Which one is the correct observation when `Br_(2)` is treated with NaF, NaCl and NaI taken in three test tubes labelled as (I), (II) and (III) ?

Options

A. `F_(2)` is liberated in (I) and `Cl_(2)` in (II)
B. Only `I_(2)` is liberated in (III)
C. Only `Cl_(2)` is liberated in (II)
D. Only `F_(2)`is liberated in (I)

Explanation

To solve the question regarding the observation when \( \text{Br}_2 \) is treated with \( \text{NaF} \), \( \text{NaCl} \), and \( \text{NaI} \) in three test tubes labeled as (I), (II), and (III), we need to consider the reactivity of halogens and their ability to displace each other.

### Step-by-Step Solution:

1. **Understanding Halogen Reactivity:**
The reactivity of halogens decreases down the group in the periodic table. The order of reactivity is:
\[ \text{F} > \text{Cl} > \text{Br} > \text{I} \]
This means that fluorine is the most reactive halogen, followed by chlorine, bromine, and iodine.

2. **Analyzing the Reactions:**
- **Test Tube I (with NaF):**
When \( \text{Br}_2 \) is added to \( \text{NaF} \), bromine cannot displace fluorine because fluorine is more reactive than bromine. Therefore, no reaction occurs in this test tube.

- **Test Tube II (with NaCl):**
Similarly, when \( \text{Br}_2 \) is added to \( \text{NaCl} \), bromine cannot displace chlorine because chlorine is also more reactive than bromine. Thus, no reaction occurs here as well.

- **Test Tube III (with NaI):**
In this case, when \( \text{Br}_2 \) is added to \( \text{NaI} \), bromine can displace iodine because iodine is less reactive than bromine. This results in the formation of \( \text{NaBr} \) and the release of iodine.

3. **Conclusion:**
Based on the analysis:
- In Test Tube I (NaF): No reaction occurs.
- In Test Tube II (NaCl): No reaction occurs.
- In Test Tube III (NaI): Iodine is released.

Therefore, the correct observation is that only iodine will be released in the third test tube (Test Tube III).

### Final Answer:
**Only iodine will be released in Test Tube III.**