For the octahedral complex $[Co{F}_6{]}^{3-}$ (Co${}^{3+}$, $d^6$), the crystal‑field splitting energy ${\Delta }_o$ is $18,000{\text{cm}}^{-1}$ and the electron‑pairing energy $P$ is $12,000{\text{cm}}^{-1}$. Calculate the crystal‑field stabilization energy (CFSE) for both the high‑spin and low‑spin configurations and determine which spin state is favoured.

Step-by-step solution

1. 1. Determine electron distribution: high‑spin $t_{2g}^4{e}_g^2$ (4 unpaired), low‑spin $t_{2g}^6{e}_g^0$ (0 unpaired).
2. 2. Compute CFSE: for high‑spin, $4(-0.4{\Delta }_o)+2(+0.6{\Delta }_o)=(-1.6+1.2){\Delta }_o=-0.4{\Delta }_o$. Plug Δ_o=18 000 ⇒ –7 200 cm⁻¹. (Note: use correct coefficients.)
3. 3. For low‑spin, $6(-0.4{\Delta }_o)=-2.4{\Delta }_o$ ⇒ –43 200 cm⁻¹. Add pairing energy for 3 pairs: $3P=36 000$ cm⁻¹, total CFSE = –43 200 – 36 000 = –49 200 cm⁻¹. (Corrected numbers.)
4. 4. Compare: –49 200 cm⁻¹ < –7 200 cm⁻¹, so low‑spin is energetically favoured.

Concept tested: Crystal‑field stabilization energy calculation

Published 10 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked