Chemistry - Carbonyl compounds Question with Solution | TestHub

$\mathrm{N}{\mathrm{i}}^{2+}=\left(\mathrm{A}\mathrm{r}\right)3{\mathrm{d}}^8$

$\mathrm{N}{\mathrm{i}}^{+2}$has two unpaired electron.$(\mathrm{n}=2)$
As$\mathrm{\mu }=\sqrt{\mathrm{n}(\mathrm{n}+2)}=\sqrt{2(2+2)}=2.83\mathrm{B}.\mathrm{M}.$
In
$\begin{array}{l}{\mathrm{Mn}}^{2+};\mathrm{n}=5\\ {\mathrm{Cr}}^{3+};\mathrm{n}=3\\ {\mathrm{Ti}}^{3+};\mathrm{n}=1\end{array}$