A) \[C{{o}^{2+}}\]
B) \[N{{i}^{2+}}\]
C) \[C{{u}^{2+}}\]
D) \[Z{{n}^{2+}}\]
Correct Answer: D
Solution :
Transition metal ion having absence of unpaired electron is diamagnetic. [a]\[C{{o}^{2+}}=[Ar]3{{d}^{7}}\](three unpaired electrons) [b]\[N{{i}^{2+}}=[Ar]3{{d}^{8}}\](two unpaired electrons) [c] \[C{{u}^{2+}}=[Ar]3{{d}^{9}}\] (one unpaired electron) [d]\[Z{{n}^{2+}}=[Ar]3{{d}^{10}}\](no unpaired electron) Thus,\[Z{{n}^{2+}}\]is diamagnetic.
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