Symmetry, distortion and the Jahn–Teller effect

Crystals of nickel(II) fluoride and copper(II) fluoride have almost the same structure: each metal ion sits in an octahedron of six $\ce{F-}$ ions. In $\ce{NiF2}$ the six Ni–F bonds are nearly equal — four of 2.02 Å and two of 1.98 Å. In $\ce{CuF2}$ four are 1.92–1.93 Å and the other two 2.30 Å, about a fifth longer. $\ce{CrF2}$ is the same: four bonds of 1.98–2.01 Å and two of 2.43 Å. Nickel, copper and chromium are close neighbours and the ligands are identical, so the difference must lie in the d electrons: $\ce{Ni^2+}$ is d⁸, $\ce{Cu^2+}$ d⁹ and $\ce{Cr^2+}$ high-spin d⁴. Why should one electron more, or a few fewer, make an octahedron deform itself?

The $t_{2g}$ and $e_g$ of the last lesson are group-theory names. Treat the d orbitals as functions — $d_{xy} \propto xy$, $d_{z^2} \propto 2z^2 - x^2 - y^2$ and so on — and how they transform under the symmetry operations is listed among the quadratic functions in the last column of the character table. Look up $O_h$: $(2z^2 - x^2 - y^2,\ x^2 - y^2)$ belong together to $E_g$ and $(xy,\ xz,\ yz)$ to $T_{2g}$, so an octahedral field must give a pair and a trio of degenerate orbitals — no energies needed. Representations take capital letters, orbitals lower-case ones: $e_g$, $t_{2g}$.

You can reach the same answer without the table. The character of the quadratic functions under an operation R depends on R alone, so the five d orbitals have one general character formula; reducing it with the reduction formula used for counting infrared bands gives the grouping in any geometry.

Stretch the octahedron along z and $O_h$ drops to $D_{4h}$. The old $e_g$ splits into $a_{1g}$ ($d_{z^2}$, down) and $b_{1g}$ ($d_{x^2-y^2}$, up); $t_{2g}$ splits into $e_g$ ($d_{xz}$, $d_{yz}$) and $b_{2g}$ ($d_{xy}$) — for weak π donors such as $\ce{F-}$ and $\ce{H2O}$ the pair falls and $d_{xy}$ rises. When a degenerate set splits its barycentre stays put: in $e_g$ one orbital falls by δ₁/2 as the other rises by δ₁/2; in $t_{2g}$ two fall by δ₂/3 each and one rises by 2δ₂/3. Because $t_{2g}$ points between the ligands, δ₂ is much smaller than δ₁. Which level splits into which in such a descent in symmetry is fixed by the correspondence between the character tables of the two groups.

Keep pulling until the axial ligands leave and you reach the square plane: $d_{x^2-y^2}$ alone at the top — the reason, met in the last lesson, that d⁸ ions favour square-planar coordination.

In 1937 H. A. Jahn and E. Teller proved that a non-linear molecule in an orbitally degenerate electronic state always has some distortion that lowers its energy and removes the degeneracy: the Jahn–Teller effect. In a regular octahedron $\ce{Cu^2+}$ (d⁹) is $t_{2g}^6e_g^3$, and the three $e_g$ electrons can be arranged as $(d_{z^2})^2(d_{x^2-y^2})^1$ or as $(d_{z^2})^1(d_{x^2-y^2})^2$ — two arrangements of equal energy, which is what orbital degeneracy means. Stretching the octahedron lowers $d_{z^2}$ by δ₁/2 and raises $d_{x^2-y^2}$ by δ₁/2: two electrons gain and one loses, a net fall of δ₁/2.

That energy falls linearly with the distortion, while stretching or squeezing bonds costs energy quadratically, like a spring — so the regular octahedron cannot be the energy minimum and the ion must distort by some finite amount. The electronic energy has a cusp at the octahedron: stretching and squashing both lower it. Real copper(II) compounds are almost always elongated.

The two common shapes of five-coordinate complexes differ little in energy. $\ce{[Ni(CN)5]^3-}$ is the classic case: in crystals of $\ce{[Cr(en)3][Ni(CN)5].1.5H2O}$ the same anion occurs both as a square pyramid and as a distorted trigonal bipyramid (Raymond, Corfield and Ibers, 1968). In the trigonal bipyramid $d_{z^2}$ points at the two axial ligands and lies highest; $(d_{x^2-y^2}, d_{xy})$ form $e'$ in the equatorial plane; $(d_{xz}, d_{yz})$ form $e''$, lowest of all. $\ce{Fe(CO)5}$ is a trigonal-bipyramidal d⁸ molecule: its 8 electrons fill $e''$ and $e'$ and leave $d_{z^2}$ empty.

The trigonal prism is $D_{3h}$ as well, with six ligands on two eclipsed triangles. Now the cone-shaped lobes of $d_{z^2}$ fit between the two layers of ligands, and it becomes the lowest level. The Mo(IV) ions of molybdenum disulfide, $\ce{MoS2}$, sit in trigonal prisms: their two d electrons pair up in $d_{z^2}$, which is why $\ce{MoS2}$ is a diamagnetic semiconductor.