A) \[C{{o}^{2+}}\]
B) \[C{{u}^{2+}}\]
C) \[M{{n}^{2+}}\]
D) \[S{{c}^{3+}}\]
Correct Answer: D
Solution :
The ion in which unpaired electrons are not present, is called diamagnetic. \[C{{o}^{2+}}=[Ar]3{{d}^{7}},4{{s}^{0}}\](three unpaired electrons) \[C{{u}^{2+}}=[Ar]3{{d}^{9}},4{{s}^{0}}\](one unpaired electron) \[S{{c}^{3+}}=[Ar]\,3{{d}^{0}},4{{s}^{0}}\](no unpaired electron) \[M{{n}^{2+}}=[Ar]\,3{{d}^{5}},4{{s}^{0}}\](Five unpaired electrons) Hence,\[S{{c}^{3+}}\]is diamagnetic ion due to absence of unpaired electrons.
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