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

Options

A. `1s^(2) 2s^(2) 2p^(6) 3s^(2) 3p^(6) 3d^(1)`
B. `1s^(2) 2s^(2) 2p^(6) 3s^(2) 3p^(6) 3d^(2)`
C. `1s^(2) 2s^(2) 2p^(6) 3s^(2) 3p^(6) 3d^(3)`
D. `1s^(2) 2s^(2) 2p^(6) 3s^(2) 3p^(6) 3d^(4)`

Answer

1

Explanation

Number of unpaired electron are given by Magnetic moment `=sqrt([n(n+2)])` where `n` is number of unpaired electrons or `1.73=sqrt([n(n+2)])` or `1.73xx1.73=n^(2)+2n`
`:. n=1`
Now vanadium atom must have one unpaired unpaired electron and thus its confiuration is:
`_(23)V^(4+): 1s^(2) 2s^(2) 2p^(6)3s^(2)3p^(6) 3d^(1)`