Calculate the crystal field stabilization energy (CFSE) in kJ mol $^{- 1}$ for the low‑spin octahedral complex [Co(CN) $_{6}$ ] $^{3 -}$ (Co $^{3 +}$ , $d^{6}$ ) given $Δ_{o} = 18000 cm^{- 1}$ . Use $1 cm^{- 1} = 0.01196 kJ mol^{- 1}$ .

medium 2 marks Class 12 · Chemistry · Coordination Compounds

Answer

- 5.17 \times 1 0^{2} kJ mol^{- 1}

(approximately)

The correct answer is $- 5.17 \times 1 0^{2} kJ mol^{- 1}$ (approximately) — The negative sign indicates stabilization relative to the spherical field.

Step-by-step solution

Determine the d‑electron count: Co $^{3 +}$ is $d^{6}$ .
In a low‑spin octahedral field the configuration is $t_{2 g}^{6} e_{g}^{0}$ , giving a CFSE of $6 \times (- 0.4 Δ_{o}) = - 2.4 Δ_{o}$ .
Calculate $- 2.4 \times 18000 cm^{- 1} = - 43200 cm^{- 1}$ .
Convert to kJ mol $^{- 1}$ : $- 43200 \times 0.01196 = - 516.7 kJ mol^{- 1}$ , i.e. about $- 5.17 \times 1 0^{2} kJ mol^{- 1}$ .
The negative sign indicates stabilization relative to the spherical field.

Common mistake: Using the high‑spin electron distribution for a strong‑field ligand like CN

^{-}

Concept tested: CFSE calculation for low‑spin octahedral complex

Published 12 September 2026 · How this solution was prepared and checked

← All Coordination Compounds questions

Calculate the crystal field stabilization energy (CFSE) in kJ mol−1 for the low‑spin octahedral complex [Co(CN)6​]3− (Co3+, d6) given Δo​=18000 cm−1. Use 1 cm−1=0.01196 kJ mol−1.

Step-by-step solution

Calculate the crystal field stabilization energy (CFSE) in kJ mol $^{- 1}$ for the low‑spin octahedral complex [Co(CN) $_{6}$ ] $^{3 -}$ (Co $^{3 +}$ , $d^{6}$ ) given $Δ_{o} = 18000 cm^{- 1}$ . Use $1 cm^{- 1} = 0.01196 kJ mol^{- 1}$ .