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In [Ni(NH₃)₄]²⁺ , nickel is present in +2 oxidation state {Ni = 3d⁸ 4s⁰}. NH₃ is a weak ligand, so it can not pair the upaired electrons. Hence, 3d orbital is partially 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.