The spin only magnetic moment of `Mn^(4+)` ion is nearly

Options

A. 3 BM
B. 6 BM
C. 3 BM
D. 5BM

Answer

C

Explanation

Electronic configuration ofMn is (Z = 25)= `[Ar]3d^(5) 4s^(2)`
`:.` Electronic configuration of `Mn^(4+)` is `Mn^(4+) = [Ar] 3d^(3)`
Thus, `Mn^(4+)` has 3-unpaired electron and its spin only magnetic moment `(mu )` is `mu = sqrt(n(n+2)) BM`
where n = number of unpaired electrons .
`mu sqrt(3(3+2)). BM implies mu =sqrt(15) = 3.87 BM~~4 ` BM