The addition of a crystal of `I_(2)` to `NaBr` turns the solution violet.

Explanation

To solve the question regarding the addition of a crystal of iodine (I₂) to sodium bromide (NaBr) and the resulting violet color, we can break down the explanation into several steps.

### Step-by-Step Solution:

1. **Understanding the Components**:
- Sodium bromide (NaBr) is an ionic compound made up of sodium ions (Na⁺) and bromide ions (Br⁻).
- Iodine (I₂) is a diatomic molecule that is known for its characteristic violet color when dissolved in a solvent.

2. **Addition of Iodine to NaBr**:
- When a crystal of iodine is added to a solution of sodium bromide, the iodine does not react with the sodium bromide. This is because iodine is less reactive compared to bromine and cannot displace bromine from its compound.

3. **Color Observation**:
- The violet color observed in the solution is due to the presence of iodine (I₂) itself. Iodine has a natural violet color, and when it is dissolved in the solution, it imparts this color to the solution.

4. **Comparison with Other Halides**:
- If sodium fluoride (NaF) or sodium chloride (NaCl) were used instead of sodium bromide, the same violet color would be observed. This is because iodine cannot displace any of the other halogens (F or Cl) from their respective salts due to its lower reactivity.

5. **Conclusion**:
- Therefore, the statement that the addition of iodine to sodium bromide turns the solution violet is true. The solution remains violet because iodine does not react with sodium bromide and retains its characteristic color.