Question

What is spin pairing energy (SPE)? How is it related to crystal field splitting energy? Relate the splitting energy for octahedral and tetrahedral complexes and briefly explain why tetrahedral complexes are almost always high spin.

Answer 1

Energy required to place two electrons in a orbital is SPE. It is endothermic in nature i.e., energy is to be supplied for this process.

If this energy can be taken from crystal field stabilization energy then spin pairing occurs. So, crystal field splitting energy have to be greater than the spin pairing energy. Hence pairing energy becomes additive with splitting energy. As an example of d^{​​​​​​4} electronic system for octahedral geometry :

High spin d^{​​​​​​4} : CFSE = {3*(2/5) - 1*(3/5)}∆_o = (3/5)∆_o

For high spin d^{​​​​​​4}​​​​​ , 3 electrons are at t_{​​​​​​2g} level and one at e_{​​​​​​g} level.

Low spin d^{​​​​​​4 :} CFSE = {4*(2/5)}∆_o - P where P is pairing energy

For low spin d^{​​​​​​4}​​​​​ , all 4 electrons are at t_{​​​​​​2g} level with one paired electrons.

Crystal field splitting energy is higher in octahedral geometry due to two factors:

I) no. Of ligands are lower in tetrahedral geometry by a factor of 2/3

II) Van Vleck operator which includes inter electronic interaction due to covalency lowers tetrahedral crystal field energy by a factor of 2/3

So, overall reduces the tetrahedral crystal field energy by a factor of 4/9 i.e., ∆_t = (4/9)∆_o

Here we see that tetrahedral crystal field stabilization energy is very less compared to octahedral and most of the cases (almost all) spin pairing energy is higher. Hence spin pairing is not possible and complexes become high spin .