Question

A compound of vanadium has magnetic moment of 1.73 BM work out electronic configuration of Vanadium ion in the compound is ${\left[Ar\right]3d}^x$. What is x?

Answer 1

Vanadium has $\left(Ar\right)4s^{2}3d^3$ configuration
magnetic moment$=\sqrt{n(n+2)}=1.73BM$
$⟹\sqrt{n(n+2)}=\sqrt{3}$
So,$n=1$
unpaired electrons =1
i.e $4s^{0}3d^1$ configuration
$3e^{-}$ removed
$⟹V^{3+}$
$⟹x=3$