Calculate magnetic moment of `Fe^(3+)` in `[Fe(CN)_(6)]^(3+)` and in `[Fe(H_(2)O)_(6)]^(3+)`
Arrange following complexes in decreasing order of magnetic moment
`[Ni(H_(2)O)_(4)]^(2+)`, `[Ni(CN)_(4)]^(2-), [Fe(CN)_(6)]^(3-), [Fe(CN)_(6)]^(4-)`

Explanation

In `[Fe(CN)_(6)]^(3-)`, `Fe^(3+)`ion has only one unpaired electron, Thus, magnetic momentof `Fe^(3+)` will be `sqrt(3)`. i.e., 1.732 B. M. `(mu_(s) = sqrt(n(n+2)))` B.M., where, nis number of unpaired electrons). In `[Fe(H_(2)O)_(6)]^(3+) Fe^(3+) ion` has 5 unpaired electrons, hence, its magnetic moment will be `sqrt(35) B.M., i.e., 5.92 B.M`
(b) `[Fe(CN)_(6)]^(4-) = `[Ni(CN)_(4)]^(2+) lt [Fe(CN)_(6)]^(3-) lt [Ni(H_(2)O)_(4)]^(2+)`
`In`[Fe(CN)_(6)]^(4-)` and `[NI(CN)_(4)]^(2)`, central atom has no unpaired electron, hence, their magnetic moment is zero. In `[Fe(CN)_(6)]^(3+)` there is only one electron with central Fetion, hence, its magnetic moment will be `1.732 B.M` In `[Ni(H_(2)O)_(4)]^(2+)` there
are two unpaired electrons with `Ni^(2+)` and hence, it.s magnetic moment will be sqrt8 B.M`