For the octahedral complex, which of the following d -electron...

{Δ}_{0} = \left(-0. 4 \times {nt}_{2g}\right) +\left(0. 6 \times {ne}_{g}\right)

where, $n =$number of electrons in ${nt}_{2g} and {ne}_{g}$.

$CFSE of$given \alpha -electrons configuration are as follows

a. For high spin,$d^{6}$;

{Δ_{{0}}\left(CFSE\right) = - 0. 4 \times4 + 0. 6 $\times$2 = - 0. 4$</p><p>b. For low spin,$d^{4}$;</p><p>${Δ_{{0}}\left(CFSE\right) = - 0. 4 \times$4 + 0. 6$\times$0 = 1. 6$

c. For low spin $d^{5}$

{Δ_{{0}}\left(CFSE\right) = - 0. 4 \times5 + 0. 6 $\times$0 = - 2. 0$</p><p>d. For high spin,$d^{7}$</p><p>${Δ_{{0}}\left(CFSE\right) = - 0. 4 \times$5 + 0. 6$\times$2 = - 0. 8$

Neglecting $- ve$ sign (as it indicates that energy is released).

Hence, maximum ${Δ}_{0} \left(CFSE\right)$ value is shown by low spin $d^{5}$ configuration.