The octahedral complex $[Co(N{H}_3{)}_6{]}^{3+}$ is a $d^6$ system. Given the octahedral crystal‑field splitting energy ${\Delta }_o=22,000{\text{cm}}^{-1}$ and the electron‑pairing energy $P=15,000{\text{cm}}^{-1}$, calculate the crystal‑field stabilization energy (CFSE) for both the high‑spin and low‑spin configurations and indicate which configuration is favoured.

Step-by-step solution

1. Determine the electron distribution for a $d^6$ ion in an octahedral field. - High‑spin: $t_{2g}^4{e}_g^2$ (four electrons in $t_{2g}$, two in $e_g$). - Low‑spin: $t_{2g}^6{e}_g^0$ (all six electrons in $t_{2g}$).
2. Calculate CFSE for the high‑spin case: CFSE = $(4\times -0.4{\Delta }_o)+(2\times +0.6{\Delta }_o)=-0.4{\Delta }_o$.
3. Insert the numerical value: $-0.4\times 22,000=-8,800{\text{cm}}^{-1}$.
4. For the low‑spin case, CFSE = $6\times -0.4{\Delta }_o=-2.4{\Delta }_o=-2.4\times 22,000=-52,800{\text{cm}}^{-1}$.
5. Account for pairing energy. High‑spin $d^6$ has two paired electrons (two $P$), low‑spin has three paired electrons (three $P$). The extra pairing cost for low‑spin is $P=15,000{\text{cm}}^{-1}$.
6. Add this cost to the low‑spin CFSE: $-52,800+15,000=-37,800{\text{cm}}^{-1}$.
7. Compare the total energies: $-8,800{\text{cm}}^{-1}$ (high‑spin) vs $-37,800{\text{cm}}^{-1}$ (low‑spin). The more negative value corresponds to greater stabilization, so the low‑spin configuration is favoured.

Concept tested: CFSE calculation and spin‑state prediction for octahedral $d^{6}$ complexes

Published 6 September 2026 · How this solution was prepared and checked