The magnetic moment of the complex anion $\left[\mathrm{Cr}(\mathrm{NO})\left(\mathrm{NH}_3\right)(\mathrm{CN})_4\right]^{2-}$ is :
$5.91 \mathrm{BM}$
$3.87 \mathrm{BM}$
$1.73 \mathrm{BM}$
$2.82 \mathrm{BM}$
Solution
$\operatorname{In}\left[\mathrm{Cr}(\mathrm{NO})\left(\mathrm{NH}_3\right)(\mathrm{CN})_4\right]^{2-}$, $\mathrm{Cr}^{2+}\left(\mathrm{d}^4\right)$ is given as :
i.e., 2 unpaired electrons
$
\mu=\sqrt{2(2+2)}=\sqrt{8}=2.82
$
