What is the correct roder of spin only magnetic moment (in BM) of `Mn^(2+), Cr^(2+) and Ti^(2+)`?

Options

A. `Mn^(2+) gt Ti^(2+) gt Cr^(2+)`
B. `Ti^(2+) gt Cr^(2+) gt Mn^(2+)`
C. `Mn^(2+) gt Cr^(2+) gt Ti^(2+)`
D. `Cr^(2+) gt Ti^(2+) gt Mn^(2+)`

Answer

C

Explanation

Spin only magnetic moment depends upon the number of unpaired electrons, more the number of unpaired electrons, greater will be the spin only magnetic moment.
`Mn = 1s^(2), 2s^(2) 2p^(6), 3s^(2) 3p^(6) 3d^(5), 4s^(2)`
`Mn^(2+) = 1s^(2), 2s^(2) 2p^(6), 3s^(2) 3p^(6) 3d^(5), 4s^(0)`
3d

Number of unpaired electrons = 5
`Cr = 1s^(2), 2s^(2) 2p^(6), 3s^(2) 3p^(6) 3d^(5), 4s^(1)`
`Cr^(2+) = 1s^(2), 2s^(2) 2p^(6), 3s^(2) 3p^(6) 3d^(4), 4s^(0)`
3d

Number of unpaired electrons =4
`Ti = 1s^(2), 2s^(2) 2p^(6), 3s^(2) 3p^(6) 3d^(2), 4s^(2)`
`overset(22)(Ti^(2+)) = 1s^(2), 2s^(2) 2p^(6), 3s^(2) 3p^(6) 3d^(2), 4s^(2)`

Number of unpaired electrons = 2
So, the correct order of spin only magnetic moment is `Mn^(2+) gt Cr^(2+) gt Ti^(2+)`
As a result of lanthanoid contraction, the atomic radii of elements of 4d and 5d becomes close and so the properties of 4d and 5d transition element show the similarities.