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)  \[{{[Fe{{(C{{N}_{6}})}_{4}}]}^{3-}}\]       \[-s{{p}^{3}}{{d}^{2}}\]

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

Correct Answer: C

Solution :

                In \[{{[Cu{{(N{{H}_{3}})}_{4}}]}^{2+}}\], \[Cu\] 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.