A compound of vanadium has a magnetic moment of 1.73 BM. Find the number of unpaired electrons of vanadium ion in the compound.

Magnetic moment = $\sqrt{n(n+2)}$ where n- number of unpaired electrons

$$\begin{gathered} \begin{array}{l}\therefore \sqrt{n(n+2)}=1.73 n^2+2n=(1.73{)}^2\end{array} \\ n=1 \end{gathered}$$

Therefore, vanadium atom must have one unpaired electron and thus its electronic configuration

${ }_{23}{\mathrm{V}}^{4+}:1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^{6}3{\mathrm{s}}^{2}3{\mathrm{p}}^{6}3{\mathrm{d}}^1$