# Crystal Field Splitting of d-Orbitals in Octahedral and Tetrahedral Ligand Fields

We're learning about the Crystal Field Theory (CFT, under coordination chemistry) at school now, and my teacher mentioned that the 5 d-orbitals split into orbital sets, namely the $\mathrm{t_{2g}}$ (triply degenerate-gerade) and the $\mathrm{e_{g}}$ (evenly degenerate-gerade) for an octahedral ligand field; whereas for the tetrahedral ligand field, the notation is the same except that the subscript 'g' is removed, indicating it is ungerade (That bit was my interpretation … I think I'm wrong though)

Now a lot of the books and sites I've looked up mention the same gerade notation for both octahedral and tetrahedral ligand fields.

So naturally my question would be "Which one's right and which one's wrong?" and a reason if possible.

• A lack of a subscripted g doesn’t mean ungerade, it means that gerade/ungerade cannot be established due to the lack of the corresponding symmetry element (centre of symmetry, if I’m not mistaken) in the compound as a whole. – Jan Aug 22 '16 at 15:42
• @Jan - yes, the lack of the 'g' in tetrahedral point groups is exactly because there is no inversion center in the compound as a whole. – Geoff Hutchison Aug 22 '16 at 17:03

The labels you are talking about refer to the orbitals’ irreducible representations. Which irreducible representations can occur is dictated by the point group of the entire compound. For octahedral compounds, the point group is the very symmetric $O_\mathrm{h}$, which basically includes every element of symmetry you can have in a cube. Tetrahedral complexes belong to the $T_\mathrm{d}$ point group with a considerably lower symmetry (i.e. a smaller number of symmetry operations).
Octahedral complexes are probably more present in the average non-coordination chemist’s brain. Since $O_\mathrm{h}$ includes a centre of symmetry, all irreducible representations have a subscripted g or u to indicate whether equal phases of an orbital are mapped onto each other upon inversion (g for gerade) or whether phases are mapped onto their opposites upon inversion (u for ungerade). Due to the d-orbitals having two nodal planes, they are automatically g, which is why they as a whole transform as $\mathrm{e_g + t_{2g}}$. Since, as I said, this is the more prevalent case in general, this is what a lot of chemists have in mind.
The tetrahedral point group $T_\mathrm{d}$ does not contain a centre of symmetry. This is easily seen when looking at a tetrahedron, which is not inversion-symmetric. Thus, the irreducible representations do not gain a subscripted g and u. The d-orbitals thus transform as $\mathrm{e + t_2}$. Adding a g to these is wrong, but common (see above).
Thus, you were correct in assuming: ‘just because everyone does it does not mean it is correct.’ The same thing can be said for $\ce{S=O}$ double bonds in sulfate.