WBJEE 2026 Physics and Chemistry Paper

Concept:Iodide ions give a brown coloration with copper sulfate solution due to liberation of iodine, and hypo removes this iodine to expose a white precipitate.Explanation:The compound $(\mathrm{X})$ is $\mathrm{KI}$.When $\mathrm{KI}$ reacts with $\mathrm{CuSO}_{4}$, cupric iodide is formed momentarily but decomposes to yield cuprous iodide and free iodine.The reaction is:$$2\mathrm{CuSO}_{4} + 4\mathrm{KI} \rightarrow \mathrm{Cu}_{2}\mathrm{I}_{2} \downarrow + \mathrm{I}_{2} + 2\mathrm{K}_{2}\mathrm{SO}_{4}$$The liberated $\mathrm{I}_{2}$ gives a brown colour to the precipitate.On adding hypo solution ($\mathrm{Na}_{2}\mathrm{S}_{2}\mathrm{O}_{3}$), the iodine reacts with thiosulfate ions:$$\mathrm{I}_{2} + 2\mathrm{Na}_{2}\mathrm{S}_{2}\mathrm{O}_{3} \rightarrow 2\mathrm{NaI} + \mathrm{Na}_{2}\mathrm{S}_{4}\mathrm{O}_{6}$$The brown colour of iodine disappears, leaving the white precipitate of $\mathrm{Cu}_{2}\mathrm{I}_{2}$.None of the other options produce a brown precipitate with $\mathrm{CuSO}_{4}$ followed by this white precipitate on adding hypo.Answer:C. $\mathrm{KI}$