A) \[C{{e}^{2+}}\]
B) \[S{{m}^{2+}}\]
C) \[E{{u}^{2+}}\]
D) \[Y{{b}^{2+}}\]
Correct Answer: D
Solution :
Lanthanoid ion with no unpaired electron is diamagnetic in nature. \[C{{e}_{58}}=[Xe]4{{f}^{2}}5{{d}^{0}}6{{s}^{2}}\] \[C{{e}^{2+}}=[Xe]4{{f}^{2}}\](Two unpaired electrons! \[S{{m}_{62}}=[Xe]4{{f}^{6}}5{{d}^{0}}6{{s}^{2}}\] \[S{{m}^{2+}}=[Xe]4{{f}^{6}}\](Six unpaired electrons) \[E{{u}_{63}}[Xe]4{{f}^{7}}5{{d}^{0}}6{{s}^{2}}\] \[E{{u}^{2+}}=[Xe]4{{f}^{7}}\](Seven unpaired electrons) \[Y{{b}_{70}}=[Xe]4{{f}^{14}}5{{d}^{0}}6{{s}^{2}}\] \[Y{{b}^{2+}}=[Xe]4{{f}^{14}}\](No unpaired electrons) Because of the absence of unpaired electrons, \[Y{{b}^{2+}}\]is diamagnetic.
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