Give reason The shape of [Ni(CO)4] is not the same as [Ni(CN)4]2-

Dear Student,

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 unpaired electrons. Hence, the 4s electrons go to the 3 d orbital. Now, 3 d orbitals are completely filled, but 4s and 4p are still available. These 4 orbitals form a degenerate set of orbitals, that means hybridization is sp3 hybridized.

In case of [Ni(CN)4 ]2-, the oxidation state of Nickel is +2. So, Ni+2 3d 8 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.

dsp 2 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.

Hope this information will clear your doubts about topic.

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