A species is diamagnetic if it has no unpaired electrons. The electronic configurations of the given lanthanoid ions are as follows:

• $\text{Ce}$ ($Z=58$): $[\text{Xe}] 4f^1 5d^1 6s^2 \implies \text{Ce}^{2+}: [\text{Xe}] 4f^2$ (2 unpaired electrons, paramagnetic)
• $\text{Eu}$ ($Z=63$): $[\text{Xe}] 4f^7 6s^2 \implies \text{Eu}^{2+}: [\text{Xe}] 4f^7$ (7 unpaired electrons, highly paramagnetic)
• $\text{Sm}$ ($Z=62$): $[\text{Xe}] 4f^6 6s^2 \implies \text{Sm}^{2+}: [\text{Xe}] 4f^6$ (6 unpaired electrons, paramagnetic)
• $\text{Yb}$ ($Z=70$): $[\text{Xe}] 4f^{14} 6s^2 \implies \text{Yb}^{2+}: [\text{Xe}] 4f^{14}$ (0 unpaired electrons, diamagnetic)

Since $\text{Yb}^{2+}$ has a completely filled $f$-subshell and no unpaired electrons, it is diamagnetic.