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

Answer

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 magnetic)
`because sqrt(1.73)=sqrt(3)`
`=> n(n+2)=3=> n=1` => Magnetic moment correspond to one unpaired electron.
=> 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)`