Jahn–Teller Distortion — When It Happens and Why

By Aniket Bhardwaj · 5 September 2026 · Advanced Chemistry

Why does copper(II) refuse to form a regular octahedron while nickel(II) is perfectly happy to? Why does one d–d band in a copper complex look like a smeared-out mess while the corresponding band in a chromium(III) complex is clean? Both answers come from a single theorem, and the theorem is short enough to state in one sentence.

The theorem

Jahn–Teller theorem: any non-linear molecule in a spatially degenerate electronic state is unstable with respect to a distortion that removes the degeneracy.

Three words in that sentence carry all the weight.

Non-linear. Linear molecules are explicitly excluded; their vibronic behaviour is described by the Renner–Teller effect instead.
Spatially degenerate. The degeneracy must be orbital. A half-filled or fully filled subshell whose electrons are distributed symmetrically is not degenerate and is not affected.
Unstable. The theorem guarantees that the symmetric geometry is not the minimum. It does not say how large the distortion is, and it does not say in which direction — only that one must occur.

Which configurations are affected

In an octahedral field, degeneracy in the e_g set produces a strong effect, because e_g orbitals (d_z² and d_x²−y²) point straight at the ligands. Degeneracy in the t_2g set produces only a weak effect, because those orbitals point between the ligands.

dⁿ	Configuration	Degeneracy	Jahn–Teller effect	Example ion
d⁰	t_2g⁰e_g⁰	none	none	Sc(III), Ti(IV)
d¹	t_2g¹	t_2g	weak	Ti(III)
d²	t_2g²	t_2g	weak	V(III)
d³	t_2g³	none	none	Cr(III)
d⁴ high spin	t_2g³e_g¹	e_g	strong	Mn(III), Cr(II)
d⁴ low spin	t_2g⁴	t_2g	weak	Mn(III) with strong field
d⁵ high spin	t_2g³e_g²	none	none	Mn(II), Fe(III)
d⁵ low spin	t_2g⁵	t_2g	weak	[Fe(CN)₆]³⁻
d⁶ high spin	t_2g⁴e_g²	t_2g	weak	Fe(II) aqua
d⁶ low spin	t_2g⁶	none	none	[Co(NH₃)₆]³⁺
d⁷ high spin	t_2g⁵e_g²	t_2g	weak	Co(II) aqua
d⁷ low spin	t_2g⁶e_g¹	e_g	strong	Ni(III), low-spin Co(II)
d⁸	t_2g⁶e_g²	none	none	Ni(II)
d⁹	t_2g⁶e_g³	e_g	strong	Cu(II)
d¹⁰	full	none	none	Zn(II), Cu(I)

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

The energy bookkeeping for a tetragonal elongation

Pull the two ligands on the z axis further away. Any orbital with z character is stabilised; any orbital in the xy plane is destabilised. Each set must keep its centre of gravity, so the splittings are fixed by that constraint.

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 e_g orbitals point at the ligands.

Copper(II), d⁹, elongated along z.

t_2g⁶: four electrons in d_xz and d_yz at −δ₂/3 each = −4δ₂/3; two electrons in d_xy at +2δ₂/3 each = +4δ₂/3. Net contribution = zero, as it must be for a filled set.

e_g³: two electrons in the stabilised d_z² at −δ₁/2 each = −δ₁; one electron in d_x²−y² at +δ₁/2.
Net = −δ₁ + δ₁/2 = −δ₁/2

So the distortion buys a stabilisation of δ₁/2. Numerically: if δ₁ = 8000 cm⁻¹, the gain is 4000 cm⁻¹. Converting with 1 cm⁻¹ = 11.96 J mol⁻¹:

4000 × 11.96 = 47 850 J mol⁻¹ = about 48 kJ mol⁻¹ — comparable to a hydrogen bond or two, which is why the effect shows up plainly in bond lengths and stability constants rather than being a subtlety.

Check the conversion: hcN_A = (6.626 × 10⁻³⁴)(2.998 × 10¹⁰ cm s⁻¹)(6.022 × 10²³) = 11.96 J mol⁻¹ per cm⁻¹. Correct.

Run the same arithmetic for high-spin d⁴: t_2g³ contributes nothing (one electron in each of the three orbitals, −δ₂/3 − δ₂/3 + 2δ₂/3 = 0) and the single e_g electron sits in d_z² at −δ₁/2. Same stabilisation, δ₁/2, from a completely different configuration — which is exactly why d⁴ and d⁹ behave alike here.

Why elongation and not compression?

To first order the theorem permits both, and both remove the degeneracy equally well. Elongation is what is almost always observed for copper(II), and the usual explanations are anharmonicity in the metal–ligand stretch and mixing of higher states; a genuinely compressed octahedron is rare. It is worth being honest in an answer: the theorem predicts that a distortion occurs, and experiment tells us which.

What you can actually observe

Bond lengths. Hexaaqua copper(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.
Electronic 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) rather than one sharp band, because the distortion has split one transition into several overlapping ones.
Distortion in the excited state. [Ti(H₂O)₆]³⁺ is d¹, so its ground state is only weakly affected — but its excited state is e_g¹ and therefore strongly Jahn–Teller active. Its absorption band, with a maximum near 20 300 cm⁻¹ and a distinct shoulder near 17 400 cm⁻¹, is the textbook demonstration that the effect operates in excited states too.
Stability constants. Copper(II) adds four ammonia ligands readily, but the fifth binds very weakly and the hexaammine is not formed in aqueous solution at all — the two axial sites are long and weak. The unusually high position of copper in the Irving–Williams stability order is attributed in part to the same effect.
Temperature dependence. A dynamic Jahn–Teller effect has the distortion hopping between the three equivalent axes fast enough that the average looks cubic; cooling freezes it into a static distortion, and EPR spectra change accordingly.

Two extensions worth knowing

Tetrahedral complexes. The theorem applies, but the effect is much smaller, because Δ_t is smaller and neither the e nor the t₂ set points directly at the ligands. Tetrahedral copper(II) species such as [CuCl₄]²⁻ are found flattened away from ideal tetrahedral geometry rather than grossly distorted.

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, and answers that use it must say so — calling it "the Jahn–Teller effect" without qualification is wrong, because the first-order theorem requires degeneracy.

Mistakes that cost marks

Claiming d⁸ octahedral is Jahn–Teller active. t_2g⁶e_g² places one electron in each e_g orbital with parallel spins. That is orbitally non-degenerate, so there is nothing to remove. The square-planar preference of Ni(II), Pd(II) and Pt(II) is a separate ligand-field argument.
Forgetting to specify spin state for d⁴, d⁶ and d⁷. High-spin d⁴ is strongly active; low-spin d⁴ is only weakly so. 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.
Adding the distortion energy to CFSE and double counting. The octahedral CFSE for 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, so state clearly which reference you are using.

Where this appears in the exam

Exam	Typical demand
CSIR-NET Chemical Sciences	Identify which of four complexes is Jahn–Teller active; explain a broad or split d–d band; account for copper(II) bond lengths and stability constants
GATE Chemistry	Configuration-to-effect matching; the tetragonal splitting pattern; CFSE with distortion
IIT-JAM / CUET-PG	Recognising d⁴, d⁷ low spin and d⁹ as the active cases
MSc coursework	Static versus dynamic distortion, EPR evidence, the pseudo Jahn–Teller effect

Convert the energies before you compare them. Ligand-field questions mix cm⁻¹, kJ mol⁻¹, eV and nanometres in the same problem, and that is where the marks go. The ABC Chemistry Calculator Suite keeps a unit converter and the scientific constants in one page beside your working.

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