In the crystal field theory (CFT), when the splitting of the d-orbital occurs, it gets divided into two parts. The upper part with higher energy is the $\mathrm{t_{2g}}$ and the lower part with lower energy is called the $\mathrm{e_g}$ as in:

t-2g and e-g

So can anyone please explain me what is this $\mathrm{t_2}$ or the $\mathrm{e}$ or the $\mathrm{_g}$ below.


As stated in both the links Geoff and Philipp have kindly commented (1, 2) they are to do with symmetry labels we chemists like to assign to orbitals. Knowing an orbitals symmetry class can lead to a lot of simplifications down the road when you use quantum mechanical calculations and even dictate the reactivity of which orbitals are "allowed" to interact together in reactions.

In this case the $\mathrm{t_2}$ groups three of the metal atom's d-orbitals into a certain class while two of the orbitals belong to the $\mathrm{e}$ class. The $\mathrm{t}$ means triply degenerate while the $\mathrm{e}$ means doubly degenerate (degenerate means have the same energy).

The $\mathrm{g}$ is not about how many energy levels are degenerate rather it is an indication of the answer to a certain operation we can perform on an orbital. It instead relates to how the orbitals behave if we hypothetically were to put a line bisecting the orbitals or "invert" them around a centre point (crudely put "going from one corner to the other". If the phase of the wavefunction (or perhaps more pragmatic) the "colour" of the orbital lobes changes while doing this we label the orbital with a $\mathrm{u}$ if we have no phase change we label it with $\mathrm{g}$. They indicate the orbitals response to an inversion about the centre of symmetry.

We can do lots more of these operations to investigate the symmetry of the orbitals which together lead us to assigning the molecule to a symmetry group. Knowing this is very important for a plethora of reasons.

| improve this answer | |
  • 16
    $\begingroup$ g and u come from the German words gerade and ungerade, meaning even and uneven, respectively. $\endgroup$ – tschoppi Oct 9 '14 at 20:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.