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Based VB theory explain why[Cr(NH_3)_6]^(3+) is paramagnetic , while [Ni(CN_4) ^(2-)is diamagnetic. |
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Answer» Solution :(a) `[Cr(NH_(3))_(6)]^(3+)` In this complex Cr is in the +3 oxidation STATE. Electronic configuration of Cr atom Electronic configuration of `Cr^(3+)` ion Hybridisation and formation of `[Cr(NH_(3))_(6)]^(3+)` complex * Due to the presence of three unpaired electrons in `[Cr(NH_(3))_(6)]^(3+)` it behaves as a PARAMAGNETIC substance * The spin magnetic moment, `mu s= sqrt(3(3+2)) = sqrt15 = 3.87BM` * `[Cr(NH_(3))_(6)]^(3+)` is an inner orbital octahedral complex.  (b) `[Ni(CN)_(4)]^(2-)` In this complex Ni is the +2 oxidation state. Electronic configuration of Ni atom Electronic configuration of `Ni^(2+)` ion  Hybridisation and formation of `[Ni(CN)_(4)]^(2-)` Complex  Since `CN^(-)` is strong field LIGAND, hence the electrons in 3d orbitals for forced to PAIR up and there is no unpaired electron in `[Ni(CN)_(4)]^(2-)`, hence it should be DIAMAGNETIC substance. |
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