Calculate the crystal field stabilization energy (CFSE) of the octahedral complex $[Fe(H_{2}O{)}_6{]}^{2+}$ given that the octahedral splitting parameter ${\Delta }_o=12000{\text{cm}}^{-1}$. Assume a high‑spin $d^6$ configuration. (Use $1{\text{cm}}^{-1}=0.01196{\text{kJmol}}^{-1}$.)

Step-by-step solution

1. Determine the oxidation state of Fe: $+2$, giving a $d^6$ configuration.
2. For a high‑spin $d^6$ octahedral ion the electron distribution is $t_{2g}^4{e}_g^2$.
3. CFSE $=(-0.4\times n_{t_{2g}}+0.6\times n_{e_g}){\Delta }_o$.
4. Substitute $n_{t_{2g}}=4$ and $n_{e_g}=2$: CFSE $=(-0.4\times 4+0.6\times 2){\Delta }_o=(-1.6+1.2){\Delta }_o=-0.4{\Delta }_o$.
5. Convert ${\Delta }_o$ to kJ mol${}^{-1}$: $12000{\text{cm}}^{-1}\times 0.01196=143.52{\text{kJmol}}^{-1}$.
6. CFSE $=-0.4\times 143.52=-57.408{\text{kJmol}}^{-1}$, rounded to $-57.4{\text{kJmol}}^{-1}$.

Concept tested: CFSE calculation for high‑spin octahedral complexes

Published 8 September 2026 · How this solution was prepared and checked