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

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 \[F{{e}^{3+}}\] \[F{{e}^{3+}}=[Ar]3{{d}^{5}}4{{s}^{0}}\] \[{{[Fe{{(CN)}_{6}}]}^{3-}}=[Ar]\] Thus, its hybridisation is \[{{d}^{2}}s{{p}^{3}}\]not \[s{{p}^{3}}{{d}^{2}}\], ie, it is an inner orbital complex.  \[{{[Co{{(en)}_{3}}]}^{3+}}\]contains total 36 electrons, ie, follows EAN rule.