Question

Which of the following ions is the strongest oxidizing agent? (Atomic Number of Ce = 58, Eu = 63, Tb = 65, Lu = 71)

• A. \( {Lu}^{3+} \)
• B. \( {Eu}^{2+} \)
• C. \( {Tb}^{4+} \)
• D. \( {Ce}^{3+} \)

Hint:

Ions with a higher charge and smaller size are generally stronger oxidizing agents because they have a higher tendency to accept electrons and achieve a stable electronic configuration.

Answer 1

The Correct Option is C

The strength of an oxidizing agent depends on its ability to accept electrons. An ion with a higher positive charge and a smaller size tends to be a stronger oxidizing agent because it is more eager to accept electrons to achieve a stable electronic configuration. The ions in the given options are:
1. Lu\(^{3+}\): This ion is a 3+ cation of lanthanum (atomic number 71). While it is a highly charged ion, it is relatively large compared to other lanthanides.
2. Eu\(^{2+}\): This ion is the 2+ cation of europium (atomic number 63). The Eu\(^{2+}\) ion is in the +2 oxidation state, and it is a stronger reducing agent compared to its +3 oxidation state.
3. Tb\(^{4+}\): This ion is the 4+ cation of terbium (atomic number 65). The Tb\(^{4+}\) ion is highly charged, and it has a very high tendency to accept electrons, making it a strong oxidizing agent.
4. Ce\(^{3+}\): This ion is the 3+ cation of cerium (atomic number 58). The Ce\(^{3+}\) ion is also a strong oxidizing agent, but not as strong as the Tb\(^{4+}\) ion.
Thus, the strongest oxidizing agent among the given ions is Tb\(^{4+}\), which is the most eager to accept electrons due to its high charge and smaller size compared to the other ions.