According to CFT, the degenerate d-orbitals split into $eg$ and $t_2 g$ groups. Are these the outermost nd orbitals which are vacant(4-d in case of 4th period) or the penultimate filled (n-1)d orbitals?

If they are the (n-1)d orbitals, how can they be degenerate if they are not completely filled? If they are (vacant) nd orbitals, how do they split into $eg$ and $t_2 g$ groups if they have no electrons to be repelled by the incoming ligands (CFT is based purely on electrostatic forces)?

  • $\begingroup$ All d orbitals. E.g. Ti$^{3+}$(aq) has only one occupied d orbital, but still splits. CFT is an approximation of Hamiltonian. It is similar with the degeneracy of hydrogen atom. zh.scribd.com/doc/188210540/… Eq. (18) around. $\endgroup$
    – user26143
    Commented Dec 21, 2013 at 21:06

1 Answer 1


The $d$ orbitals that split are the $(n-1)$ $d$-orbitals because they are the relevant ones that do chemistry (because they have the electrons).

If the orbitals are degenerate, it doesn't matter whether they're filled or not, the energy level is the same for all the degenerate states. So it also doesn't matter if they're filled or not, the energy states will split up according to CFT.

The compound only "feels" this raising or lowering of energy depending on how many electrons are filled in into the different orbitals after splitting.

To maybe clarify some confusion: The existence of an orbital doesn't mean that the electron has to be in there, it is only a mathematical construct.

  • $\begingroup$ If (n-1)d orbitals have electrons, how can the ligands fill it up with its own electrons? According to VBT, vacant d-orbitals(hybrid?) are needed. Does CFT use hybrid orbitals like VBT? $\endgroup$
    – user80551
    Commented Dec 22, 2013 at 15:52
  • $\begingroup$ Using MO-LCAO you see that you mix the ligand group orbitals (LGO's) with the $d$ orbitals with the same symmetry. This gives you twice as many orbitals, where the ligand electrons go into the strongly stabilized ones and the metal electrons go into the destabilized ones (which in turn is of the form that we know from CFT). $\endgroup$
    – tschoppi
    Commented Dec 22, 2013 at 21:59
  • $\begingroup$ In CFT, the ligands are approximated as external fields, without no electron in its description. $\endgroup$
    – user26143
    Commented Dec 23, 2013 at 1:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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