The most commonly used reducing agent is

Options

A. `AlCl_(3)`
B. `PbCl_(2)`
C. `SnCl_(4)`
D. `SnCl_(2)`

Explanation

To determine the most commonly used reducing agent from the given options, we need to analyze each compound based on its ability to get oxidized. A reducing agent is a substance that donates electrons to another substance, thereby reducing it while itself getting oxidized.

### Step-by-Step Solution:

1. **Identify the Compounds**: The options given are AlCl3, PbCl2, SnCl4, and SnCl2.

2. **Analyze AlCl3**:
- **Oxidation State**: In AlCl3, aluminum (Al) has an oxidation state of +3 (since each Cl is -1, and there are three Cl atoms).
- **Oxidation Potential**: Aluminum cannot be oxidized further because +3 is its maximum oxidation state. Therefore, AlCl3 cannot act as a reducing agent.
- **Conclusion**: AlCl3 is not a good reducing agent.

3. **Analyze PbCl2**:
- **Oxidation State**: In PbCl2, lead (Pb) has an oxidation state of +2.
- **Inert Pair Effect**: As we move down the group, the inert pair effect makes the +2 state more stable than the +4 state. Thus, Pb prefers to stay in the +2 oxidation state and does not easily get oxidized to +4.
- **Conclusion**: PbCl2 is not a good reducing agent.

4. **Analyze SnCl4**:
- **Oxidation State**: In SnCl4, tin (Sn) has an oxidation state of +4.
- **Oxidation Potential**: This is the maximum oxidation state for tin, and it cannot be oxidized further. Therefore, SnCl4 cannot act as a reducing agent.
- **Conclusion**: SnCl4 is not a good reducing agent.

5. **Analyze SnCl2**:
- **Oxidation State**: In SnCl2, tin (Sn) has an oxidation state of +2.
- **Oxidation Potential**: Sn can be oxidized from +2 to +4, which means it can donate electrons to reduce another substance.
- **Conclusion**: SnCl2 can act as a reducing agent.

6. **Final Answer**: Among the given options, SnCl2 is the most commonly used reducing agent.