A) \[M{{n}^{2+}}\]
B) \[F{{e}^{2+}}\]
C) \[C{{o}^{2+}}\]
D) \[N{{i}^{2+}}\]
Correct Answer: A
Solution :
\[M{{n}^{2+}}(Z=25):[Ar]\,\,3{{d}^{5}}4{{s}^{0}};5\]unpaired electrons \[F{{e}^{2+}}(Z=26):[Ar]\,\,3{{d}^{6}}4{{s}^{0}};4\]unpaired electrons \[C{{o}^{2+}}(Z=27):[Ar]\,\,3{{d}^{7}}4{{s}^{0}};3\]unaired electrons \[N{{i}^{2+}}(Z=28):[Ar]\,\,3{{d}^{8}}4{{s}^{0}};2\]unpaired electrons Hence,\[M{{n}^{2+}}\]cation Has the maximum number of unpaired electrons. .
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