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The magnetic moment of [Mn(CN)_(6)]^(3-) is 2.8 B.M and that of [MnBr_(4)]^(2-)is 5.9 B.M. What are the geometries of these complex ions |
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Answer» Solution :For complex `[Mn(CN)_(6)]^(3-)`, the NUMBER of unpaired electrons is calculated as, `2.8 = sqrt(n(n+2)) IMPLIES n = 2` `[Mn(CN)_(6)]^(3-)` has TWO unpaired electrons. Hence the geometry is octahedral with dsp hybridisation. For complex `[MnBr_(4)]^(2-)` the number of unpaired electrons is calculated as `5.9 = sqrt(n(n+2)) implies n=5` `[MnBr_(4)]^(2-)`has 5 unpaired electrons. Hence the geometry is tetrahedral with `sp^(3)` hybridisation |
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