question_answer

There hybridizations of \[Ni{{(CN)}^{4}}\] and \[Cr{{({{H}_{2}}O)}_{6}}^{2+},\] respectively =

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

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

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

D) \[ds{{p}^{2}}\] 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 field ligand, hence it will cause pairing of electrons. Hence, \[ds{{p}^{2}}\]

\[Cr{{({{H}_{2}}O)}_{{{6}^{+2}}}}C{{r}^{2}}:[Ar]3{{d}^{4}}4s{}^\circ \]

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