A transition metal ion is $d^{6}$ in a high‑spin octahedral complex. The octahedral crystal‑field splitting energy $Δ_{o}$ is 18,000 $c m^{- 1}$ and the electron‑pairing energy $P$ is 12,000 $c m^{- 1}$ . Calculate the crystal‑field stabilization energy (CFSE) for the high‑spin and low‑spin possibilities and decide which spin state is favoured.

medium 3 marks Class 12 · Chemistry · d- and f-Block Elements

Answer

Low‑spin complex is favoured; CFSE = –19,200

c m^{- 1}

The correct answer is Low‑spin complex is favoured; CFSE = –19,200 $c m^{- 1}$ — Since –19,200 $c m^{- 1}$ (low‑spin) is more negative (lower energy) than –7,200 $c m^{- 1}$ (high‑spin), the low‑spin arrangement is energetically favoured.

Step-by-step solution

For a high‑spin $d^{6}$ octahedral ion the electron distribution is $t_{2 g}^{4} e_{g}^{2}$ .
CFSE (high‑spin) = (4 $\times$ -0.4 $Δ_{o}$ ) + (2 $\times 0$ .6 $Δ_{o}$ ) = -0.4 $Δ_{o}$ .
Substituting $Δ_{o} = 18, 000$ $c m^{- 1}$ , CFSE = -0.4 $\times$ 18,000 = -7,200 $c m^{- 1}$ .
In a low‑spin $d^{6}$ configuration the electrons pair in the $t_{2 g}$ set: $t_{2 g}^{6} e_{g}^{0}$ .
CFSE (low‑spin) = (6 $\times$ -0.4 $Δ_{o}$ ) = -2.4 $Δ_{o}$ .
Add the pairing energy cost: low‑spin has two extra paired electrons compared with high‑spin, so add 2P.
CFSE (low‑spin) = -2.4 $Δ_{o}$ + 2P = -2.4 $\times 18$ ,000 + 2 $\times 12$ ,000 = -43,200 + 24,000 = -19,200 $c m^{- 1}$ .
Since –19,200 $c m^{- 1}$ (low‑spin) is more negative (lower energy) than –7,200 $c m^{- 1}$ (high‑spin), the low‑spin arrangement is energetically favoured.

Common mistake: Omitting the pairing energy term when evaluating the low‑spin configuration

Concept tested: CFSE calculation for d^6 octahedral complexes

Published 2 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked

Step-by-step solution

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