Sc^{3+}$
2. $\ce{TiO2}={2{(-2)}}+{x}={0}⟶{-4}+{x}={0}⟶{x}={Ti}={+4}$ $Ti^{4+}$
3. $\ce{V2O5}={5{(-2)}}+{2{x}}={0}⟶{-10}+{2{x}}={0}⟶{x}={V}={+5}$ $V^{5+}$
4. $\ce{CrO3}={3{(-2)}}+{x}={0}⟶{-6}+{x}={0}⟶{x}={Cr}={+6}$ $Cr^{6+}$
5. $\ce{MnO2}={2{(-2)}}+{x}={0}⟶{-4}+{x}={0}⟶{x}={Mn}={+4}$ $Mn^{4+}$
6. $\ce{Fe3O4}=\ce{FeO}·\ce{Fe2O3}=$
$\ce{FeO}={-2}+{x}={0}⟶{x}={Fe}={+2}$ $Fe^{2+}$
$\ce{Fe2O3}={3{(-2)}}+{2{x}}={0}⟶{-6}+{2x}={0}⟶{x}={Fe}={+3}$ $Fe^{3=}$
(One Fe Atom has an oxidation state of +2 and the other 2 Fe atoms have an oxidation state of +3)
7. $\ce{Co3O4}=\ce{CoO}·\ce{Co2O3}=$
$\ce{CoO}={-2}+{x}={0}⟶{x}={Co}={+2}$ $Co^{2+}$
$\ce{Co2O3}={3{(-2)}}+{2{x}}={0}⟶{-6}+{2x}={0}⟶{x}={Co}={+3}$ $Co^{3+}$
(One Co Atom has an oxidation state of +2 and the other 2 Co atoms have an oxidation state of +3)
8. $\ce{NiO}={-2}+{x}={0}⟶{x}={Ni}={+2}$ $Ni^{2+}$
9. $\ce{Cu2O}={-2}+{2{x}}={0}⟶{-2}+{2x}={0}⟶{x}={Cu}={+1}$ $Cu^{1+}$
A19.1.25
Sc3+; Ti4+; V5+; Cr6+; Mn4+; Fe2+ and Fe3+; Co2+ and Co3+; Ni2+; Cu+
19.2: Coordination Chemistry of Transition Metals
Q19.2.1
Indicate the coordination number for the central metal atom in each of the following coordination compounds:
1. [Pt(H2O)2Br2]
2. [Pt(NH3)(py)(Cl)(Br)] (py = pyridine, C5H5N)
3. [Zn(NH3)2Cl2]
4. [Zn(NH3)(py)(Cl)(Br)]
5. [Ni(H2O)4Cl2]
6. [Fe(en)2(CN)2]+ (en = ethylenediamine, C2H8N2)
S19.2.1
First we must identify whether or not the ligand has more than one bonded atom (bidentate/polydentate). Using the table below we are able to do this.
Ligand Number of bonded atoms
Ammine (NH3) monodentate
Aqua (H2O) monodentate
Bromo (Br) monodentate
Chloro (Cl) monodentate
Cyano (CN) monodentate
Pyridine (C5H5N) monodentate
Ethylenediamine (C2H8N2) bidentate
Now that we have identified the number of bonded atoms from each ligand, we can find the total number of atoms bonded to the central metal ion, giving us the coordination number.
1. $[Pt(H_2O)_2Br_2]$: We can identify the metal ion in the complex as Pt, platinum, as the other two are listed as ligands above and are nonmetallic. We can now use the number of ligands and their bonding atoms to find its coordination number. From the table above we see that H2O has only one bonding atom and Br as well. So for each Br atom we have one bonding atom, and we have two of these, yielding 2 bonding atoms; this is the same for H2O, giving us a total number of 4 bonding atoms, and therefore a coordination number of 4.
2. $[Pt(NH_3)(py)(Cl)(Br)]$ (py = pyridine, C5H5N): The metal ion in this complex, similarly to the first one, can be identified as Pt, platinum. The ligands can be identified as NH3, pyridine, Cl, and Br, which are all monodentate ligands and have one bonding atom each. Since we have four ligands, each with one bonding atom, the total number of bonding atoms on the metal ion is 4, therefore the complex has a coordination number of 4.
3. $[Zn(NH_3)_2Cl_2]$: The metal ion in this complex can be identified as Zn, zinc, and the ligands can be identified as NH3 and Cl. Since these two are both monodentate ligands they have one bonding atom each. Since we have a total of two NH3 and two Cl ligands, we get a total of four monodentate ligands, giving us 4 bonding atoms and a coordination number of 4.
4. $[Zn(NH_3)(py)(Cl)(Br)]$: The metal ion in this complex can be identified as Zn, zinc, and the ligands can be identified as NH3, pyridine, Cl, and Br, which are all monodentate ligands and have one bonding atom each. Since we have four ligands, each with one bonding atom, the total number of bonding atoms on the metal ion is 4, therefore the complex has a coordination number of 4.
5. $[Ni(H_2O)_4Cl_2]$: The metal ion in this complex can be identified as Ni, nickel, and we can now use the number of ligands and their bonding atoms to find its coordination number. From the table above we see that H2O has only one bonding atom and Cl as well. So for each Cl atom we have one bonding atom, and we have two of these, yielding 2 bonding atoms. H2O is the same, having only one bonding atom, but there are four of these. So this gives us a total number of 6 bonding atoms, and therefore a coordination number of 6.
6. $[Fe(en)_2(CN)_2]^+$ (en = ethylenediamine, C2H8N2): The metal ion in this complex can be identified as Fe, iron, and the ligands can be identified as (en) and CN. Since (en) is bidentate, meaning it has 2 bonding atoms, and there are two of these, the total number of bonding atoms from (en) is four. Since CN is monodentate, meaning it has one bonding atom, and there are two of these, the total number of bonding atoms from CN ligand is two. So, the total number of bonding atoms is 6, therefore the complex has a coordination number of 6.
A19.2.1
1. The 2 aqua and the 2 bromo ligands form a total of 4 coordinate covalent bonds and as a result the coordination number is 4.
2. The ammine, pyridine, chloro and bromo each form one coordinate covalent bond that gives a total of 4 and hence CN=4.
3. Two ammine and two chloro ligands give a total of 4 coordinate covalent bonds and a CN = 4.
4. One ammine, a pyrimidine, a chloro and a bromo ligand give a total of 4 covalent bonds, resulting in CN = 4.
5. Four aqua ligands and two chloro ligands form a total of 6 coordinate covalent bonds and a CN =6.
6. Ethylenediamine is a bidentate ligand that forms two coordinate covalent bonds; along with two cyano ligands, it forms a total of 6 bonds, and hence has a CN=6.
Q19.2.2
Give the coordination numbers and write the formulas for each of the following, including all isomers where appropriate:
1. tetrahydroxozincate(II) ion (tetrahedral)
2. hexacyanopalladate(IV) ion
3. dichloroaurate ion (note that aurum is Latin for "gold")
4. diamminedichloroplatinum(II)
5. potassium diamminetetrachlorochromate(III)
6. hexaamminecobalt(III) hexacyanochromate(III)
7. dibromobis(ethylenediamine) cobalt(III) nitrate
S19.2.2
To determine coordination numbers we must count the total number of ligands bonded to the central metal and distinguish monodentate and polydentate ligands. To determine the formulas, we use the nomenclature rules and work backwards.
1. "tetrahydroxo" = 4 hydroxide ligands; since hydroxide is a monodentate ligand, we have a total of 4 bonds to the central metal.
Coordination Number: 4
We review the basics of nomenclature and see that "tetra" = 4 and "hydroxo" = OH-. Since the charge on zinc is 2+, which is given in the nomenclature by the Roman numerals, we can calculate the total charge on the complex to be 2-.
Formula: [Zn(OH)4]2−
2. "hexacyano" = 6 cyanide ligands; since cyanide is a monodentate ligand, we have a total of 6 bonds to the central metal.
Coordination Number: 6
We review the basics of nomenclature and see that "hexa" = 6 and "cyano" = CN-. Since the charge on Pd is 4+, which is given in the nomenclature by the Roman numerals, we can calculate the total charge on the complex to be 2-.
Formula: [Pd(CN)6]2−
3. "dichloro" = 2 chloride ligands; since chloride is a monodentate ligand, we have a total of 2 bonds to the central metal.
Coordination Number: 2
We review the basics of nomenclature and see that "di" = 2 and "chloro" = Cl-. Since the charge on Au is always 1+, we can calculate the total charge on the complex to be 1-.
Formula: [AuCl2]
4. "diammine" = 2 ammonia ligands and "dichloro" = 2 chloride ligands; since both ammonia and chloride ligands are monodentate, we have a total of 4 bonds to the central metal.
Coordination Number: 4
We review the basics of nomenclature and see that "di" = 2, "chloro" = Cl- and "ammine" = NH3. Since the charge on Pt is 2+, which is given in the nomenclature by the Roman numerals, we can calculate that the total charge is 0, so the complex is neutral.
Formula: [Pt(NH3)2Cl2]
5. "diammine" = 2 ammonia ligands and "tetrachloro" = 4 chloride ligands; since both ammonia and chloride ligands are monodentate, we have a total of 6 bonds to the central metal.
Coordination Number: 6
We review the basics of nomenclature and see that "di" = 2, "ammine" = NH3, "tetra" = 4 and "chloro" = Cl-. Since the charge on the central metal, Cr, is 3+, which is given in the nomenclature by the Roman numerals, we can calculate that the total charge of the complex is 1-. The "potassium" at the front of the nomenclature indicates that it is the corresponding cation to this anionic complex.
Formula: K[Cr(NH3)2Cl4]
6. Both of the metal complexes have "hexa" monodentate ligands, which means both have coordination numbers of 6.
Coordination Number: 6
We review the basics of nomenclature and see that "hexa" = 6, "ammine" = NH3, and "cyano" = CN-. Since the charge on Cr is 3+ and Co is 3+, which is given in the nomenclature by the Roman numerals, we find that these complexes' charges balance out.
Formula: [Co(NH3)6][Cr(CN)6]
7. "dibromo" = 2 bromide ligands "bis(ethylenediamine)" = 2 (en) ligands; bromide is a monodentate ligand while (en) is a bidentate ligand. Therefore, we have a coordination number of 6.
Coordination Number: 6
We review the basics of nomenclature and see that "di" = 2, "bromo" = Br-, "bis" = 2, and "ethylenediamine" = en. Since the charge on Co is 3+, which is given in the nomenclature by the Roman numerals, we find that the total charge of the complex is 1+. Nitrate is the corresponding anion to this cationic complex.
