When `Br_2` is treated with aqueous solutions of NaF, NaCl and NaI separately

Options

A. `F_(2),Cl_(2) ` and `I_(2)` are liberated
B. Only `F_(2)` and `Cl_(2)` are liberated
C. Only `I_(2)` is liberated
D. Only `Cl_(2)` is liberated

Explanation

To solve the problem of what happens when \( \text{Br}_2 \) is treated with aqueous solutions of \( \text{NaF} \), \( \text{NaCl} \), and \( \text{NaI} \) separately, we need to consider the reactivity of bromine with these halide ions.

### Step 1: Understanding the Reactivity of Halogens
Halogens are known to exhibit a trend in reactivity. In general, a more reactive halogen can displace a less reactive halogen from its compound. The reactivity trend of halogens is as follows:
\[ \text{F} > \text{Cl} > \text{Br} > \text{I} \]

### Step 2: Reaction of \( \text{Br}_2 \) with \( \text{NaF} \)
When \( \text{Br}_2 \) is treated with \( \text{NaF} \):
- Bromine (\( \text{Br}_2 \)) is less reactive than fluorine (\( \text{F}_2 \)).
- Therefore, bromine cannot displace fluorine from sodium fluoride.
- No reaction occurs.

**Result:** No reaction with \( \text{NaF} \).

### Step 3: Reaction of \( \text{Br}_2 \) with \( \text{NaCl} \)
When \( \text{Br}_2 \) is treated with \( \text{NaCl} \):
- Bromine is more reactive than chlorine.
- Thus, bromine can displace chlorine from sodium chloride.
- The reaction can be represented as:
\[ \text{Br}_2 + 2 \text{NaCl} \rightarrow 2 \text{NaBr} + \text{Cl}_2 \]

**Result:** \( \text{Cl}_2 \) is produced, and \( \text{NaBr} \) is formed.

### Step 4: Reaction of \( \text{Br}_2 \) with \( \text{NaI} \)
When \( \text{Br}_2 \) is treated with \( \text{NaI} \):
- Bromine is less reactive than iodine.
- Therefore, bromine cannot displace iodine from sodium iodide.
- No reaction occurs.

**Result:** No reaction with \( \text{NaI} \).

### Summary of Reactions:
1. **With \( \text{NaF} \)**: No reaction.
2. **With \( \text{NaCl} \)**: Reaction occurs, producing \( \text{NaBr} \) and \( \text{Cl}_2 \).
3. **With \( \text{NaI} \)**: No reaction.

### Final Answer:
- \( \text{Br}_2 + \text{NaF} \) → No reaction
- \( \text{Br}_2 + \text{NaCl} \) → \( 2 \text{NaBr} + \text{Cl}_2 \)
- \( \text{Br}_2 + \text{NaI} \) → No reaction