A) \[M{{n}^{2+}}\]
B) \[F{{e}^{2+}}\]
C) \[T{{i}^{2+}}\]
D) \[C{{r}^{2+}}\]
Correct Answer: A
Solution :
[a] \[M{{n}^{2+}}\] \[_{25}Mn=[Ar]{{\,}^{18}}\,3{{d}^{5}}4{{s}^{2}}\] \[M{{n}^{2+}}=[Ar]{{\,}^{18}}\,3{{d}^{5}}\] \[_{20}Fe=[Ar]{{\,}^{18}}\,3{{d}^{6}}4{{s}^{2}}\] \[F{{e}^{2+}}=[Ar]{{\,}^{18}}\,3{{d}^{5}}\] \[_{22}T{{i}^{2+}}=[Ar]{{\,}^{18}}\,3{{d}^{2}}4{{s}^{2}}\] \[T{{i}^{2+}}=[Ar]{{\,}^{18}}\,3{{d}^{2}}\] \[_{24}C{{r}^{2+}}=[Ar]{{\,}^{18}}\,3{{d}^{5}}4{{s}^{1}}\] \[C{{r}^{2+}}=[Ar]{{\,}^{18}}\,3{{d}^{4}}\] Thus \[M{{n}^{2+}}\] has maximum number of unpaired electrons.
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