A) \[Z{{n}^{+}}\]
B) \[F{{e}^{2+}}\]
C) \[{{N}^{3+}}\]
D) \[C{{u}^{+}}\]
Correct Answer: B
Solution :
\[Zn_{(At.\,No.\,=30)}^{+}=1{{s}^{2}},\,2{{s}^{2}}2{{p}^{6}},\,3{{s}^{2}}3{{p}^{6}}3{{d}^{10}},\,4{{s}^{1}}\] \[Fe_{(At.\,No.\,26)}^{2+}=1{{s}^{2}},\,2{{s}^{2}}2{{p}^{6}},\,3{{s}^{2}}3{{p}^{6}}3{{d}^{6}}\] \[N_{(At.\,No.=7)}^{3+}=1{{s}^{2}},2{{s}^{2}}\,2{{p}^{0}}\] \[Cu_{(At.\,No=29)}^{+}=1{{s}^{2}},2{{s}^{2}}2{{p}^{6}},3{{s}^{2}}3{{p}^{6}}3{{d}^{10}}\] So, maximum number of unpaired electrons are present in \[F{{e}^{2+}}\].You need to login to perform this action.
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