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
Labeling the orbitals: when the symmetry drops below Oh labeling the d orbitals as t2g and eg is no longer appropriate or correct. Identifying the correct point group and then using the corresponding character table quickly gives the correct irreducible representations to label the orbitals.
A second way to do this is to use a Correlation Table, which shows the connection between the labels of various point groups.
a1g + b1g
a1 + b1
a1 + b1
b2g + eg
b2 + e
b2 + e
a1 + e
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
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?
either, (nearly always is elongation, often to CN = 4)
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.