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

Consider a tetragonal case:

Need to introduce additional parameters, δ_{1} and δ_{2}

The situation is such that E(x^{2}–y^{2}) – 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(x

^{2}–y^{2}) + δ_{2}] – [E(xy) + 2δ_{1}] = 10Dq[E(x

^{2}–y^{2}) – E(xy)] + δ_{2}- 2δ_{1}= 10Dq10Dq + δ

_{2}-2δ_{1}= 10 Dqδ

_{2}= 2δ_{1}

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 d^{1}

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 – 2δ_{1}

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.