24.5 Crystal-Field Theory

Although the ability to form complexes is common to all metal ions, the most numerous and interesting complexes are formed by the transition elements. Scientists have long recognized that the magnetic properties and colors of transition-metal complexes are related to the presence of d electrons in metal orbitals. In this section we will consider a model for bonding in transition-metal complexes, called the crystal-field theory, that accounts for many of the observed properties of these substances. (The name crystal field arose because the theory was first developed to explain the properties of solid, crystalline materials, such as ruby. The same theoretical model applies to complexes in solution.)

We have already noted that the ability of a metal ion to attract ligands such as water around itself can be viewed as a Lewis acid-base interaction. The base—that is, the ligand—can be considered to donate a pair of electrons into a suitable empty orbital on the metal, as shown in Figure 24.27. However, we can assume that much of the attractive interaction between the metal ion and the surrounding ligands is due to the electrostatic forces between the positive charge on the metal and negative charges on the ligands. If the ligand is ionic, as in the case of Cl or SCN, the electrostatic interaction occurs between the positive charge on the metal center and the negative charge on the ligand. When the ligand is neutral, as in the case of H2O or NH3, the negative ends of these polar molecules, containing an unshared electron pair, are directed toward the metal. In this case the attractive interaction is of the ion-dipole type. In either case the result is the same; the ligands are attracted strongly toward the metal center. The assembly of metal ion and ligands is lower in energy than the fully separated charges, as illustrated on the left side of Figure 24.28.

Figure 24.27 Representation of the metal-ligand bond in a complex as a Lewis acid-base interaction. The ligand, which acts as a Lewis base, donates charge to the metal via a metal hybrid orbital. The bond that results is strongly polar, with some covalent character. It is often sufficient to assume that the metal-ligand interaction is entirely electrostatic in character, as is done in the crystal-field model.

Figure 24.28 In the crystal-field model the bonding between metal ion and donor atoms is considered to be largely electrostatic. The energy of the metal ion plus coordinated ligands is lower than that of the separated metal ion plus ligands because of the electrostatic attraction. At the same time, the energies of the metal d electrons are increased by the repulsive interaction between these electrons and the electrons of the ligands. These repulsive interactions give rise to the splitting of the metal d-orbital energies.

In a 6-coordinate octahedral complex we can envision the ligands approaching along the x, y, and z axes, as shown in Figure 24.30(a). Using the physical arrangement of ligands and metal ion shown in this figure as our starting point, let's consider what happens to the energies of electrons in the metal d orbitals as the ligands approach the metal ion. Keep in mind that the d electrons are the outermost electrons of the metal ion. We know that the overall energy of the metal ion plus ligands will be lower (more stable) when the ligands are drawn toward the metal center. At the same time, however, there is a repulsive interaction between the outermost electrons on the metal and the negative charges on the ligands. This interaction is called the crystal field. The crystal field causes the energies of the d electrons on the metal ion to increase, as shown in Figure 24.28. The d orbitals of the metal ion, however, do not all behave in the same way under the influence of the crystal field.

Figure 24.30 (a) An octahedral array of negative charges approaching a metal ion. (b-f) The orientations of the d orbitals relative to the negatively charged ligands. Notice that the lobes of the dz2 and dx2 - y2 orbitals (b and c) point toward the charges. The lobes of the dxy, dyz, and dxz orbitals (d-f) point between the charges.

A key feature of Figure 24.28, shown in red, is the fact that the d orbitals of the metal ion do not all have the same energy. To see why, we must consider the shapes of the d orbitals and how their lobes are oriented relative to the ligands (Figure 24.30). In the isolated metal ion the five d orbitals are equal in energy. However, the dz2 and dx2 - y2 orbitals [Figure 24.3 (b) and (c)] have lobes directed along the x, y, and z axes pointing toward the approaching ligands, whereas the dxy, dyz and dxz orbitals (Figure 24.30d, e, and f) have lobes that are directed between the axes along which the ligands approach. Thus, electrons in the dx2 - y2 and dz2 orbitals experience stronger repulsions than those in the dxy, dxz, and dyz orbitals. As a result, an energy separation, or splitting, occurs between the three lower-energy d orbitals and the two higher-energy ones. In the material that follows we will concentrate on this splitting of the d orbital energies by the crystal field, which is depicted in Figure 24.31. Notice that the energy gap between the two sets of d orbitals is labeled . (The energy gap, , is sometimes referred to as the crystal-field splitting energy.)

