Question

A chloride of vanadium has a magnetic moment of 1.732 BM. It's formula could be:

• A. $VC{l}_2$
• B. $VC{l}_3$
• C. $VC{l}_5$
• D. $VC{l}_4$

Answer 1

The correct option is D $VC{l}_4$

Using the spin only formula,$\mu =\sqrt{n(n+2)}=1.732$
$\Rightarrow n=1$.
V has a configuration of $\left[Ar\right]3d^{3}4s^2$. For one unpaired electron to be present, its oxidation state must be +4.
In $VC{l}_4$, V is in this state and hence, this is the answer.