Jahn-Teller distortion: Worked examples — Chemistry deep dive

This is the practice arena for Jahn–Teller distortion. The parent note told you the rule; here we run every kind of input through it, one worked example per case, so no configuration can ever surprise you in an exam.

Before we start, a quick promise: every symbol we use is defined the moment it appears. If the parent used a shorthand, we re-earn it here.

Look at the figure below: it lays out these two sets on one energy ladder, with the coral $e_{g}$ pair sitting above the mint $t_{2 g}$ trio, split by the octahedral gap $Δ_{o}$ . Every example on this page is really just asking "how are electrons parked on these two shelves, and is either shelf lopsided?"

The scenario matrix

Every case this topic can throw at you falls into one of these cells. Our examples below are labelled by cell so you can see the coverage is complete.

Cell	What makes it special	Covered by
A. Even $e_{g}$ , even $t_{2 g}$	both degenerate sets balanced → zero distortion	Ex 1 ( $d^{3}$ ), Ex 6 ( $d^{8}$ )
B. Uneven $e_{g}$ (strong)	the textbook big distortion	Ex 2 ( $d^{9}$ ), Ex 3 (hs $d^{4}$ ), Ex 4 (ls $d^{7}$ )
C. Uneven $t_{2 g}$ only (weak)	tiny, usually ignored distortion	Ex 5 ( $d^{1}$ ), Ex 5b ( $d^{2}$ )
D. Spin-state changes the answer	same $d$ -count, high-spin vs low-spin differ	Ex 7 ( $d^{6}$ ), Ex 7b ( $d^{5}$ )
E. Degenerate / boundary inputs	$d^{0}$ , $d^{10}$ — nothing to distort	Ex 8
F. Real-world word problem	reading distortion from bond-length data	Ex 9 (Cu²⁺ crystal)
G. Exam twist	direction of distortion (elongate vs compress)	Ex 10

The energy bookkeeping tool we reuse everywhere:

Why these particular fractions? Because a split conserves the total energy — nobody creates energy for free. Two orbitals drop $δ_{2} /3$ , one rises $2 δ_{2} /3$ , and $- 2 (δ_{2} /3) + 2 δ_{2} /3 = 0$ . ✓ The centre of gravity stays put. The next figure shows this split happening for both shelves at once — trace each coloured arrow from the grey un-split level to see who moves up (coral) and who moves down (mint).

Worked examples

Worked example Ex 1 —

$d^{3}$ (Cr³⁺): Cell A Statement: Does $[Cr (H_{2} O)_{6}]^{3 +}$ , a $d^{3}$ ion, undergo Jahn–Teller distortion?

Forecast: Guess now — will the octahedron stay perfect or squash? (Hint: count electrons in each set.)

Write the config. In an octahedral field $d^{3} = t_{2 g}^{3} e_{g}^{0}$ . Why this step? We must see how the electrons spread across the two degenerate sets before we can judge evenness.
Check the $e_{g}$ set. It holds 0 electrons — empty. Why this step? $e_{g}$ is the strong-distortion set; empty means it can't drive anything.
Check the $t_{2 g}$ set. It holds 3 electrons, one in each of the three orbitals — perfectly even (half-filled). Why this step? Evenness is exactly what kills distortion; each orbital identical → no energetic reason to break symmetry.
Apply $E_{JT}$ . Lowered and raised orbitals carry equal electrons in a balanced set, so $E_{JT} = 0$ . Why this step? Plugging into the bookkeeping formula turns "even" into a hard number, $0$ , confirming the verdict.

Answer: no distortion — symmetric octahedron.

Verify: With three equal orbitals each holding one electron, any distortion that lowers one raises another by the same amount and moves an equal electron count → net zero. Sanity: $Cr^{3 +}$ complexes are indeed regular octahedra. ✓

Worked example Ex 2 —

$d^{9}$ (Cu²⁺): Cell B, the classic Statement: Predict the JT behaviour and stabilization energy of $Cu^{2 +}$ ( $d^{9}$ ).

Forecast: Strong or weak distortion? By how much, in units of $δ_{1}$ ?

