# 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

They are equivalent in energy because symmetry says so, and because the calculations give that as an answer. Okay, that wasn’t exactly satisfactory, was it?

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.

• If they 'actually form a group of three' then the d-orbital manifold would be able to hold 12 electrons. These images, are they mathematical representations or empirically determined, such as in doi:10.1038/nature03183? I'm pretty sure even the more complicated molecular orbitals people have imaged look surprisingly like we thought they should. – StevieD Mar 25 '16 at 8:15
• @StevieD No to your first comment, because QC tells us that there is only space for ten. Just because something can mathematically extend in a beautiful way doesn’t mean it does. Partial yes to your second comment. But a professor of mine actually used that explanation and showed us how to look at a specific orbital representation. I have forgotten where to find it so I cannot reproduce it, unfortunately =C. – Jan Mar 30 '16 at 23:08
• Yes Jan, I meant that the fact that they only accommodate 10 is evidence that there can only be 5 subshells of d orbitals. Likewise the ml quantum number agrees. The waveform of these orbitals may be represented as the sum of other waveforms, but these don't have a physical representation....I.e. it's just an artifact of the math. When in a ligand field the orbitals split in such a way that the net sum of the orbital energies are 0, that is they maintain the barycenter. If orbitals go up in energy due to repulsion between ligand orbitals and metal orbitals (spacial overlap) then – StevieD Mar 30 '16 at 23:39
• Others must go down to maintain an overall conservation of energy. – StevieD Mar 30 '16 at 23:44