A) \[C{{u}^{+}}\]
B) Zn
C) \[F{{e}^{2+}}\]
D) \[N{{i}^{3+}}\]
Correct Answer: C
Solution :
(i) \[C{{u}^{+}}(Z=29)\,\,\,\,[Ar]\,\,3{{d}^{10}}4{{s}^{0}}\] It has no unpaired electron. (ii) \[Zn\,\,(Z=30)\,\,\,\,[Ar]\,\,3{{d}^{10}}4{{s}^{2}}\] It has no unpaired electron. (iii) \[F{{e}^{2+}}\,\,(Z=26)\,\,\,\,[Ar]\,\,3{{d}^{6}}4{{s}^{0}}\] It has four unpaired electrons. (iv) \[N{{i}^{3+}}\,\,(Z=28)\,:\,\,\,[Ar]\,\,3{{d}^{7}}4{{s}^{0}}\] It has three unpaired electrons Therefore \[F{{e}^{2+}}\] has maximum number of unpaired electrons Note - A paramagnetic species has unpaired electrons. A diamagnetic species has no unpaired electrons.
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