A) \[\text{C}{{\text{e}}^{\text{2+}}}\]
B) \[\text{S}{{\text{m}}^{\text{2+}}}\]
C) \[\text{E}{{\text{u}}^{\text{2+}}}\]
D) \[\text{Y}{{\text{b}}^{\text{2+}}}\]
Correct Answer: D
Solution :
Lanthanide 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) \[E{{u}_{63}}=[Xe]4{{f}^{7}}5{{d}^{0}}6{{s}^{2}}\] \[E{{u}_{63}}=[Xe]4{{f}^{7}}\] (seven unpaired electrons) \[Y{{b}_{70}}=[Xe]4{{f}^{7}}5{{d}^{0}}6{{s}^{2}}\] \[Y{{b}^{2+}}=[Xe]4{{f}^{14}}\] (no unpaired electrons) Because of the absence of unpaired electrons, \[Y{{b}^{2+}}\] is diamagnetic.
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