Identify lanthanoids known to show the +4 oxidation state

In the lanthanoid series, the most common oxidation state is $+3$. However, a few lanthanoids can also show $+4$ due to extra stability associated with certain electronic configurations.

Common examples of lanthanoids showing $+4$ are:

• $\mathrm{Ce}^{4+}$
• $\mathrm{Tb}^{4+}$
• $\mathrm{Pr}^{4+}$ in some compounds

Check each option

Option A: $\mathrm{Tb}$

Terbium can show the $+4$ oxidation state, for example in $\mathrm{TbO_2}$.

So, Tb does show $+4$.

Option B: $\mathrm{Dy}$

Dysprosium generally shows the $+3$ oxidation state. It is not known to show a stable $+4$ oxidation state in the standard lanthanoid chemistry discussed for JEE.

So, Dy does not show $+4$.

Option C: $\mathrm{Ce}$

Cerium commonly shows the $+4$ oxidation state, for example in $\mathrm{CeO_2}$.

So, Ce does show $+4$.

Option D: $\mathrm{Eu}$

Europium commonly shows $+2$ and $+3$ oxidation states. It is not known to show $+4$.

So, Eu does not show $+4$.

Conclusion

Among the given options, both $\mathrm{Dy}$ and $\mathrm{Eu}$ do not show the $+4$ oxidation state in standard JEE-level chemistry.

Therefore, the question as a single-correct MCQ is problematic because it has more than one valid answer.

Comparison with stored answer

The stored correct answer is D (Eu).

But $\mathrm{Dy}$ also does not show $+4$ oxidation state. Hence I do not fully agree with the stored answer as the question appears ambiguous or incorrectly framed.