Octahedral Crystal Field Splitting Diagram

How are the $\mathrm{e_g}$ orbitals degenerate with each other?

Note: This isn't a homework question. After the semester ended (I don't go to MIT), I ended up on MIT open course-ware to watch some videos about areas of chemistry I haven't covered yet or haven't covered well. I am asking the question here because I have no other avenue in which to ask questions to other people. I'm just trying to use my time during the pandemic to build my knowledge of chemistry. I don't have a lot of understanding about this, I'm just looking for some help


2 Answers 2


It is because perfect octahedral symmetry is normally assumed; the two $\mathrm{e_g}$ levels and three $\mathrm{t_{2g}}$ are degenerate. If there was a distortion, say by lengthening both $z$-axis ligand positions then the $\mathrm{e_g}$ degeneracy would be removed as the $\mathrm d_{z^2}$ becomes more stable than $\mathrm d_{x^2-y^2}$. This happens because the $z$-axis ligand has more effect on $\mathrm d_{z^2}$ than on $\mathrm d_{x^2-y^2}$ orbitals simply due to its position. The $\mathrm d_{xy}$ orbital also increases in energy removing the degeneracy of the $\mathrm{t_{2g}}$ and the $\mathrm d_{yz},$ $\mathrm d_{zx}$ lowered.


The crystal field splitting is based on where the ligands (modelled as point charges) are in relation to the orbitals. In an octahedral complex, the ligands are all at 90° from each other and are placed on each of the $x,$ $y,$ $z$ axes. The orbitals that lie on these axes will experience the most repulsion and will rise in energy, while the orbitals between the axes $(\mathrm{t_{2g}})$ will lower in energy as they experience less repulsion from the ligands, and the average overall energy is maintained.

I can only assume that the degree by which the $\mathrm{e_g}$ orbitals are raised is the same due to the $\mathrm d_{z^2}$ orbital technically being a linear combination of what would have been the $\mathrm d_{z^2-x^2}$ and $\mathrm d_{z^2-y^2}$ orbitals. This means that the $\mathrm{e_g}$ orbitals lie on the axes to the same extent as each other, so experience the same overall repulsion from the ligands.


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