A) octahedral, tetrahedral and square planar
B) tetrahedral, square planar and octahedral
C) square planar, tetrahedral and octahedral
D) octahedral, square planar and octahedral
Correct Answer: B
Solution :
[b] (i) \[N{{i}^{2+}}+4C{{l}^{-}}\to {{[Ni{{(Cl)}_{4}}]}^{2-}}\](Tetrahedral) \[Ni(Z=28)=3{{d}^{8}}4{{s}^{2}},N{{i}^{2+}}=3{{d}^{8}}\] Since \[C{{l}^{-}}\] ion is a weak ligand so pairing does not occur, thus \[s{{p}^{3}}\]hybridisation and is tetrahedralYou need to login to perform this action.
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