A compound of vanadium has magnetic moment of 1.73 BM. What is the electronic configuration of the vanadium ion in the compound ?

Explanation

`mu=sqrt(n(n+2)) BM or 1.73=sqrt(n(n+2)) or 3.0=n(n+2) rArr n=1`
Magnetic moment corresponds to one unpaired electron.
According to electronic configuration `(""_(23)V: 1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)3d^(3)4s^(2))`, vanadium has 3 unpaired electrons in the 3d level. To convert it to a species containing only 1 unpaired electron, 4 electrons should be removed.
Thus electronic configuration of `V^(4+): 1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)3d^(1)4s^(0)`.