A) \[Z{{n}^{2+}}\]
B) \[F{{e}^{2+}}\]
C) \[N{{i}^{3+}}\]
D) \[C{{u}^{+}}\]
Correct Answer: B
Solution :
\[Zn(30)=[Ar]\,\,3{{d}^{10}},4{{s}^{2}}\] \[Z{{n}^{2+}}=[Ar]\,3{{d}^{10}}\](no unpaired electron) \[Fe(26)=[Ar]\,3{{d}^{8}},4{{s}^{2}}\] \[F{{e}^{2+}}=[Ar]3{{d}^{6}}\] (four unpaired d electrons) \[Ni(28)=[Ar]3{{d}^{8}},4{{s}^{2}}\] \[N{{i}^{3+}}=[Ar]3{{d}^{7}}\](three unpaired d electrons) \[Cu(29)=[Ar]3{{d}^{10}},4{{s}^{1}}\] \[C{{u}^{+}}=[Ar]3{{d}^{10}}\](no unpaired electron)You need to login to perform this action.
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