The transition metal ion that has 'spin-only' magnetic moment value of 5.92 is

Options

A. `Mn^(2+)`
B. `Fe^(2+)`
C. `V^(2+)`
D. `Cu^(2+)`

Answer

A

Explanation

`Mn^(2-)` with electronic configuration `3d^5 4s^0` has 5 unpaired electrons.
Spin - only magnetic moment `(mu_s) = sqrt(n(n 12))`
For `Mn^(2-) : (mu_s) = sqrt(5(5+2)) = 5.916 = 5.92 B.M`
`mu+s` for `Fe^(2+)` = 4.90 B.M `V^(2+)` = 3.87 B. M `Cu^(2+)` = 1.73 B.M
So the correct option is (a) ie. `Mn^(2+)`