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NEET 2013
Exam
NEET
Year
2013
Subject
Chemistry
Chapter
d and f Block Elements

NEET 2013 · Chemistry · d and f Block Elements

Which of the following lanthanoid ions is diamagnetic? (At. nos....

Which of the following lanthanoid ions is diamagnetic? (At. nos. Ce=58,Sm=62,Eu=63\mathrm{Ce}=58, \mathrm{Sm}=62, \mathrm{Eu}=63, Yb=70\mathrm{Yb}=70 )

Options

  1. ACe2+\mathrm{Ce}^{2+}
  2. BSm2+\mathrm{Sm}^{2+}
  3. CEu2+\mathrm{Eu}^{2+}
  4. DYb2+Yb^{2+}

Correct answer

D. Yb2+Yb^{2+}

Step-by-step solution

Lanthanoid ion with no unpaired electron is diamagnetic in nature. Ce58=[Xe]4f25d06s2Ce2+=[Xe]4f2 (two unpaired electrons) \begin{array}{ll} \mathrm{Ce}_{58} & =[\mathrm{Xe}] 4 f^2 5 d^0 6 s^2 \\ C e^{2+} & =[\mathrm{Xe}] 4 f^2 \quad \text{ (two unpaired electrons) } \end{array} Sm62=[Xe]4f65d06s2Sm2+=[Xe]4f6(six unpaired electrons) Eu63=[Xe]4f75d06s2Eu2+=[Xe]4f7 (seven unpaired electrons) Yb70=[Xe]4f145d06s2Yb2+=[Xe]4f14 (no unpaired electrons) \begin{aligned} \mathrm{Sm}_{62} & =[\mathrm{Xe}] 4 f^6 5 d^0 6 s^2 \\ \mathrm{Sm}^{2+} & =[\mathrm{Xe}] 4 f^6 \quad(s i x \text{ unpaired electrons) } \\ \mathrm{Eu}_{63} & =[\mathrm{Xe}] 4 f^7 5 d^0 6 s^2 \\ \mathrm{Eu}^{2+} & =[\mathrm{Xe}] 4 f^7 \quad \text{ (seven unpaired electrons) } \\ \mathrm{Yb}_{70} & =\left[\mathrm{X}_e\right] 4 f^{14} 5 d^0 6 s^2 \\ \mathrm{Yb}^{2+} & =[\mathrm{Xe}] 4 f^{14} \quad \text{ (no unpaired electrons) } \end{aligned} Because of the absence of unpaired electrons, Yb2+\mathrm{Yb}^{2+} is diamagnetic.
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