A) Zn
B) \[F{{e}^{2+}}\]
C) \[N{{i}^{3+}}\]
D) \[C{{u}^{+}}\]
Correct Answer: B
Solution :
\[Zn:1{{s}^{2}},2{{s}^{2}}2{{p}^{6}},3{{s}^{2}}3{{p}^{6}}3{{d}^{10}},4{{s}^{2}};\] no unpaired electron \[F{{e}^{2+}}:1{{s}^{2}},2{{s}^{2}}2{{p}^{6}},3{{s}^{2}}3{{p}^{6}}3{{d}^{6}};\] four unpaired electron \[N{{i}^{3+}}:1{{s}^{2}},\text{ }2{{s}^{2}}2{{p}^{6}},\text{ }3{{s}^{2}}3{{p}^{6}}3{{d}^{7}};\] three unpaired electrons \[C{{u}^{+}}:1{{s}^{2}},2{{s}^{2}}2{{p}^{6}},3{{s}^{2}}3{{p}^{6}}3{{d}^{10}};\] no unpaired electron Hence,\[F{{e}^{2+}}\]has maximum number of unpaired electrons.
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