Saved Bookmarks
| 1. |
A solution of [Ni(H_2 O)_6]^(2+) is green but a solution of [Ni(CN)_4]^(2-) is colourless . Explain. |
|
Answer» Solution :In `[NI(H_2 O)_6]^(2-)` Ni is in +2 state with the CONFIGURATION `2d^8`,i.e., it has two unpaired electrons which do not pair up in the presence of the weak `H_2 O` ligand. So, 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^-` ligand, the two unpaired electrons in the 3d-orbitals pair up. Hence, there is no unpaired ELECTRON present, Hence, it is colourless. |
|