Jahn-Teller distortion

WHAT is it?

The key trigger word is degeneracy that is unevenly occupied.

In an octahedral field the five d-orbitals split into:

== $t_{2g}$ == ( $d_{xy}, d_{yz}, d_{xz}$ ) — lower, point between ligands.
== $e_g$ == ( $d_{z^2}, d_{x^2-y^2}$ ) — higher, point at ligands.

A JT distortion occurs when either set is degenerate but unequally filled.

WHY does it happen?

The strong vs weak rule:

$e_g$ orbitals point directly at the ligands → uneven filling here causes a strong, observable distortion.
$t_{2g}$ orbitals point between ligands → uneven filling causes only a weak, often unobservable distortion.

HOW does the energy actually split? (Derivation from scratch)

Consider an elongation along the $z$ -axis: the two axial ligands move away from the metal.

Step 1 — split the $e_g$ set.

$d_{z^2}$ has its lobes along $z$ → repulsion drops → energy lowers by $\delta_1/2$ .
$d_{x^2-y^2}$ lies in the $xy$ plane → repulsion increases → energy raises by $\delta_1/2$ .

Why this step? Because the perturbation must conserve the centre of energy ("barycentre"): one goes down as much as the other goes up.

Step 2 — split the $t_{2g}$ set.

$d_{xz}, d_{yz}$ (have $z$ component) → lower by $\delta_2/3$ .
$d_{xy}$ (purely in-plane) → raise by $2\delta_2/3$ (keeping barycentre).

Why barycentre weighting? Two orbitals go down, one goes up; the signed total must be zero: $-2(\delta_2/3) + (2\delta_2/3)=0$ . ✓

Step 3 — net stabilization. For a configuration, the JT stabilization energy is the sum of (electrons in lowered orbitals)×(drop) minus (electrons in raised orbitals)×(rise):

WHICH configurations distort?

The distortion is strong when the $e_g$ set is unevenly occupied:

Configuration	$e_g$ occupation	JT?
$d^9$ (Cu²⁺)	$e_g^3$ ( $t_{2g}^6 e_g^3$ )	Strong
high-spin $d^4$ (Cr²⁺, Mn³⁺)	$e_g^1$	Strong
low-spin $d^7$ (Co²⁺ ls, Ni³⁺)	$e_g^1$	Strong
$d^3$ , $d^6$ ls, $d^8$ , $d^{10}$	$e_g$ evenly filled (0,0,2)	None

Weak (usually ignored) when only $t_{2g}$ is uneven: e.g. $d^1, d^2, t_{2g}$ low-spin cases.

Flashcards

Jahn–Teller theorem statement

A non-linear molecule in a degenerate electronic ground state distorts geometrically to remove the degeneracy and lower its energy.

Trigger condition for JT distortion

Uneven (asymmetric) occupation of a degenerate set of orbitals.

Which orbital set gives STRONG distortion and why

e_g

— its lobes point directly at the ligands, coupling strongly to bond lengths.

Which gives only WEAK distortion

t_{2g}

— lobes point between ligands, weak coupling.

Classic strong-JT ion

Cu²⁺ (

d^9

t_{2g}^6 e_g^3

d

counts with strong JT (high spin)

d^9

, high-spin

d^4

, low-spin

d^7

(uneven

e_g

d

counts with NO JT

d^3

d^8

d^{10}

, low-spin

d^6

(evenly filled sets).

Effect on Cu²⁺ octahedron geometry

Elongation along z → 4 short + 2 long M–L bonds.

Why elongation lowers

d_{z^2}

Axial ligands move away → less repulsion on the z-pointing orbital.

Does barycentre stay fixed during splitting

Yes — total energy of a split set is conserved; rises balance drops.

Recall Feynman: explain to a 12-year-old

Picture a metal ball wearing a cage of 6 marbles, perfectly even. The ball has little electron "pockets." If the electrons sit lopsidedly — more on one side than the other — the cage is uncomfortable. So it stretches: two marbles slide a bit farther away. This makes the crowded electrons more relaxed and the whole thing happier (lower energy). That stretching is the Jahn–Teller effect. It happens a lot for copper, which is why copper compounds look "squashed."

Connections

Crystal Field Theory — provides the $t_{2g}/e_g$ splitting JT acts on.
Octahedral Splitting and $\Delta_o$ — the parent energy diagram.
High-spin vs Low-spin Complexes — decides $e_g$ occupation, hence JT.
Colour of Transition Metal Complexes — JT splitting adds extra absorption bands (e.g. broad Cu²⁺ band).
Stability and Distortion in $d^9$ Cu(II) — direct application.
Tetrahedral vs Octahedral Geometry — where distortions are strong vs weak.

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, Jahn-Teller distortion ka basic idea simple hai. Octahedral complex mein metal ke d-orbitals do groups mein split hote hain: niche wale $t_{2g}$ (teen orbital) aur upar wale $e_g$ (do orbital). Ab agar koi degenerate set (jaha orbitals same energy ke hain) mein electrons uneven baithe hain — matlab ek orbital mein zyada, doosre mein kam — to molecule "uncomfortable" feel karta hai. Energy bachane ke liye wo apni shape thodi distort kar leta hai (usually z-axis pe stretch ho jaata hai), jisse degeneracy toot jaati hai aur total energy kam ho jaati hai.

Sabse important rule: strong distortion sirf tab hoti hai jab $e_g$ set uneven ho. Kyun? Kyunki $e_g$ orbitals ( $d_{z^2}$ , $d_{x^2-y^2}$ ) seedha ligands ki taraf point karte hain, to bond length pe inka effect bada hota hai. $t_{2g}$ orbitals ligands ke beech mein point karte hain, isliye unki unevenness se sirf "teeny" weak distortion aati hai jo aksar ignore kar dete hain.

Classic example hai Cu²⁺ ( $d^9$ ): configuration $t_{2g}^6 e_g^3$ hai, to $e_g$ mein 3 electrons hai — ye toh uneven hona hi pad gaya (ek orbital full, ek half). Result: octahedron z-axis pe stretch ho jaata hai, 4 short + 2 long bonds banti hain. Isi tarah high-spin $d^4$ (Mn³⁺, Cr²⁺) aur low-spin $d^7$ bhi strong distortion dikhate hain. Lekin $d^3$ , $d^8$ , $d^{10}$ mein sets evenly filled hote hain, to koi distortion nahi.

Exam tip: pehle d-count likho, fir $t_{2g}/e_g$ filling likho, fir dekho $e_g$ uneven hai ya nahi. Agar haan → strong JT. Yahi 20% concept poore questions ka 80% cover kar deta hai.

WHAT is it?

WHY does it happen?

HOW does the energy actually split? (Derivation from scratch)

WHICH configurations distort?

Common mistakes

Flashcards

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Coordination Chemistry

Connections