Calculate the crystal field stabilization energy (CFSE) in kJ mol⁻¹ of the high‑spin octahedral complex [Fe(H₂O)₆]³⁺ (d⁵) given Δ₀ = 14 000 cm⁻¹. Use 1 cm⁻¹ = 0.01196 kJ mol⁻¹.
Answer
0 kJ mol⁻¹
The correct answer is 0 kJ mol⁻¹ — Convert to kJ mol⁻¹: 0 cm⁻¹ × 0.01196 kJ mol⁻¹ cm⁻¹ = 0 kJ mol⁻¹.
Step-by-step solution
- Determine the electronic configuration of high‑spin d⁵: t₂g³e_g².
- Count the number of electrons in the t₂g and e_g sets: N_t₂g = 3, N_e_g = 2.
- Apply the CFSE formula for an octahedral field: CFSE = (−0.4·N_t₂g + 0.6·N_e_g)·Δ₀.
- Substitute the numbers: CFSE = (−0.4·3 + 0.6·2)·Δ₀ = (−1.2 + 1.2)·Δ₀ = 0.
- Multiply by Δ₀ to obtain the energy: 0·Δ₀ = 0 cm⁻¹.
- Convert to kJ mol⁻¹: 0 cm⁻¹ × 0.01196 kJ mol⁻¹ cm⁻¹ = 0 kJ mol⁻¹.
Common mistake: Using the low‑spin formula for a high‑spin complex or forgetting that the CFSE of a high‑spin d⁵ is zero.
Concept tested: CFSE calculation
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