The compound of vanadium has magnetic
moment of 1.73 BM. The vanadium chloride
has the formula:

We know that Electronic configuration of vanadium is 

$\mathrm{V}\left(23\right):1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^{6}3{\mathrm{s}}^{2}3{\mathrm{p}}^{6}3{\mathrm{d}}^{3}4{\mathrm{s}}^2$

Now, we also know that 

Magnetic moment $=\sqrt{\mathrm{n}(\mathrm{n}+2)}\mathrm{BM}=\sqrt{\mathrm{n}(\mathrm{n}+2)}=1.73\mathrm{BM}$ (Given)

Thus, n=1 (one unpaired electron)

Now Electronic configuration of vanadium with one unpaired electron is 

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

Therefore,its chloride is ${\mathrm{VCl}}_4$.

Hence, option(c) is the answer.