Config: $d^{9} = t_{2 g}^{6} e_{g}^{3}$ . Why this step? We need the $e_{g}$ count — that's what decides "strong".
$e_{g}$ holds 3 electrons in 2 orbitals → forced unevenness: one orbital doubly filled, one singly filled. Why this step? Three electrons cannot spread evenly over two boxes → the degeneracy MUST break → strong distortion.
Assign on elongation: put 2 electrons in the lowered $d_{z^{2}}$ (drops $δ_{1} /2$ ), 1 electron in the raised $d_{x^{2} - y^{2}}$ (rises $δ_{1} /2$ ). Why this step? Elongation lowers $d_{z^{2}}$ ; nature fills the cheaper orbital more.
Compute: $E_{JT} = 2 (\frac{δ _{1}}{2}) - 1 (\frac{δ _{1}}{2}) = + \frac{δ _{1}}{2} > 0$ . Why this step? Positive means real energy saved → distortion happens.

Answer: strong distortion, elongation along $z$ , gain $= δ_{1} /2$ .

Verify: $t_{2 g}^{6}$ is full and even → contributes 0. Only $e_{g}$ matters. $2 (δ_{1} /2) - 1 (δ_{1} /2) = δ_{1} /2$ . ✓ Cu²⁺ salts really show 4 short + 2 long bonds.

Worked example Ex 3 — high-spin

$d^{4}$ (Mn³⁺, Cr²⁺): Cell B Statement: Does high-spin $[Mn (H_{2} O)_{6}]^{3 +}$ ( $d^{4}$ ) distort, and by how much?

Forecast: One electron in $e_{g}$ — enough for a strong distortion?

Config (high spin): $d^{4} = t_{2 g}^{3} e_{g}^{1}$ . Why this step? Weak-field $H_{2} O$ → electrons spread out to maximise unpaired spins.
$e_{g}$ holds 1 electron in 2 orbitals → uneven (one orbital occupied, one empty). Why this step? One electron in a degenerate pair is the simplest possible unevenness → still strong.
Assign on elongation: the single $e_{g}$ electron drops into the lowered $d_{z^{2}}$ ( $- δ_{1} /2$ ); nothing sits in the raised $d_{x^{2} - y^{2}}$ . Why this step? Elongation lowers $d_{z^{2}}$ , so the one electron parks in the cheaper orbital.
Compute: $E_{JT} = 1 (\frac{δ _{1}}{2}) - 0 = + \frac{δ _{1}}{2}$ . Why this step? $t_{2 g}^{3}$ is evenly half-filled → 0 contribution, so only the $e_{g}$ electron counts.

Answer: strong distortion, gain $= δ_{1} /2$ .

Verify: Same $δ_{1} /2$ magnitude as Cu²⁺ from the $e_{g}$ side. $MnF_{6}^{3 -}$ and $[Cr (H_{2} O)_{6}]^{2 +}$ are experimentally distorted. ✓

Worked example Ex 4 — low-spin

$d^{7}$ (Ni³⁺, ls Co²⁺): Cell B Statement: Show that low-spin $d^{7}$ undergoes strong JT distortion.

Forecast: $d^{7}$ is more electrons — does that change the $e_{g}$ count from $d^{4}$ ?

Config (low spin): strong field fills $t_{2 g}$ first → $t_{2 g}^{6} e_{g}^{1}$ . Why this step? Strong-field ligand → $Δ_{o}$ large → pairing in $t_{2 g}$ beats occupying $e_{g}$ .
$e_{g}$ holds 1 electron → uneven → strong distortion (same shape of argument as Ex 3). Why this step? Uneven $e_{g}$ is the strong-distortion trigger, regardless of how many electrons sit below in $t_{2 g}$ .
Compute: $t_{2 g}^{6}$ full & even → 0. $E_{JT} = 1 (\frac{δ _{1}}{2}) = + \frac{δ _{1}}{2}$ . Why this step? Only the lone $e_{g}$ electron in the lowered orbital contributes.

Answer: strong distortion, gain $= δ_{1} /2$ .

Verify: Compare to high-spin $d^{7} = t_{2 g}^{5} e_{g}^{2}$ : here $e_{g}^{2}$ is even (one each) → only $t_{2 g}^{5}$ is uneven → weak. So the spin state flips the answer — see High-spin vs Low-spin Complexes. ✓

Worked example Ex 5 —

$d^{1}$ (Ti³⁺): Cell C (weak) Statement: Does $[Ti (H_{2} O)_{6}]^{3 +}$ ( $d^{1}$ ) distort? How strongly, and what is $E_{JT}$ ?

