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. The electronic configuration of `V^(4+) : 1s^(2) 2s^(2) 2p^(6) 3s^(2) 3p^(6) 3d^(1) 4s^(0)`.