Give reason for the following:
The structure of [Ni(CO)4] (tetrahedral) is different from the structure of [Ni (CN)4]-2 (square planar)
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 dsp2 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 dsp2 hybridization, one d -orbital [which is d (x2–y2)] is involved in hybridization with one s and two p -orbital. This gives rise to the square planar geometry.