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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.
now let us compair with
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.
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