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).