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Assertion: A solution of `[Ni(H_(2)O)_(6)]^(2+)` is green but a solution of `[Ni(CN)_(4)]^(2-)` is colourless Reason: `[Ni(CN)_(4)]^(2-)` is square planar complex . |
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Answer» In `[Ni(H_2O)_6]^(2+)`, Ni is in `+2` state with the configuration `3d^(8)`, i.e., it has two unpaired electrons which do not pair up in the presence of the weak `H_2O` ligand. Hence, it is coloured. The d-d transition, absorbs red light and the complementary light emitted is green. In case of `[Ni(CN)_4]^(2-)`, Ni is again in `+2` state with the configuration `3d^(8)` but in presence of the strong `CN^(ɵ)` ligand, the two unpaired electrons in the `3d` orbitals pair up. Thus, there is no unpaired electron present. Hence, it is colourless. |
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