Amongst the following, the lowest degree of paramgnetism per mole of the compound at `298 K` will be shown by

Options

A. `MnSO_4. 4H_2O`
B. `CuSO_4. 5H_2O`
C. `FeSO_4 . 6H_2O`
D. `NiSO_4. 6H_2O`

Answer

B

Explanation

`Mn^(2+)` (in `MnSO_4 .2H_2O`) has the configuration `1s^2 2s^2 2p^6 3s^2 3p^6 3d^5` , i.e. , it has 5 unpaired electrons in 3d -orbitals. Similarly the number of unpaired electrons in other species are `Cu^(+2)` (in `CuSO_4 . 5H_2O`):`1s^2 2s^2 2p^6 3s^2 3p^2 3p^6 3d^9` , i.e., one unpaired electron.
`Fe^(2+)` ( in `FeSO_4 . 6H_2O)`: `1s^2 2s^2 2p^6 3s^2 3p^6 3d^6` , i.e. four unpaired electrons.
`Ni^(2+)` (in`NiSO_4 . 6H_2O`) : `1s^2 2s^2 2p^6 3s^2 3p^6 3d^8` , i.e., two unpaired electrons.
So we find minimum number of unpaired electrons (i.e. 1 ) in `CuSO_4 . 5H_2O` so it has the lowest degree of paramagnetism