Question:
A compound (X) when treated with \( \text{CuSO}_4 \) solution yields a brown precipitate. On adding hypo solution the precipitate turns white. The compound (X) is
A compound (X) when treated with \( \text{CuSO}_4 \) solution yields a brown precipitate. On adding hypo solution the precipitate turns white. The compound (X) is
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This is a standard volumetric/qualitative test for copper. Remember that \( \text{Cu}^{2+} \) does not form a stable \( \text{CuI}_2 \) because \( \text{I}^- \) is a strong reducing agent and easily reduces \( \text{Cu}^{2+} \) to \( \text{Cu}^+ \) while being oxidized to \( \text{I}_2 \).
Updated On: May 28, 2026
- \( \text{KBr} \)
- \( \text{K}_2\text{CrO}_3 \)
- \( \text{KI} \)
- \( \text{K}_3\text{PO}_4 \)
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Question:
The question describes a qualitative analysis test where an unknown compound X reacts with copper(II) sulfate to form a brown precipitate, which then decolors/turns white upon addition of sodium thiosulfate (hypo) solution. We need to identify X.
Step 2: Key Formula or Approach:
The reactions of halide ions with transition metal cations are highly characteristic:
- Copper(II) ions (\( \text{Cu}^{2+} \)) oxidize iodide ions (\( \text{I}^- \)) to elemental iodine (\( \text{I}_2 \)) while being reduced to insoluble white copper(I) iodide (\( \text{CuI} \)).
- Sodium thiosulfate (\( \text{Na}_2\text{S}_2\text{O}_3 \), hypo) is a reducing agent that reduces brown \( \text{I}_2 \) to colorless \( \text{I}^- \).
Step 3: Detailed Explanation:
1. Reaction with \( \text{CuSO}_4 \):
When \( \text{CuSO}_4 \) is mixed with potassium iodide (\( \text{KI} \)), cupric ions (\( \text{Cu}^{2+} \)) oxidize the iodide ions:
\[ 2\text{Cu}^{2+} + 4\text{I}^- \rightarrow 2\text{CuI} \downarrow \text{(white precipitate)} + \text{I}_2 \text{(brown solution)} \]
The mixture of the white \( \text{CuI} \) precipitate and the liberated brown iodine (\( \text{I}_2 \)) appears visually as a brown precipitate/suspension.
2. Reaction with Hypo:
When sodium thiosulfate (hypo) is added, it reduces the brown iodine to colorless iodide ions, forming tetrathionate ions:
\[ \text{I}_2 + 2\text{S}_2\text{O}_3^{2-} \rightarrow 2\text{I}^- + \text{S}_4\text{O}_6^{2-} \text{(colorless)} \]
Once the brown color of iodine is discharged, the insoluble white precipitate of copper(I) iodide (\( \text{CuI} \)) becomes clearly visible. This explains why the precipitate "turns white".
Step 4: Final Answer:
The compound (X) is potassium iodide, \( \text{KI} \), which corresponds to option (C).
The question describes a qualitative analysis test where an unknown compound X reacts with copper(II) sulfate to form a brown precipitate, which then decolors/turns white upon addition of sodium thiosulfate (hypo) solution. We need to identify X.
Step 2: Key Formula or Approach:
The reactions of halide ions with transition metal cations are highly characteristic:
- Copper(II) ions (\( \text{Cu}^{2+} \)) oxidize iodide ions (\( \text{I}^- \)) to elemental iodine (\( \text{I}_2 \)) while being reduced to insoluble white copper(I) iodide (\( \text{CuI} \)).
- Sodium thiosulfate (\( \text{Na}_2\text{S}_2\text{O}_3 \), hypo) is a reducing agent that reduces brown \( \text{I}_2 \) to colorless \( \text{I}^- \).
Step 3: Detailed Explanation:
1. Reaction with \( \text{CuSO}_4 \):
When \( \text{CuSO}_4 \) is mixed with potassium iodide (\( \text{KI} \)), cupric ions (\( \text{Cu}^{2+} \)) oxidize the iodide ions:
\[ 2\text{Cu}^{2+} + 4\text{I}^- \rightarrow 2\text{CuI} \downarrow \text{(white precipitate)} + \text{I}_2 \text{(brown solution)} \]
The mixture of the white \( \text{CuI} \) precipitate and the liberated brown iodine (\( \text{I}_2 \)) appears visually as a brown precipitate/suspension.
2. Reaction with Hypo:
When sodium thiosulfate (hypo) is added, it reduces the brown iodine to colorless iodide ions, forming tetrathionate ions:
\[ \text{I}_2 + 2\text{S}_2\text{O}_3^{2-} \rightarrow 2\text{I}^- + \text{S}_4\text{O}_6^{2-} \text{(colorless)} \]
Once the brown color of iodine is discharged, the insoluble white precipitate of copper(I) iodide (\( \text{CuI} \)) becomes clearly visible. This explains why the precipitate "turns white".
Step 4: Final Answer:
The compound (X) is potassium iodide, \( \text{KI} \), which corresponds to option (C).
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