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A solution of [Ni(H_(2)O_(6)]^(2+) is green, whereas a solution of [Ni(CN)_(4)]^(2-) is colourless-Explain. |
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Answer» Solution :`(i)` In `[Ni(H_(2)O)_(6)]^(2+)` , `Ni` is in `+2` oxidation state with the configuration `3d^(8)`.i.e., it has two unpaired electrons which do not pair up in the presence of WEAK `H_(2)O` ligand. HENCE, it is coloured. The `d-d` transtion ABSORBS red light and the complementary light emitted is green. `(ii)` In the case of `[Ni(CN)_(4)]^(2-)`, `Ni` is again in `+2` oxidation state with the configuration `3d^(8)`, but in the presence of strong `CN^(-)` ligand the two unpaired electrons in the `3d` orbitals pair up. THUS there is no unpaired electron present. Hence it is colourless. Therefore, a solution of `[Ni(H_(2)O)_(6)]^(2+)` is green, whereas a solution of `[Ni(CN)_(4)]^(2-)` is colourless. |
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