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The octahedral complex [Co(NH₃)₆]³⁺ has a high–spin d⁶ configuration and an octahedral field splitting parameter Δ₀ = 15 000 cm⁻¹. Calculate the crystal field stabilization energy (CFSE) in kJ mol⁻¹ for this complex. (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 — State the magnitude as the stabilization energy.

Step-by-step solution

  1. Determine electronic configuration: high‑spin d⁶ → t₂g⁴e_g².
  2. Compute CFSE: (4 × –0.4Δ₀) + (2 × 0.6Δ₀) = (–1.6 + 1.2)Δ₀ = –0.4Δ₀.
  3. Insert Δ₀ = 15 000 cm⁻¹: CFSE = –0.4 × 15 000 cm⁻¹ = –6 000 cm⁻¹.
  4. Convert to kJ mol⁻¹: 6 000 × 0.01196 = 71.76 ≈ 71.8 kJ mol⁻¹.
  5. State the magnitude as the stabilization energy.
Common mistake: Using low‑spin electron count for d⁶ instead of high‑spin

Concept tested: Crystal field stabilization energy (CFSE)

Published · How this solution was prepared and checked

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