Why `Sm^(2+) , Eu^(2+)` and `yb^(2+)` ions in solution are good reducing agent but an aqeius solution of `Ce^(4+)` is a good oxidising agent?

Answer

To understand why `Sm^(2+)`, `Eu^(2+)`, and `Yb^(2+)` ions in solution are good reducing agents while an aqueous solution of `Ce^(4+)` is a good oxidizing agent, we can analyze the oxidation states and their tendencies to gain or lose electrons.

### Step-by-Step Solution:

1. **Identify the Oxidation States**:
- The ions in question are `Sm^(2+)`, `Eu^(2+)`, `Yb^(2+)`, and `Ce^(4+)`.
- The most common oxidation state for lanthanides is +3.

2. **Understanding Reducing Agents**:
- A reducing agent is a substance that donates electrons and gets oxidized in the process.
- `Sm^(2+)`, `Eu^(2+)`, and `Yb^(2+)` can easily lose an electron to convert to their +3 oxidation state (`Sm^(3+)`, `Eu^(3+)`, `Yb^(3+)`).
- This tendency to lose an electron makes them good reducing agents.

3. **Understanding Oxidizing Agents**:
- An oxidizing agent is a substance that accepts electrons and gets reduced in the process.
- `Ce^(4+)` has a strong tendency to gain an electron to convert to its +3 oxidation state (`Ce^(3+)`).
- This ability to gain an electron makes `Ce^(4+)` a good oxidizing agent.

4. **Conclusion**:
- `Sm^(2+)`, `Eu^(2+)`, and `Yb^(2+)` are good reducing agents because they can easily lose an electron to achieve a more stable +3 state.
- `Ce^(4+)` is a good oxidizing agent because it readily accepts an electron to achieve the +3 state.

### Summary:
- `Sm^(2+)`, `Eu^(2+)`, and `Yb^(2+)` → Good reducing agents (they oxidize to +3).
- `Ce^(4+)` → Good oxidizing agent (it reduces to +3).