JEE Chemistry 2025 Question with Solution

Given: The ions are $Lu^{3+}$, $Eu^{2+}$, $Tb^{4+}$ and $Ce^{3+}$.

Find: The strongest oxidizing agent.

An oxidizing agent accepts electrons and gets reduced. So the strongest oxidizing agent will be the ion that has the greatest tendency to gain an electron and move to a more stable oxidation state.

From the solution analysis:

• $Lu^{3+}$ has a stable filled $4f^{14}$ configuration, so it has low tendency to accept electrons.
• $Eu^{2+}$ tends to go to the more stable $Eu^{3+}$ state, so it behaves as a reducing agent rather than an oxidizing agent.
• $Tb^{4+}$ readily gains an electron to form the more stable $Tb^{3+}$ state, so it acts as a strong oxidizing agent.
• $Ce^{3+}$ is not as strong an oxidizing agent as $Tb^{4+}$.

Thus, among the given ions, $Tb^{4+}$ has the highest tendency to accept an electron.

Therefore, the correct option is C.

Given: Strength of an oxidizing agent depends on how readily the species accepts electrons.

Find: Which ion among the given lanthanide ions is the strongest oxidizing agent.

A higher positive charge generally increases electron-accepting tendency. Also, if reduction leads to a more stable oxidation state, the oxidizing character becomes stronger.

Evaluate the ions one by one:

1. $Lu^{3+}$: This ion corresponds to a very stable filled $4f^{14}$ arrangement, so it is not strongly inclined to accept an electron.
2. $Eu^{2+}$: This ion prefers oxidation to $Eu^{3+}$, so it is a reducing agent, not the strongest oxidizing agent.
3. $Tb^{4+}$: This highly charged ion can accept an electron and convert to the more stable $Tb^{3+}$ state. This makes it strongly oxidizing.
4. $Ce^{3+}$: It is weaker as an oxidizing agent than $Tb^{4+}$.

Hence, the ion with the greatest tendency to be reduced is $Tb^{4+}$.

Therefore, the strongest oxidizing agent is $Tb^{4+}$, so the correct option is C.