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.

Two words carry the weight. Non-linear: linear molecules are excluded by the theorem itself (their vibronic behaviour is described by the Renner–Teller effect). Degenerate means orbitally (spatially) degenerate: a half-filled or completely filled set whose electrons are spread symmetrically is not degenerate and is not affected.

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.

Three rows carry the strong effect and they are the three worth memorising: high-spin d⁴, low-spin d⁷ and d⁹ — exactly the configurations with an odd number of electrons in the e_g 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.

The energy bookkeeping for a tetragonal elongation

Pull the two ligands on the z axis further away. Every orbital with z character is stabilised and every orbital in the xy plane is destabilised. Each set keeps its centre of gravity, which fixes the splittings:

• e_g splits by δ₁: d_z² falls by δ₁/2, d_x²−y² rises by δ₁/2.
• t_2g splits by δ₂: d_xz and d_yz fall by δ₂/3 each, d_xy rises by 2δ₂/3.
• δ₁ is much larger than δ₂, because the e_g orbitals point at the ligands.

The same arithmetic for high-spin d⁴: t_2g³ contributes nothing (−δ₂/3 − δ₂/3 + 2δ₂/3 = 0) and the single e_g electron sits in d_z² at −δ₁/2 — the same δ₁/2 from a different configuration, which is why d⁴ and d⁹ behave alike. To check a configuration quickly, use the electron configuration calculator.

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. To first order the theorem permits both; the usual explanations for the preference are anharmonicity in the metal–ligand stretch and mixing with higher states, and a genuinely compressed octahedron is rare. So the theory predicts that a distortion occurs, and experiment tells us which — 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.

In numbers:

• Bond lengths. Hexaaquacopper(II) in crystals typically shows four short equatorial Cu–O bonds near 1.95 Å and two long axial bonds near 2.3–2.4 Å, instead of six equal bonds.
• Spectra. The d–d absorption of [Cu(H₂O)₆]²⁺ is a single very broad, asymmetric band near 800 nm (about 12 500 cm⁻¹).
• Excited states too. [Ti(H₂O)₆]³⁺ is d¹, so its ground state is only weakly affected — but its excited state is e_g¹ and strongly Jahn–Teller active: the band (maximum near 20 300 cm⁻¹) has a distinct shoulder near 17 400 cm⁻¹.
• Stepwise binding. Copper(II) adds four ammonia ligands readily, but the fifth binds very weakly and the hexaammine is not formed in aqueous solution — the two axial sites are long and weak.
• Dynamic vs static. In a dynamic Jahn–Teller effect the distortion hops between the three equivalent axes fast enough that the average looks cubic; cooling freezes it into a static distortion, and EPR spectra change accordingly.

Square planar: the limit of elongation — but not a Jahn–Teller case for d⁸

Take the elongation far enough and the two axial ligands are effectively removed, leaving a square planar complex; many copper(II) complexes sit somewhere along this path. Square planar d⁸ complexes (Ni(II) with strong-field ligands, Pd(II), Pt(II)) are a different story: octahedral d⁸ (t_2g⁶e_g², one electron in each e_g orbital) is not degenerate, so it is not Jahn–Teller active. Their square planar preference is a separate ligand-field argument: in a strong field the eight electrons fill the four lower orbitals and leave the strongly antibonding d_x²−y² empty.

Second-order (pseudo) Jahn–Teller effect

A molecule with a non-degenerate ground state can still distort if a low-lying excited state of the right symmetry mixes into it strongly enough. This is a distinct mechanism: an answer that uses it must say so, because the first-order theorem requires degeneracy.

Mistakes that cost marks

• Calling octahedral d⁸ Jahn–Teller active. t_2g⁶e_g² puts one electron in each e_g orbital with parallel spins — orbitally non-degenerate, nothing to remove.
• Forgetting the spin state for d⁴, d⁶ and d⁷. High-spin d⁴ is strongly active; low-spin d⁴ only weakly. The answer changes with the ligand field.
• Treating high-spin d⁵ as active. One electron in every orbital is the most symmetric arrangement possible.
• Saying the theorem predicts elongation. It predicts distortion.
• Applying it to linear molecules. They are excluded by the theorem itself.
• Double counting with CFSE. The octahedral CFSE of d⁹ is 6(−0.4Δ_o) + 3(+0.6Δ_o) = −0.6Δ_o; the δ₁/2 from the distortion is an additional stabilisation measured from the undistorted octahedron — state which reference you use.

Where this appears in the exam

Convert the energies before you compare them: ligand-field questions mix cm⁻¹, kJ mol⁻¹, eV and nanometres in the same problem.

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. Δ_t is also smaller. Tetrahedral copper(II) species such as [CuCl₄]²⁻ are found flattened away from the ideal tetrahedron rather than grossly distorted.