Why is the pairing energy not considered when calculating the crystal field splitting energy?

Question

Consider the low-spin octahedral crystal field splitting diagram for a d4 system.

When we calculate the crystal field splitting energy we simply observe the absorbed color of light and calculate the frequency of a photon required to promote an electron from the t_2g to the e_g orbitals. Why is the spin pairing energy not considered (PE) in these calculations? If we had a simple d3 system, I would understand why the energy of the photon absorbed is equal to the crystal field splitting energy since we are not dealing with any paired electrons; however, in the case of the d4 system, doesn't "removing" one of the paired electrons actually release energy? Why do we say the energy of the photon absorbed is equal to Δ_o and not Δ_o - PE?

Answer 1

It is actually the two single-paired electrons are more likely to be promoted than the two paired electrons.

Promoting the electrons always cost energy whether the electrons in question are paired or not. In the classical examples of hybridization (where they might actually promote a paired electron), this energy cost is compensated by the formation of stable covalent bonds, but in this case, a complex, no bonds formed here.

So why the single electrons are promoted before the paired ones?

The spin rule in the coordination complexes tell that the allowed d-d transitions must not involve the change in the total spin of the electrons.

In your case of a low-spin d4, promoting the the paired electrons will leave the d-orbitals with single electrons with opposite spin, which is highly unfavorable. The changing of the spin of the opposite-spin electron is forbidden per the spin rule since it will change the total spin of the system.