Figure 24.31 Energies of the d orbitals in an octahedral crystal field.

Let's examine how the crystal-field model accounts for the observed colors in transition-metal complexes. The energy gap between the d orbitals, labeled , is of the same order of magnitude as the energy of a photon of visible light. It is therefore possible for a transition-metal complex to absorb visible light, which excites an electron from the lower-energy d orbitals into the higher-energy ones. The [Ti(H2O)6]3+ ion provides a simple example because titanium(III) has only one 3d electron. As shown in Figure 24.26, [Ti(H2O)6]3+ has a single absorption peak in the visible region of the spectrum. The maximum absorption is at 510 nm (235 kJ/mol). Light of this wavelength causes the d electron to move from the lower-energy set of d orbitals into the higher-energy set, as shown in Figure 24.32. The absorption of 510-nm radiation that produces this transition causes substances containing the Ti(H2O)63+ ion to appear purple.

Figure 24.32 The 3d electron of [Ti(H2O)6]3+ is excited from the lower-energy d orbitals to the higher-energy ones when irradiated with light of 510-nm wavelength.

The magnitude of the energy gap, , and consequently the color of a complex depend on both the metal and the surrounding ligands. For example, [Fe(H2O)6]3+ is light violet, [Cr(H2O)6]3+ is violet, and [Cr(NH3)6]3+ is yellow. Ligands can be arranged in order of their abilities to increase the energy gap, . The following is an abbreviated list of common ligands arranged in order of increasing :

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This list is known as the spectrochemical series.

Ligands that lie on the low end of the spectrochemical series are termed weak-field ligands; those on the high end are termed strong-field ligands. Figure 24.33 shows schematically what happens to the crystal-field splitting when the ligand is varied in a series of chromium(III) complexes. (This is a good place to remind you that when a transition metal is ionized, the valence s electrons are removed first. Therefore, the outer electron configuration for chromium is [Ar]3d54s1; that for Cr3+ is [Ar]3d3.) Notice that as the field exerted by the six surrounding ligands increases, the splitting of the metal d orbitals increases. Because the absorption spectrum is related to this energy separation, these complexes vary in color.

Figure 24.33 Crystal-field splitting in a series of octahedral chromium(III) complexes.

SAMPLE EXERCISE 24.7

Which of the following complexes of Ti3+ exhibits the shortest wavelength absorption in the visible spectrum: [Ti(H2O)6]3+; [Ti(en)3]3+; [TiCl6]3–?

SOLUTION The wavelength of the absorption is determined by the magnitude of the splitting between the d-orbital energies in the field of the surrounding ligands. The larger the splitting, the shorter the wavelength of the absorption corresponding to the transition of the electron from the lower- to the higher-energy orbital. The splitting will be largest for ethylenediamine, en, the ligand that is highest in the spectrochemical series. Thus, the complex with the shortest wavelength absorption is [Ti(en)3]3+.

PRACTICE EXERCISE

The absorption spectrum of [Ti(NCS)6]3– shows a band that lies intermediate in wavelength between those for [TiCl6]3– and [TiF6]3–. What can we conclude about the place of NCS in the spectrochemical series? Answer: It lies between Cl and F; that is, Cl < NCS < F.

Electron Configurations in Octahedral Complexes

The crystal-field model also helps us understand the magnetic properties and some important chemical properties of transition-metal ions. From our earlier discussion of electronic structure in atoms, we expect that electrons will always occupy the lowest-energy vacant orbitals first and that they will occupy a set of degenerate orbitals one at a time with their spins parallel (Hund's rule). Thus, if we have one, two, or three electrons to add to the d orbitals in an octahedral complex ion, the electrons will go into the lower-energy set of orbitals, with their spins parallel, as shown in Figure 24.34. When we wish to add a fourth electron, a problem arises. If the electron is added to the lower-energy orbital, an energy gain of magnitude is realized, as compared with placing the electron in the higher-energy orbital. However, there is a penalty for doing this, because the electron must now be paired up with the electron already occupying the orbital. The energy required to do this, relative to putting it in another orbital with parallel spin, is called the spin-pairing energy. The spin-pairing energy arises from the greater electrostatic repulsion of two electrons that share an orbital as compared with two that are in different orbitals.

