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

Options

A. `[Ar]3d^(2)`
B. `[Ar]3d^(1)4s^(0)`
C. `[Ar]3d^(3)`
D. `[Ar]3d^(0)4s^(1)`

Answer

B

Explanation

Magnetic moment`=sqrt(n(n+1))BM`
`1.73=sqrt(n(n+2))`
`sqrt(3)=sqrt(n(n+2))`
n=1 (Number of unparied electrons)
`V_(22)to3d^(2)4s^(2)`
`V^(3+)to3d^(1)4s^(0)` (has one unpaired electron)