Formula: [Co(en)2Br2]NO3
A19.2.2
1. 4, [Zn(OH)4]2−;
2. 6, [Pd(CN)6]2−;
3. 2, [AuCl2];
4. 4, [Pt(NH3)2Cl2];
5. 6, K[Cr(NH3)2Cl4];
6. 6, [Co(NH3)6][Cr(CN)6];
7. 6, [Co(en)2Br2]NO3
Q19.2.3
Give the coordination number for each metal ion in the following compounds:
1. [Co(CO3)3]3− (note that CO32− is bidentate in this complex)
2. [Cu(NH3)4]2+
3. [Co(NH3)4Br2]2(SO4)3
4. [Pt(NH3)4][PtCl4]
5. [Cr(en)3](NO3)3
6. [Pd(NH3)2Br2] (square planar)
7. K3[Cu(Cl)5]
8. [Zn(NH3)2Cl2]
S19.2.3
You can determine a compound's coordination number based on how many ligands are bound to the central atom.
1) In this compound, Cobalt is the central atom, and it has 3 CO32- molecules attached to it. However, CO32- is a bidentate ligand, which means it binds to the central atom in two places rather than one. This means that the coordination number of [Co(CO3)3]3- is 6. A coordination number of 6 means that the structure is most likely octahedral.
2) In this compound, Copper is the central atom. 4 ammonia molecules are attached to it. This means the coordination number is 4, and the structure is likely tetrahedral.
3) For this compound, we can ignore the (SO4)3 because it is not bound to the central atom. The central atom is cobalt, and it has 4 ammonia molecules and 2 bromine molecules bound to it. The coordination number is 6.
4) There are two compounds here, indicated by the brackets. The central atom for both is platinum. One of them has 4 ammonia molecules attached, and the other has 4 chlorine atoms attached. Both complexes have a coordination number of 4.
5) We can ignore (NO3)3 for this compound. The central atom is Chromium. There are 3 ethylenediamine molecules attached to the chromium. Ethylenediamine is a bidentate ligand, so the coordination number is 6.
6) Palladium is the central atom. 2 ammonia molecules and 2 bromine atoms are bound to the palladium atom. The coordination number is 4.
7) We can ignore the K3 structure. Copper is the central atom, and there are 5 chlorine molecules attached to it. The coordination number is 5, so the structure is either trigonal bipyramidal or square pyramidal.
8) In this compound, zinc is the central atom. There are 2 ammonia molecules and 2 chlorine atoms attached. This means that the coordination number is 4.
Q19.2.4
Sketch the structures of the following complexes. Indicate any cis, trans, and optical isomers.
1. [Pt(H2O)2Br2] (square planar)
2. [Pt(NH3)(py)(Cl)(Br)] (square planar, py = pyridine, C5H5N)
3. [Zn(NH3)3Cl]+ (tetrahedral)
4. [Pt(NH3)3Cl]+ (square planar)
5. [Ni(H2O)4Cl2]
6. [Co(C2O4)2Cl2]3− (note that $\ce{C2O4^2-}$ is the bidentate oxalate ion, $\ce{−O2CCO2-}$)
S19.2.4
Cis and trans are a type of geometric isomer, meaning there is a difference in the orientation in which the ligands are attached to the central metal. In cis, two of the same ligands are adjacent to one another and in trans, two of the same ligands are directly across from one another.
Optical isomers → have the ability to rotate light, optical isomers are also chiral. Only chiral complexes have optical isomers
Chiral → asymmetric, structure of its mirror image is not superimposable
Enantiomers: chiral optical isomers (compound can have multiple enantiomers)
Tetrahedral complex with 4 distinct ligands → always chiral
• For tetrahedral, if 2 ligands are the same, then it cannot be chiral, has a plane of symmetry
Solutions:
a. $[Pt(H_2O)_2Br_2]$ (square planar)
This complex has 2 kinds of ligands. The matching ligands can either be adjacent to each other and be cis, or they can be across from each other and be trans.
b. $[Pt(NH_3)(py)(Cl)(Br)]$ (square planar, py = pyridine, $C_5H_5N$)
This complex has 4 different ligands. There is no plane of symmetry in any of the enantiomers, making the structures chiral and therefore has optical isomers.
c. $[Zn(NH_3)_3Cl]^+$ (tetrahedral)
There is a plane of symmetry from $NH_3$ through $Zn$ to the other $NH_3$, therefore it is not chiral.
d. $[Pt(NH_3)_3Cl]^+$ (square planar)
There is a plane of symmetry from $NH_3$ through $Pt$ to the other $NH_3$, therefore it is not chiral.
e. $[Ni(H_2O)_2Cl_2]$
The $Cl$ ligands can either be right next to each other, or directly across from one another allowing for both cis and trans geometries.
f. $[Co(C_2O_4)_2Cl_2]^-3$ (note that $C_2O_4^-2$ is the bidentate oxalate ion, $^−O_2CCO_2^-$
There is a plane of symmetry from $Cl$ through Co to the other $Cl$ in a "trans" chlorine configuration, therefore it is not chiral in a chlorine "trans" configuration. However, there is no symmetry in the chlorine "cis" configuration, indicating multiple "cis" isomers.
A19.2.4
a. [Pt(H2O)2Br2]:
b. [Pt(NH3)(py)(Cl)(Br)]:
c. [Zn(NH3)3Cl]+ :
d. [Pt(NH3)3Cl]+ :
e. [Ni(H2O)4Cl2]:
f. [Co(C2O4)2Cl2]3−:
Q19.2.5
Draw diagrams for any cis, trans, and optical isomers that could exist for the following (en is ethylenediamine):
1. [Co(en)2(NO2)Cl]+
2. [Co(en)2Cl2]+
3. [Pt(NH3)2Cl4]
4. [Cr(en)3]3+
5. [Pt(NH3)2Cl2]
S19.2.5
We are instructed to draw all geometric isomers and optical isomers for the specified compound. Optical isomers exist when an isomer configuration is not superimposable on its mirror image. This means there are two distinct molecular shapes. Often a left and right hand are cited as an example; if you were to take your right hand and place it upon your left, you cannot make the major parts of your hand align on top of one another. The basic idea when deciding whether something is optically active is to look for a plane of symmetry--if you are able to bisect a compound in a manner that establishes symmetry, then the compound does not have an optical isomer.
Cis isomers exist when there are 2 ligands of the same species placed at 90 degree angles from each other. Trans isomers exist when there are 2 ligands of the same species placed at 180 degree angles from each other.
Problem 1
This compound is an octahedral molecule, so the six ligands (atoms in the complex that are not the central transition metal) are placed around the central atom at 90 degree angles. Two optical isomers exist for [Co(en)2(NO2)Cl]+. The second isomer is drawn by taking the mirror image of the first.
Problem 2
This compound is also an octahedral molecule. Two cis (optical) isomers and one trans isomer exist for [Co(en)2Cl2]+. The trans isomer can be drawn by placing the chlorine ligands in positions where they form a 180 degree angle with the central atom. The first cis isomer can be drawn by placing the chlorine ligands in positions where they form a 90 degree angle with the central atom. The second cis isomer can be found by mirroring the first cis isomer, like we did in problem 1.
Problem 3
This compound is also an octahedral molecule. One trans isomer and one cis isomer of [Pt(NH3)2Cl4] exist. The trans isomer can be drawn by placing the ammonia ligands in positions where they form a 180 degree angle with the central atom. The cis isomer can be drawn by placing the ammonia ligands in positions where they form a 90 degree angle with the central atom.
Problem 4
This compound is also an octahedral molecule. Two optical isomers for [Cr(en)3]3+ exist. The second optical isomer can be drawn by taking the mirror image of the first optical isomer.
Problem 5
This compound is a square planar complex, so the ligands are placed around the central atom in a plane, at 90 angles. A trans isomer and a cis isomer exist for the complex [Pt(NH3)2Cl2]. The trans isomer can be drawn by placing the ammonia ligands in positions where they form a 180 degree angle in the plane with the central atom. The cis isomer can be drawn by placing the ammonia ligands in positions where they form a 90 degree angle in the plane with the central atom.
Q19.2.6
Name each of the compounds or ions given in Exercise Q19.2.3, including the oxidation state of the metal.
S19.2.6
Rules to follow for coordination complexes
1. Cations are always named before the anions.
2. Ligands are named before the metal atom or ion.
3. Ligand names are modified with an ‐o added to the root name of an anion. For neutral ligands the name of the molecule is used, with the exception of OH2, NH3, CO and NO.
4. The prefixes mono‐, di‐, tri‐, tetra‐, penta‐, and hexa‐ are used to denote the number of simple ligands.
5. The prefixes bis‐, tris‐, tetrakis‐, etc., are used for more complicated ligands or ones that already contain di‐, tri‐, etc.
6. The oxidation state of the central metal ion is designated by a Roman numeral in parentheses.
7. When more than one type of ligand is present, they are named alphabetically. Prefixes do not affect the order.
8. If the complex ion has a negative charge, the suffix –ate is added to the name of the metal.
9. In the case of complex‐ion isomerism the names cis, trans, fac, or mer may precede the formula of the complex‐ion name to indicate the spatial arrangement of the ligands. Cis means the ligands occupy adjacent coordination positions, and trans means opposite positions just as they do for organic compounds. The complexity of octahedral complexes allows for two additional geometric isomers that are peculiar to coordination complexes. Fac means facial, or that the three like ligands occupy the vertices of one face of the octahedron. Mer means meridional, or that the three like ligands occupy the vertices of a triangle one side of which includes the central metal atom or ion.
A19.2.6
1. tricarbonatocobaltate(III) ion;
2. tetraaminecopper(II) ion;
3. tetraaminedibromocobalt(III) sulfate;
4. tetraamineplatinum(II) tetrachloroplatinate(II);
5. tris-(ethylenediamine)chromium(III) nitrate;
6. diaminedibromopalladium(II);
7. potassium pentachlorocuprate(II);
8. diaminedichlorozinc(II)
Q19.2.7
Name each of the compounds or ions given in Exercise Q19.2.5.
S19.2.7
Given:
1. [Co(en)2(NO2)Cl]+
2. [Co(en)2Cl2]+
3. [Pt(NH3)2Cl4]
4. [Cr(en)3]3+
5. [Pt(NH3)2Cl2]
Wanted:
Names of the above compounds.
1. [Co(en)2(NO2)Cl]+
Step 1: Attain the names of the ligands and metal cation. Names can be found here.
Co: Cobalt
en: Ethylenediamine
NO2: Nitro
Cl: Chloro
Step 2: Add the appropriate pre-fixes to each ligand depending on the number. Pre-fixes can be found here.
(en)2: bis(Ethylenediamine)
Step 3: Find the charges of the ligands. Charges can be found here.
en: 0
NO2: -1
Cl: -1
Step 4: Algebraically attain the charge of the metal cation using the overall charge of the complex ion and the individual ligand charges.
Co + 2(en) + (NO2) + Cl = 1
Co +2(0) + (-1) + (-1) = 1
Co = 3
Step 5: For the name alphabetically place the ligands, pre-fixes should not be accounted, and use roman numerals for the metal cation which should be placed last.
