question_answer

The hybridization involved in complex \[{{[Ni{{(CN)}_{4}}]}^{2-}}\]is (Atomic number of Ni = 28)                                               

A)  \[ds{{p}^{2}}\]             

B)  \[s{{p}^{3}}\]                   

C)  \[{{d}^{2}}s{{p}^{2}}\]           

D)  \[{{d}^{2}}s{{p}^{3}}\]

Correct Answer: A

Solution :

\[{{[Ni(CN)4]}^{2-}}\] Let oxidation state of Ni in \[{{[Ni(CN)4]}^{2-}}\]is x. \[\therefore \]                 \[x-4=2\] or                            x = 2 Now,   \[N{{i}^{2+}}=[Ar],3{{d}^{8}},4{{s}^{0}}\] \[\because \]\[C{{N}^{-}}\] is a strong field ligand. Hence, all unpaired electrons are paired up. \[\therefore \] Hybridisation of \[{{[Ni{{(CN)}_{2}}]}^{2-}}\]is \[ds{{p}^{2}}\]