Consider the octahedral complex $[Fe(CN{)}_6{]}^{4-}$ (Fe${}^{2+}$, $d^6$). Given the octahedral crystal‑field splitting energy ${\Delta }_o=21,000;{\text{cm}}^{-1}$ and the electron‑pairing energy $P=12,000;{\text{cm}}^{-1}$, calculate the crystal‑field stabilization energy (CFSE) for the high‑spin and low‑spin configurations. Which spin state is favoured?

Step-by-step solution

1. High‑spin ($t_{2g}^4{e}_g^2$): CFSE $=(4\times -0.4+2\times +0.6){\Delta }_o=(-1.6+1.2){\Delta }_o=-0.4{\Delta }_o$.
2. Insert ${\Delta }_o$: $-0.4\times 21,000=-8,400;{\text{cm}}^{-1}$.
3. Low‑spin ($t_{2g}^6{e}_g^0$): CFSE $=(6\times -0.4){\Delta }_o=-2.4{\Delta }_o$.
4. Insert ${\Delta }_o$: $-2.4\times 21,000=-50,400;{\text{cm}}^{-1}$.
5. Compare total energies including pairing: high‑spin has 2 paired electrons → add $2P=24,000$. Low‑spin has 3 paired electrons → add $3P=36,000$.
6. Total energy high‑spin: $-8,400+24,000=15,600$.
7. Total energy low‑spin: $-50,400+36,000=-14,400$.
8. Since $-14,400<15,600$, the low‑spin state is lower in energy and hence favoured.

Concept tested: CFSE and spin state selection

Published 11 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked