Question

Arrange the following complexes in decreasing orger of magnetic moment:
`[Ni(H_2O)_4]^(2+),[Ni(CN)_4]^(2-),[Fe(CN)_6]^(3-),[Fe(CN)_6]^(4-)`

Answer 1

`[Fe(CN)_6]^(4-)=[Ni(CN)_4]^(2-)lt[Fe(CN_6]^(3-)lt[Ni(H_2O)_4]^(2+)`
In `[Fe(CN)_6]^(4-)` and `[Ni(CN)_4[^(2-)`, central atom has no unpaired electron, hence, their magentic moment is zero. In `[Fe(CN)_6]^(3-)` there is only one electron with central `Fe^(3+)` ion, hence, its magentic moment will be 1.732 BM. In `[Ni(H_2O)_4]^(2+)`, there are two unpaired electrons with `Ni^(2+)` and hence, its magnetic moment will be `sqrt8` BM.