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Low spin complex of $d^6$-cation in an octahedral field will have the following energy: ($\Delta_0$ = Crystal Field Splitting Energy in an octahedral field, P = Electron pairing energy)
A
$-\frac{2}{5}\Delta_0 + P$
B
$-\frac{12}{5}\Delta_0 + P$
C
$-\frac{12}{5}\Delta_0 + 3P$
D
$-\frac{2}{5}\Delta_0 + 2P$
Detailed Solution
In a low spin $d^6$ octahedral complex all six electrons are paired in the $t_{2g}$ orbitals: $t_{2g}^6e_g^0$.
Each $t_{2g}$ electron is stabilised by $-\frac{2}{5}\Delta_0$ ($= -0.4\Delta_0$) and each $e_g$ electron is destabilised by $+\frac{3}{5}\Delta_0$.
Energy from splitting $= 6\times\left(-\frac{2}{5}\Delta_0\right) + 0 = -\frac{12}{5}\Delta_0$
There are three electron pairs, so the pairing energy is 3P.
Total energy $= -\frac{12}{5}\Delta_0 + 3P$
Each $t_{2g}$ electron is stabilised by $-\frac{2}{5}\Delta_0$ ($= -0.4\Delta_0$) and each $e_g$ electron is destabilised by $+\frac{3}{5}\Delta_0$.
Energy from splitting $= 6\times\left(-\frac{2}{5}\Delta_0\right) + 0 = -\frac{12}{5}\Delta_0$
There are three electron pairs, so the pairing energy is 3P.
Total energy $= -\frac{12}{5}\Delta_0 + 3P$