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) \[Fe(26)=[Ar]\,3{{d}^{6}},4{{s}^{2}}\] \[F{{e}^{2+}}=[Ar]\,\,3{{d}^{6}}\] (four unpaired d electrons)You need to login to perform this action.
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