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 Oh 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 d1 yes compression d2 yes elongation d3 no d4 (hs) yes either d4 (ls) yes compression d5 (hs) no d5 (ls) yes elongation d6 (hs) yes compression d6 (ls) no d7 (hs) yes elongation d7 (ls) yes either d8 no d9 yes either, (nearly always is elongation to CN = 4) d10 no

Tetrahedral Complexes

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

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

The splitting in Td complexes is always less than the splitting in Oh complexes with the same ligands (Δt < Δo). (Fewer ligands give a smaller electrostatic field; in the exact ionic limit .)

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