### Correct Option: D. The mixture turns reddish brown ### Detailed Explanation: When we mix aqueous solutions of tetraoxosulphate (VI) acid (H₂SO₄), potassium trioxoiodate (V) (KI₃), and potassium iodide (KI), we need to consider the chemical reactions that occur between these substances. 1. **Understanding the Components**: - **Tetraoxosulphate (VI) Acid (H₂SO₄)**: This is a strong acid that can donate protons (H⁺ ions) in solution. - **Potassium Trioxoiodate (V) (KI₃)**: This compound contains the triiodide ion (I₃⁻), which is formed when potassium iodide (KI) is mixed with iodine (I₂). The triiodide ion is known for its characteristic reddish-brown color. - **Potassium Iodide (KI)**: This is a salt that dissociates in solution to give potassium ions (K⁺) and iodide ions (I⁻). 2. **Chemical Reactions**: - When H₂SO₄ is added to a solution containing KI, it can oxidize the iodide ions (I⁻) to iodine (I₂). The reaction can be represented as: \[ 2 \text{I}^- + \text{H}_2\text{SO}_4 \rightarrow \text{I}_2 + \text{H}_2\text{O} + \text{SO}_4^{2-} \] - The iodine (I₂) can then react with more iodide ions to form the triiodide ion (I₃⁻): \[ \text{I}_2 + \text{I}^- \rightarrow \text{I}_3^- \] - The presence of the triiodide ion (I₃⁻) in solution gives the mixture a reddish-brown color. 3. **Observation**: - As a result of these reactions, the solution will turn reddish-brown due to the formation of the triiodide ion (I₃⁻). This is the key observation when these solutions are mixed. ### Why Other Options Are Incorrect: - **Option A: A white precipitate is formed**: - This option is incorrect because there are no insoluble salts formed in this reaction. The components involved do not lead to the formation of a white precipitate under the conditions described. - **Option B: A green precipitate is formed**: - This option is also incorrect. A green precipitate is not a characteristic observation in the reactions involving H₂SO₄, KI, and KI₃. The expected products do not include any green compounds. - **Option C: The mixture remains colourless**: - This option is incorrect because the reaction produces iodine and subsequently triiodide, which imparts a distinct reddish-brown color to the solution. Therefore, the mixture will not remain colorless. ### Summary of Key Points: - Mixing H₂SO₄ with KI and KI₃ leads to the oxidation of iodide ions to iodine. - Iodine can further react with iodide ions to form the triiodide ion (I₃⁻), which is reddish-brown. - The correct observation when these solutions are mixed is that the mixture turns reddish-brown. - Other options are incorrect as they do not reflect the actual chemical behavior of the substances involved. ### Revision Summary: - H₂SO₄ oxidizes I⁻ to I₂, which can form I₃⁻. - The triiodide ion (I₃⁻) is responsible for the reddish-brown color. - No precipitates are formed in this reaction. - The correct observation is that the mixture turns reddish-brown (Option D).