Chlorobis(ethylenediamine)nitrocobalt(III)
2. [Co(en)2Cl2]+
Step 1: Attain the names of the ligands and metal cation. Names can be found here.
Co: Cobalt
en: Ethylenediamine
Cl: Chloro
Step 2: Add the appropriate pre-fixes to each ligand depending on the number. Pre-fixes can be found here.
(en)2: bis(Ethylenediamine)
Cl2: dichloro
Step 3: Find the charges of the ligands. Charges can be found here.
en: 0
Cl: -1
Step 4: Algebraically attain the charge of the metal cation using the overall charge of the complex ion and the individual ligand charges.
Co + 2(en) +2(Cl) = 1
Co + 2(0) + 2(-1) = 1
Co = 3
Step 5: For the name alphabetically place the ligands, pre-fixes should not be accounted, and use roman numerals for the metal cation which should be placed last.
Dichlorobis(Ethylenediamine)cobalt(III)
3. [Pt(NH3)2Cl4]
Step 1: Attain the names of the ligands and metal cation. Names can be found here.
Pt: Platinum
NH3: Ammine
Cl: Chloro
Step 2: Add the appropriate pre-fixes to each ligand depending on the number. Pre-fixes can be found here.
(NH3)2: diammine
Cl4: tetrachloro
Step 3: Find the charges of the ligands. Charges can be found here.
NH3: 0
Cl: -1
Step 4: Algebraically attain the charge of the metal cation using the overall charge of the complex ion and the individual ligand charges.
Pt + 2(NH3) + 4(Cl) = 0
Pt + 2(0) + 4(-1) = 0
Pt = 4
Step 5: For the name alphabetically place the ligands, pre-fixes should not be accounted, and use roman numerals for the metal cation which should be placed last.
Diamminetetrachloroplatinum(IV)
4. [Cr(en)3]3+
Step 1: Attain the names of the ligands and metal cation. Names can be found here.
Cr: Cromium
en: ethylenediamine
Step 2: Add the appropriate pre-fixes to each ligand depending on the number. Pre-fixes can be found here.
(en)3: tris(ethylenediamine)
Step 3: Find the charges of the ligands. Charges can be found here.
en: 0
Step 4: Algebraically attain the charge of the metal cation using the overall charge of the complex ion and the individual ligand charges.
Cr + 3(en) = 3
Cr + 3(0) = 3
Cr = 3
Step 5: For the name alphabetically place the ligands, pre-fixes should not be accounted, and use roman numerals for the metal cation which should be placed last.
Tris(ethylenediamine)cromium(III)
5. [Pt(NH3)2Cl2]
Step 1: Attain the names of the ligands and metal cation. Names can be found here.
NH3: Ammine
Cl: Chloro
Pt: Platinum
Step 2: Add the appropriate pre-fixes to each ligand depending on the number.
(NH3)2: diammine
Cl2: dichloro
Step 3: Find the charges of the ligands. Charges can be found here.
NH3: 0
Cl: -1
Step 4: Algebraically attain the charge of the metal cation using the overall charge of the complex ion and the individual ligand charges.
Pt + 2(NH3) + 2(Cl) = 0
Pt + 2(0) + 2(-1) = 0
Pt = 2
Step 5: For the name alphabetically place the ligands, pre-fixes should not be accounted, and use roman numerals for the metal cation which should be placed last.
Diamminedichloroplatinum(II)
A19.2.7
1. Chlorobis(ethylenediamine)nitrocobalt(III)
2. Dichlorobis(Ethylenediamine)cobalt(III)
3. Diamminetetrachloroplatinum(IV)
4. Tris(ethylenediamine)cromium(III)
5. Diamminedichloroplatinum(II)
Q19.2.8
Specify whether the following complexes have isomers.
1. tetrahedral [Ni(CO)2(Cl)2]
2. trigonal bipyramidal [Mn(CO)4NO]
3. [Pt(en)2Cl2]Cl2
S19.2.8
Isomers are compounds that have the same number of atoms, but have different structures. Structural isomers (linkage, ionization, coordination) and stereoisomers (geometric and optical) can occur with several compounds.
1. tetrahedral $\mathrm{[Ni(CO)_2(Cl)_2]}$
(Fig 1.) In this model, nickel is the dark green central atom, carbonyl ligands are the pink atom, and chloro ligands are the light green atoms.
Immediately, we can cancel out the possibility of linkage, ionization, and coordination isomers. There are no other coordination complexes for coordination isomerism, there is no ligand that can bond to the atom in more than one way for it to exhibit linkage isomerism, and there are no ions outside the coordination sphere for ionization isomerism.
This is a tetrahedral structure which immediately rules out any geometric isomers since they require 90° and/or 180° bond angles. Tetrahedral structures have 109.5° angles.
To confirm that the structure has no optical isomer, we must determine if there is a plane of symmetry. Structures that have no plane of symmetry are considered chiral and would have optical isomers.
(Fig 2.) We can rotate the structure and find that there is indeed a plane of symmetry through the two chloro ligands and central atom and between the carbonyl ligands.
Since there is a plane of symmetry, we can conclude that there are no optical isomers.
Overall, there are no isomers that exist for this compound.
2. trigonal byprimidal $\mathrm{[Mn(CO)_4(NO)]}$
(Fig 3.) The central purple atom is manganese, the carbonyl ligands are the pink atoms, and the nitrosyl ligand is the fuschia atom.
There are no ions, other coordination complex, and ambidentate ligands. Therefore, no structural isomers exist for this structure.
Geometric isomers do not exist for this compound because there is only one nitrosyl ligand.
(Fig 4.) Dashed line bisects molecule and shows plane of symmetry. The molecule is rotated in this image.
In the image above, after the structure has been rotated, we can see that there is a plane of symmetry. Thus, there are no optical isomers.
No isomers (the ones mentioned above) exist for this compound.
3. $\mathrm{[Pt(en)_{2}Cl_2]Cl_2}$
(Fig 5.) The green atoms are the chloro ligands, the the central atom is platinum, and the grey/blue atoms are ethyldiamine ligands.
Coordination isomerism cannot exist for this complex because there are no other complexes. There are no linkage isomers because there are no ambidentate ligands. Ionization isomers cannot exist in this complex either, even though there is a neutral molecule outside the coordination sphere. If we exchange $\mathrm{Cl_2}$ with one ethyldiamine molecule, There would be 5 ligands in the coordination sphere instead of 4. This difference in the ratio of metal atom to ligands means that an ionization isomer cannot exist.
(Fig 6.) Here, one chloro ligand exchanged places with the ethyldiamine so that it can be at a 90° angle with the other chloro ligand.
The image above, shows the chloro and ethyldiamine ligands at a 90° angle with its other identical ligand. This is the cis isomer, while Fig. 5 shows the trans isomer.
Fig 5. shows that there is a plane of symmetry in the trans isomer. Therefore, that structure does not have an optical isomer. On the other hand, the cis isomer does not have a plane of symmetry and therefore has an optical isomer.
A19.2.8
none; none; The two Cl ligands can be cis or trans. When they are cis, there will also be an optical isomer.
Q19.2.9
Predict whether the carbonate ligand $\ce{CO3^2-}$ will coordinate to a metal center as a monodentate, bidentate, or tridentate ligand.
S19.2.9
$\ce{CO3^2−}$ can be either monodentate or bidentate, since two of its oxygen atoms have lone pairs as shown above and can form covalent bonds with a transition metal ion. In most cases carbonate is monodentate because of its trigonal planar geometry (there is 120 degrees between the oxygens so it's hard for both to bind to the same metal). However, in some cases it will bind to two different metals, making it bidentate.
A19.2.9
CO3-2 will coordinate to a metal center as a monodentate ligand.
Q19.2.10
Draw the geometric, linkage, and ionization isomers for [CoCl5CN][CN].
S19.2.10
Isomers are compounds with same formula but different atom arrangement. There are two subcategories: structural isomers, which are isomers that contain the same number of atoms of each kind but differ in which atoms are bonded to one another, and stereoisomers, isomers that have the same molecular formula and ligands, but differ in the arrangement of those ligands in 3D space.
There are three subcategories under structural isomers: ionization isomers, which are isomers that are identical except for a ligand has exchanging places with an anion or neutral molecule that was originally outside the coordination complex; coordination isomers, isomers that have an interchange of some ligands from the cationic part to the anionic part; and linkage isomers, in two or more coordination compounds in which the donor atom of at least one of the ligands is different.
There are also two main kinds of stereoisomers: geometric 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, and optical isomers, which occurs when the mirror image of an object is non-superimposable on the original object.
Some of the isomers look almost identical, but that is because the CN ligand can be attached by both (but not at the same time) the C or N.
19.3: Spectroscopic and Magnetic Properties of Coordination Compounds
Q19.3.1
Determine the number of unpaired electrons expected for [Fe(NO2)6]3−and for [FeF6]3− in terms of crystal field theory.
S19.3.1
• The crystal field theory is is a model that describes the breaking of degeneracies of electron orbital states, usually d or f orbitals, due to a static electric field produced by a surrounding charge distribution.
• The degenerate d-orbitals split into two levels, e$_{g}$ and t$_{2g}$, in the presence of ligands.
• The energy difference between the two levels is called the crystal-field splitting energy, $\Delta_{\circ}$.
• After 1 electron each has been filled in the three t$_{2g}$ orbitals, the filling of the fourth electron takes place either in the e$_{g}$ orbital or in the t$_{2g}$, where the electrons pair up. depending on whether the complex is high spin or low spin.
• If the $\Delta_{\circ}$ value of a ligand is less than the pairing energy (P), then the electrons enter the e$_{g}$ orbital, but if the $\Delta_{\circ}$ value of a ligand is more than the pairing energy (P), then the electrons enter the t$_{2g}$ orbital.
• when the $\Delta_o$ is less than the pairing energy, the electrons prefer then eg orbitals because there is not enough energy to pair the electrons together. It will be high spin
• when the $\Delta_o$ is more then the pairing energy, the electrons prefer the t2g because there is enough energy to pair the electrons. It will be low spin.
Step 1: Determine the oxidation state of the Fe
For $[Fe(NO_{2}){_{6}}]^{3-}$ and $[FeF_{6}]^{3-}$, both $NO_{2}$ and $F_{6}$ have a charge of -1. Since there is 6 of them then that means the charge is -6 and in order for there to be an overall charge of -3, Fe has to have a +3 charge.
Step 2: Determine type of ligand
Based on the spectrochemical series we can see that $NO_{2}^{-}$ is a stronger field ligand than F$^{-}$, and therefore is a low spin complex because it has a high $\Delta_{\circ}$ unlike F$^{-}$ which is a high spin.
