JEE Chemistry 2023 Question with Solution

Given: $Cu^{2+}$ is treated with KI, producing a white precipitate X. The resulting iodine-containing solution is titrated with sodium thiosulphate to form Y.

Find: The identities of X and Y.

From the extracted solution, iodide first reacts with copper(II):

The solution also states that iodide acts as a strong reducing agent and reduces $Cu^{2+}$ to $Cu^+$.

The unstable intermediate then changes as shown in the solution:

So the white precipitate X is copper(I) iodide, i.e. $CuI$.

The liberated iodine forms triiodide in presence of excess iodide:

During titration with sodium thiosulphate, the solution gives:

Hence Y is $Na_2S_4O_6$.

Therefore, according to the primary solution working, X = CuI and Y = Na_2S_4O_6. Among the given options, this corresponds to Option D.

However, the extracted the solution explicitly marks The Correct Option is C, which conflicts with both the working and the listed options. Since the working is the primary source, the defensible choice from the options is D.