Forecast: One lonely electron in $t_{2 g}$ — big distortion or barely-there?

Config: $d^{1} = t_{2 g}^{1} e_{g}^{0}$ . Why this step? We locate the single electron to see which shelf is lopsided.
$e_{g}$ empty → no strong contribution. Why this step? The strong-distortion shelf is untouched, so any effect must come from $t_{2 g}$ .
$t_{2 g}$ holds 1 electron in 3 orbitals → uneven → a distortion is predicted. Why this step? The theorem guarantees some distortion for any uneven degenerate set.
Compute the size. On elongation the electron parks in a lowered $t_{2 g}$ orbital ( $d_{x z}$ or $d_{y z}$ , drop $δ_{2} /3$ ), with nothing in the raised $d_{x y}$ : $E_{JT} = 1 (\frac{δ _{2}}{3}) - 0 = \frac{δ _{2}}{3}$ . Why this step? Writing the actual number shows the effect is not zero — it's real, just governed by the small $δ_{2}$ .
Judge strength. $t_{2 g}$ lobes point between ligands → they barely couple to bond lengths → $δ_{2} ≪ δ_{1}$ → $E_{JT} = δ_{2} /3$ is tiny. Why this step? A formula alone doesn't say "weak"; the geometry of $t_{2 g}$ is what makes $δ_{2}$ small.

Answer: weak (essentially unobservable) distortion, $E_{JT} = δ_{2} /3$ .

Verify: This is exactly why the famous $Ti^{3 +}$ absorption band shows only a slight shoulder, not a big split — see Colour of Transition Metal Complexes. ✓

Worked example Ex 5b —

$d^{2}$ (V³⁺): Cell C (weak) Statement: Does $[V (H_{2} O)_{6}]^{3 +}$ ( $d^{2}$ ) distort?

Forecast: Two electrons now — do they fill $t_{2 g}$ evenly, or is it still lopsided?

Config: $d^{2} = t_{2 g}^{2} e_{g}^{0}$ . Why this step? Count first, judge second.
$e_{g}$ empty → no strong contribution. Why this step? Rules out the big textbook distortion right away.
$t_{2 g}$ holds 2 electrons in 3 orbitals → uneven (two orbitals occupied, one empty). Why this step? Two electrons cannot fill three boxes evenly → the degenerate set is lopsided.
Compute the size. On elongation both electrons sit in the two lowered orbitals ( $d_{x z}, d_{y z}$ , each $- δ_{2} /3$ ), leaving raised $d_{x y}$ empty: $E_{JT} = 2 (\frac{δ _{2}}{3}) - 0 = \frac{2 δ _{2}}{3}$ . Why this step? Placing electrons in the cheapest orbitals maximises the gain and gives the actual number.

Answer: weak distortion, $E_{JT} = 2 δ_{2} /3$ (still governed by tiny $δ_{2}$ , so unobservable in practice).

Verify: $t_{2 g}$ -only unevenness → $δ_{2} ≪ δ_{1}$ → practically undetectable, matching real $V^{3 +}$ octahedra looking regular. ✓

Worked example Ex 7 —

$d^{6}$ , both spins (Fe²⁺, Co³⁺): Cell D Statement: Compare JT behaviour of high-spin vs low-spin $d^{6}$ .

Forecast: Will the two spin states agree or disagree?

Low-spin $d^{6}$ : $t_{2 g}^{6} e_{g}^{0}$ . $t_{2 g}$ full & even, $e_{g}$ empty → no distortion. Why this step? Strong field packs all six into $t_{2 g}$ , filling it completely and evenly.
High-spin $d^{6}$ : $t_{2 g}^{4} e_{g}^{2}$ . $e_{g}^{2}$ = even (one each), but $t_{2 g}^{4}$ = one orbital doubly filled, two singly filled → uneven $t_{2 g}$ . Why this step? Four electrons over three $t_{2 g}$ boxes cannot be even.
Judge strength: unevenness is in $t_{2 g}$ only → weak distortion. Why this step? $t_{2 g}$ couples weakly to bond lengths, so the effect is small.

