Which of the following lanthanoid ions is diamagnetic? (At. nos. $\mathrm{Ce}=58, \mathrm{Sm}=62,…

Which of the following lanthanoid ions is diamagnetic? (At. nos. $\mathrm{Ce}=58, \mathrm{Sm}=62, \mathrm{Eu}=63$, $\mathrm{Yb}=70$ )
  1. $\mathrm{Ce}^{2+}$
  2. $\mathrm{Sm}^{2+}$
  3. $\mathrm{Eu}^{2+}$
  4. $Yb^{2+}$

Solution

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

Asked in: NEET 2013 (All India)

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