Figure 24.34 Electron configurations associated with one, two, and three electrons in the 3d orbitals in octahedral complexes.

The ligands that surround the metal ion and the charge on the metal ion often play major roles in determining which of the two electronic arrangements arises. Consider the [CoF6]3– and [Co(CN)6]3– ions. In both cases the ligands have a 1- charge. However, the F ion, on the low end of the spectrochemical series, is a weak-field ligand. The CN ion, on the high end of the spectrochemical series, is a strong-field ligand. It produces a larger energy gap than does the F ion. The splittings of the d-orbital energies in the complexes are compared in Figure 24.35.

Figure 24.35 Population of d orbitals in the high-spin [CoF6]3– ion (small ) and low-spin [Co(CN)6]3– ion (large ).

A count of the electrons in cobalt(III) tells us that we have six electrons to place in the 3d orbitals. Let's imagine that we add these electrons one at a time to the d orbitals of the CoF63– ion. The first three will go into the lower-energy orbitals with spins parallel. The fourth electron could go into a lower-energy orbital, pairing up with one of those already present. This would result in an energy gain of as compared with putting it in one of the higher-energy orbitals. However, it would cost energy in an amount equal to the spin-pairing energy. Because F is a weak-field ligand, is small, and the more stable arrangement is the one in which the electron is placed in the higher-energy orbital. Similarly, the fifth electron we add goes into a higher-energy orbital. With all of the orbitals containing at least one electron, the sixth must be paired up, and it goes into a lower-energy orbital. In the case of the [Co(CN)6]3– complex, the crystal-field splitting is much larger. The spin-pairing energy is smaller than , so electrons are paired in the lower-energy orbitals, as illustrated in Figure 24.35.

The [CoF6]3– complex is referred to as a high-spin complex; that is, the electrons are arranged so that they remain unpaired as much as possible. On the other hand, the [Co(CN)6]3– ion is referred to as a low-spin complex. These two different electronic arrangements can be readily distinguished by measuring the magnetic properties of the complex, as described earlier. The absorption spectrum also shows characteristic features that indicate the electronic arrangement.

SAMPLE EXERCISE 24.8

Predict the number of unpaired electrons in 6-coordinate high-spin and low-spin complexes of Fe3+.

SOLUTION The Fe3+ ion possesses five 3d electrons. In a high-spin complex these are all unpaired. In a low-spin complex the electrons are confined to the lower-energy set of d orbitals, with the result that there is one unpaired electron:

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PRACTICE EXERCISE

For which d electron configurations does the possibility of a distinction between high-spin and low-spin arrangements exist in octahedral complexes? Answer: d4, d5, d6, d7

Tetrahedral and Square-Planar Complexes

Thus far we have considered the crystal-field model only for the complexes of octahedral geometry. When there are only four ligands about the metal, the geometry is tetrahedral, except for the special case of metal ions with a d8 electron configuration, which we will discuss in a moment. The crystal-field splitting of the metal d orbitals in tetrahedral complexes differs from that in octahedral complexes. Four equivalent ligands can interact with a central metal ion most effectively by approaching along the vertices of a tetrahedron. It turns out—and this is not easy to explain in just a few sentences—that the splitting of the metal d orbitals in a tetrahedral crystal is just the opposite of that for the octahedral case. That is, three of the metal d orbitals are raised in energy, and the other two are lowered, as illustrated in Figure 24.36. Because there are only four ligands instead of six, as in the octahedral case, the crystal-field splitting is much smaller for tetrahedral complexes. Calculations show that for the same metal ion and ligand set, the crystal-field splitting for a tetrahedral complex is only four ninths as large as for the octahedral complex. For this reason all tetrahedral complexes are high spin; the crystal field is never large enough to overcome the spin-pairing energies.

Figure 24.36 Energies of the d orbitals in a tetrahedral crystal field.

Square-planar complexes, in which four ligands are arranged about the metal ion in a plane, represent a common geometric form. You can envision the square-planar complex as formed by removing two ligands from along the vertical z axis of the octahedral complex. As this happens, the four ligands in the plane are drawn in more tightly. The changes that occur in the energy levels of the d orbitals are illustrated in Figure 24.37.

Figure 24.37 Effect on the relative energies of the d orbitals caused by removing the two negative charges from the z axis of an octahedral complex. When the charges are completely removed, the square-planar geometry results.

