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

Explanation

Vanadium belongs to 3d series with Z=23. The magnetic moment of 3d series metal is given by spin only formula.
`mu=sqrt(n(n+2))` BM (BM=Bohr's magneton)
`thereforesqrt(1.73)=sqrt3`
`rArr` n(n+2)=3 `rArr n=1` ltbr `rArr` Magnetic moment correspond to one unpaired electron.
`rArr` Electronic configuration of vanadium atom `1s^(2) 2s^(2) 2p^(6)3s^(2)3p^(6)4s^(2)3d^(3)`.
For one unpaired electron 4 electron must be removed in which first 2 electron are lost from 4s orbital (outermost).
Electronic configuration of `V^(+4)`
`1s^(2) 2s^(2) 2p^(6) 3s^(2) 3p^(6) 4s^(0) 3d^(1)`