question_answer

Among \[{{[Ni{{(CN)}_{4}}]}^{2-}},\,{{[NiC{{l}_{4}}]}^{2-}}\] and \[[Ni\,\,{{(CO)}_{4}}]\]

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 \[Ni{{(CO)}_{4}}-\,s{{p}^{3}}\to \] Tetrahedral