Which of the following statements are correct with respect to ${[N i C l_{4}]}^{2 -} and [N i {(P P h_{3})}_{2} C l_{2}]$ ?

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${[N i C l_{4}]}^{- 2} i s square planar$

In both cases $N i$ undergoes $d s p^{2}$ hybridisation

$[N i {(P P h_{3})}_{2} C l_{2}] i s square planar$

$[N i {(P P h_{3})}_{2} C l_{2}] i s t e t r a h e d r a l$

answer is B.

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Detailed Solution

$C l^{-}$ is weak ligand so no pairing of electrons take place. so ${[N i C l_{4}]}^{- 2}$ is tetrahedral. ${pph}_{3}$ group is bulkier one so it favours tetrahedral geometry, through ${pph}_{3}$ is strong field ligand.

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