Question

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

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

Answer 1

Answer: (C)

$VC{l}_4$

As we know,
Magnetic moment $=\sqrt{n(n+2)}BM$ $=\sqrt{n(n+2)}=1.73BM$

Thus, $n=1$ (one unpaired electron)

Electronic configuration:

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

$V^{4+}:1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^1$ with one unpaired electron. So its chloride is $VC{l}_4$