A) tetrahedral
B) square planar
C) square planar and tetrahedral respectively
D) tetrahedral and square planar respectively
Correct Answer: C
Solution :
The geometry of \[{{[Ni{{(CN)}_{4}}]}^{2-}}\] is square planar and\[{{[NiC{{l}_{4}}]}^{2-}}\] is tetrahedral \[\text{C}{{\text{N}}^{-}}\]is a strong ligand (crystal field theory) hence it cause the pairing of electron in \[\text{N}{{\text{i}}^{\text{2+}}}\]while \[\text{C}{{\text{l}}^{-}}\]is a weak ligand so it does not cause pairing of electronsYou need to login to perform this action.
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