A) \[C{{e}^{4+}}\]
B) \[Y{{b}^{2+}}\]
C) \[L{{u}^{3+}}\]
D) \[E{{u}^{2+}}\]
Correct Answer: D
Solution :
Lanthanoid ions having the unpaired electrons are paramagnetic. \[C{{e}^{4+}}={{[Xe]}_{54}}\] (No unpaired electron) \[Y{{b}^{2+}}=[Xe]4{{f}^{14}}\] (No unpaired electron) \[L{{u}^{3+}}=[Xe]4{{f}^{14}}\] (No unpaired electron) \[E{{u}^{2+}}=[Xe]4{{f}^{7}}\](Seven unpaired electrons) Hence,\[E{{u}^{2+}}\]paramagnetic lanthanoid ion.
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