Answer: low-spin $d^{6}$ → none; high-spin $d^{6}$ → weak. The spin state changes the verdict.

Verify: $[Fe (CN)_{6}]^{4 -}$ (ls) is a clean octahedron; high-spin $Fe^{2 +}$ shows only slight, weak distortion. ✓

Worked example Ex 7b —

$d^{5}$ , both spins (Mn²⁺, Fe³⁺): Cell D Statement: Compare JT behaviour of high-spin vs low-spin $d^{5}$ .

Forecast: High-spin $d^{5}$ is the famous "all five singly filled" case — even or uneven?

High-spin $d^{5}$ : $t_{2 g}^{3} e_{g}^{2}$ . $t_{2 g}^{3}$ = one electron each (even), $e_{g}^{2}$ = one each (even) → both sets even → no distortion. Why this step? Five electrons over five orbitals, one apiece, is the most symmetric filling possible.
Low-spin $d^{5}$ : $t_{2 g}^{5} e_{g}^{0}$ . $e_{g}$ empty, but $t_{2 g}^{5}$ = two orbitals doubly filled, one singly filled → uneven $t_{2 g}$ . Why this step? Five electrons over three $t_{2 g}$ boxes cannot be even.
Judge strength: low-spin unevenness is in $t_{2 g}$ only → weak distortion. Why this step? Again $t_{2 g}$ couples weakly, so even the uneven low-spin case is barely visible.

Answer: high-spin $d^{5}$ → none; low-spin $d^{5}$ → weak. Opposite pattern to $d^{6}$ — always check the spin state.

Verify: High-spin $Mn^{2 +}$ / $Fe^{3 +}$ complexes are undistorted (half-filled shell, maximally symmetric); low-spin $d^{5}$ shows only weak $t_{2 g}$ distortion. ✓

Worked example Ex 8 —

$d^{0}$ and $d^{10}$ : Cell E (boundary/degenerate) Statement: What happens for the "edge" ions $Sc^{3 +}$ ( $d^{0}$ ) and $Zn^{2 +}$ ( $d^{10}$ )?

Forecast: Nothing in the shell, or everything in the shell — either way?

$d^{0}$ : $t_{2 g}^{0} e_{g}^{0}$ — no $d$ -electrons at all. Why this step? We check the electron count first; here there is literally nothing to be uneven.
$d^{0}$ has no degenerate occupation to be uneven → empty sets are trivially even. Why this step? The trigger is uneven occupation; zero electrons cannot be uneven, so no distortion.
$d^{10}$ : $t_{2 g}^{6} e_{g}^{4}$ — every orbital doubly filled. Why this step? We check the top boundary the same way, by counting occupation.
$d^{10}$ is a completely full set → perfectly even. Why this step? Lowering one orbital and raising another moves equal electron counts → net zero, so no distortion.
$E_{JT} = 0$ in both. Why this step? Both boundaries plug into the formula as exact cancellation — the natural "off" ends of the effect.

Answer: no JT distortion for $d^{0}$ or $d^{10}$ . These are the natural "off" boundaries of the whole effect.

Verify: $Zn^{2 +}$ and $Sc^{3 +}$ complexes are undistorted and colourless (no partly-filled $d$ set to absorb). ✓

Worked example Ex 9 — Word problem: reading bond lengths (Cell F)

Statement: X-ray data for a copper(II) complex give four M–L bonds of $1.95 \overset{˚}{A}$ and two M–L bonds of $2.30 \overset{˚}{A}$ . Which axis distorted, and does this match the $d^{9}$ prediction?

Forecast: Are the longer bonds the axial ( $z$ ) ones or the equatorial ones?

Identify the odd pair. Four bonds are equal ( $1.95$ ), two are different ( $2.30$ ) — the two longer bonds define the unique axis. Why this step? JT elongation stretches exactly two collinear (axial) bonds, leaving four equatorial bonds equal.
Longer = pulled away = along $z$ . So the $z$ -axis is elongated by $2.30 - 1.95 = 0.35 \overset{˚}{A}$ per bond. Why this step? Elongation increases axial bond length; less repulsion → $d_{z^{2}}$ lowers.
Match to theory. $Cu^{2 +} = d^{9} = t_{2 g}^{6} e_{g}^{3}$ → strong elongation predicted (Ex 2). Data agree. Why this step? Linking data to the config confirms this is genuine Jahn–Teller and not measurement noise.

