Question

Crystal field stabilisation energy for high spin $d^4$ octahedral complex is

• A. $-1.8\Delta_0$
• B. $-1.6\Delta_0+P$
• C. $-1.2\Delta_0$
• D. $-0.6\Delta_0$

Answer 1

The Correct Option is D

 To determine the Crystal Field Stabilization Energy (CFSE) for a high-spin octahedral complex with a \(d^4\) electronic configuration, let's go step by step through the concept of crystal field theory and its application:

1. The crystal field theory describes how the degeneracy of the d-orbitals in a transition metal ion is lifted in the presence of a ligand field, leading to a stabilization or destabilization of the electron configuration. In an octahedral field, the d-orbitals split into two sets: the lower energy set \((t_{2g})\) and the higher energy set \((e_g)\).
2. In a high-spin octahedral complex, weak field ligands cause minimal pairing of electrons leading to maximization of unpaired electrons. For a \(d^4\) ion:
   • The electronic configuration will be \(t_{2g}^3 e_g^1\).
3. The energy difference between the \(t_{2g}\) and \(e_g\) orbitals is denoted by \(\Delta_0\). The \(t_{2g}\) orbitals are lower in energy by \(-\frac{2}{5}\Delta_0\) each, and the \(e_g\) orbitals are higher in energy by \(+\frac{3}{5}\Delta_0\) each.
4. To calculate the Crystal Field Stabilization Energy (CFSE) for \(t_{2g}^3 e_g^1\):
   • Contribution by \(t_{2g}\) electrons: \(3 \times \left(-\frac{2}{5}\Delta_0\right) = -\frac{6}{5}\Delta_0\)
   • Contribution by \(e_g\) electron: \(1 \times \left(+\frac{3}{5}\Delta_0\right) = +\frac{3}{5}\Delta_0\)
5. Thus, the total CFSE is: \(= -\frac{6}{5}\Delta_0 + \frac{3}{5}\Delta_0 = -\frac{3}{5}\Delta_0\)

The CFSE calculated is approximately \(-0.6\Delta_0\), which matches the correct answer. Thus, the correct option is \(-0.6\Delta_0\).