Square-planar complexes are characteristic of metal ions with a d8 electron configuration. They are nearly always low spin; that is, the eight d electrons are spin-paired to form a diamagnetic complex. Such an electronic arrangement is particularly common among the ions of heavier metals, such as Pd2+, Pt2+, Ir+, and Au3+.

SAMPLE EXERCISE 24.9

Four-coordinate nickel(II) complexes exhibit both square-planar and tetrahedral geometries. The tetrahedral ones, such as [NiCl4]2–, are paramagnetic; the square-planar ones, such as [Ni(CN)4]2–, are diamagnetic. Show how the d electrons of nickel(II) populate the d orbitals in the appropriate crystal-field splitting diagram in each case.

SOLUTION Nickel(II) has an electron configuration of [Ar]3d8. The population of the d electrons in the two geometries is given here:

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PRACTICE EXERCISE

How many unpaired electrons do you predict for the tetrahedral [CoCl4]2– ion?

Answer: three

We have seen that the crystal-field model provides a basis for explaining many features of transition-metal complexes. In fact, it can be used to explain many observations in addition to those we have discussed. However, many lines of evidence show that the bonding between transition-metal ions and ligands must have some covalent character. Molecular orbital theory (Sections 9.7 and 9.8) can also be used to describe the bonding in complexes. However, the application of molecular orbital theory to coordination compounds is beyond the scope of our discussion. The crystal-field model, although not entirely accurate in all details, provides an adequate and useful description.

SAMPLE INTEGRATIVE EXERCISE 24: Putting Concepts Together

The oxalate ion has the Lewis structure shown in Figure 24.5. (a) Show the geometrical structure of the complex formed by coordination of oxalate to cobalt(II), forming [Co(Ox)(H2O)4]. (b) Write the formula for the salt formed upon coordination of three oxalate ions to Co(II), assuming that the charge-balancing cation is Na+. (c) Sketch all of the possible geometric isomers for the cobalt complex formed in part (b). Are any of these isomers chiral? Explain. (d) The formation constant for formation of the cobalt(II) complex formed by coordination of three oxalate anions, as in part (b), is 5.0 109. By comparison, the formation constant for formation of the cobalt(II) complex with three molecules of ortho-phenanthroline (Figure 24.5) is 9 1019. From these results, what conclusions can you draw regarding the relative Lewis base properties of the two ligands toward cobalt(II)? (e) Using the approach described in Sample Exercise 17.14, calculate the concentration of free aqueous Co(II) ion in a solution initially containing 0.040 M oxalate ion and 0.0010 M Co2+(aq).

SOLUTION (a) The complex formed by coordination of one oxalate ion is octahedral (see figure below).

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(b) Because the oxalate ion has a charge of 2-, the net charge of a complex with three oxalate anions and one Co2+ ion is 4-. Therefore, the coordination compound has the formula Na4[Co(C2O4)3]. (c) There is only one geometric isomer. However, the complex is chiral, in the same way as is the [Co(en)3]3+ complex, Figure 24.20(b). Notice that these two mirror-image forms are not superimposable. Thus, there are two enantiomers:

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(d) The ortho-phenanthroline ligand is bidentate in the same way as oxalate, so the two ligands should enjoy the same benefits of the chelate effect. Thus, we can conclude that ortho-phenanthroline is a stronger Lewis base toward Co2+ than is oxalate. This is consistent with what we learned about bases in Section 16.7. Nitrogen bases are generally stronger than oxygen bases. For example, ammonia, NH3, is a stronger base than water, H2O. In addition, the carbonyl oxygens on each of the oxalate carbons, Figure 24.5, tend to draw electron density away from the carbons, thus causing the oxygen atoms that are coordinated to the metal to be weaker bases. (e) The equilibrium we must consider involves three moles of oxalate ion (represented as Ox2–):

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The formation constant expression is:

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Because the oxalate concentration is so much higher than that of Co2+, we can assume that [Ox2–] is 0.040 M even after the complex has formed. We can also make the assumption that nearly all of the Co2+ has been converted to complex, since the formation constant is pretty large. Let's denote the small concentration of Co2+ that remains uncomplexed by x. Then we have:

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Inserting these values into the equilibrium-constant expression, we have:

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Solving for x, we obtain x = 3.1 10–9 M. From this we can see that the oxalate has complexed all but a tiny fraction of the Co2+ present in solution.