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Crystal field stabilization energy for high spin $d^4$ octahedral complex is
A
$-0.6\Delta_0$
B
$-1.8\Delta_0$
C
$-1.6\Delta_0 + P$
D
$-1.2\Delta_0$
Detailed Solution
In an octahedral field the d orbitals split into $t_{2g}$ (each electron stabilised by $-0.4\Delta_0$) and $e_g$ (each electron destabilised by $+0.6\Delta_0$).
High spin $d^4$ means the fourth electron goes to $e_g$ instead of pairing: configuration $t_{2g}^3e_g^1$.
CFSE = $3\times(-0.4\Delta_0) + 1\times(+0.6\Delta_0)$
CFSE = $-1.2\Delta_0 + 0.6\Delta_0$
CFSE = $-0.6\Delta_0$
(No pairing energy term appears, because no electrons are paired.)
High spin $d^4$ means the fourth electron goes to $e_g$ instead of pairing: configuration $t_{2g}^3e_g^1$.
CFSE = $3\times(-0.4\Delta_0) + 1\times(+0.6\Delta_0)$
CFSE = $-1.2\Delta_0 + 0.6\Delta_0$
CFSE = $-0.6\Delta_0$
(No pairing energy term appears, because no electrons are paired.)