A compound of vanadium has a magneitc moment `(mu)` of `1.73 BM`. If the vanadium ion in the compound is present as `V^(x+)`, then, the value of `x` is ?

Options

A. 1
B. 2
C. 3
D. 4

Answer

4

Explanation

Given `mu=sqrt(n(n+2))=1.73BM`
where `n` is number of unpaired electrons,
`:. n=1,._(23)V=1s^(2)2S^(2)2P^(6)3s^(2)3p^(2)3d^(3)`

for `n=1` it must release 4 electrons, first two from `4s-` orbital and then next two electrons from `3d-` orbital. So, `x=4`