A) \[{{[NiC{{l}_{4}}]}^{2-}}\] is square planar and \[{{[Ni{{(CN)}_{4}}]}^{2-}},\,Ni{{(CO)}_{4}}\] are tetrahedral
B) \[Ni\,{{(CO)}_{4}}\] is square planar and \[{{[Ni\,{{(CN)}_{4}}]}^{2-}},\,{{[NiC{{l}_{4}}]}^{2-}}\] are tetrahedral
C) \[{{[Ni\,{{(CN)}_{4}}]}^{2-}}\] is square planar and \[[Ni\,{{(C{{l}_{4}}]}^{2-}},Ni{{(CO)}_{4}}\] are tetrahedral
D) None of the above
Correct Answer: C
Solution :
\[{{[Ni{{(CN)}_{4}}]}^{2}}-\,ds{{p}^{2}}\to \] Square planar \[{{[NiC{{l}_{4}}]}^{2-}}-\,s{{p}^{3}}\to \] Tetrahedral \[Ni{{(CO)}_{4}}-\,s{{p}^{3}}\to \] TetrahedralYou need to login to perform this action.
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