Which of the following does not liberate `Br_(2)` form KBr ?

Options

A. `I_(2)`
B. `CI_(2)`
C. conc. `H_(2)SO_(4)`
D. `F_(2)`

Explanation

To determine which of the following does not liberate \( \text{Br}_2 \) from \( \text{KBr} \), we need to analyze the reactivity of the given substances with potassium bromide (KBr).

### Step-by-Step Solution:

1. **Identify the Options**:
We need to evaluate the given options, which typically include halogens and possibly other substances. Let's assume the options are:
- A) Iodine (I2)
- B) Chlorine (Cl2)
- C) Fluorine (F2)
- D) Concentrated H2SO4

2. **Reactivity of Iodine (I2)**:
- Iodine is a weaker oxidizing agent compared to bromine. Therefore, it cannot displace bromine from potassium bromide.
- **Conclusion**: Iodine does not liberate \( \text{Br}_2 \) from \( \text{KBr} \).

3. **Reactivity of Chlorine (Cl2)**:
- Chlorine is a stronger oxidizing agent than bromine. It can displace bromine from potassium bromide.
- **Reaction**:
\[ 2 \text{KBr} + \text{Cl}_2 \rightarrow 2 \text{KCl} + \text{Br}_2 \]
- **Conclusion**: Chlorine does liberate \( \text{Br}_2 \) from \( \text{KBr} \).

4. **Reactivity of Fluorine (F2)**:
- Fluorine is the strongest oxidizing agent among the halogens and can also displace bromine from potassium bromide.
- **Reaction**:
\[ 2 \text{KBr} + \text{F}_2 \rightarrow 2 \text{KF} + \text{Br}_2 \]
- **Conclusion**: Fluorine does liberate \( \text{Br}_2 \) from \( \text{KBr} \).

5. **Reactivity of Concentrated H2SO4**:
- Concentrated sulfuric acid reacts with potassium bromide to form hydrogen bromide (HBr), which is a strong reducing agent. HBr can further reduce sulfuric acid to produce bromine.
- **Reaction**:
\[ \text{KBr} + \text{H}_2\text{SO}_4 \rightarrow \text{HBr} + \text{KHSO}_4 \]
\[ \text{HBr} + \text{H}_2\text{SO}_4 \rightarrow \text{SO}_2 + \text{H}_2\text{O} + \text{Br}_2 \]
- **Conclusion**: Concentrated H2SO4 does liberate \( \text{Br}_2 \) from \( \text{KBr} \).

6. **Final Conclusion**:
- Among the options, only iodine does not liberate \( \text{Br}_2 \) from \( \text{KBr} \). Therefore, the correct answer is **A) Iodine (I2)**.