The Jahn–Teller Effect: Why Some Complexes Distort

The statement

Any non-linear molecule in a degenerate electronic state is unstable and will distort to lower its symmetry, thereby removing the degeneracy. The distortion costs some energy in strain but gains more in electronic stabilisation, so the net effect is favourable.

Linear molecules are excluded, which is worth remembering as a stated condition rather than an afterthought.

Which configurations are affected

In an octahedral field, degeneracy arises whenever a set of orbitals is unevenly occupied. Whether the resulting distortion is significant depends on which set.

Identifying which of a given set of complexes will show strong distortion is the most common question, and it is answered by writing the configuration and checking the e_g occupation.

Elongation or compression

The usual outcome is elongation along one axis, giving four short bonds and two long ones. Elongation lowers the energy of the orbital pointing along that axis, and if the odd electron occupies that orbital, the system gains stabilisation.

Compression is possible in principle and would stabilise the other orbital instead, but elongation is observed far more often. The reason is that removing ligand repulsion along one axis is generally easier than increasing it, so the theory predicts the distortion but not reliably its direction — a limitation worth stating.

Observable consequences

• Bond lengths differ in a complex that would otherwise be regular, which crystallography shows directly.
• Absorption bands broaden or split, because the degenerate levels are no longer degenerate. A broad or shouldered d–d band in a d⁹ complex is the classic spectroscopic signature.
• Stability constants are enhanced for distorted complexes relative to what the regular geometry would give, because the distortion provides extra stabilisation.

The Irving–Williams series shows a maximum at the d⁹ ion partly for this reason, which connects this topic to complex stability — a link questions sometimes ask for.

Square planar as the limit

Take the elongation far enough and the two axial ligands are effectively removed, leaving a square planar complex. This is why d⁸ complexes of certain metals are square planar rather than octahedral: the extreme distortion is so favourable that the axial ligands leave entirely.

Presenting square planar geometry as the extreme case of Jahn–Teller distortion is a good higher-order answer, and it explains why square planar geometry is so strongly associated with d⁸.

Frequently asked questions

Why does a half-filled or filled set not distort?

Because there is no degeneracy to remove — every orbital in the set is equally occupied, so distorting gains nothing electronically while still costing strain energy.

Why is t_2g distortion weak?

Because those orbitals point between the ligands, so changing their occupation has little effect on the metal–ligand interaction and produces only a small energy gain.

Does the theorem predict the size of the distortion?

No. It predicts that distortion will occur but not its magnitude or direction. Those require more detailed calculation or experiment.

Does it apply to tetrahedral complexes?

Yes in principle, but the effects are much smaller because no tetrahedral d orbital points directly at a ligand, so the degeneracies are less strongly coupled to geometry.