Spin only magnetic moment of an octahedral complex of $\mathrm{Fe}^{2+}$ in the presence of a strong field ligand in BM is :
Correct Option: , 3
In presence of SFL $\Delta_{0}>\mathrm{P}$ means pairing occurs therefore
$\therefore$ No of unpaired $\mathrm{e}^{-}(\mathrm{s})=0$
$\therefore \mu=\sqrt{n(n+2)} B M=0$
$\left[n=\right.$ No of unpaired $\left.e^{-}(s)\right]$
In $\mathrm{NiCl}_{2} \mathrm{Ni}^{+2}$ is having configuration $3 \mathrm{~d}^{8}$
$\therefore \quad$ Number of unpaired electron $=2$
After formation of oxidised product
$\left[\mathrm{Ni}(\mathrm{CN})_{6}\right]^{-2} \mathrm{Ni}^{+4}$ is obtained
$\mathrm{Ni}^{+4} \Rightarrow 3 \mathrm{~d}^{6}$ and $\mathrm{CN}^{-}$is strong field ligand
$\therefore$ number of unpaired electrons $=0$
$\therefore$ The charge is $2-0=2$