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InterviewSolution
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What can be inferred from the magnetic moment of the followingcomplex 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):} |
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Answer» Solution :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.  
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