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.You need to login to perform this action.
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