Step 3: Draw the crystal field
$[Fe(NO_{2}){_{6}}]^{3-}$
$[FeF_{6}]^{3-}$
There is 1 unpaired electron for $[Fe(NO_{2}){_{6}}]^{3-}$, and 5 for $[FeF_{6}]^{3-}$ based on the crystal field theory.
A19.3.1
[Fe(NO2)6]3−:1 electron
[FeF6]3−:5 electrons
Q19.3.2
Draw the crystal field diagrams for [Fe(NO2)6]4− and [FeF6]2−. State whether each complex is high spin or low spin, paramagnetic or diamagnetic, and compare Δoct to P for each complex.
S19.3.2
a) $[Fe(NO_2)_6]^{_{-4}}$
• NO2- has a -1 charge. The overall ion has a -4 charge, therefore Fe must be +2 charge. (The math: $x+(6)(-1)=-4, x+-6=-4, x=+2$ or 2+(6*-1)=-4)
• Fe2+ has 6 valence electrons.
• Next we look at the ligand bonded to Fe, which is NO2- . Based on the spectrochemical Series, NO2- is a strong field ligand meaning that it has a large DELTAo large splitting energy in comparison to the pairing energy, P. So the electrons would rather pair up, as it takes the least amount of energy.
• So [Fe(NO2)6]4− is low spin.
• All 6 electrons are paired up, so it is diamagnetic.
b) $[FeF_6]^{_{-3}}$
• F- has a -1 charge. The overall ion has a -3 charge, therefore Fe must be +3 charge. (The math: $x+(-1)(6)=-3, x+-6=-3, x=+3$ or 3+(6*-1)=-3)
• F3+ has 5 valence electrons.
• Next we look at the ligand bonded to Fe, which is F- . Based on the Spectrochemical Series, F- is a weak field ligand, meaning that it has a small DELTAo or small splitting energy in comparison to the pairing energy, P. So the electrons would rather split up and move up to the higher energy level, rather than pairing up, as it takes the least amount of energy.
• So [Fe(NO2)6]4− is high spin.
• [FeF6]3− is paramagnetic because it has unpaired electrons.
Q19.3.3
Give the oxidation state of the metal, number of d electrons, and the number of unpaired electrons predicted for [Co(NH3)6]Cl3.
S19.3.3
The oxidation state of the metal can be found by identifying the charge of one of each molecule in the coordinate compound, multiplying each molecule's charge by the respective number of molecules present, and adding the products. This final sum represents the charge of the overall coordination compound. You can then solve for the oxidation state of the metal algebraically. In this case, one chloride anion Cl- has a charge of -1. So three chloride anions have a total charge of -3. One ammine ligand NH3 has no charge so six ammine ligands have a total charge of zero. Finally, we are trying to solve for the oxidation state of a cobalt ion. Now we can write the equation that adds the total charges of each molecule or ion and is equal to the total charge of the overall coordinate compound.
$(\text{oxidation}\text{ state}\text{ of}\text{ Co})+(-3)+0=0$
$\text{oxidation}\text{ state}\text{ of}\text{ Co}=+3$
So the oxidation state of Co is +3. Now we need to identify the number of d-electrons in the Co3+ ion. The electron configuration for cobalt that has no charge is
$[\text{Ar}]4\text{s}^23\text{d}^7$
However, a Co3+ ion has 3 less electrons than its neutral counterpart and has an electron configuration of
$[\text{Ar}]3\text{d}^6$
For transition metals, the $\text{s}$ electrons are lost first. So cobalt loses its two $4\text{s}$ electrons first and then loses a single $3\text{d}$ electron meaning Co3+ ion has 6 $\text{d}$ electrons. To predict the number of unpaired electrons, we must first determine if the complex is high spin or low spin. Whether the complex is high spin or low spin is determined by the ligand in the coordinate complex. Specifically, the ligand must be identified as either a weak-field ligand or a strong-field ligand based on the spectrochemical series. Weak-field ligands induce high spin while strong-field ligands induce low spin. We can then construct the energy diagram or crystal field diagram of the designated spin that has the proper electron placings. The geometric shape of the compound must also be identified to construct the correct diagram. Finally, from this crystal field diagram we can determine the number of unpaired electrons. The number of electrons in the diagram is equal to the number of $\text{d}$ electrons of the metal. The ligand in this case is NH3, which is a strong field ligand according to the spectrochemical series. This means that the complex is low spin. Additionally, six monodentate ligands means the ligand field is octahedral. The number of electrons that will be in the diagram is 6 since the metal ion Co3+ has 6 $\text{d}$ electrons. Now the proper crystal field diagram can be constructed.
From the crystal field diagram, we can tell that the complex has no unpaired electrons.
A19.3.3
a) 3+
b) 6 d electrons
c) No unpaired electrons
Q19.3.4
The solid anhydrous solid CoCl2 is blue in color. Because it readily absorbs water from the air, it is used as a humidity indicator to monitor if equipment (such as a cell phone) has been exposed to excessive levels of moisture. Predict what product is formed by this reaction, and how many unpaired electrons this complex will have.
S19.3.4
From our knowledge of ligands and coordination compounds (or if you need a refresher Coordination Compounds), we can assume the product of CoCl2 in water. H2O is a common weak field ligand that forms six ligand bonds around the central Cobalt atom while the Chloride stays on the outer sphere. We can use this to determine the complex:
$[Co(H_2O)_6]Cl_2$
From this formation, we can use the Crystal Field Theory (CFT)(Crystal Field Theory) to determine the number of unpaired electrons. This coordination compound has six ligand bonds attached to the central atom which means the CFT model will follow the octahedral splitting. Keep in mind that we know H2O is a weak field ligand and will produce a high spin. High spin is when the electrons pairing energy (P) is greater than the octahedral splitting energy. Thus, the electrons spread out and maximize spin.
In order to fill out our crystal field diagram, we need to determine the charge of cobalt. Because the H2O ligand is neutral, and there are two chlorine ions, we can deduce the charge of cobalt is plus two in order to make the coordination complex neutral. From here, we can use the electron configuration of Co2+ is [Ar]4s23d7. The electrons that are taken away from the cobalt atom in order to form the plus two charge will from the 4s orbital and leave the 3d orbital untouched. Thus, there will be 7 electrons in the crystal field diagram and appear as:
We can see here that there are 3 unpaired electrons.
A19.3.4
[Co(H2O)6]Cl2 with three unpaired electrons.
Q19.3.5
Is it possible for a complex of a metal in the transition series to have six unpaired electrons? Explain.
A19.3.5
Is it possible for a complex of a metal in the transition series to have six unpaired electrons? Explain.
It is not possible for a metal in the transition series to have six unpaired electrons. This is because transition metals have a general electron configuration of (n-1)d1-10 ns1-2 where n is the quantum number. The last electron will go into the d orbital which has 5 orbitals that can each contain 2 electrons, yielding 10 electrons total. According to Hund's Rule, electrons prefer to fill each orbital singly before they pair up. This is more energetically favorable. Since there are only 5 orbitals and due to Hund's Rule, the maximum number of unpaired electrons a transition metal can have is 5. Therefore, there cannot be a complex of a transition metal that has 6 unpaired electrons.
For example, lets look at iron's electron configuration. Iron has an electron configuration of 1s22s22p63s23p64s23d6. Now the most important orbital to look at is the d orbital which has 6 electrons in it, but there are only 4 unpaired electrons as you can see by this diagram:
3d: [↿⇂] [↿] [↿][↿][↿]
Each [ ] represents an orbital within the d orbital. This diagram follows Hund's rule and shows why no transition metal can have 6 unpaired electrons.
Q19.3.6
How many unpaired electrons are present in each of the following?
1. [CoF6]3− (high spin)
2. [Mn(CN)6]3− (low spin)
3. [Mn(CN)6]4− (low spin)
4. [MnCl6]4− (high spin)
5. [RhCl6]3− (low spin)
S19.3.6
1. For [CoF6]3-, we first found the oxidation state of Co, which is 3+ since F has a 1- charge and since there is 6 F, Co's charge has to be 3+ for the overall charge to be 3-.
$\textrm{charge of Co} + \textrm{-6} = \textrm{-6}$ $\textrm{charge of Co} = \textrm{+3}$
After finding the oxidation state, I then go to the periodic table to find its electron configuration: [Ar]3d6
We distribute the 6 d-orbital electrons along the complex and since it is high spin, the electrons is distributed once in each energy level before it is paired. There is only one pair and the other 4 electrons are unpaired, making the answer 4.
2. The same process is repeated. We find the charge of Mn, which is 3+, making the electron configuration: [Ar]3d4
$\textrm{charge of Mn} + \textrm{-6} = \textrm{-3}$ $\textrm{charge of Mn} = \textrm{+3}$
There is a difference between this and number 1. This is low spin so instead of distributing one electron in each level before pairing it, I must distribute one electron on the bottom and then pair them all up before I'm able to move to the top portion. So since there is 4, there is only a pair at dyz and the other two electrons are unpaired. Making the number of unpaired electrons 2.
3.The same process as number 2 is applied. The only difference is that the charge of Mn is now 2+ so the electron configuration: [Ar]3d5.
$\textrm{charge of Mn} + \textrm{-6} = \textrm{-4}$ $\textrm{charge of Mn} = \textrm{+2}$
Since is it low spin like number 2, I only need to add an extra electron to the next level, making that 2 pairs of electron and only 1 electron unpaired.
4. Since Cl has a -1 charge like CN, Mn's charge is also 2+ with the same electron configuration as number 3, which is 5.
$\textrm{charge of Mn} + \textrm{-6} = \textrm{-4}$ $\textrm{charge of Mn} = \textrm{+2}$
With 5 electrons, this is high spin instead of low. So as stated in number 1, we pair distribute the electrons on all levels first. Since there are 5 electrons and 5 levels and they are al distributed, there are zero pairs, making that 5 unpaired electrons.
5. Using the same process as the problems above, Rh's charge is 3+, with the electron configuration: [Kr]4d6.
$\textrm{charge of Rh} + \textrm{-6} = \textrm{-3}$ $\textrm{charge of Mn} = \textrm{+3}$
With a low spin and 6 electrons, all electrons are paired up, making it 0 electrons that are unpaired.
4; 2; 1; 5; 0
Q19.3.7
Explain how the diphosphate ion, [O3P−O−PO3]4−, can function as a water softener that prevents the precipitation of Fe2+ as an insoluble iron salt.
S19.3.7
The diphosphate ion, [O3P−O−PO3]4− can function as a water softener keeping the iron in a water soluble form because of its more negative electrochemical potential than water's. This is similar to the way plating prevents metals from reacting with oxygen to corrode.
Mineral deposits are formed by ionic reactions. The Fe2+ will form an insoluble iron salt of iron(III) oxide-hydroxide when a salt of ferric iron hydrolyzes water. However, with the addition of [O3P−O−PO3]4−, the Fe2+ cations are more attracted to the PO3 group, forming a Fe(PO3) complex.
