Jahn-Teller distortion: Exercises — Chemistry deep dive

Before any problem, here is the entire toolkit, built from zero.

Look at the figure below: the $e_{g}$ orbitals aim their lobes straight into the ligands (red), while the $t_{2 g}$ lobes slip into the gaps between them. This single picture is why moving a ligand barely nudges $t_{2 g}$ but strongly shifts $e_{g}$ — the whole page rests on it.

Where $δ_{1}$ and $δ_{2}$ come from (defining the symbols)

Why the $\frac{1}{2}$ in $\frac{δ _{1}}{2}$ ? The $e_{g}$ set has 2 orbitals. When a gap $δ_{1}$ opens, the barycentre (average energy) must stay put, so the two orbitals sit symmetrically about it: one at $- \frac{δ _{1}}{2}$ , the other at $+ \frac{δ _{1}}{2}$ . Their gap is $\frac{δ _{1}}{2} - (- \frac{δ _{1}}{2}) = δ_{1}$ . ✓ The denominator "2" is literally the number of orbitals in the set.

Why the $\frac{1}{3}$ and $\frac{2}{3}$ in $t_{2 g}$ ? The $t_{2 g}$ set has 3 orbitals. Under $z$ -elongation two of them ( $d_{x z}, d_{y z}$ , which have a $z$ -component) go down and one ( $d_{x y}$ , purely in-plane) goes up. To keep the barycentre fixed with a 2-down/1-up split, the lone rising orbital must move twice as far as each falling one: two orbitals at $- \frac{δ _{2}}{3}$ balance one at $+ \frac{2 δ _{2}}{3}$ . The denominator "3" is the number of $t_{2 g}$ orbitals; the "2" in $\frac{2 δ _{2}}{3}$ counts how many orbitals share the opposite side.

The geometry behind the $t_{2 g}$ signs

Where the $E_{JT}$ formula comes from

Formula Deriving the stabilization energy

A molecule's electronic energy is just the sum, over every electron, of the energy of the orbital it sits in. Before distortion, take the barycentre as the zero of energy — every electron sits at $0$ , so the total is $0$ . After distortion, each orbital has shifted by some $\pm Δ$ . The new total is $E_{after} = \sum_{electrons} (shift of its orbital) .$ The energy we saved is $E_{JT} = E_{before} - E_{after} = - \sum (shifts)$ . Writing "down-shifts" as positive savings and "up-shifts" as costs: $E_{JT} = energy saved \sum (electrons that dropped) \times (their drop) - energy spent \sum (electrons that rose) \times (their rise) .$ Distortion happens only when this is positive — i.e. you parked more electrons in the lowered orbitals than in the raised ones. That requires an uneven occupation of a degenerate set.

Why do the drops and rises balance when the set is evenly filled? Because the barycentre (centre of energy) is conserved: whatever one orbital gains, its partner loses. So a set that is filled symmetrically has zero net change — no reason to distort.

Level 1 — Recognition

Recall Solution L1·Q1

Fill $t_{2 g}$ first (it is lower), obeying high-spin rules where stated.

(a) $d^{4}$ high-spin: $t_{2 g}^{3} e_{g}^{1}$ → $e_{g}$ has 1 electron in a 2-orbital set → uneven.
(b) $d^{3}$ : $t_{2 g}^{3} e_{g}^{0}$ → $e_{g}$ empty (0,0) → even.
(c) $d^{9}$ : $t_{2 g}^{6} e_{g}^{3}$ → $e_{g}$ has 3 in a 2-orbital set → one orbital 2, one orbital 1 → uneven.
(d) $d^{8}$ : $t_{2 g}^{6} e_{g}^{2}$ → $e_{g}$ has 2 → one each → even.

Recall Solution L1·Q2

Rule: strong distortion ⇔ the == $e_{g}$ set is unevenly filled== (because $e_{g}$ lobes point directly at ligands). From Q1: uneven $e_{g}$ only in high-spin $d^{4}$ and $d^{9}$ . Strong JT: high-spin $d^{4}$ and $d^{9}$ . ( $d^{3}, d^{8}$ : even $e_{g}$ → none.)

Level 2 — Application

Recall Solution L2·Q1

After elongation the single $e_{g}$ electron drops into the lowered $d_{z^{2}}$ .

Electrons that dropped: 1, drop $= \frac{δ _{1}}{2}$ .
Electrons that rose: 0. $E_{JT} = 1 \cdot \frac{δ _{1}}{2} - 0 = \frac{δ _{1}}{2} .$ Positive → distortion is favourable. ✓

Recall Solution L2·Q2

Optimal filling: put 2 electrons in the lowered $d_{z^{2}}$ , 1 in the raised $d_{x^{2} - y^{2}}$ .

