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

Explanation

The magnitude of moment `(mu)` of a compound species ion is given by
`mu = sqrt(n(n+ 2))BM`
(n is the number of unpaired electron BM is the unit of magnitude moment in bohr magneton)
`:. 1.73 = sqrt(n(n + 2))`
ON solving for r we get `n = 1` This mean that vandium ion `(Z = 23)` in the compound has one ampaired electromnm `3d `

So electronic configuration must be
`1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)3d^(6)`
i.e. vanadium exies `V^(3+)` ion in the compound since the ground state electronic configuration of `_(23)V` is
`1s^(2)2s^(2)2p^(2)3s^(2)3p^(6)3d^(8)4s^(2)`
`3d