Key idea: For lanthanoids, the divalent state $\text{Ln}^{2+}$ is especially stable when it gives a particularly stable $4f$ electronic configuration such as:

• half-filled: $4f^7$
• fully filled: $4f^{14}$

Check the electronic configurations of the given lanthanoids:

• Ce ($Z=58$): $\text{Ce} = [Xe]4f^15d^16s^2$ $\text{Ce}^{2+} \approx [Xe]4f^25d^0$ This is not especially stable.

• Sm ($Z=62$): $\text{Sm} = [Xe]4f^66s^2$ $\text{Sm}^{2+} = [Xe]4f^6$ This is somewhat stable, but not half-filled or fully filled.

• Eu ($Z=63$): $\text{Eu} = [Xe]4f^76s^2$ $\text{Eu}^{2+} = [Xe]4f^7$ This gives a half-filled $4f^7$ configuration, which is highly stable.

• Yb ($Z=70$): $\text{Yb} = [Xe]4f^{14}6s^2$ $\text{Yb}^{2+} = [Xe]4f^{14}$ This gives a fully filled $4f^{14}$ configuration, also highly stable.

Compare Eu$^{2+}$ and Yb$^{2+}$: Both are unusually stable in the divalent state. However, in standard JEE chemistry, Eu$^{2+}$ is regarded as the most stable/common divalent lanthanoid because of the exceptional stability of the half-filled $4f^7$ configuration.

Evaluate options:

• A: Ce — incorrect
• B: Sm — incorrect
• C: Eu — correct
• D: Yb — very stable, but not the best answer here

Final answer: $\boxed{\text{Eu (Option C)}}$

Comparison with stored correct answer: Stored answer is C, which matches the derived answer.