Assertion : `Sb_(2)S_(3)` is not soluble in yellow ammonium sulphide.
Reason : The common ion effect due to `S^(2)` ions reduces the solubility of `Sb_(2)S_(3)`

Options

A. if both Assertion and Reason are true and reason is the correct explanation of the assertion.
B. if both assertion and reason are true but reason is not the correct expalanation of the assertion
C. if assertion is true, reason is false
D. both assertion and reason are false statements.

Explanation

To solve the question, we need to analyze both the assertion and the reason provided.

### Step-by-Step Solution:

1. **Understanding the Assertion:**
- The assertion states that \( \text{Sb}_2\text{S}_3 \) is not soluble in yellow ammonium sulfide.
- Yellow ammonium sulfide is a solution that contains \( \text{S}^{2-} \) ions.

2. **Solubility of \( \text{Sb}_2\text{S}_3 \):**
- \( \text{Sb}_2\text{S}_3 \) (antimony sulfide) is known to have limited solubility in water but can dissolve in ammonium sulfide due to the formation of complex ions.
- The reaction can be represented as:
\[ \text{Sb}_2\text{S}_3 + 3 \text{NH}_4\text{S} \rightarrow 2 \text{NH}_4\text{SbS}_3 \]
- This indicates that \( \text{Sb}_2\text{S}_3 \) is indeed soluble in ammonium sulfide.

3. **Understanding the Reason:**
- The reason states that the common ion effect due to \( \text{S}^{2-} \) ions reduces the solubility of \( \text{Sb}_2\text{S}_3 \).
- The common ion effect implies that the presence of a common ion (in this case, \( \text{S}^{2-} \)) in a solution will shift the equilibrium of a sparingly soluble salt to the left, thus reducing its solubility.
- However, since \( \text{Sb}_2\text{S}_3 \) is soluble in ammonium sulfide, the presence of \( \text{S}^{2-} \) does not reduce its solubility in this context.

4. **Conclusion:**
- Both the assertion and the reason are incorrect:
- The assertion is false because \( \text{Sb}_2\text{S}_3 \) is soluble in yellow ammonium sulfide.
- The reason is also false because the common ion effect does not apply in this case as \( \text{Sb}_2\text{S}_3 \) can dissolve in the presence of \( \text{S}^{2-} \).

### Final Answer:
Both the assertion and the reason are false.