Question

A compound of vanadium has a magnetic moment of $1.73$ $BM$. The electronic configuration of vanadium ion in the compound is:

• A. $\left[Ar\right]3d^2$
• B. $\left[Ar\right]3d^{1}4s^0$
• C. $\left[Ar\right]3d^3$
• D. $\left[Ar\right]3d^{0}4s^1$

Answer 1

The correct option is B $\left[Ar\right]3d^{1}4s^0$

The electronic configuration of $V(Z=23)=1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^3$

If the number of unpaired electrons be n, then the Magnetic moment = $\sqrt{n(n+2)}$ BM

If n=1, then the magnetic momentum = $\sqrt{1(1+2)}=\sqrt{3}=1.73$ BM

Therefore vanadium has one unpaired electron and the electronic configuration of the ion is $\left[Ar\right]3d^{1}4s^0$. since the electrons are first removed from the outermost shell.