Dropped: 2 electrons × $\frac{δ _{1}}{2}$ = $δ_{1}$ .
Risen: 1 electron × $\frac{δ _{1}}{2}$ = $\frac{δ _{1}}{2}$ . $E_{JT} = δ_{1} - \frac{δ _{1}}{2} = \frac{δ _{1}}{2} .$ Same magnitude as $d^{4}$ — both are "one net electron of $e_{g}$ imbalance". This is why $d^{4}$ and $d^{9}$ show comparable strong distortions.

The energy-level diagram below shows exactly this $z$ -elongation: watch the red $d_{z^{2}}$ level drop and the lone $e_{g}$ electron settle into it — the geometric source of the $+ \frac{δ _{1}}{2}$ saving.

Recall Solution L2·Q3

$d_{x z} - \frac{δ _{2}}{3} + d_{y z} - \frac{δ _{2}}{3} + d_{x y} + \frac{2 δ _{2}}{3} = - \frac{2 δ _{2}}{3} + \frac{2 δ _{2}}{3} = 0. ✓$ Two orbitals fall by one unit each; one rises by two units — the centre of energy does not move.

Level 3 — Analysis

Recall Solution L3·Q1

The size of a distortion depends on how strongly an orbital couples to bond lengths — i.e. how much moving a ligand changes that orbital's energy. This is exactly the difference between the two splitting sizes: $δ_{1}$ (large) vs $δ_{2}$ (small).

$e_{g}$ lobes ( $d_{z^{2}}, d_{x^{2} - y^{2}}$ ) point directly at the ligands. Move an axial ligand and the $d_{z^{2}}$ energy changes a lot → large $δ_{1}$ → the molecule can gain a lot by distorting → large, X-ray-visible distortion.
$t_{2 g}$ lobes point between ligands. Moving a ligand barely changes their energy → tiny $δ_{2}$ → small gain → negligible distortion. So even though both sets can be "uneven", only $e_{g}$ imbalance pays enough to bend real bonds.

Recall Solution L3·Q2

$e_{g}^{1}$ : one electron in a two-orbital degenerate set → uneven $e_{g}$ → strong JT. It mirrors high-spin $d^{4}$ (also $e_{g}^{1}$ ): the lone $e_{g}$ electron drops into $d_{z^{2}}$ under $z$ -elongation. $E_{JT} = 1 \cdot \frac{δ _{1}}{2} = \frac{δ _{1}}{2} > 0.$ Prediction: strong distortion, elongation along $z$ (4 short + 2 long bonds).

Recall Solution L3·Q3

Low-spin $d^{6}$ : $t_{2 g}^{6} e_{g}^{0}$ → $t_{2 g}$ fully filled (2,2,2), $e_{g}$ empty → both even → no JT.
High-spin $d^{6}$ : $t_{2 g}^{4} e_{g}^{2}$ → $e_{g}$ even (1,1); $t_{2 g}$ uneven (2,1,1) → only weak $t_{2 g}$ JT (usually ignored, since $δ_{2} ≪ δ_{1}$ ). So $d^{6}$ shows no strong JT in either case. The claim confuses " $e_{g}$ imbalance" (strong) with "any imbalance". Corrected: strong JT needs an odd $e_{g}$ population — $d^{6}$ never has one.

Level 4 — Synthesis

Worked example L4·Q1

Build a general rule: for a given octahedral $d^{n}$ , the strong-JT $e_{g}$ imbalance equals $∣ electrons in d_{z^{2}} - electrons in d_{x^{2} - y^{2}} ∣$ at the symmetric point (0, 1, or 2 possible). Fill this table and mark strong JT (HS = high-spin, LS = low-spin):

$d^{n}$	spin	$t_{2 g}$	$e_{g}$	$e_{g}$ imbalance	strong JT?
$d^{1}$	—	1	0	?	?
$d^{4}$	HS	3	1	?	?
$d^{7}$	LS	6	1	?	?
$d^{9}$	—	6	3	?	?
$d^{10}$	—	6	4	?	?

Recall Solution L4·Q1

$e_{g}$ imbalance = 1 whenever $e_{g}$ holds an odd number (1 or 3); 0 when it holds 0, 2, or 4.

