question_answer

Which one of the following is wrongly matched?

A) \[{{\text{ }\!\![\!\!\text{ Cu(N}{{\text{H}}_{\text{3}}}{{\text{)}}_{\text{4}}}\text{ }\!\!]\!\!\text{ }}^{\text{2+}}}\] -Square planar

B) \[\text{ }\!\![\!\!\text{ Ni(CO}{{\text{)}}_{\text{4}}}\text{ }\!\!]\!\!\text{ }\] - Neutral ligand

C) \[\text{ }\!\![\!\!\text{ Fe(CN)}_{6}^{3-}\text{ }\!\!]\!\!\text{ }\]   - \[\text{ }\!\![\!\!\text{ Fe(CN)}_{6}^{3-}\text{ }\!\!]\!\!\text{ s}{{\text{p}}^{3}}{{d}^{2}}\]

D) \[{{\text{ }\!\![\!\!\text{ Co(en}{{\text{)}}_{3}}\text{ }\!\!]\!\!\text{ }}^{3+}}\]- Follows EAN rule

Correct Answer: C

Solution :

(a)\[{{[Cu{{(N{{H}_{3}})}_{4}}]}^{2+}},Cu\] is present as \[C{{u}^{2+}}C{{u}^{2+}}=[Ar]3{{d}^{9}}4{{s}^{0}}\] \[{{[Cu{{(N{{H}_{3}})}_{4}}]}^{2+}}=[Ar]\] (\[N{{H}_{3}}\] being a strong field ligand shifts one electron from 3d orbital to 4p orbital.)  In \[[Ni{{(CO)}_{4}}],CO\]is a neutral ligand. In\[{{[Fe{{(CN)}_{6}}]}^{3-}},\] Fe is present as\[F{{e}^{3+}}\] \[F{{e}^{3+}}=[Ar]3{{d}^{5}}4{{s}^{0}}\] \[{{[Fe(CN)6]}^{3-}}=[Ar]\] Thus. its hybridisation is \[{{d}^{2}}s{{p}^{3}}\]not\[s{{p}^{3}}{{d}^{2}},\] i.e., it is an inner orbital complex. \[{{[Co{{(en)}_{3}}]}^{3+}}\] contains total 36 electrons, i.e., follows EAN rule.