Jahn–Teller Distortion: When It Happens and Why
An octahedral complex looks perfectly symmetric on paper, but many real complexes are not. [Cu(H2O)6]2+ has four short Cu–O bonds and two long ones. The reason is the Jahn–Teller effect, and it is a favourite topic in inorganic chemistry for IIT-JAM, GATE, CSIR-NET and CUET-PG. This article gives you a method to decide, for any complex, whether to expect a strong distortion, a weak one, or none.
The Jahn–Teller theorem
The theorem states that a non-linear molecule in an electronically degenerate ground state is unstable. It will distort to a lower-symmetry shape that removes the degeneracy and lowers the energy. In practice, for octahedral complexes it means: if the d-electron configuration leaves the d orbitals unevenly filled in a degenerate set, the complex distorts.
The theorem tells you that a distortion will occur. It does not tell you its size or direction. That needs extra reasoning, given below.
Step 1: Count d electrons and find the configuration
In an octahedral field the d orbitals split into t2g (dxy, dxz, dyz; lower) and eg (dz², dx²−y²; higher). Fill the electrons using the ligand field (weak field = high spin, strong field = low spin).
Step 2: Look for an uneven degenerate set
| Configuration | Uneven set | Jahn–Teller effect | Typical example |
|---|---|---|---|
| d1 (t2g1) | t2g | Weak | Ti(III) |
| d2 (t2g2) | t2g | Weak | V(III) |
| d3 (t2g3) | None (half-filled t2g) | None | Cr(III) |
| d4 high spin (t2g3 eg1) | eg | Strong | Cr(II), Mn(III) |
| d4 low spin (t2g4) | t2g | Weak | strong-field Mn(III) |
| d5 high spin (t2g3 eg2) | None | None | Mn(II), Fe(III) with weak ligands |
| d5 low spin (t2g5) | t2g | Weak | [Fe(CN)6]3− |
| d6 high spin (t2g4 eg2) | t2g | Weak | Fe(II) with weak ligands |
| d6 low spin (t2g6) | None | None | [Co(NH3)6]3+ |
| d7 high spin (t2g5 eg2) | t2g | Weak | Co(II) with weak ligands |
| d7 low spin (t2g6 eg1) | eg | Strong | Ni(III) with strong ligands |
| d8 (t2g6 eg2) | None | None | Ni(II) |
| d9 (t2g6 eg3) | eg | Strong | Cu(II) |
| d10 | None | None | Zn(II) |
The rule of thumb: an uneven eg set (eg with 1 or 3 electrons, and no other filling) gives a strong distortion, because the eg orbitals point straight at the ligands and respond strongly to them. An uneven t2g set gives only a weak distortion, because t2g orbitals point between the ligands. The weak effect is often too small to see in the structure.
Step 3: Elongation or compression?
Take d9. The two eg orbitals hold three electrons: dz² gets 2 and dx²−y² gets 1 (or the reverse).
- Tetragonal elongation (the two axial bonds stretch): dz² drops by δ1/2 and dx²−y² rises by δ1/2. With two electrons in dz² and one in dx²−y², the net energy change is 2(−δ1/2) + 1(+δ1/2) = −δ1/2. The complex is stabilised.
- Tetragonal compression (the two axial bonds shorten): the splitting reverses. dx²−y² drops by δ1/2 and dz² rises by δ1/2. Now the two electrons sit in dx²−y² and one in dz², so the net change is again 2(−δ1/2) + 1(+δ1/2) = −δ1/2. On this simple energy count, elongation and compression stabilise d9 equally.
The filled t2g set (6 electrons) gives no net energy change in either case, because the splitting inside t2g is balanced (shifts of −δ2/3 for two orbitals and +2δ2/3 for one add up to zero when filled). In practice, elongation is observed far more often than compression. Do not overstate why: the common textbook explanation is that elongation weakens the two axial bonds while keeping four strong equatorial bonds, which usually costs less bonding energy than compressing two bonds. Textbooks differ in how fully they explain this, so give the observed trend in your exam answer.
Worked examples
Step 1: Cu(II) is d9.
Step 2: Octahedral d9 is t2g6 eg3. The eg set is unevenly filled (3 electrons in 2 orbitals).
Step 3: Uneven eg means a strong distortion, usually elongation.
Step 4: Extra stabilisation from the distortion = δ1/2 (where δ1 is the splitting of the eg set). The regular octahedral CFSE is 6(−0.4Δo) + 3(+0.6Δo) = −0.6Δo.
Answer: yes, strongly distorted; four short and two long Cu–O bonds are expected.
Step 1: Co(III) is d6. NH3 is a strong-field ligand toward Co(III), so the complex is low spin: t2g6 eg0. No uneven degenerate set. No Jahn–Teller effect.
Step 2: Mn(III) is d4. H2O is a weak-field ligand, so the complex is high spin: t2g3 eg1. The single eg electron is uneven.
Step 3: High-spin d4 gives a strong distortion. The CFSE of regular geometry is 3(−0.4Δo) + 1(+0.6Δo) = −0.6Δo. The one eg electron enters the lowered orbital, giving an extra −δ1/2.
Answer: the Co(III) complex is regular; the Mn(III) complex is strongly Jahn–Teller distorted.
Step 1: High-spin d7 is t2g5 eg2. The eg set is evenly filled (one electron in each orbital), but the t2g set is uneven. Weak effect.
Step 2: Low-spin d7 is t2g6 eg1. The single eg electron is uneven. Strong effect.
Answer: the spin state decides the strength. Low-spin d7 is strongly distorted; high-spin d7 only weakly.
Common mistakes
- Forgetting to decide spin state first. d4, d5, d6 and d7 change their answer between high and low spin.
- Saying "d9 and d4 are the same." They are strong cases for the same reason (eg uneven), but only when d4 is high spin.
- Assuming a distortion needs the ligands to be different. Jahn–Teller distortion occurs even with six identical ligands. It comes from the electrons, not from the ligands.
- Treating distortion as extreme. It lowers symmetry but does not break the complex. The complex remains six-coordinate.
- Claiming compression never happens. Elongation is more common, but compression is known.
- Quoting bond lengths from memory. Do not invent numbers. Say "two longer axial bonds" unless the question gives data.
Exam relevance
| Question type | What to do |
|---|---|
| Which complex shows Jahn–Teller distortion? | Find d count and spin state, then check eg first |
| Predict structure of a Cu(II) or Mn(III) complex | Say tetragonal elongation: four short, two long bonds |
| Link to spectra | Splitting of levels gives broad or split absorption bands |
| CFSE with distortion | Regular CFSE plus the extra stabilisation |
Check the current official syllabus of your exam to see how deep your paper goes. Many papers ask only the identification step.
Use the suite for the electron-count and CFSE arithmetic while you practise these classifications.
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