Jahn-Teller distortion: Question bank — Chemistry deep dive

This is a reasoning gym for the JT topic. Every item below is a conceptual trap: the "obvious" answer is often wrong. Cover the answer, commit to a reason, then reveal. If your reasoning differs from the answer's reasoning, that's the trap catching you.

Prerequisites you may want open: Crystal Field Theory, Octahedral Splitting and $\Delta_o$, High-spin vs Low-spin Complexes.

True or false — justify

True or false: Every complex with a partially filled

d

shell shows a Jahn–Teller distortion.

False. What matters is whether a degenerate set is unevenly occupied.

d^{3}

(

t_{2 g}^{3}

) and

d^{8}

(

t_{2 g}^{6} e_{g}^{2}

) are partially filled yet evenly spread within each degenerate set, so

E_{JT} = 0

True or false: A

d^{3}

ion in an octahedral field distorts because it has unpaired electrons.

False.

t_{2 g}^{3}

means each of the three degenerate

t_{2 g}

orbitals holds exactly one electron — that is even occupation of the degenerate set. Unpaired ≠ uneven; the set is symmetric, so no distortion.

True or false:

t_{2 g}

unevenness produces a large, X-ray-observable distortion.

False.

t_{2 g}

lobes point between the ligands, so they couple weakly to bond lengths — the distortion is tiny and usually unobservable. Only uneven $e_{g}$ gives the strong textbook 4-short-2-long geometry.

True or false: The Jahn–Teller theorem tells you the distortion will be an elongation.

False. The theorem guarantees only that some symmetry-lowering distortion occurs. Both elongation and compression lift the degeneracy; elongation is usually slightly more stable but the theorem never predicts direction.

True or false: A

d^{10}

ion like Zn²⁺ shows strong Jahn–Teller distortion.

False.

t_{2 g}^{6} e_{g}^{4}

fills every degenerate orbital completely and evenly. There is no way to gain electronic energy by splitting a full set, so

E_{JT} = 0

True or false: The barycentre (centre of energy) of a set shifts downward when the set splits under distortion.

False. The barycentre is conserved. If one orbital drops by

δ /2

its partner rises by

δ /2

; the net energy of the set is unchanged. The stabilization comes only from how electrons populate the split levels, not from the set moving down.

True or false: Jahn–Teller distortion changes the total number of

d

electrons available for the metal.

False. Distortion only rearranges orbital energies and hence occupation within the same electron count. Electron number is fixed by oxidation state; JT never adds or removes electrons.

True or false: Low-spin

d^{6}

(e.g.

[Co(NH_{3})_{6}]^{3 +}

) is a strong Jahn–Teller ion.

False. Low-spin

d^{6}

t_{2 g}^{6} e_{g}^{0}

: the

t_{2 g}

set is completely full (even) and

e_{g}

is empty (even). Both degenerate sets are evenly occupied → no distortion.

True or false: Jahn–Teller distortion applies to a linear molecule such as a two-coordinate metal centre.

False. The theorem explicitly requires a non-linear geometry. Linear systems are excluded from the JT theorem (their degeneracies are handled by the separate Renner–Teller effect).

Spot the error

"Cu²⁺ (

d^{9}

) distorts because it has one unpaired electron." — find the flaw.

The unpaired electron is a symptom, not the cause. The real cause is that the

e_{g}

set holds 3 electrons (

e_{g}^{3}

), which cannot be split evenly between two orbitals — one is doubly filled, one singly filled. That uneven

e_{g}

occupation is the trigger.

d^{8}

is like

d^{9}

minus one electron, so it should also distort strongly." — find the flaw.

d^{8}

t_{2 g}^{6} e_{g}^{2}

— the

e_{g}

pair sits one in each

e_{g}

orbital, i.e. evenly. Removing that even pairing to make

d^{9}

(

e_{g}^{3}

) is exactly what breaks the symmetry. Counting electrons off

d^{9}

hides that the

e_{g}

occupation flips from even to uneven.

"During elongation the

d_{z^{2}}

orbital rises because the metal–ligand bonds get longer." — find the flaw.

Longer axial bonds mean the axial ligands move away, so the

z

-pointing

d_{z^{2}}

feels less repulsion and drops. The claim inverts the physics; less repulsion always lowers, not raises, the orbital energy.

"Elongation lowers

d_{z^{2}}

, so both

e_{g}

orbitals move down and the complex is stabilized." — find the flaw.

The barycentre is conserved: if

d_{z^{2}}

drops by

δ_{1} /2

, then

d_{x^{2} - y^{2}}

must rise by

δ_{1} /2

. Both cannot fall. Stabilization comes from placing more electrons in the lowered

d_{z^{2}}

than in the raised

d_{x^{2} - y^{2}}

"High-spin

d^{4}

distorts because its

t_{2 g}^{3}

half-fill is uneven." — find the flaw.

t_{2 g}^{3}

is evenly half-filled (one electron per orbital) → no

t_{2 g}

contribution. The distortion is driven by the single

e_{g}

electron (

e_{g}^{1}

), which occupies one

e_{g}

orbital and leaves the other empty — that is the uneven set.

"Tetrahedral complexes distort just as strongly as octahedral ones by the same argument." — find the flaw.

