Question

A compound of vanadium has a spin magnetic moment of 1.73 BM. Work out the electronic configuration of the vanadium ion in the compound.

• A. $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{1}4s^2$
• B. $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^2$
• C. $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{3}4s^1$
• D. $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{2}4s^2$

Answer 1

The correct option is A $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{1}4s^2$

We know,
spin magnetic moment $=\sqrt{n(n+2)}$
Where $n$ is the number of unpaired electrons.
$1.73=\sqrt{n(n+2)}$ $\therefore n=1$
$\therefore$ Vanadium atom must have one unpaired electron hence its configuration is ${}_{23}{V}^{4+}=1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^1$