Question

A solution of \[\ce{[Ni(H2O)6]^2+}\] is green, whereas a solution of \[\ce{[Ni(CN)4]^2-}\] is colorless – Explain.

Answer 1

1. In \[\ce{[Ni(H2O)6]^2+}\], Ni is in +2 oxidation state with the configuration 3d⁸, i.e., it has two unpaired electrons which do not pair up in the presence of weak H₂O ligand. Hence, it is coloured. The d – d transition absorbs red light and the complementary light emitted is green.

2. In the case of \[\ce{[Ni(CN)4]^2-}\], Ni is again in +2 oxidation state with the configuration 3d⁸, but in the presence of strong CN^– ligand the two impaired electrons in the 3d orbitals pair up. Thus there is no unpaired electron present. Hence it is colourless. Therefore, a solution of \[\ce{[Ni(H2O)6]^2+}\] is green, whereas a solution of \[\ce{[Ni(CN)4]^2-}\] is colourless.