JT applies to any non-linear geometry, but in tetrahedral fields the

e

t_{2}

splitting is small and the orbital orientations relative to ligands differ, so coupling to bond length is weak → distortions are generally small. The principle is general; the magnitude is not.

"In the split

t_{2 g}

set,

d_{x y}

rises by

δ_{2} /3

to match the two orbitals that fell." — find the flaw.

To conserve the barycentre with two orbitals falling by

δ_{2} /3

each, the single rising orbital must go up by

2 δ_{2} /3

- 2 (δ_{2} /3) + 2 (δ_{2} /3) = 0

. Using

δ_{2} /3

for the rise would break the balance.

Why questions

Why does uneven

e_{g}

occupation cause a stronger distortion than uneven

t_{2 g}

occupation?

The

e_{g}

orbitals (

d_{z^{2}}, d_{x^{2} - y^{2}}

) point their lobes directly at the ligands, so changing bond lengths changes their energies a lot.

t_{2 g}

lobes point between ligands, coupling weakly to bond lengths, so the energy gain from distorting is tiny.

Why must the distortion lower the total energy rather than just move orbitals around?

Distorting costs elastic/strain energy, so it only happens if the electronic energy drop is larger. That drop exists only when you can move electrons from a now-raised orbital into a now-lowered one — which requires an unevenly occupied degenerate set.

Why is elongation along

z

usually preferred over compression for Cu²⁺?

Both remove the degeneracy, but elongation moves the two axial ligands away, letting the doubly-occupied

d_{z^{2}}

relax the most and giving a slightly larger net stabilization. Hence the observed 4 short + 2 long M–L bonds.

Why does the barycentre rule (

- 2 (δ /3) + 2 (δ /3) = 0

etc.) have to hold at all?

A pure geometric distortion redistributes repulsion among orbitals but does not change the average electron–ligand interaction of a set. So the signed sum of energy shifts within a degenerate set is zero — the set's centre of energy is fixed.

Why does Jahn–Teller distortion add extra features to the absorption spectrum of Cu²⁺?

Splitting the

e_{g}

(and

t_{2 g}

) levels creates new, closely spaced energy gaps, so instead of one clean

d

–

d

transition you get several overlapping ones. This broadens or splits the band — visible as the characteristic broad Cu²⁺ absorption. See Colour of Transition Metal Complexes.

Why does high-spin vs low-spin choice matter for predicting JT in

d^{7}

The spin state sets the

e_{g}

occupation. Low-spin

d^{7}

t_{2 g}^{6} e_{g}^{1}

— uneven

e_{g}

→ strong JT. High-spin

d^{7}

t_{2 g}^{5} e_{g}^{2}

—

e_{g}

evenly filled (no strong JT). Same

d

count, opposite prediction. See High-spin vs Low-spin Complexes.

Edge cases

Edge case: What does JT predict for

d^{0}

(e.g. Ti⁴⁺, Sc³⁺)?

No distortion. With zero

d

electrons there is no degenerate occupied set to split, so there is nothing to gain — the octahedron stays symmetric.

Edge case:

d^{5}

high-spin (Mn²⁺, Fe³⁺) — does it distort?

No. High-spin

d^{5}

t_{2 g}^{3} e_{g}^{2}

: each of all five orbitals holds exactly one electron. Every degenerate set is evenly half-filled →

E_{JT} = 0

, a symmetric octahedron.

Edge case:

d^{5}

low-spin (e.g.

[Fe(CN)_{6}]^{3 -}

) — does it distort?

Only weakly. Low-spin

d^{5}

t_{2 g}^{5} e_{g}^{0}

: the

e_{g}

set is empty (even), but

t_{2 g}^{5}

is unevenly filled. Since only

t_{2 g}

is uneven, the distortion is small and usually unobserved.

Edge case: Does a spin-degenerate ground state that is orbitally non-degenerate trigger JT?

No. The JT theorem concerns orbital (spatial) degeneracy of the electronic ground state. Pure spin degeneracy does not couple to nuclear geometry the same way, so it does not force a distortion.

Edge case: If two different distortions give the same energy gain, what does the theorem say?

The theorem guarantees that some symmetry-lowering distortion occurs but does not uniquely pick one. When paths are near-degenerate the molecule may show a dynamic JT effect, hopping between equivalent distorted geometries rather than freezing into one.

Edge case:

d^{10}

d^{0}

— do they distort for the same reason of "no distortion"?

Both give

E_{JT} = 0

but for mirror-image reasons:

d^{0}

has an empty set (nothing to lower),

d^{10}

has every degenerate set completely full (splitting a full set gains nothing). Either way, no uneven occupation exists.

Recall One-line master test

For any $d^{n}$ : write the octahedral configuration, look at each degenerate set ( $t_{2 g}$ then $e_{g}$ ). Strong JT ⇔ uneven $e_{g}$ . Weak JT ⇔ uneven $t_{2 g}$ only. No JT ⇔ every degenerate set is empty, evenly half-full, or completely full.

Connections

Jahn-Teller distortion — the parent concept these traps drill.
High-spin vs Low-spin Complexes — fixes $e_{g}$ occupation, hence JT prediction.
Stability and Distortion in $d^9$ Cu(II) — the flagship strong-JT case.
Colour of Transition Metal Complexes — why JT splitting broadens $d$ – $d$ bands.
Tetrahedral vs Octahedral Geometry — why magnitude differs by geometry.
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