Question

A compound of vanadium has a magnetic moment $\left(\mu \right)$ of $1.73BM$. If the vanadium ion in the compound is present as $V^{x+}$, then the value of $x$ is ?

Answer 1

Given,
$\mu =\sqrt{n(n+2)}=1.73BM$
(where $n$ is number of unpaired electrons)
$\therefore n=1;{}_{23}V=1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^3$
For $n=1$, it must release 4 electrons, first two from $4s-orbital$ and then next two electrons from $3rd-orbital$. So, $x=4$.