`CuFeS_(2)+O_(2)("excess")overset(Delta)toX(s)+Y(s)`
Note: Corresponding sulphates of copper and iron are formed respectively.
which is correct. choose for X and Y?

Options

A. X liberates iodine from soluble iodides like Kl
B. Y on heating liberates `SO_(3)` only
C. X forms bluish white precipitate with sodium hydroxide which redissolves in excess of sodium hydroxide.
D. Y on reaction with potassium ferricyanide gives brown precipitate.

Explanation

To solve the question, we need to identify the products \(X\) and \(Y\) formed when \(CuFeS_2\) (chalcopyrite) reacts with excess oxygen upon heating. The note indicates that the corresponding sulfates of copper and iron are formed.

### Step-by-Step Solution:

1. **Write the Reaction**:
The initial reaction involves copper iron sulfide (\(CuFeS_2\)) reacting with oxygen (\(O_2\)). The balanced equation for the reaction is:
\[ CuFeS_2 + 4O_2 \rightarrow 2CuSO_4 + FeSO_4 \]
Here, \(CuSO_4\) is the sulfate of copper and \(FeSO_4\) is the sulfate of iron.

2. **Identify \(X\) and \(Y\)**:
From the balanced equation, we can identify:
- \(X = CuSO_4\) (copper sulfate)
- \(Y = FeSO_4\) (iron(II) sulfate)

3. **Verify the Statements**:
- **Statement 1**: "X liberates iodine from soluble iodides like KI."
- This is correct. Copper sulfate reacts with potassium iodide to release iodine.
- **Statement 2**: "Y on heating liberates only sulfur trioxide."
- This is incorrect. Iron(II) sulfate decomposes to form both sulfur dioxide and sulfur trioxide upon heating.
- **Statement 3**: "X forms bluish white precipitate with sodium hydroxide which re-dissolves in excess of sodium hydroxide."
- This is incorrect. Copper sulfate forms a blue precipitate of copper(II) hydroxide with sodium hydroxide, which does not dissolve in excess NaOH.
- **Statement 4**: "Y on reaction with potassium ferricyanide gives brown precipitate."
- This is incorrect. The reaction of iron(II) sulfate with potassium ferricyanide produces a deep blue precipitate, not brown.

4. **Conclusion**:
The only correct statement is that \(X\) liberates iodine, confirming that \(X = CuSO_4\) and \(Y = FeSO_4\).

### Final Answer:
- \(X = CuSO_4\) (copper sulfate)
- \(Y = FeSO_4\) (iron(II) sulfate)

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