The excess minerals in this type of water is considered hard thus its name hard water.
Q19.3.8
For complexes of the same metal ion with no change in oxidation number, the stability increases as the number of electrons in the t2g orbitals increases. Which complex in each of the following pairs of complexes is more stable?
1. [Fe(H2O)6]2+ or [Fe(CN)6]4−
2. [Co(NH3)6]3+ or [CoF6]3−
3. [Mn(CN)6]4− or [MnCl6]4−
S19.3.8
The Spectrochemical Series is as follows $I^{-}<Br^{-}<SCN^{-}\approx Cl^{-}<F^{-}<OH^{-}<ONO^{-}<ox<H_{2}O<SCN^{-}<EDTA<NH_{3}<en<NO_{2}^{-}<CN$
The strong field ligands (on the right) are low spin which fills in more electrons in the t2g orbitals. The weak field ligands (on the left) are high spin so it can fill electrons in the t2g orbitals and eg orbitals. In conclusion, more electrons are filled up from the strong field ligands because the electrons don't move up to the eg orbitals.
a. $[Fe(CN)_{6}]^{4-}$
$CN$ is a stronger ligand than $H_{2}O$ so it is low spin, which fills up the t2g orbitals.
b. $[Co(NH_{3})_{6}]^{3+}$
$NH_{3}$ is a stronger ligand than $F$.
c. $[Mn(CN)_{6}]^{4-}$
$CN$ is a stronger ligand than $Cl^{-}$.
For more information regarding the shape of the complex and d-electron configuration, libretext provides more information on how to classify high and low spin complexes.
A19.3.8
[Fe(CN)6]4−; [Co(NH3)6]3+; [Mn(CN)6]4−
Q19.3.9
Trimethylphosphine, P(CH3)3, can act as a ligand by donating the lone pair of electrons on the phosphorus atom. If trimethylphosphine is added to a solution of nickel(II) chloride in acetone, a blue compound that has a molecular mass of approximately 270 g and contains 21.5% Ni, 26.0% Cl, and 52.5% P(CH3)3 can be isolated. This blue compound does not have any isomeric forms. What are the geometry and molecular formula of the blue compound?
S19.3.9
1)Find the empirical formula. There is a total of 270 grams. To find out how many grams of each element/compound there are, multiply the percentage by the mass (270).
$(270g)(0.215)=58.05gNi$
$(270g)(0.26)=70.2gCl$
$(270g)(0.525)=141.75gP(CH_3)_3$
Now that we have the grams of each element/compound, we can convert them to moles by using their molar mass.
$(58.055gNi)(\frac{1mol.}{58.69gNi})=0.989mol.Ni$
$(70.2gCl)(\frac{1mol.}{35.45gCl})=1.98mol.Cl$
$(141.75gP(CH_3)_3)(\frac{1mol.}{76.07gP(CH_3)_3})=1.86mol.P(CH_3)_3$
Now that we have the moles of all elements/compounds, we can find the ratio of all them to each other. To do this, we take the element/compound with the least amount of moles and divide all element/compound moles by this amount. In this case, Ni has the least number of moles.
$\frac{0.989mol.Ni}{0.989mol.Ni}=1$
$\frac{1.98mol.Cl}{0.989mol.Ni}=approx.2$
$\frac{1.86mol.P(CH_3)_3}{0.989mol.Ni}=approx.2$
We now know the ratio of all element/compounds in the blue compound.
The empirical formula is: NiCl(P(CH3)3)2
This formula shows us there are 4 ligands. There are 2 chlorine ligands and 2 trimethylphosphine ligands. This means that the blue compound has either a tetrahedral or square planar shape, where tetrahedral shapes are capable of different isomeric forms when all ligands are different (because if not, there is only 1 way for them to be arranged), and square planar shapes are capable of cis/trans forms. In the problem, it states this compound does not have any isomeric forms, therefore this has a tetrahedral shape.
A19.3.9
a) NiCl(P(CH3)3)2
b) Tetrahedral
Q19.3.10
Would you expect the complex [Co(en)3]Cl3 to have any unpaired electrons? Any isomers?
S19.3.10
Assign oxidation states to each element. Cl- has a -1 oxidation state. En is neutral, so 0. The entire complex is also neutral, so in order to balance the charges out, Co must be +3 because there are 3 chlorides, which gives a -3 charge.
STEP 2:
Write the electron configuration for $Co^{3+}$. $[Ar]3d^6$. There are 6 electrons.
STEP 3:
Check where en lies on the spectrochemical series. Does it have a strong field strength? It does, so these electrons will exist at the d-level with high splitting energy because the magnitude of the pairing energy is less than the crystal field splitting energy in the octahedral field.
You will the notice that there aren't any unpaired electrons when you draw the Crystal Field Theory (CFT) diagram.
This complex does not have any geometric isomers because cis-trans structures cannot be formed. The mirror image is nonsuperimpoasable, which means the enantiomers are chiral molecules; if the mirror image is placed on top on the original molecule, then they will never be perfectly aligned to give the same molecule.
A19.3.10
The complex does not have any unpaired electrons. The complex does not have any geometric isomers, but the mirror image is nonsuperimposable, so it has an optical isomer.
Q19.3.11
Would you expect the Mg3[Cr(CN)6]2 to be diamagnetic or paramagnetic? Explain your reasoning.
S19.3.11
The first step to determine the magnetism of the complex is to calculate the oxidation state of the transition metal. In this case, the transition metal is Cr.
Before doing so, we need to find charge of the of the complex ion [Cr(CN)6]2 given that the oxidation state of Mg3 is 2+. Using the subscripts of the $\displaystyle Mg^{2+}$ ion and the [Cr(CN)6]2 complex, we find that the oxidation state of [Cr(CN)6]2 , x, to be:
$\displaystyle 3(+2)+2(x)=0$
$\displaystyle x=3$
Now that we found the charge of the coordination complex, we are able to find the charge of the transition metal Cr given that the charge of CN is -1. Again, using the subscripts we find the oxidation state of Cr, y, to be:
$\displaystyle y + 6(-1)=-3$
$\displaystyle y=3$
Therefore, the oxidation state of the transition metal Cr is $\displaystyle Cr^{3+}$
Next, using the transition metal $\displaystyle Cr^{3+}$ and the periodic table as reference, we can determine the electron configuration of $\displaystyle Cr^{3+}$ to be $\displaystyle [Ar]d^{3}$. This means that $\displaystyle Cr^{3+}$ has 3 unpaired electrons in the 3d sublevel. Therefore, we find that since at least one electron is unpaired(in this case all 3 electrons are unpaired), Mg3[Cr(CN)6]2 is paramagnetic.
A19.3.11
a) Paramagnetic
Q19.3.12
Would you expect salts of the gold ion, Au+, to be colored? Explain.
S19.3.12
No. Colored ions have unpaired electrons in their outmost orbital. A partially filled d orbital, for example, can yield various colors. After completing the noble gas configuration, we see that Au+ has a configuration of [Xe] 4f145d10. Since Au+ has a completely filled d sublevel, we are certain that any salts of the gold ion, Au+ will be colorless.
*An example of a colored ion would be copper(II). Cu2+ has an electron configuration of [Ar]3d9. It has one unpaired electron. Copper(II) appears blue.
A19.3.12
No. Au+ has a complete 5d sublevel.
Q19.3.13
[CuCl4]2− is green. [Cu(H2O)6]2+is blue. Which absorbs higher-energy photons? Which is predicted to have a larger crystal field splitting?
S19.3.13
Although a color might appear a certain way, it actual absorbs a different color, opposite of it on the color wheel.
In this case;
[CuCl4]2- appears green but is opposite of red on the color wheel which is absorbed and is characterized by wavelengths 620-800 nanometers.
[Cu(H2O)6]2+ appears blue but is opposite of orange on the color wheel which is absorbed and is characterized by wavelengths 580-620 nanometers.
When determining which absorbs the higher energy photons, one must look at the complex itself. A higher energy indicates a high energy photon absorbed and a lower energy indicates a lower energy photon absorbed. How can we determine this? By looking at the complex and more specifically the ligand attached and its location in the spectrochemical series.
The ligands attached are Water and Chlorine and since Water is a stronger ligand than Chlorine according to the series, it also has larger energy, indicating a higher energy. This means that the complex [Cu(H2O)6]2+ absorbs a higher energy photon because of its a stronger ligand than chlorine.
Part 2 of this question also asks which complex is predicted to have a larger crystal field splitting. To determine this you also use the spectrochemical series and see which ligand is stronger. Since H2O is stronger than Cl- on the spectrochemical series, we can say [Cu(H2O)6]2+ has a higher crystal field splitting.
A19.3.13
a) [Cu(H2O)6]2+
b) [Cu(H2O)6]2+ has a higher crystal field splitting | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/09%3A_Coordination_Chemistry_I_-_Structure_and_Isomers/9.P%3A_Problems.txt |
This section describes the types of experimental data that give us hints about the electronic structure of coordination complexes (ie the molecular orbitals, their energy levels, and electron configurations). The types of experimental observations discussed here have been used to generate the theories of bonding that are described later in this chapter.
Introduction
The frontier (valence) electrons and orbitals define the chemical and physical properties of any compound. In the transition metals and their coordination compounds, the \(d\) orbitals and \(d\) electrons are the ones of interest. Thus, the \(d\) orbitals and electrons will be the focus of this chapter.
One of the first things you might notice about valence \(d\) orbitals is the order of orbital energy levels of atoms and ions in the transition metals. For example, in the first row of transition metals, the 4s orbital fills before the \(3d\) in the atoms. This indicates that \(4s\) is lower in energy than \(3d\) in these atoms. However, in the ions of the same elements, the \(3d\) orbital is occupied while the \(4s\) orbital is empty. This indicates the opposite, that the \(3d\) orbital is lower in energy than \(4s\) in these ions. The change in the relative energy levels of the \(3d\) and \(4s\) orbitals in atoms compared to ions is a result of the combined effects of shielding and penetration, as discussed in a previous chapter. The result is that the 3d and 4s orbital energy levels are close enough in energy that subtle changes in the environment can reverse the order.
Furthermore, you are going to see that once the atom is surrounded by ligands, the five \(d\) orbitals are no longer degenerate. In a coordination compound, with ligands bound in a specific geometry, the \(d\) orbitals become split into slightly different energy levels. The splitting pattern depends on where the ligands are in space; in other words it depends on the coordination geometry. The patterns of splitting for tetrahedral and octahedral geometries are shown below. Notice that the splitting is reversed between metals in a tetrahedral coordination environment compared to an octahedral environment.
When the orbitals of a coordination compound are filled with electrons, the electron configuration depends on the energy difference between the lower and higher energy levels. For example, given five \(d\) electrons in an octahedral environment, there are two possible electron configurations. All five electrons may occupy the lower level of \(d\) orbitals (Fig. \(3\), left), so that some electrons are paired. Another possibility is that all the electrons will be unpaired if they populate both levels of the d orbitals (Fig. \(3\), right).
