Idea: A species is a good oxidizing agent if it gets reduced easily, i.e. it has a strong tendency to gain electrons.

In lanthanoids, the common stable oxidation state is $+3$. So:

• Species in higher oxidation states like $+4$ tend to get reduced to $+3$ and hence can act as oxidizing agents.
• Species in lower oxidation states like $+2$ tend to get oxidized to $+3$ and hence usually act as reducing agents, not oxidizing agents.

Now examine each species:

A. $\mathrm{Sm}^{2+}$

• $\mathrm{Sm}^{2+}$ tends to get oxidized to $\mathrm{Sm}^{3+}$.
• Therefore, it behaves as a reducing agent, not an oxidizing agent.
• Not correct.

B. $\mathrm{Ce}^{2+}$

• $\mathrm{Ce}^{2+}$ is an unusually low oxidation state and tends to get oxidized to $\mathrm{Ce}^{3+}$.
• So it is also a reducing agent.
• Not correct.

C. $\mathrm{Ce}^{4+}$

• $\mathrm{Ce}^{4+}$ readily gets reduced to $\mathrm{Ce}^{3+}$.
• Hence, it is a good oxidizing agent.
• Correct.

D. $\mathrm{Tb}^{4+}$

• $\mathrm{Tb}^{4+}$ also tends to get reduced to $\mathrm{Tb}^{3+}$.
• Therefore, it acts as an oxidizing agent.
• Correct.

Thus, the good oxidizing agents are: $\mathrm{Ce}^{4+} \text{ and } \mathrm{Tb}^{4+}$

So the correct option is: $\boxed{\text{A: C and D only}}$

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