Low spin complex of $d^6$-cation in an octahedral field will have the following energy $\left(\Delta_o=\right.$ crystal field splitting energy in an octahedral field, $P=$ electron pairing energy)

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Aicharya · Accepted Answer

$\text {C.F.S.E. }=(-0.4 x+0.6 y) \Delta_o+z P$ where $\mathrm{x}=$ number of electrons occupying $\mathrm{t}_{2 g}$ orbital $y=$ number of electrons occupying $e_g$ orbital $z=$ number of pairs of electrons For low spin $\mathrm{d}^6$ complex electronic configuration $\begin{aligned} & =t_{2 g}^6 e_g^0 \text { or } t_{2 g}^{2,2,2} e_g^0 \ & \therefore \quad x=6, y=0, z=3 \ & \text { C.F.S.E. }=(-0.4 \times 6+0 \times 0.6) \Delta_o+3 P \ & =\frac{-12}{5} \Delta_o+3 P \end{aligned}$

Low spin complex of $d^6$-cation in an octahedral field will have the following energy…

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