# Energy of orbitals under a crystal field

It is shown in octahedral crystal field splitting that both the $e_g$ orbitals have equal energy. But how can that be possible ?

The $d_{x^2-y^2}$ is directed along both the axes (x and y) whereas the $d_{z^2}$ is directed along only the z-axis, shouldn't the later have less energy than the former ?

• You envisioned the orbitals wrong. For example see: chemed.chem.purdue.edu/genchem/topicreview/bp/ch12/… – MaxW Mar 6 '16 at 6:57
• @MaxW I have read what is given in the link, but couldn't find where I am wrong. Would be helpful if you can point out. – Abhirikshma Mar 7 '16 at 15:08

The next complicated answer on the road from over-simplified to reality is that the $\mathrm{d_{z^2}}$ extends to both ligands, too: Do not forget the ‘ring’ in the $xy$ plane. By extending towards the ligands on the $z$ axis further than the $\mathrm{d_{x^2 - y^2}}$ orbitals do to their ligands, it is less stabilised than it looks, and the interaction with the equatorial ligands also destabilises.
The really sophisticated and probably most correct answer is that we generally have a wrong view of those two orbitals. They are mathematically equivalent and would actually form a group of three: $\mathrm{d_{x^2 - y^2}}$, $\mathrm{d_{z^2 - x^2}}$ and $\mathrm{d_{y^2 - z^2}}$ — but because we can only have two, wee need to add two together and commonly take the two that give ‘$\mathrm{d_{z^2}}$’. In some images of molecular orbitals you can see that they are much more identical than they initially look.