Question

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

• A. $V{Cl}_2$
• B. $V{Cl}_3$
• C. $V{Cl}_4$
• D. $V{Cl}_5$

Answer 1

The correct option is C $V{Cl}_4$

$\sqrt{n(n+2)}=1.73$ Solving this we get the number of unpaired electrons as 1. $V^{4+}$ will result in 1 unpaired electron. Hence the solution is $VC{l}_4$