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 objective is to identify the most potent oxidizing agent among the provided ions. This requires an understanding of oxidizing agents.

An oxidizing agent facilitates the oxidation of other substances by accepting electrons, thereby undergoing reduction itself. The strength of an oxidizing agent is dictated by its electron affinity.

Analysis of each option:

• \({Lu}^{3+}\): Lutetium's completed f-electron shells (4f¹⁴) confer stability and a low propensity for electron acceptance.
• \({Eu}^{2+}\): Europium in the +2 oxidation state readily oxidizes to the more stable +3 state (Eu³⁺), classifying it as a moderate, not strong, oxidizing agent.
• \({Tb}^{4+}\): Terbium in the +4 oxidation state readily accepts an electron to achieve the stable +3 state (Tb³⁺), designating it as a strong oxidizing agent.
• \({Ce}^{3+}\): Cerium in the +3 oxidation state can be further oxidized to Ce⁴⁺ under specific conditions, acting as a weak oxidizing agent.

\({Tb}^{4+}\) is identified as the strongest oxidizing agent among the given ions due to its strong tendency to accept an electron, driven by the stability gained upon reduction to the +3 state.

Consequently, \({Tb}^{4+}\) is the correct answer.