Magnetic moments `2.84 B.M` is given by :
(At. nos. ni = 28, Ti = 22, Cr = 24, Co = 27).

Options

A. `Co^(2+)`
B. `Ni^(2+)`
C. `Ti^(3+)`
D. `Cr^(2+)`

Explanation

For the ions of the first series of transition metals, the magnetic moment is determinef by the number of unpaired electrons and is calculated by using the 'spin-only' formula:
`mu = sqrt(n (n+2)) B.M.`
Magnetic moment `2.84 B.M.` (which lies between 2 and 3) corresponds to 2 unpaired electrons:
`Ni (Z = 21) : 3d^(8) 4s^(2) rarr Ni^(2+) (3d^(8))`, 2 unpaired electrons
`Ti (Z = 22) : 3d^(2) 4s^(2) rarr Ti^(3+) (3d^(1))`, 1 unpaired electron
`Cr (Z = 24) : 3d^(5) 4s^(1) rarr Cr^(2+) (3d^(4))`, 4 unpaired electrons
`Co (Z = 27) : 3d^(7) 4s^(2) rarr Co^(2+) (3d^(7))`, 3 unpaired electrons
The similar question was asked in 2014 with a slight difference : `Cr^(3+)` in place of `Cr^(2+)` , `Mn^(2+)` in place of `Co^(2+)`