The geometry of $N i {(C O)}_{4} a n d N i {({P H}_{3})}_{2} {C l}_{2}$ are
(a) both square planer (b) tetrahedral and square planer
(c) both tetrahedral (d) square planar and tetrahedral

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Solution

The correct option is B
the geometry of $N i (c o) 4$ is tetrahedral, and the geometry of $N i (P h_{3})_{2} c l_{2}$ are square planer.
The reason $- n i (c o)_{4}$ is an unusual complex, it has ni in 0 oxidation state, thus there are 8 electrons in 3d an ND 2 electrons in 4s. co is the strongest ligand and does cause pairing by leaving 1 empty d orbital. now, the pair electron is the 4s shift to 3d leaving 4s orbital empty. and free to accommodate electrons from co. so, there is no free orbital as its a weak ligand, so, it has $s p^{3}$ hybridisation . meaning tetrahedral shape.

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