10.2.1: Crystal Field Theory

As the name implies, crystal field theory was developed as a way of explaining phenomena in ionic crystalline solids. Bethe and van Vleck wished to provide a rationale for the magnetic properties of these materials, as well as their colors. The latter resulted from the wavelength of light absorbed, revealing something about differences in electronic energy levels in the ions of the crystalline solid. The magnetic properties also depended on electronic energy levels, because magnetism depended upon whether electrons were paired, and whether or not electrons are paired depends on the presence or absence of orbitals at the same energy level.

They first considered a transition metal ion in an octahedral hole formed between layers of counterions packed in a crystalline lattice. The metal ion would have six near neighbors: three above and three below. Of course, the ions would be held together by electrostatics: the negative charges on the counterions would be attracted to the positive charge on the metal ion.

Figure \(\PageIndex{1}\): Illustration of a positively charged metal ion (blue) surrounded by six near neighbors with negative charges (red). (CC-BY-NC; Chris Schaller)

But in addition to that positive charge, the metal ion also has electrons of its own. How did the electrons of those neighboring counterions affect the energy of those metal electrons? "...If we’re going to work this out carefully, we need to consider three different situations", thought the physicists. "We need to start by considering the energy of the metal valence orbitals in the absence of counterions. Since these are transition metal ions, we will pay close attention to the five \(d\) orbitals. What happens when those \(d\) orbitals are placed in a field of surrounding electrons, if the field is evenly distributed around them? In other words, what happens to them in a spherical field of negative charge? And then, what happens if those electrons are not evenly distributed? What happens if the surrounding electrons approach only along the Cartesian axes?"

Figure \(\PageIndex{2}\): Illustration of the electrostatic charges to consider in idealized situations. (Left) imaginary situation of a "lone" metal ion with five d electrons in its valence shell. (Middle) Imaginary situation where a \(d^5\) metal ion is surrounded by a spherical distribution of donor (Lewis base) electrons from a set of ligands. (Right) Imaginary situation where a \(d^5\) metal ion is surrounded by a set of ligand donor electrons that are located along Cartesian coordinates (in an octahedral geometry). (CC-BY-NC; Chris Schaller)

The Cartesian axes are important here because, in an octahedron, the six ligands are ninety degrees (\(90^{\circ}\)) apart from each other in space. We can think of two of those ligand electrons as lying in either direction from the metal ion along the z axis; two of them along the x axis; and the last two along the y axis.

Figure \(\PageIndex{3}\): Illustration of a positively charged metal ion (blue) surrounded by six ligands (red) in an octahedral geometry where each ligand lies along the Cartesian axes. (CC-BY-NC; Chris Schaller)

Of course, when these metal electrons are placed in a field of negative charge, repulsion results, and the electrons on the metal go up in energy. They all increase in energy by the same amount.

Figure \(\PageIndex{4}\): The relative energy levels of the valence \(d\) orbitals corresponding to the imaginary situations pictured in Figure \(\PageIndex{2}\) above. (Left) The five valence \(d\) orbitals of a "lone" metal ion are degenerate. (Middle) The electrons in the five valence \(d\) orbitals of a metal ion would be uniformly repelled by a spherical distribution of negative charge, and thus would be higher in energy than a "lone" metal ion. (Right) The electrons in each of the five valence \(d\) orbitals of a metal ion would interact to different extents with ligands located along each of the Cartesian coordinates, and thus their energies would be raised or lowered compared to the situation where charge was distributed in a uniform sphere. (CC-BY-NC; Chris Schaller)

The octahedral field is a different question, however, because the five \(d\) orbitals have different spatial distributions. Two of them (the \(d_{z^2}\) and \(d_{x^2-y^2}\)) lie along the axes. These orbitals are referred to as having \(e_g\) symmetry. The other three (the \(d_{xy}, d_{xz}\), and \(d_{yz}\)) lie in between the axes. These three are referred to as having \(t_{2g}\) symmetry.

Figure \(\PageIndex{5}\): Illustration of the boundary surfaces of the five \(d\) orbitals. (CC-BY-NC; Chris Schaller)

Those two orbitals along the axes experience repulsion from the neighboring anions. Because all of the negative charge is focused in six positions rather than being uniformly distributed, the repulsion experienced by these orbitals is greater than it would be in a spherical field. On the other hand, the three orbitals between the axes experience very little repulsion; their energy is actually much lower than it would be in a spherical field of negative charge.

These two sets of orbitals are therefore found at two different energy levels. The energy difference between them is called the octahedral field splitting, \(\Delta_o\). Because the total amount of repulsion is the same in the octahedral field and the spherical field – it is just distributed differently – the average energy level of the orbitals should be the same in the two fields. This average energy of the orbitals is called the barycenter, a term borrowed from astronomy. Because two of the orbitals are raised above the barycenter in the octahedral field and three are lowered below the barycenter, then the \(e_g\) set must be \(\frac{3}{5}\Delta_o\) (\(=0.6\Delta_o\)) above the barycenter and the \(t_{2g}\) set must be \(\frac{2}{5} \Delta_o\) (\(=0.4\Delta_o\)) below the barycenter. That means the average energy lies at the barycenter.

A parameter called the crystal field stabilization energy (CFSE) was developed to assess the energy difference between the metal ion in an octahedral coordination geometry and a spherical field. To determine CFSE, we simply add up the energy of all the electrons relative to the energy level of electrons in the spherical field, in units of \(\Delta_o\). For an octahedral case:

\[ \text{CFSE}= \left[ \frac{2}{5} \left( \text{# of electrons in } t_{2g}\right) + \frac{3}{5} \left( \text{# of electrons in } e_g \right) \right] \times \Delta_o \nonumber \]

Problems

Exercise \(\PageIndex{1}\)

Determine the crystal field stabilization energy (CFSE) in the following octahedral ions:

a) \(\ce{V^3+}\) b) \(\ce{Ni^2+}\) c) \(\ce{Cr^3+}\) d) \(\ce{Zn^2+}\)

Answer a

\(\ce{V^3+}\) is \(d^2\), with two electrons in the \(t_{2g}\) and zero electrons in \(e_g\).

CFSE = \([2(-0.4) + 0(0.6)]\times \Delta_o = -0.8\Delta_o\)

Answer b

\(\ce{Ni^2+}\) is \(d^8\), with six electrons in \(t_{2g}\) and two electrons in \(e_g\).

CFSE = \([6(-0.4) + 2(0.6)]\times \Delta_o = -1.2\Delta_o\)

Answer c

\(\ce{Cr^3+}\) is \(d^3\), with three electrons in \(t_{2g}\) and zero electrons in \(e_g\).

CFSE = \([3(-0.4) + 0(0.6)]\times \Delta_o = -1.2\Delta_o\)

Answer d

\(\ce{Zn^2+}\) is \(d^{10}\), with six electrons in \(t_{2g}\) and four electrons in \(e_g\).

CFSE = \([6(-0.4) + 4(0.6)]\times \Delta_o = 0\Delta_o\).