`I_(2)` cannot liberate `Cl_(2)` form aqueous KCI but it can liberate `Cl_(2)` form `KCIO_(3)`.

Explanation

To understand why \( I_2 \) cannot liberate \( Cl_2 \) from aqueous \( KCl \) but can do so from \( KClO_3 \), we need to analyze the oxidation states and the oxidizing power of the involved species.

### Step-by-Step Solution:

1. **Understanding the Reactants**:
- \( KCl \) (potassium chloride) contains \( Cl^- \) ions, where chlorine is in the -1 oxidation state.
- \( KClO_3 \) (potassium chlorate) contains \( Cl \) in the +5 oxidation state.

2. **Oxidizing Agent Comparison**:
- Chlorine (\( Cl_2 \)) is a strong oxidizing agent, meaning it can easily gain electrons and oxidize other substances.
- Iodine (\( I_2 \)) is a weaker oxidizing agent compared to chlorine. It cannot oxidize \( Cl^- \) (in \( KCl \)) to \( Cl_2 \) because it does not have sufficient oxidizing power.

3. **Reaction with \( KCl \)**:
- When \( I_2 \) is mixed with \( KCl \), the reaction would be:
\[ I_2 + 2KCl \rightarrow \text{(no reaction)} \]
- Since iodine is a weaker oxidizing agent, it cannot oxidize \( Cl^- \) to \( Cl_2 \).

4. **Reaction with \( KClO_3 \)**:
- In the case of \( KClO_3 \), the chlorine is in a higher oxidation state (+5) and can act as a stronger oxidizing agent.
- The reaction can be represented as:
\[ KClO_3 + I_2 \rightarrow KIO_3 + Cl_2 \]
- Here, iodine is oxidized to \( IO_3^- \) and chlorine is reduced from +5 to 0, resulting in the liberation of \( Cl_2 \).

5. **Conclusion**:
- The key reason \( I_2 \) cannot liberate \( Cl_2 \) from \( KCl \) but can from \( KClO_3 \) is due to the relative oxidizing strengths of the species involved. \( KClO_3 \) is a stronger oxidizing agent than \( I_2 \), allowing the reaction to proceed.