question_answer

The hybridisation involved in complex \[{{[Ni{{(CN)}_{4}}]}^{2-}}\]is (Atomic number of Ni = 28)  [NEET 2015 (Re)]

A) \[ds{{p}^{2}}\]

B) \[s{{p}^{3}}\] 

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

D) \[{{d}^{2}}s{{p}^{3}}\]

Correct Answer: A

Solution :

[a] \[{{[Ni(CN)4]}^{2-}}\]

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 \[ds{{p}^{2}}\]