The experimental hydration enthalpy of Ni²⁺ is −494 kcal mol⁻¹. In the absence of crystal field splitting, the hydration enthalpy would be −464 kcal mol⁻¹. Use 1 kcal mol⁻¹ = 350 cm⁻¹. Find Δₒ for [Ni

Question

The experimental hydration enthalpy of Ni²⁺ is −494 kcal mol⁻¹. In the absence of crystal field splitting, the hydration enthalpy would be −464 kcal mol⁻¹. Use 1 kcal mol⁻¹ = 350 cm⁻¹. Find Δₒ for [Ni(H₂O)₆]²⁺ in cm⁻¹.

TestHub · Accepted Answer

The CFSE is −494 − (−464) = −30 kcal mol⁻¹ = −30 × 350 = −10500 cm⁻¹. Ni²⁺ is d⁸, t 2 g ​ ⁶ e g ​ ², so CFSE = 6(−0.4Δₒ) + 2(0.6Δₒ) = −1.2Δₒ. Hence Δₒ = 10500 ÷ 1.2 = 8750 cm⁻¹.

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