A number of factors determine which case results, including the identity of the metal ion and the nature of the ligands involved. These different filling orders have consequences, too. The magnetic properties of the complex, its propensity to undergo reactions, and its optical properties all depend on the configuration of these frontier electrons.
This chapter is an exploration of the nature of the metal-ligand bond and how both geometry and the mode of bonding influence the properties of coordination complexes.
10.01: Evidence for Electronic Structures
We can get some insight into the stability of coordination complexes by looking at their formation constants. The formation constant is the equilibrium constant for the reaction leading to the formation of a coordination complex. The greater the formation constant, the more thermodynamically stable the product complex compared to the reactant complex.
Many complexation reactions take place in aqueous solution by dissolving a metal ion in water to produce the "aqueous metal ion". An aqueous metal ion is one that is coordinated by a number of water molecules, and since it is charged, there must be counter ions to balance that charge. When an aqueous metal ion reacts to form a new complex, its formation constant indicates the stability of the product compared to the aqueous complex. Depending on the number of ligands that are replaced in the product compared to the reactant, the formation complex is defined as either a stepwise formation constant ($K$) or an overall formation constant ($\beta$).
For example, the formation of [Cu(NH3)(OH2)5]2+ complex ion from Cu2+(aq) can be represented by the following stepwise reaction:1
$\ce{[Cu(OH2)6]^2+ + NH3⇌[Cu(NH3)(OH2)5]^2+ + H2O} \qquad \qquad K_1=1.9 \times 10^4\nonumber$
The formation constant, $K_1$, is a stepwise formation constant because it represents the substitution of a single aquo ligand (a coordinated water) by an ammine ligand. Another example is the reaction below that represents the substitution of four aquo ligands for four ammine ligands. The overall formation of the $\ce{[Cu(NH3)4(OH2)2]^2+}$ complex ion from Cu2+(aq) would be represented by this equation:
$\ce{[Cu(OH2)6]^2+ + 4NH3⇌[Cu(NH3)4(OH2)2]^2+ + 4H2O} \qquad \qquad \beta_4=1.1 \times 10^{13}\nonumber$
This time, the formation constant represents the substitution of four ammine ligands for four aquo ligands. This transformation does not happen in one step, so the formation constant, $\beta_4$ is a composite of the equilibrium constants for four individual steps:
$\begin{array}{cl} \ce{[Cu(OH2)6]^2+ + NH3⇌[Cu(NH3)(OH2)5]^2+ + H2O} & K_1=\frac{\ce{[Cu(NH3)(OH2)5^2+]}}{\ce{[Cu(OH2)6^2+][NH3]}}=1.9 \times 10^4 \ \ce{[Cu(NH3)(OH2)5]^2+ + NH3⇌[Cu(NH3)2(OH2)4]^2+ + H2O} & K_2=\frac{\ce{[Cu(NH3)2(OH2)4^2+]}}{\ce{[Cu(NH3)(OH2)5^2+][NH3]}} =3.9 \times 10^3 \ \ce{[Cu(NH3)2(OH2)4]^2+ + NH3⇌[Cu(NH3)3(OH2)3]^2+ + H2O} & K_3=\frac{\ce{[Cu(NH3)3(OH2)3^2+]}}{\ce{[Cu(NH3)2(OH2)4^2+][NH3]}}=1.0 \times 10^3 \ \ce{[Cu(NH3)3(OH2)3]^2+ + NH3⇌[Cu(NH3)4(OH2)2]^2+ + H2O} & K_4=\frac{\ce{[Cu(NH3)4(OH2)2^2+]}}{\ce{[Cu(NH3)3(OH2)3^2+][NH3]}} =1.5 \times 10^2\ \end{array}\nonumber$
$\beta_4 = K_1\times K_2\times K_3 \times K_4 =(1.9\times 10^4)(3.9\times 10^3)(1.0 \times 10^3)(1.5 \times 10^2)=1.1\times 10^{13}\nonumber$
(\beta\) notation
The $\beta$ notation is used to distinguish an overall formation constant from a stepwise formation constant. However, sometimes alternative notation will be used. You may see other texts, the literature, or your instructor using alternative ways to distinguish an overall formation constant like $\beta_4$ from the stepwise formation constant, like $K_4$. The important thing is to understand the conventions used for whatever source you are reading.
In both the examples above, the equilibrium lies to the right (in other words, the formation constant is >> 1). Another way to interpret this is that Cu(II) has a higher affinity for ammonia than for water, and thus the ammine complex is more stable than the aqueous metal ion. In the case of the copper tetraammine complex, replacement of each individual water by ammonia is favorable, so the aggregate formation constant $\beta_4$ becomes very large. The preference of Cu(II) for ammonia can be explained by the increased basicity of ammonia compared to water.
Formation constants can provide valuable insight into factors that contribute to the stability of coordination complexes, but we may need a more extensive set of data to build a more complete picture. If we look at formation constants in a series of complexes that have something in common, we may get an idea about additional factors that play a role in complex stability. Take a look at these examples:2
$\begin{array}{cl} \ce{Cu+(aq) +2 Cl- ⇌ [CuCl2]-} & K = 3.0 \times 10^5 \ \ce{Cu+(aq) + 2 Br- ⇌[CuBr2]-} & K = 8.0\times 10^5\ \ce{Cu+(aq) + 2 I- ⇌[CuI2]-} & K = 8.0\times 10^8\ \end{array}\nonumber$
The reactions above are simplified by abbreviating the aqueous metal ion, $\ce{[Cu(OH2)n]^+}$ as $\ce{Cu+(aq)}$. This time, we are dealing with Cu(I) rather than Cu(II). Note that basicity does not seem to play a role in this series. The pKa's of the conjugate acids HCl, HBr, and HI are -6.7, -8.7, and -9.3, respectively. If basicity were important, chloride should bind most tightly, followed by bromide, and then iodide; that trend is reversed here. Instead, this case can be explained by an application of Pearson’s hard-soft acid-base theory (HSAB). HSAB theory predicts that small, charge-dense bases will bind more tightly with small, charge-dense acids. Archetypically, hard bases have N, O, or F donor atoms. Conversely, larger, more polarizable bases bind more tightly with larger, more polarizable acids. Soft bases archetypically have P or S donor atoms. In this series of copper complexes, the iodide is the softest, most polarizable of the three halide ions shown, whereas the chloride is the hardest (although not as hard as fluoride). Cu(I) has a relatively low charge density, so it can be considered a soft acid. Soft acids and soft bases form strong bonds because of the pronounced covalency between the donor and acceptor. Of the three halides, iodide is the most capable of providing that soft-soft interaction with the Cu(I) ion.
Let’s look at another major contributor to complex stability. In this case, aqueous nickel ion reacts with ammonia to produce the hexaammine complex, $\ce{[Ni(NH3)6]^2+}$. That’s similar to the reaction of aqueous copper ion with ammonia; ammonia is a better donor in this case than water. However, ammine is easily displaced by ethylenediamine ($\ce{H2NCH2CH2NH2}$, abbreviated en) to provide an ethylenediamine complex, $\ce{[Ni(en)3]2+}$.2,3
$\begin{array}{cl} \ce{[Ni(OH2)6]^2+ + 6 NH3 ⇌[Ni(NH3)6]^2+} & K = 2.0\times10^8 \ \ce{[Ni(NH3)6]^2+ + 3en ⇌[Ni(en)3]^2+ + 6 NH3} & K = 4.7\times10^9\ \end{array}\nonumber$
Once again, we see that an ammine ligand is a much more effective donor than an aquo ligand. The ethylenediamine ligand, however, binds exceptionally tightly, and can even displace ammine ligands. To gain insight into why a metal ion would have higher affinity toward $\ce{H2NCH2CH2NH2}$ (en) than $\ce{NH3}$, despite the fact that the two ligands have similar donor atoms, we can look at the thermodynamic parameters (the entropy and enthalpy) of the reaction for the formation of $\ce{[Ni(en)3]^2+}$ from $\ce{[Ni(NH3)6]^2+}$:
$\begin{array}{rcl} \Delta H^{\circ} & =& -12.1 \text{ kJ mol}^{-1}\ \Delta S^{\circ} & = & 184.9 \text{ J mol}^{-1} K^{-1}\ \text{and at 298 K,} -T\Delta S^{\circ} & = & -55.1 \text{ kJ mol}^{-1} \end{array}\nonumber$
That last term, $-T\Delta S^{\circ}$, is included for easier comparison to the enthalpy contribution, since these two factors are evaluated via the Gibbs free energy equation, $\Delta G = \Delta H - T\Delta S$. We see that the formation of $\ce{[Ni(en)3]^2+}$ from $\ce{[Ni(NH3)6]^2+}$ is slightly favored enthalpically (by $\Delta H^{\circ}=-12.1 \text{kJ mol}^{-1}$), even though there are nitrogen atom donor ligands in both reactant and product complexes. However, this small preference is not likely the most important driving force for the large formation constant. The entropy factor is much larger in comparison ($T\Delta S^{\circ}=-55.1 \text{ kJ mol}^{-1}$), indicating entropy is driving the huge formation constant for the ethylenediamine complex.
How do we interpret that large $T\Delta S^{\circ}$? The simplest approach to interpreting a change in entropy of a reaction is to evaluate the number of molecules in the reactants compared to the products. If there is an increase in the number of molecules, there is an increase in disorder (entropy).
In the replacement of aquo ligands with ammine ligands, there are seven species on either side of the equation:
$\begin{array}{rcl} \ce{[Ni(OH2)6]^2+ + 6 NH3} & \ce{⇌} & \ce{[Ni(NH3)6]^2+} \ \text{7 molecules} & & \text{7 molecules}\ \end{array}\nonumber$
We might not expect this reaction to have a large entropic driving force because the number of molecules is the same in the reactants and products.
On the other hand, in the replacement of ammine ligands with ethylenediamine ligands, there are four molecules on the reactant side, and seven on the products side.
$\begin{array}{rcl} \ce{[Ni(NH3)6]^2+ + 3en} & \ce{⇌} & \ce{[Ni(en)3]^2+ + 6 NH3} \ \text{4 molecules} & & \text{7 molecules}\ \end{array}\nonumber$
That means there is a net increase in the number of molecules in solution when the ethylenediamine replaces the ammines. An increase in the number of molecules represents an increase in the partitioning of energy, which is entropically favorable. The underlying reason for this difference is that each ethylenediamine ligand has two nitrogen donors, allowing one ethylenediamine to replace two ammines. The high formation constant for ligands that contain multiple donor atoms is called the chelate effect, from the Greek word for "crab" (which has two claws with which it can grab things). We call donors capable of binding through multiple atoms “polydentate” ligands (for “many-toothed”, indicating they can bite more strongly into the metal and hold on).
