For the octahedral complex $[Cr(N{H}_3{)}_6{]}^{3+}$ (Cr${}^{3+}$, $d^3$), the octahedral crystal‑field splitting energy is ${\Delta }_o=15,000;{\text{cm}}^{-1}$ and the electron‑pairing energy is $P=8,000;{\text{cm}}^{-1}$. 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. For a $d^3$ ion, the electronic configuration in octahedral field is $t_{2g}^3{e}_g^0$ for both high‑spin and low‑spin.
2. Compute CFSE: $(n_{t_{2g}}\times (-0.4)+n_{e_g}\times (+0.6)){\Delta }_o=(3\times -0.4+0\times 0.6){\Delta }_o=-1.2{\Delta }_o$.
3. Insert ${\Delta }_o=15,000;{\text{cm}}^{-1}$: $-1.2\times 15,000=-18,000;{\text{cm}}^{-1}$? Wait: $-0.4\times 3=-1.2$ so $-1.2\times 15,000=-18,000$.
4. Correct calculation: CFSE = $-1.2{\Delta }_o=-1.2\times 15,000=-18,000;{\text{cm}}^{-1}$. This is the same for both spin states because no electrons are paired.
5. Thus both high‑spin and low‑spin CFSE are $-18,000;{\text{cm}}^{-1}$; no energetic preference for one spin state.

Concept tested: CFSE calculation for d3 complex

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