Question

Sol. [Key Concept: Diamagnetic f⁰ and f¹⁴ ions]

Ce⁴⁺ has the closed shell [Xe], equivalently 4f⁰, and Yb²⁺ has [Xe]4f¹⁴. Both have no unpaired electrons and are diamagnetic, so option (3) is correct. Ce³⁺ is [Xe]4f¹, not 4f³. Quantitative magnetic moments of other lanthanoid ions generally require orbital and spin–orbit contributions; the spin only formula should not be used as a general measured-moment formula for them.

Magnetic moment $\mu = \sqrt{n(n + 2)}$

$$\begin{matrix} & n \rightarrow \text{~number of unpaired electron~} \\ & {Ce}^{4 +} \Rightarrow (Xe)4f^{0} \\ & \mu = 0 \\ & {Yb}^{2 +} \Rightarrow (Xe)4f^{14} \\ & \mu = 0 \\ & {Ce}^{3 +} \Rightarrow (Xe)4f^{(3)} \\ & \mu = \sqrt{3} \\ & {Eu}^{2 +} \Rightarrow (Xe)4f^{7} \\ & \mu = \sqrt{63} \\ & {Gd}^{3 +} \Rightarrow (Xe)4f^{7} \\ & \mu = \sqrt{63} \\ & {Eu}^{3 +} \Rightarrow (Xe)4f^{6} \\ & \mu = \sqrt{48} \\ & \text{~Paramagnetic~} \end{matrix}$$

Hence ${Ce}^{4 +}$ and ${Yb}^{2 +}$ are only diamagnetic.