Oxidation States of Lanthanides
The oxidation states of lanthanide elements are generally less predictable compared to the transition metals. However, there are some trends that can be observed.

Explanation:
- The oxidation states of the lanthanide elements are primarily determined by the number of valence electrons they possess.
- The lanthanide elements have partially filled 4f orbitals, which can accommodate up to 14 electrons.
- The most common oxidation state of the lanthanides is +3, where they lose all of their valence electrons.
- However, some lanthanides can exhibit other oxidation states as well.
- In the case of the elements given in the question, the stability of their +4 oxidation state can be analyzed.

Stability of +4 Oxidation State
To determine the stability of the +4 oxidation state, we need to consider the electronic configurations of the elements and their ability to lose four electrons.

La (Z = 57)
- The electron configuration of La is [Xe] 5d1 6s2.
- To achieve a +4 oxidation state, La would need to lose four electrons, resulting in a configuration of [Xe].
- However, losing four electrons would require a significant amount of energy, making the +4 oxidation state of La relatively unstable.

Eu (Z = 63)
- The electron configuration of Eu is [Xe] 4f7 6s2.
- To achieve a +4 oxidation state, Eu would need to lose four electrons, resulting in a configuration of [Xe] 4f3.
- The 4f orbitals are relatively stable, and losing electrons from these orbitals would require a significant amount of energy.
- Therefore, the +4 oxidation state of Eu is relatively unstable.

Sm (Z = 62)
- The electron configuration of Sm is [Xe] 4f6 6s2.
- To achieve a +4 oxidation state, Sm would need to lose four electrons, resulting in a configuration of [Xe] 4f2.
- The 4f orbitals are relatively stable, and losing electrons from these orbitals would require a significant amount of energy.
- Therefore, the +4 oxidation state of Sm is relatively unstable.

Ce (Z = 58)
- The electron configuration of Ce is [Xe] 4f1 5d1 6s2.
- To achieve a +4 oxidation state, Ce would need to lose four electrons, resulting in a configuration of [Xe].
- The 4f orbitals are relatively stable, and losing electrons from these orbitals would not require a significant amount of energy.
- Therefore, the +4 oxidation state of Ce is relatively stable.

Conclusion
Based on the analysis of the electronic configurations and stability of the +4 oxidation state, the correct answer is option 'B' - Ce (Z = 58). Ce is the only element among the given options that can form a stable +4 oxidation state.