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

Explanation

Step I. Calculation of the number of unpaired electrons.
Magnetic moment ` (mu)=sqrt(n(n+2)) " " ` (Here n=number of unpaired electrons)
As `" " mu =1.73` BM, therefore, `1.73 = sqrt(n(n+2))`
`(1.73)^(2)=n=(n+2) or 3=n^(2)+2n`
`n^(2)+2n-3 =0 or (n-1)(n+3)=0. ` Therefore, `n=1`
Step II. Configuration of vanadium ion.
The electronic configuration of vanadium `(Z=23) ` is `: 1s^(2) 2s^(2) 2p^(6) 3s^(2) 3p^(6) 3d^(3) 4s^(2)`
Since the ion has only one unpaired electron, therefore its configuration is `: 1s^(2) 2s^(2)2p^(6) 3s^(2) 3p^(6) 3d^(1)`
Thus, vanadium is present as ` V^(4+)` ion or V (IV)