Foundations — Jahn-Teller distortion

This page is a toolbox. The parent note on Jahn-Teller distortion throws around symbols like $t_{2g}$, $e_g$, $d_{z^2}$, $\delta$, "degenerate", "barycentre". If any of those are just squiggles to you, start here. We build every one from a picture, in an order where each rung of the ladder rests on the one below it.

1. The metal ion and its ligands — the physical stage

Before any orbital talk, picture the actual object.

The 6 ligands sit at the 6 corners of an octahedron: two on the $x$-axis, two on the $y$-axis, two on the $z$-axis (one up, one down). That is the "perfectly symmetric octahedron" the intuition callout keeps mentioning.

2. Axes and the coordinate frame

We need names for directions, so we set up three perpendicular axes: $x$, $y$, $z$. Think of them as three lines through the metal at right angles, like the edges of a room meeting at a corner.

• $x$ and $y$ span the flat floor (the "equatorial plane" or $xy$-plane).
• $z$ points straight up and down (the "axial" direction).

Why bother? Because "stretch the octahedron along $z$" and "the orbital points along $z$" are the two sentences the whole derivation hinges on. Without axis names those sentences are meaningless.

3. What an orbital is — and its shape

For transition metals the important ones are the five $d$-orbitals. Their shapes are the entire reason Jahn–Teller happens, so look at them carefully.

Two families by where the lobes point:

The subscript is just a label for the direction the cloud reaches:

• $d_{xy}$ ::: lobes in the $xy$-plane, between the $x$ and $y$ axes.
• $d_{z^2}$ ::: lobes along the $z$-axis (with a small doughnut around the middle).
• $d_{x^2-y^2}$ ::: lobes along the $x$ and $y$ axes, straight at those four ligands.

4. Repulsion → energy: the key link

Now connect the two pictures above.

That is the whole engine. "Energy goes up" just means "this arrangement is less stable / more strained". We measure energies so we can compare arrangements and pick the lowest.

5. Splitting into $t_{2g}$ and $e_g$, and the gap ${\Delta }_o$

Because three orbitals point between ligands and two point at them, the five $d$-orbitals split into two groups of equal-energy orbitals. This is the result of Crystal Field Theory, and it is where the parent's $t_{2g}$ and $e_g$ come from.

The labels $t$ and $e$ come from symmetry theory: $t$ tags a group of three equal orbitals, $e$ tags a group of two. The little "$g$" just means the orbital looks the same if you flip it through the centre — you can treat it as decoration for now. The gap ${\Delta }_o$ itself is set by how strongly the ligands push; see Octahedral Splitting and $\Delta_o$ for the full picture. Keep ${\Delta }_o$ clearly separate from the tiny splits inside each group that we meet in §8 — ${\Delta }_o$ is the big gap between the groups; ${\delta }_1,{\delta }_2$ are small splits within a group caused by distortion.

6. Degeneracy — the star word

Everything above builds to this one term, which the parent uses in its very definition.

Why does the topic need this word? Because Jahn–Teller only fires when a degenerate group is filled unevenly. So we must be able to say precisely "these orbitals are at the same height" — that is degenerate — before we can say "and the electrons in them are lopsided".

7. Even vs uneven occupation

Examples using the $e_g$ pair ($d_{z^2}$, $d_{x^2-y^2}$):

• $e_g^0$ = $(0,0)$ → even.
• $e_g^2$ = $(1,1)$ → even (one each).
• $e_g^1$ = $(1,0)$ → uneven → triggers JT.
• $e_g^3$ = $(2,1)$ → uneven → triggers JT (this is copper's case).

8. What the distortion looks like in 3D: elongation vs compression

Now watch the octahedron actually deform along $z$.

9. Symbols ${\delta }_1$, ${\delta }_2$, and the barycentre rule

Once we distort, each orbital's energy shifts by a small amount. The parent writes such a shift with the Greek letter $\delta$ ("delta", meaning "a small change").

Why is ${\delta }_1>{\delta }_2$ (why is the $e_g$ split "strong")? Go back to §4. The $e_g$ orbitals point straight at the ligands, so when a ligand moves even a little, the repulsion on an $e_g$ orbital changes a lot → a big energy shift → large ${\delta }_1$. The $t_{2g}$ orbitals point between the ligands, so moving a ligand barely changes their repulsion → a small energy shift → tiny ${\delta }_2$. Strong coupling to bond length = strong split.

Under an elongation along $z$, the figure shows exactly which orbital goes where:

Notice why $d_{xz},d_{yz}$ drop: both reach along $z$, so they benefit when the $z$-ligands retreat; $d_{xy}$ lies purely in-plane, so it does not benefit and relatively rises.

The barycentre rule.

10. Putting it together — $E_{\text{JT}}$

Now the parent's stabilization formula reads in plain words: add up (electrons that dropped)×(how far they dropped), subtract (electrons that rose)×(how far they rose). If that sum is positive, distorting saved energy, so the molecule does it.

Because a dropped orbital can hold more electrons than the risen one only when the group was unevenly filled, this quantity is positive exactly in the uneven case — which is why uneven occupation is the trigger.

A worked example — $d^9$ (Cu²⁺) in numbers

Take an elongation and put in numbers. Say the $e_g$ split is ${\delta }_1=1.0$ energy unit (so each $e_g$ orbital moves ${\delta }_1/2=0.5$). Config $t_{2g}^6{e}_g^3$: the $t_{2g}$ is full (6 electrons, evenly spread → no $t_{2g}$ contribution), and the $e_g$ holds 3: 2 electrons in the lowered $d_{z^2}$, 1 in the raised $d_{x^2-y^2}$.

$E_{\text{JT}}=\underset{\text{dropped}}{\underbrace{2\times \frac{{\delta }_1}{2}}}-\underset{\text{risen}}{\underbrace{1\times \frac{{\delta }_1}{2}}}=\frac{{\delta }_1}{2}=0.5>0$

Positive → elongation lowers energy → Cu²⁺ distorts. A full walk-through lives in Stability and Distortion in $d^9$ Cu(II).

A weak-JT example — $d^1$ (one $t_{2g}$ electron)

A lone $t_{2g}$ electron ($t_{2g}^1$) is an uneven $t_{2g}$ occupation, so a weak JT is expected. Put the single electron in a lowered $d_{xz}$ (drop ${\delta }_2/3$), nothing risen:

$E_{\text{JT}}=1\times \frac{{\delta }_2}{3}-0=\frac{{\delta }_2}{3}>0$

Still positive, so a distortion is favoured — but because ${\delta }_2\ll {\delta }_1$, the saving is tiny and the geometric distortion is usually too small to see. Uneven $t_{2g}$ occupations (e.g. $t_{2g}^1$, $t_{2g}^2$, $t_{2g}^4$, $t_{2g}^5$) give only these weak effects; the strong textbook distortions all come from uneven $e_g$.

Prerequisite map

Each box is a symbol or idea you now own; together they feed straight into the topic. For the same content in Hinglish, see 3.4.10 Jahn-Teller distortion (Hinglish).

Equipment checklist

Self-test — can you answer each before revealing?