An octahedral complex of a metal ion has a $d^4$ configuration. The crystal‑field splitting energy is ${\Delta }_o=18,000{\text{cm}}^{-1}$ and the pairing energy is $P=11,000{\text{cm}}^{-1}$. Calculate the crystal‑field stabilization energy (CFSE) for the low‑spin configuration and decide whether the low‑spin or high‑spin state is favoured.

Step-by-step solution

1. Write the electron arrangement for low‑spin $d^4$: $t_{2g}^4{e}_g^0$.
2. In this arrangement there are two paired electrons, so the pairing energy contribution is $2P$.
3. Calculate the crystal‑field term: $4\times (-0.4{\Delta }_o)=-1.6{\Delta }_o$.
4. Insert the numerical values: $-1.6\times 18,000=-28,800{\text{cm}}^{-1}$.
5. Add the pairing‑energy cost: $-28,800+2\times 11,000=-6,800{\text{cm}}^{-1}$.
6. For the high‑spin $d^4$ configuration ($t_{2g}^3{e}_g^1$) there is no pairing energy.
7. Calculate its crystal‑field term: $3(-0.4{\Delta }_o)+1(0.6{\Delta }_o)=-0.6{\Delta }_o$.
8. Numerically: $-0.6\times 18,000=-10,800{\text{cm}}^{-1}$.
9. Compare the two CFSE values: $-10,800{\text{cm}}^{-1}$ (high‑spin) is more negative than $-6,800{\text{cm}}^{-1}$ (low‑spin).
10. Therefore the high‑spin configuration is energetically favoured.

Concept tested: CFSE calculation and spin‑state preference

Published 14 September 2026 · How this solution was prepared and checked