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A solution of [Ni(H_(2)O_(6)]^(2+) is green but a solution of [Ni(CN)_(4)]^(2-)is colourless. Explain. |
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Answer» Solution :In `[NI(H_(2)O_(2))_(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_(2)O` 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^(@)` but in presence of the strong CN 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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