How does Crystal Field theory change if the complexes are not octahedral?

Consider a tetragonal case:

Need to introduce additional parameters, δ₁ and δ₂

The situation is such that E(x²–y²) – E(xy) = 10Dq (moving the ligand along the z axis should have no effect on the relative energies of the orbitals in the xy plane).

For compression case :

[E(x²–y²) + δ₂] – [E(xy) + 2δ₁] = 10Dq

[E(x²–y²) – E(xy)] + δ₂ - 2δ₁ = 10Dq

10Dq + δ₂ -2δ₁ = 10 Dq

δ₂ = 2δ₁

Can we predict when this will happen? Yes, using the Jahn-Teller theorem

Jahn-Teller Theorem: In a nonlinear molecule a degenerate electronic state will distort to remove the degeneracy and to increase the stability

Consider d¹

In an O_h geometry, the electronic state is triply degenerate (the single electron can be in one of three orbitals of identical energy).

Axial elongation gives a state that is still degenerate (doubly) so would need to further distort.

Axial compression leads to a singly degenerate state and increased stability.

LFSE = –4Dq – 2δ₁

This should occur even if all the ligands are the same!

Which configurations should be J-T active?

configuration	active?	distortion geometry
d^1	yes	compression
d^2	yes	elongation
d^3	no
d^4 (hs)	yes	either
d^4 (ls)	yes	compression
d^5 (hs)	no
d^5 (ls)	yes	elongation
d^6 (hs)	yes	compression
d^6 (ls)	no
d^7 (hs)	yes	elongation
d^7 (ls)	yes	either
d^8	no
d^9	yes	either, (nearly always is elongation to CN = 4)
d^10	no

Tetrahedral Complexes

Tetrahedral symmetry is fairly common but can not be treated as a distortion from O_h

Ligands between axes are destabilized, ligands along axes are stabilized.

The splitting in T_d complexes is always less than the splitting in O_h complexes with the same ligands (Δ_t < Δ_o). (Fewer ligands give a smaller electrostatic field; in the exact ionic limit .)

This means that T_d complexes are always high spin and usually bluer.