question_answer

Which one of the following is wrongly matched?

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

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

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

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

Correct Answer: C

Solution :

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