Answer: $z$ -axis elongation of $0.35 \overset{˚}{A}$ , consistent with $d^{9}$ Jahn–Teller.

The figure below draws exactly this geometry — the four mint equatorial bonds are short and equal, the two dashed coral axial bonds are the long pair defining the elongated $z$ -axis.

Verify: $2.30 - 1.95 = 0.35$ ; "4 short + 2 long" is the textbook Cu²⁺ signature — see Stability and Distortion in $d^9$ Cu(II). ✓

Worked example Ex 10 — Exam twist: elongate or compress? (Cell G)

Statement: For $d^{9}$ , both elongation and compression remove the degeneracy. Using the $e_{g}$ occupation, argue which is usually preferred.

Forecast: Which distortion lets the doubly-filled orbital relax the most?

List the two options. Elongation (axial out) lowers $d_{z^{2}}$ , raises $d_{x^{2} - y^{2}}$ . Compression (axial in) does the reverse: lowers $d_{x^{2} - y^{2}}$ , raises $d_{z^{2}}$ . Why this step? Either one splits the $e_{g}$ pair, so either gives $E_{JT} > 0$ — the theorem alone can't pick.
Put the electrons in. With $e_{g}^{3}$ you have 2 electrons to place in the lowered orbital and 1 in the raised one.
- Elongation: the doubly-filled orbital is $d_{z^{2}}$ (a single lobe along $z$ + a ring) → relaxing two axial bonds gives those 2 electrons a lot of room.
- Compression: the doubly-filled orbital would be $d_{x^{2} - y^{2}}$ , needing four equatorial bonds to move — a stiffer, costlier deformation. Why this step? The preferred distortion is the one giving the biggest net (electronic gain − strain cost).
Pick the cheaper stretch. Relaxing 2 axial bonds (elongation) is easier than pushing 4 equatorial bonds → elongation usually wins. Why this step? Moving fewer bonds costs less elastic strain, so the net energy saving is larger.

Answer: elongation is the common outcome, but compression is not forbidden — the theorem guarantees distortion, not its direction. Overwhelmingly Cu²⁺ shows 4 short + 2 long bonds (elongation), with only rare compressed exceptions.

Verify: The "4 short + 2 long" elongation signature dominates real Cu²⁺ crystallography, matching "some distortion, direction not guaranteed". ✓

Coverage check

Recall Did we hit every matrix cell?

A (even/even) — Ex 1, Ex 6, Ex 7b(hs) ::: $d^{3}$ , $d^{8}$ , hs $d^{5}$ : no distortion. B (uneven $e_{g}$ , strong) — Ex 2, 3, 4 ::: $d^{9}$ , hs $d^{4}$ , ls $d^{7}$ : gain $δ_{1} /2$ . C (uneven $t_{2 g}$ only, weak) — Ex 5, 5b ::: $d^{1}$ gives $δ_{2} /3$ , $d^{2}$ gives $2 δ_{2} /3$ . D (spin-state dependent) — Ex 7, 7b ::: $d^{6}$ (ls none, hs weak); $d^{5}$ (hs none, ls weak). E (boundary/degenerate) — Ex 8 ::: $d^{0}$ , $d^{10}$ : nothing to distort. F (word problem) — Ex 9 ::: reading 4-short/2-long bond data. G (exam twist) — Ex 10 ::: elongate vs compress.

Connections

Jahn-Teller distortion — the parent theory these examples drill.
Crystal Field Theory — source of the $t_{2 g} / e_{g}$ split.
Octahedral Splitting and $\Delta_o$ — the energy diagram we perturb.
High-spin vs Low-spin Complexes — decides $e_{g}$ count (Ex 4, 7, 7b).
Stability and Distortion in $d^9$ Cu(II) — Ex 2 and Ex 9 in depth.
Colour of Transition Metal Complexes — why weak $d^{1}$ distortion barely shows (Ex 5).
Tetrahedral vs Octahedral Geometry — where distortions turn weak.