Calculate the crystal field stabilization energy (CFSE) in kJ mol⁻¹ for the high‑spin octahedral complex [MnCl $_{6}$ ] $^{3 -}$ where Mn $^{3 +}$ (d $^{4}$ ) has an octahedral field splitting $D e l t a_{0} = 10, 000$ cm $^{- 1}$ . Use 1 cm $^{- 1} = 0.01196$ kJ mol $^{- 1}$ .

hard 4 marks Class 12 · Chemistry · Coordination Compounds

Answer

-71.8 kJ mol

^{- 1}

The correct answer is -71.8 kJ mol $^{- 1}$ — Convert to kJ mol $^{- 1}$ : $- 6, 000 \times 0.01196 = - 71.76$  kJ mol $^{- 1}$ , rounded to $- 71.8$  kJ mol $^{- 1}$ .

Step-by-step solution

Determine the electron distribution for high‑spin d $^{4}$ in an octahedral field: 3 electrons in $t_{2 g}$ , 1 in $e_{g}$ .
Apply the CFSE formula: CFSE = $(n_{t_{2 g}} \times - 0.4 Δ_{0}) + (n_{e_{g}} \times 0.6 Δ_{0})$ .
Substitute $n_{t_{2 g}} = 3$ , $n_{e_{g}} = 1$ , $Δ_{0} = 10, 000$  cm $^{- 1}$ : CFSE = $(3 \times - 0.4 + 1 \times 0.6) \times 10, 000 = - 0.6 \times 10, 000 = - 6, 000$  cm $^{- 1}$ .
Convert to kJ mol $^{- 1}$ : $- 6, 000 \times 0.01196 = - 71.76$  kJ mol $^{- 1}$ , rounded to $- 71.8$  kJ mol $^{- 1}$ .

Common mistake: Using the wrong electron distribution or forgetting the negative sign

Concept tested: Crystal field stabilization energy

Published 7 October 2026 · How this solution was prepared and checked

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Calculate the crystal field stabilization energy (CFSE) in kJ mol⁻¹ for the high‑spin octahedral complex [MnCl6​]3− where Mn3+ (d4) has an octahedral field splitting Delta0​=10,000 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 high‑spin octahedral complex [MnCl $_{6}$ ] $^{3 -}$ where Mn $^{3 +}$ (d $^{4}$ ) has an octahedral field splitting $D e l t a_{0} = 10, 000$ cm $^{- 1}$ . Use 1 cm $^{- 1} = 0.01196$ kJ mol $^{- 1}$ .