CBSE Class 12-science Answered

In [Ni(CO)₄ ] , nickel is present in zero oxidation state {Ni = 3d⁸ 4s²}. CO is a very strong ligand, so it can pair the upaired electrons. Hence, the 4s electrons goes to the 3 d orbital. Now, 3 d orbital is completely filled, but 4s and 4p are still available. These 4 orbitals form a degenerate set of orbitals, that means hybridisation is sp3 hybridised.

In case of [Ni(CN)₄ ]^2-, oxidation state of Nickel is +2. So, Ni²⁺ : 3d⁸ 4s⁰ . Now, cyanide also causes pairing of unpaired electrons, in 3d orbital, all the 8 electrons will get paired, so now, 1 more orbital is left.... and there are 4 ligands to bond with. Hence, the hybridization will be dsp² so hence, it is a square planar complex because all dsp^2 complexes are square planar. The singly unpaired electron will pair up only if the ligand field is very strong and that too only in the lower energy orbitals.

In dsp ² hybridization, one d-orbital [which is d(x ²–y²)] is involved in hybridization with one s and two p-orbital. This gives rise to the square planar geometry.