$d^{n}$	spin	$t_{2 g}$	$e_{g}$	imbalance	strong JT?
$d^{1}$	—	1	0	0	No (only weak $t_{2 g}$ )
$d^{4}$	HS	3	1	1	Yes
$d^{7}$	LS	6	1	1	Yes
$d^{9}$	—	6	3	1	Yes
$d^{10}$	—	6	4	0	No

General rule: Strong octahedral JT ⇔ $e_{g}$ holds an odd number of electrons (1 or 3).

Recall Solution L4·Q2

$t_{2 g}^{3}$ evenly filled → contribution $= 1 (- \frac{δ _{2}}{3}) + 1 (- \frac{δ _{2}}{3}) + 1 (+ \frac{2 δ _{2}}{3}) = 0.$
$e_{g}^{1}$ in lowered $d_{z^{2}}$ → $+ \frac{δ _{1}}{2}$ stabilization. $E_{JT,total} = 0 + \frac{δ _{1}}{2} = \frac{δ _{1}}{2} > 0.$ Positive ⇒ the elongated structure is lower in energy ⇒ distortion occurs. The $t_{2 g}$ term vanishing shows why we may safely ignore it for the "strong" verdict.

Level 5 — Mastery

Recall Solution L5·Q1

Colour arises from a $d$ – $d$ electron jump, whose energy sets the absorbed wavelength.

In an undistorted octahedron a $d^{9}$ ion has essentially one $e_{g} \leftarrow t_{2 g}$ -type transition energy → one band.
JT elongation splits both $e_{g}$ (into $d_{z^{2}}$ low, $d_{x^{2} - y^{2}}$ high) and $t_{2 g}$ . Now there are several slightly-different transition energies (e.g. into $d_{z^{2}}$ vs $d_{x^{2} - y^{2}}$ ).
These closely-spaced transitions overlap → one broad, asymmetric/split absorption band. That is exactly why aqueous Cu²⁺ shows a wide pale-blue band rather than a sharp line — a direct fingerprint of the JT split.

Recall Solution L5·Q2

Octahedral $d^{4}$ (HS): uneven $e_{g}^{1}$ , and $e_{g}$ lobes point at ligands → strong JT.
Tetrahedral: the labels flip — the lower set is $e$ (two orbitals), the upper is $t_{2}$ (three). Crucially, in a tetrahedron no $d$ -orbital points straight at a ligand (ligands sit in "gaps"), and the overall splitting $Δ_{t} \approx \frac{4}{9} Δ_{o}$ is much smaller.
So even when a degenerate tetrahedral set is uneven, the coupling to bond length is weak and the energy gain is small → weak, generally unobservable distortion. Conclusion: same electron count, but geometry decides whether the imbalance can "pay" for a visible distortion.

Recall Solution L5·Q3

"Strong JT" ⇔ odd $e_{g}$ (imbalance 1). High-spin $e_{g}$ populations:

$d^{4}$ : $t_{2 g}^{3} e_{g}^{1}$ → odd ✔
$d^{9}$ : $t_{2 g}^{6} e_{g}^{3}$ → odd ✔
(all others high-spin have $e_{g}$ = 0, 2, or 4 → even ✘) Answer: high-spin $d^{4}$ and $d^{9}$ . Shared feature: the $e_{g}$ set — whose lobes aim straight at ligands — holds an odd electron, forcing an uneven, strongly-coupled distortion (elongation ⇒ 4 short + 2 long).

Connections

Crystal Field Theory — supplies the $t_{2 g} / e_{g}$ split these exercises perturb.
Octahedral Splitting and $\Delta_o$ — the baseline diagram before distortion.
High-spin vs Low-spin Complexes — sets the $e_{g}$ population (odd vs even).
Colour of Transition Metal Complexes — L5·Q1's broad Cu²⁺ band.
Stability and Distortion in $d^9$ Cu(II) — the flagship strong-JT ion.
Tetrahedral vs Octahedral Geometry — why L5·Q2's tetrahedral case stays weak.

Recall Self-test map

Trigger for strong JT ::: odd number of electrons in the $e_{g}$ set (1 or 3). What is $δ_{1}$ ::: the total energy gap opened inside the $e_{g}$ set by the distortion. What is $δ_{2}$ ::: the total energy gap opened inside the $t_{2 g}$ set; $δ_{2} ≪ δ_{1}$ . $E_{JT}$ for $d^{4}$ HS from $e_{g}$ ::: $+ δ_{1} /2$ . $E_{JT}$ for $d^{9}$ from $e_{g}$ ::: $+ δ_{1} /2$ (2 dropped, 1 risen). Strong-JT high-spin counts ::: $d^{4}$ and $d^{9}$ . Why $t_{2 g}$ imbalance is weak ::: its lobes point between ligands → small $δ_{2}$ → weak coupling to bond length.