question_answer

The compound having tetrahedral geometry is

A)  \[{{[Ni\,{{(CN)}_{4}}]}^{2-}}\]   

B)  \[{{[Pd\,{{(CN)}_{4}}]}^{2-}}\]

C)  \[{{[NiC{{l}_{4}}]}^{2-}}\]                            

D)  \[{{[PdC{{l}_{4}}]}^{2-}}\]

E)  None of these

Correct Answer: C

Solution :

                \[{{[NiC{{l}_{4}}]}^{2-}}\] \[N{{i}^{2+}}={{d}^{8}}\]           In \[NiC{{l}_{4}},\] pairing will not occur because \[C{{l}^{-}}\] is not a strong ligand. Therefore, it shows\[(s{{p}^{3}})\]tetrahedral geometry.