What can be inferred from the magnetic moment values of the following complex species?
`{:("Example","Magnetic moment(BM)"),(K_(4)[Mn(CN)_(6)],2.2),([Fe(H_(2)O)_(6)]^(2+),5.3),(K_(2)[MnCl_(4)],5.9):}`

Explanation

Magnetic moment `(mu) = sqrt(n(n+2))BM`
When `n= 1 , mu = sqrt(1(1+1)) = sqrt(3) = 1.73`
When `n=2, mu = sqrt(2(2+2))= sqrt(8) = 2.83`
When `n = 3 , mu = sqrt( 3( 3+2)) = sqrt(15) = 3.87`
When `n=4 , mu = sqrt( 4( 4+2)) = sqrt( 32) = 5.66`,
When `n = 5 , mu= sqrt( 5 ( 5+2)) = sqrt( 35) 5.92`
(i) `K_(4)[Mn(CN)_(6)]`. In this complex, Mn is in `+2` oxidation state, i.e., as `Mn^(2+)`. `mu = 2.2` BM shows that it has only on unpaired electron. Thus, when `CN^(-)` ligands approach `Mn^(2+)` ion, the electron in 3d pairup as shown in Figure.
Thus, `CN^(-)` is a strong ligand.
The hybridisation involved is `d^(2) sp^(3)` forming inner orbital octahedral complex.
(ii) `[Fe(H_(2)O)_(6) ]^(2+)`. In this complex, Fe is in `+2` state, i.e., as `Fe^(2+)` .
`mu = 5.3` BM shows that there are four unpaired electrons. This means that the electrons in 3d do not pair up when the ligands, `H_(2)O` molecules, approach. Thus, `H_(2)O` is a weak ligand. To accomodate the electrons donated by six `H_(2)O` molecule, the hybridisation will be `sp^(3) d^(2)`. Hence, it will be an outer orbital octahedral complex .
(iii) `K_(2)[MnCl_(4)]` . In this complex, Mn is in `+2` state, i.e., as `Mn^(2+), mu = 5.92` BM shows that there are five unpaired electrons. Thus, the hybridisation involved will be `sp^(3)` and the complex will be tetrahedral.