A) 5
B) 4
C) 3 a
D) 2
Correct Answer: B
Solution :
\[{}_{26}Fe=1{{s}^{2}},2{{s}^{2}}2{{p}^{6}},3{{s}^{2}}3{{p}^{6}}3{{d}^{6}},4{{s}^{2}}\] \[F{{e}^{2+}}=[Ar]\,3{{d}^{6}}\] \[\] Hence, unpaired electrons in \[F{{e}^{2+}}\] (ferrous ion) = 4
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