question_answer

Ther hydridizations of \[Ni\,{{(CN)}_{4}}\] and \[Cr\,({{H}_{2}}O)_{6}^{2+},\]respectively.

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

B) \[d\,s{{p}^{2}}\] and \[{{d}^{2}}s{{p}^{3}}\]

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

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

Correct Answer: C

Solution :

\[Ni\,{{(CN)}_{4}}Ni:[Ar]\,3{{d}^{8}}4{{s}^{2}}=t_{2g}^{2,\,2,\,2}e_{g}^{2,\,2}\]

\[C{{N}^{-}}\]is a strong filed ligand, hence it will cause pairing of electrons. Hence, \[ds{{p}^{2}}.\]

octahedral complex: \[{{d}^{2}}s{{p}^{3}}\]