Crystal field stabilization energy for high spin $\mathrm{d}^4$ octahedral complex is
Crystal field stabilization energy for high spin $\mathrm{d}^4$ octahedral complex is
$-1.8 \Delta_0$
$-1.6 \Delta_0+P$
$-1.2 \Delta_{\circ}$
$-0.6 \Delta_0$
Solution
Key Idea:- In case of high spin complex, $\Delta_0$ is small. Thus, the energy required to pair up the fourth electron with the electrons of lower energy d-orbitals would be higher than that required to place the electrons in the higher d-orbital. Thus, pairing does not occur.
For high spin $\mathrm{d}^4$ octahedral complex,
$\therefore$ Crystal field stabilisation energy
$\begin{aligned}
& =(-3 \times 0.4+1 \times 0.6) \Delta_{\circ} \\
& =(-1.2+0.6) \Delta_0 \\
& =-0.6 \Delta_0
\end{aligned}$
$\therefore$ Crystal field stabilisation energy
$\begin{aligned}
& =(-3 \times 0.4+1 \times 0.6) \Delta_{\circ} \\
& =(-1.2+0.6) \Delta_0 \\
& =-0.6 \Delta_0
\end{aligned}$