Formation constants of coordination complexes affect important processes ranging from industrial catalysis to biology. For example, most people are aware of carbon monoxide poisoning. Carbon monoxide leads to suffocation by displacing molecular oxygen from hemoglobin (Hb), the oxygen-carrying protein in blood cells. Hemoglobin has an Fe(II) ion bound in its active site which in turn coordinates molecular oxygen, increasing the oxygen-carrying capacity of a blood cell. The two relevant equilibria are:4
$\begin{array}{cl} \ce{Hb + O2 ⇌HbO2} & K = 3.2 \; \mu M^{-1} \ \ce{Hb + CO ⇌HbCO} & K = 750 \; \mu M^{-1} \ \end{array}\nonumber$
Here, Hb is just a shorthand for hemoglobin. The formation constants shown above indicate that carbon monoxide binds to hemoglobin hundreds of times more tightly than does oxygen. This difference can’t be caused by a chelate effect; both $\ce{CO}$ and $\ce{O2}$ are monodentate donors in this case. The origin for this difference comes from more subtle differences in metal-ligand binding that we will develop later in this chapter.
Problems
Exercise $1$
Show how we can combine the formation constants given above for $\ce{[Ni(NH3)6]^2+}$ and $\ce{[Ni(en)3]^2+}$ from $\ce{[Ni(HO2)6]^2+}$ to determine the formation constant for the ethylenediamine complex from aqueous nickel ion.
$\ce{[Ni(OH2)6]^2+ + 3 en ⇄ [Ni(en)3]^2+ + 6 H2O} \qquad K = ?\nonumber$
Answer
We know the formation constants of $\ce{[Ni(NH3)6]^2+}$ and $\ce{[Ni(en)3]^2+}$. Let's call them "reactions $i$ and $j$":
$\begin{array}{lcr} \text{Reaction }i: & \ce{[Ni(OH2)6]^2+ + 6 NH3 ⇌[Ni(NH3)6]^2+} & K_i = \frac{\ce{[Ni(NH3)6]}}{\ce{[Ni(OH2)6][NH3]}} = 2.0\times10^8\ \text{Reaction }j: & \ce{[Ni(NH3)6]^2+ + 3en ⇌[Ni(en)3]^2+ + 6 NH3} & K_j = \frac{\ce{[Ni(en)3][NH3]^6}}{\ce{[Ni(NH3)6][en]^3}}= 4.7\times10^9\ \end{array}\nonumber$
and we want to find the equilibrium constant for the formation of $\ce{[Ni(en)3]^2+}$ from $\ce{[Ni(HO2)6]^2+}$. We'll call this "reaction $k$".
$\begin{array}{lcr} \text{Reaction }k: & \ce{[Ni(OH2)6]^2+ + 3 en ⇄ [Ni(en)3]^2+ + 6 H2O} & K_k = ?\ \end{array}\nonumber$
Reaction "$k$" is the sum of reaction "$i$" followed by reaction "$j$".
$\begin{array}{rcl} K_k & = & K_i\times K_j \ &=& \frac{\ce{[Ni(NH3)6]}}{\ce{[Ni(OH2)6][NH3]}} \times \frac{\ce{[Ni(en)3][NH3]^6}}{\ce{[Ni(NH3)6][en]^3}} = \frac{\ce{[Ni(en)3]}}{\ce{[Ni(OH2)6][en]^3}}\ &=& (2.0\times 10^8 )(4.7\times 10^9 ) = 9.4\times 10^{17}\ \end{array}\nonumber$
Exercise $2$
The formation constants shown below correspond to the substitution of one ligand on aqueous Cu(II) ion.5 Propose a trend that explains the different stabilities of the complexes.
$\begin{array}{cl} \ce{NH3} & K = 2.0 \times 10^3 \ \ce{F-} & K = 8.0 \ \ce{Cl-} & K = 1.2 \ \ce{Br-} & K = 0.9 \ \end{array}\nonumber$
Answer
This trend could be explained by basicity, since ammonium ion is a weak base and the acidity of the hydrogen halides increase in the order HF < HCl < HBr. The ammine ligand is a better donor than any of the halides to the nickel ion, but fluoride is better than chloride and chloride is better than bromide. The reason for the acidity trend varies in the midst of this series. Among the halides, fluoride is the best donor because it is the least polarizable and least stable ion. However, in the second row of the periodic table, the nitrogen in ammonia is a better donor than fluoride because nitrogen is less electronegative than fluorine. This difference is enough to compensate for the fact that ammine is a neutral ligand, presumably with less Coulombic attraction to the copper ion than fluoride.
Note that this trend is very different for Cu(II) than for a similar series of ligands with Cu(I). Cu(II), with greater charge density, is harder in character than Cu(I). Thus, it doesn’t necessarily bind more tightly to softer donors the way Cu(I) does.
Exercise $3$
Ethylenediamine can be displaced from nickel by tren, (NH2CH2CH2)3N, another polydentate ligand.3
$\ce{[Ni(en)2(OH2)2]^2+ + tren ⇄ [Ni(tren)(H2O)2]^2+ + 2 en } \qquad K = 76\nonumber$
1. What is the denticity of tren?
2. It is reported that ΔH° = +13.0 kJ mol-1 and ΔS° = 79.5 J mol-1 K-1. Propose a reason for the moderately large formation constant.
Answer
a) Tren has a denticity of 4; it is sometimes described as a tetradentate ligand.
b) The enthalpy of reaction ΔH° = +13.0 kJ mol-1 and at 298 K, -TΔS° = -23.7 kJ mol-1. That means there are no arguments based on the donor-acceptor strength, such as HSAB or basicity, because the enthalpy of reaction is positive. The tightness of binding must result from the positive entropy, which comes from the net increase in the number of molecules as the reaction proceeds (the chelate effect).
The moderately positive enthalpy change is probably related to ring strain in the multidentate complex. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/10%3A_Coordination_Chemistry_II_-_Bonding/10.01%3A_Evidence_for_Electronic_Structures/10.1.01%3A_Thermodynamic_Data.txt |
Electronic Structure
The electronic structure of coordination complexes can lead to several different properties that involve different responses to magnetic fields. These properties can vary between related compounds because of differences in electron counts, geometry, or donor strength. As a result, magnetic measurement of these materials can be used as a tool to provide insight into the structure of a coordination complex.
Diamagnetic and paramagnetic compounds are two common categories defined by interaction with a magnetic field. Diamagnetic compounds display a very slight repulsion from magnetic fields. The interaction is quite subtle and requires the proper instrumentation if it is to be observed; it’s not something you would normally notice by waving a magnet in front of a vial of coordination complex. In contrast, paramagnetic compounds display a very slight attraction to magnetic fields. In terms of electronic structure, diamagnetic compounds have all of their electrons in pairs whereas paramagnetic compounds have one or more unpaired electrons. These definitions are not limited to coordination complexes. Atmospheric oxygen, O2, is a common example of a paramagnetic compound. Atmospheric nitrogen, N2, is a common example of a diamagnetic compound.
These differences arise because of a quantum mechanical property of electrons and other subatomic particles: spin. Spin does not have a direct macroscopic analogy; physical chemists often urge caution about equating it to a phenomenon on the level at which we observe the universe. However, we do know that spin is related to magnetic phenomena. We can think of an electron as having a magnetic moment. There are two allowed values of this magnetic moment ($+\frac{1}{2}$ and $-\frac{1}{2}$) that differ in orientation; spin is a vector quantity. Normally, there is no preference for one orientation over the other. If a coordination complex has one unpaired electron, and we have a million of those coordination complexes in a very tiny sample, about 500,000 of the unpaired electrons would have spin = $+\frac{1}{2}$ and about 500,000 of them would have spin = $-\frac{1}{2}$. The spin of each electron is randomly oriented. As a result, the material would have no net magnetic moment. In the presence of a magnetic field, however, there is a small energy difference between those two spin states. As a result, a majority of the spins adopt the energetically more favorable orientation. Although the material by itself has no net magnetic field, one has temporarily been induced by the influence of an external magnetic field. This is what we mean by magnetic susceptibility.
Measuring Magnetic Susceptibility
All diamagnetic compounds have a roughly similar response to a magnetic field; either their electrons are all paired or they are not. Paramagnetic compounds, however, can respond quite differently to a magnetic field, depending on the number of unpaired electrons. The more unpaired electrons there are, the stronger the magnetic susceptibility will be, and so the stronger the attraction between the compound and the magnetic field. For this reason, measuring the strength of attraction of a compound for a magnetic field can reveal the number of unpaired electrons in the compound. This relationship is most straightforward for complexes of first row transition metals ($3d$ metals), becoming a little more complicated for the $4d$ and $5d$ series.
A Guoy balance is probably the simplest example of a method for determining the presence of unpaired electrons in a coordination complex.1 It uses a balance, a sample holder and a magnet. Essentially, the magnet is weighed in the presence and absence of the sample. The discrepancy between the two measurements arises because of the interaction of the sample with the magnetic field. Any diamagnetic sample causes slight downward repulsion, registering as a heavier weight. A paramagnetic sample suspended between the poles of the magnet causes a slight upward pull on the magnet, which registers as a lighter weight. The more unpaired electrons there are, the greater the induced magnetic moment, and so the lighter the weight becomes.
From that measurement, we can extract a parameter that we normally refer to as the effective magnetic moment of the material, $\mu_{eff}$. The magnetic moment is expressed in units called Bohr magnetons. A Bohr magneton (BM or $\mu_{B}$) is defined as:
$1 \text{BM} =\frac{e h}{4 \pi m c} \nonumber$
in which e is the charge on an electron; $h$ is Planck’s constant; $m$ is the mass of the electron; $c$ is the speed of light.
This experimental quantity is fitted to a simple relationship that depends on the number of unpaired electrons. This mathematical model gives a prediction of the “spin-only” magnetic moment. If the magnetic moment arises purely from the number of unpaired electrons without additional complicating factors, the fit to the experimental data is pretty good.
$\mu_{ eff } \approx \mu_{ so }=g \sqrt{S(S+1)} \nonumber$
Here, g is the gyromagnetic ratio, which is a proportionality constant between the angular momentum of the electron and the magnetic moment. It has a value of 2.00023, or approximately 2.0. The term $\sqrt{S(S+1)}$ is the value of the angular momentum, which depends on the number of unpaired electrons; S is the absolute value of the sum of the individual spins of the valence electrons. Of course, the spins would cancel out in electrons that were paired, because one would have $m_s = \frac{1}{2}$ and the other would have $m_s = -\frac{1}{2}$. For unpaired electrons, Hund’s rule states that they would have parallel spins; for example, two unpaired electrons gives $S=1$.
