Jahn-Teller distortion: Visual walkthrough — Chemistry deep dive

We only need one physical fact to start: like charges repel. A d-electron is negative. A ligand's lone pair is negative. So an electron cloud that points straight at a ligand feels a big shove; one that points between ligands feels a small shove. Hold that thought — everything follows from it.

Step 1 — Draw the octahedron and name the axes

WHAT. We place a metal ion at the origin and six ligands (the electron-donating groups) at fixed positions: two on the $z$ -axis (top and bottom, called axial), and four in the $x y$ -plane (called equatorial).

WHY. Before any orbital talk, we must agree where things are. The words "axial" and "equatorial" and the labels $x, y, z$ are the coordinate frame every later step points back to. Pick them wrong and every sign flips.

PICTURE.

Step 2 — Sort the five d-orbitals by where they point

WHAT. A d-orbital is just a region of space where an electron is likely to be. There are five of them. We split them into two families by a single yes/no question: does this orbital point at a ligand?

WHY. From Step 1's physical fact, orbitals that point at ligands cost more energy (strong repulsion) than orbitals that point between them. So this one question already sorts the five orbitals into a high group and a low group — this is exactly the crystal-field splitting.

PICTURE.

The gap between these two groups is written $Δ_{o}$ (see Octahedral Splitting and $\Delta_o$). Within each group, the orbitals are so far exactly equal in energy — this equality is called degeneracy.

Step 3 — The uncomfortable case: a degenerate group filled unevenly

WHAT. Suppose we must place 3 electrons into the two-member $e_{g}$ group. Two orbitals, three electrons → one orbital is doubly filled, the other singly filled. The group is degenerate but occupied unevenly.

WHY. This is the only trigger for Jahn–Teller. If a degenerate group is filled evenly (like $t_{2 g}^{3}$ : one electron in each) there is no preferred direction and nothing happens. Uneven filling secretly picks a direction — and picking a direction is exactly what "distortion" means.

PICTURE.

Recall

Why does an evenly filled degenerate set not distort? ::: Both orbitals hold the same number of electrons, so lowering one and raising the other cancels out — no net energy is gained.

Step 4 — Distort: pull the axial ligands away (elongation)

WHAT. We move the two $z$ -axis ligands slightly outward. The $M$ – $L$ bond length along $z$ grows; the four equatorial bonds stay put (or shrink a touch).

WHY. By Step 1's physical fact, any orbital that points along $z$ now feels less repulsion (its ligands ran away) → it drops in energy. Any orbital living in the $x y$ -plane feels relatively more repulsion → it rises. So the distortion is a lever that pushes some orbitals down and others up. We chose the $z$ -axis, but by symmetry any axis would do — the molecule just picks one.

PICTURE.

Step 5 — Split the $e_{g}$ set (the strong one)

WHAT. The two $e_{g}$ orbitals were on one line. Elongation splits them: $d_{z^{2}}$ (lobes along $z$ ) drops, $d_{x^{2} - y^{2}}$ (lobes in the $x y$ -plane) rises.

WHY this split, and this amount. The total energy of the group cannot appear from nowhere — a perturbation only redistributes energy. So the average (the barycentre) stays fixed: whatever one orbital gains, the other loses. With two orbitals splitting symmetrically, each moves by half the total split $δ_{1}$ .

PICTURE.

Because $e_{g}$ points straight at ligands, $δ_{1}$ is large — this is why $e_{g}$ unevenness gives the strong, visible distortion.

Step 6 — Split the $t_{2 g}$ set (the weak one, with barycentre bookkeeping)

WHAT. The three $t_{2 g}$ orbitals also split. Two of them ( $d_{x z}, d_{y z}$ ) have a $z$ -component → they drop. One ( $d_{x y}$ ) is purely in-plane → it rises.

WHY the uneven split here. Now it is two down, one up. Barycentre still demands the signed total be zero. If the two down-movers each fall by $\frac{δ _{2}}{3}$ , the single up-mover must rise by $\frac{2 δ _{2}}{3}$ to balance. Weighting by how many orbitals move is the key subtlety.

PICTURE.

Because $t_{2 g}$ points between ligands, $δ_{2}$ is tiny → the $t_{2 g}$ distortion is usually unobservable.

Step 7 — Add it up: the Jahn–Teller stabilization energy

WHAT. For any electron configuration we now total the payoff: every electron in a lowered orbital earns its drop; every electron in a raised orbital pays its rise.

WHY. The molecule distorts only if it saves energy overall. Summing "electrons × how far they moved" gives that net saving. If it's positive, distort; if zero, stay symmetric.

PICTURE.

Step 8 — Worked cases (all d-counts covered)

WHAT. Plug real configurations into Step 7's formula. We include the distorting cases and the flat cases so no scenario is left unshown.

WHY. A formula you can't run on examples is decoration. These are the cases you'll be tested on.

The one-picture summary

Read it left to right: symmetric octahedron → uneven degenerate filling → elongate along $z$ → $e_{g}$ splits (big), $t_{2 g}$ splits (tiny), barycentres held → count electrons into lowered vs raised → $E_{JT} > 0$ ⇒ the distortion is real.

Recall Feynman retelling of the whole walkthrough

Six marbles sit around a metal ball in a perfect cage (Step 1). The ball's electron pockets come in two flavours: pockets that stare right at a marble (expensive, the $e_{g}$ family) and pockets that peek between marbles (cheap, the $t_{2 g}$ family) — Step 2. Trouble starts when a set of equal pockets gets filled lopsidedly, like two people crammed onto one of two identical benches (Step 3). The ball fixes this by sliding the top and bottom marbles outward (Step 4). Now the up-down pockets relax and drop; the sideways pockets tighten and rise — and because energy is only ever shuffled, the drops and rises exactly balance around the middle (Steps 5–6). Finally we tally: electrons in the dropped pockets bank energy, electrons in the raised pockets spend it (Step 7). If the bank comes out ahead — which happens whenever that expensive $e_{g}$ family is filled unevenly, as in copper's $d^{9}$ — the stretch is worth it and stays (Step 8). Even filling? The ledger reads zero and the cage stays perfectly square.

Connections

Crystal Field Theory — supplies the $t_{2 g} / e_{g}$ split we distort.
Octahedral Splitting and $\Delta_o$ — the $Δ_{o}$ gap of Step 2.
High-spin vs Low-spin Complexes — sets the $e_{g}$ occupation that triggers Step 3.
Colour of Transition Metal Complexes — the extra split adds absorption bands.
Stability and Distortion in $d^9$ Cu(II) — the Step 8 flagship case.
Tetrahedral vs Octahedral Geometry — where distortions are strong vs weak.