A compound of vanadium has a magnetic moment ($\mathrm{\mu }$) of 1.73 BM. If the vanadium ion in the compound is present as V ^x+, then, the value of x is?

$\text{ Given }\mathrm{\mu }=\sqrt{\mathrm{n}(\mathrm{n}+2)}=1.73\mathrm{BM}$
(where n is number of unpaired electrons)
$\therefore \mathrm{n}=1;{}_{23}\mathrm{V}=1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^{6}3{\mathrm{s}}^{2}3{\mathrm{p}}^{6}4{\mathrm{s}}^{2}3{\mathrm{d}}^3$

for n = 1 it must release 4 electrons, first two from 4s-orbital and then next two electrons from 3d-orbital. So, x = 4.