$\begin{array}{|c|c|c|} \hline \text{Number of} & \text{Maximum total} & \text{Spin-only magnetic moment, } \ \text{unpaired electrons, } n & \text{spin, S} & \mu_{so} (BM) \ \hline 1 & \frac{1}{2} & 1.73 \ 2 & 1 & 2.83 \ 3 & \frac{3}{2} & 3.87 \ 4 & 2 & 4.90 \ 5 & \frac{5}{2} & 5.92\ \hline \end{array} \nonumber$
The table above shows how the magnetic moment changes with the number of unpaired electrons. The maximum number of unpaired electrons given is five, because for a transition metal a sixth electron would have to pair up in a previously occupied orbital. In f-block elements, there could be seven unpaired electrons before pairing occurs because there are seven f orbitals rather than just five d orbitals.
Note that the expression for spin-only magnetic moment is sometimes written in an alternative way, based directly on the number of unpaired electrons, $n$, rather than $S$.
$\mu_{ eff } \approx \mu_{ so }=\sqrt{n(n+2)} \nonumber$
There is an additional approximation in these cases, based on how the values of the spin-only magnetic moment correlate with the number of unpaired electrons. If we always round the value of $\mu_{so}$, then it is one greater than the number of unpaired electrons. Thus, $\mu_{so} \approx n + 1$, provided of course that there are any unpaired electrons at all; the relationship doesn’t hold if n = 0 because then $\mu_{so}=0$, not 1.
In reality, observed magnetic moments are slightly different than spin-only magnetic moments. In some cases, the observed magnetic moment is smaller than expected, but those cases are more complicated and we won’t consider them here. Very often, the observed magnetic moments are larger than predicted because orbital angular momentum also plays a role in determining the magnitude of the overall magnetic moment. We may use a modified expression that takes this part into account.
$\mu_{ eff } \approx \mu_{ s + L }=g \sqrt{S(S+1)+\frac{1}{4} L(L+1)} \nonumber$
Just as S is the absolute value of the sum of the spin quantum numbers in the ion, $m_s$, L is the absolute value of the sum of the orbital quantum numbers, $m_l$. There are five d orbitals, with $m_l= 2, 1, 0, -1, -2$, and the value of L has to be maximized according to Hund’s rule. That means the value of L can be 3, 2, or 0.
As an example, suppose we have a $\ce{Co^2+}$ ion. $\ce{Co^2+}$ ion has a $d^7$ configuration. It has five $d$ orbitals, so four of these electrons are paired, leaving only three unpaired electrons with parallel spins, so $S = 3/2$. In order to maximize the orbital quantum number, $L$, two electrons will be in an orbital with $m_l= 2$, two electrons will be in an orbital with $m_l=1$, and one electron will be in each orbital with $m_l$ = 0, -1, and -2. When we take the sum, we find $L = 2 + 2 + 1 + 1 + 0 -1 -2 = 3$. That gives us:
$\mu_{ s + L }=g \sqrt{3 / 2(3 / 2+1)+\frac{1}{4} 3(3+1)}=5.20 \nonumber$
In comparison, $m_{so} = 3.87$ in this case. Observed values of $\mu_{eff}$ vary between different Co2+ complexes but are generally in the range 4.1 – 5.2 BM.2 The orbital contribution is often smaller than expected, so magnetic susceptibilities frequently fall somewhere between $\mu_{so}$ and $\mu_{S+L}$.
The use of a simple Guoy balance is an historically important method of determining magnetic susceptibility, and it is sometimes used in undergraduate laboratory experiments. Other methods are frequently used to measure magnetic susceptibility in the research laboratory. A magnetic susceptibility balance operates on a similar principle to the one behind the Guoy balance. A sample is placed within the poles of an electromagnet, causing the electromagnet to move very slightly. The current in the electromagnet is adjusted, changing its magnetic field, until the magnet comes back to its initial position. The magnitude of the current adjustment is proportional to the magnetic susceptibility of the sample.
A superconducting quantum interference device (SQUID) uses a superconducting loop in an external magnetic field to measure magnetic susceptibility of a sample.3 The sample is mechanically moved though the superconducting loop, inducing a change in current and magnetic field that are proportional to the magnetic susceptibility of the sample. Commercially produced SQUID magnetometers often have variable temperature controls, allowing magnetic susceptibility to be measured across a range of temperatures.
The Evans NMR method is quite common because of the widespread availability of high-field NMR spectrometers in research labs. A capillary containing the paramagnetic sample and a reference analyte is placed in an NMR tube containing the same reference analyte but without the paramagnetic sample. The analyte in the presence of the paramagnetic material will experience a local, induced magnetic field owing to the effect of the superconducting magnet of the NMR instrument on the paramagnetic sample. Its NMR signals will shift as a result. The NMR signal of the analyte outside the capillary will undergo no such shift, and the difference in the two signals is an indicator of the magnetic susceptibility of the sample.
There are other types of magnetic behavior in addition to diamagnetism and paramagnetism. Ferromagnetic materials have long-range order with spins oriented parallel to each other, even in the absence of an external magnetic field. Common, permanent magnets are made from ferromagnetic materials. Antiferromagnetic materials also have long-range order, but the magnetic moments are arranged in opposing pairs. These three different magnetic behaviors are diagnosed by the temperature dependence of the magnetic susceptibility. Paramagnetic materials display magnetic susceptibility ($\chi_M$, related to $\mu_{eff}$_) that increases with the inverse of temperature. Ferromagnetic materials have a critical temperature below which magnetic susceptibility rapidly rises. Antiferromagnetic materials have a critical temperature below which magnetic susceptibility rapidly falls. However, these behaviors are largely beyond the scope of the current discussion.
Problems
Exercise $1$
Show that, given g = 2.0, then $g \sqrt{S(S+1)}=\sqrt{n(n+2)}$.
Answer
\begin{aligned} \mu_{ so } &=g \sqrt{S(S+1)} \ &=2 \sqrt{S(S+1)} \ &=\sqrt{4 S(S+1)} \ &=\sqrt{2 S(2 S+2)} \end{aligned} \nonumber
but $S=n(1 / 2)$ or $n=2 S$ then
$\mu_{ so }=\sqrt{n(n+2)} \nonumber$
Exercise $2$
Calculate the value of $\mu_{so}$ in the following cases:
a) V4+ b) Cr2+ c) Ni2+ d) Co3+ e) Mn2+ f) Fe2+
Answer
a) $V ^{4+} \text{ is } d ^{1} ; n=1 ; \sqrt{n(n+2)}=1.73$
b) $Cr ^{2+} \text{ is } d ^{4} ; n=4 ; \sqrt{n(n+2)}=4.90$
c) $Ni ^{2+} \text{ is } d ^{8} ; n=2 ; \sqrt{n(n+2)}=2.83$
d) $Co ^{3+} \text{ is } d ^{6} ; n=0 ; \sqrt{n(n+2)}=0$
e) $Mn ^{2+} \text{ is } d ^{5} ; n=5 ; \sqrt{n(n+2)}=5.92$
f) $Fe ^{2+} \text{ is } d ^{6} ; n=4 ; \sqrt{n(n+2)}=4.90$
g) $Cr ^{3+} \text{ is } d ^{3} ; n=3 ; \sqrt{n(n+2)}=3.87$
h) $V ^{3+} \text{ is } d ^{2} ; n=2 ; \sqrt{n(n+2)}=2.83$
Exercise $3$
Calculate the value of $\mu_{eff}$ in the following cases:
a) V4+ b) Cr2+ c) Ni2+ d) Co3+ e) Mn2+ f) Fe2+
g) Cr3+ h) V3+
Answer
a) $V ^{4+}\text{ is } d ^{1} ; S=1 / 2 ; L=2 ; \mu_{ s + L }=g \sqrt{1 / 2(1 / 2+1)+\frac{1}{4} 2(2+1)}=3.00$
b) $Cr ^{2+}\text{ is } d ^{4} ; S=2 ; L=2+1+0-1=2 ; \mu_{ s + L }=g \sqrt{2(2+1)+\frac{1}{4} 2(2+1)}=5.48$
c) $Ni ^{2+}\text{ is } d ^{8} ; S=1 ; L=2+2+1+1+0+0-1-2=3 ; \mu_{ s + L }=g \sqrt{1(1+1)+\frac{1}{4} 3(3+1)}=2.24$
d) $Co ^{3+}\text{ is } d ^{6} ; S=2 ; L=2+2+1+0-1-2=2 ; \mu_{ s + L }=g \sqrt{(2+1)+\frac{1}{4} 2(2+1)}=5.48$
e) $Mn ^{2+}\text{ is } d ^{5} ; S=5 / 2 ; L=2+1+0-1-2=0 ; \mu_{ s + L }=g \sqrt{5 / 2(5 / 2+1)+\frac{1}{4} 0(0+1)}=5.92$
f) $Fe ^{2+}\text{ is } d ^{6} ; S=2 ; L=2+2+1+0-1-2=2 ; \mu_{ s + L }=g \sqrt{2(2+1)+\frac{1}{4} 2(2+1)}=5.48$
g) $Cr ^{3+}\text{ is } d ^{3} ; S=3 / 2 ; L=2+1+0=3 ; \mu_{ s + L }=g \sqrt{3 / 2(3 / 2+1)+\frac{1}{4} 3(3+1)}=5.20$
h) $V ^{3+}\text{ is } d ^{2} ; S=1 ; L=2+1=3 ; \mu_{ s + L }=g \sqrt{1(1+1)+\frac{1}{4} 3(3+1)}=2.24$
10.1.03: Electronic Spectra
The interaction of different wavelengths of electromagnetic radiation with matter provides a vast amount of information about material structure. The absorption of UV-visible light is connected to the excitation of electrons to higher orbitals. Inherently, this spectroscopy reveals information about the electronic structure of compounds, including coordination complexes. This topic is important enough that it is the subject of a later chapter.
10.1.04: Coordination Number and Molecular Shapes
Examination of physical properties, such as electronic spectra or magnetic susceptibility, can often be used to distinguish between possible molecular geometries of coordination complexes. It can be difficult to predict the coordination number of a complex formed from a specific metal ion and a given set of ligands, to say nothing of its geometry. Steric crowding and valence electron count around the metal in the complex are just two of the factors that influence coordination number. Even for a given coordination number, there are sometimes different possible coordination geometries. For example, five-coordinate can adopt square pyramidal or trigonal bipyramidal geometry. In most cases, it is especially difficult to predict which geometry prevails. In fact, many complexes adopt a geometry somewhere between the two. Four-coordinate geometry offers a similar choice between square planar and tetrahedral geometry, although in this case it can be easier to predict which one is likely to occur. The energetic difference between these two possible geometries can often be explained based on the electronic structure of the complex. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/10%3A_Coordination_Chemistry_II_-_Bonding/10.01%3A_Evidence_for_Electronic_Structures/10.1.02%3A_Magnetic_Susceptibility.txt |
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