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Calculate the crystal field stabilization energy (CFSE) in kJ mol⁻¹ for the high‑spin octahedral complex [Cr(H₂O)₆]³⁺ (Cr³⁺, d³) given Δ₀ = 12 000 cm⁻¹. Use 1 cm⁻¹ = 0.01196 kJ mol⁻¹.

medium 2 marks Class 12 · Chemistry · Coordination Compounds
Answer
-172.2 kJ mol⁻¹

The correct answer is -172.2 kJ mol⁻¹ — Round to one decimal: -172.2 kJ mol⁻¹.

Step-by-step solution

  1. For high‑spin d³, all 3 electrons occupy t₂g orbitals: configuration t₂g³ e_g⁰.
  2. CFSE = (n_t2g)(-0.4Δ) + (n_eg)(+0.6Δ) = 3·(-0.4Δ) = -1.2Δ.
  3. Insert Δ = 12 000 cm⁻¹: CFSE = -1.2×12 000 = -14 400 cm⁻¹.
  4. Convert to kJ mol⁻¹: -14 400 × 0.01196 = -172.224 kJ mol⁻¹.
  5. Round to one decimal: -172.2 kJ mol⁻¹.
Common mistake: Using low‑spin electron configuration

Concept tested: CFSE calculation for high‑spin d³

Published · How this solution was prepared and checked

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