Q24 / 155 · WBJEE 2026 · Chemistry
A compound (X) when treated with $CuSO_4$ solution yields a brown precipitate. On adding hypo solution the precipitate turns white. The compound (X) is
- A$KBr$
- B$K_2CrO_3$
- Correct answerC$KI$
- D$K_3PO_4$
Solution
When potassium iodide ($KI$) is treated with copper sulphate ($CuSO_4$) solution, it forms unstable cupric iodide ($CuI_2$), which immediately decomposes to give a white precipitate of cuprous iodide ($Cu_2I_2$) and liberates iodine ($I_2$).
$2CuSO_4 + 4KI \rightarrow Cu_2I_2 + I_2 + 2K_2SO_4$
The liberated iodine imparts a brown colour to the mixture, making the precipitate appear brown.
When hypo solution ($Na_2S_2O_3$) is added to this mixture, it reacts with the liberated iodine to form soluble sodium tetrathionate ($Na_2S_4O_6$) and sodium iodide ($NaI$).
$I_2 + 2Na_2S_2O_3 \rightarrow Na_2S_4O_6 + 2NaI$
As the brown iodine is consumed in the reaction, the white precipitate of cuprous iodide ($Cu_2I_2$) becomes visible. Therefore, the compound (X) is $KI$.
Answer: $KI$
$2CuSO_4 + 4KI \rightarrow Cu_2I_2 + I_2 + 2K_2SO_4$
The liberated iodine imparts a brown colour to the mixture, making the precipitate appear brown.
When hypo solution ($Na_2S_2O_3$) is added to this mixture, it reacts with the liberated iodine to form soluble sodium tetrathionate ($Na_2S_4O_6$) and sodium iodide ($NaI$).
$I_2 + 2Na_2S_2O_3 \rightarrow Na_2S_4O_6 + 2NaI$
As the brown iodine is consumed in the reaction, the white precipitate of cuprous iodide ($Cu_2I_2$) becomes visible. Therefore, the compound (X) is $KI$.
Answer: $KI$
Hence, the correct answer is option: 'C'.