Jahn-Teller Effect | Coordination Chemistry Class 12

Jahn-Teller Effect (JTE)

Distortion in Octahedral Complexes | Coordination Chemistry

1. The Jahn-Teller Theorem (1937)

Statement: "Any non-linear molecular system in a degenerate electronic state will be unstable and will undergo some sort of distortion to lower its symmetry and remove the degeneracy."

Simply put: If the d-orbitals ($t_{2g}$ or $e_g$) are unsymmetrically filled, the regular octahedron will distort (stretch or compress) to become more stable.

2. When does JTE occur?

The extent of distortion depends on the electronic configuration of the metal ion.

A. Asymmetrical Filling (Distortion Occurs)

• $t_{2g}$ Asymmetry (Weak Distortion): Since $t_{2g}$ orbitals point *between* ligands, unequal filling causes only slight repulsion changes.
  Examples: $d^1, d^2, d^4$ (Low Spin), $d^5$ (Low Spin).
• $e_g$ Asymmetry (Strong Distortion): Since $e_g$ orbitals point *directly at* ligands, unequal filling causes massive repulsion differences.
  Examples: $d^4$ (High Spin), $d^9$, $d^7$ (Low Spin).

B. Symmetrical Filling (No Distortion)

If orbitals are empty, half-filled, or fully filled, the octahedron remains regular.

Examples: $d^3, d^5$ (High Spin), $d^6$ (Low Spin), $d^8, d^{10}$.

3. Types of Distortion (Tetragonal)

To remove degeneracy in the $e_g$ level ($d_{x^2-y^2}, d_{z^2}$), the bonds along the Z-axis change length relative to X and Y.

Z-out (Tetragonal Elongation)

• Most Common (Observed in $Cu^{2+}$).
• The two ligands on the Z-axis move away.
• $Z$-bonds > $X,Y$-bonds.
• Stabilization: The orbital with a Z-component ($d_{z^2}$) lowers in energy because ligands are further away.

Z-in (Tetragonal Compression)

• Rare.
• The two ligands on the Z-axis move closer.
• $Z$-bonds < $X,Y$-bonds.
• The orbital with a Z-component ($d_{z^2}$) rises in energy due to increased repulsion.

4. Case Study: Copper(II) $d^9$

Config: $t_{2g}^6 e_g^3$.

The $e_g$ level contains 3 electrons. This is asymmetric (one orbital has 2, the other has 1).

• Without JTE: $d_{x^2-y^2}$ and $d_{z^2}$ are degenerate.
• With Z-out JTE: The system lowers energy by placing the paired electrons in the lower energy $d_{z^2}$ orbital and the single electron in the higher energy $d_{x^2-y^2}$ orbital.
• Result: $[Cu(H_2O)_6]^{2+}$ is not a perfect octahedron; it has 4 short equatorial bonds and 2 long axial bonds.

Practice Quiz

Test your ability to predict JTE distortions.

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