WBJEE 2024 Physics & Chemistry Paper

Let's analyze each option to determine its correctness: Option A: Solid $\mathrm{I}_2$ is freely soluble in waterThis statement is incorrect. lodine $(\mathrm{I}_2)$ is only slightly soluble in water. It does not dissolve freely due to its nonpolar nature, which does not interact well with polar solvent molecules like water. Option B: Solid $\mathrm{I}_2$ is freely soluble in water but only in presence of excess KIThis statement is correct. When KI (potassium iodide) is present in excess, iodine forms a soluble complex, $\mathrm{KI}_3$ (potassium triiodide), which is much more soluble in water. The excess iodide ions from KI react with iodine:$\mathrm{I}_2 + \mathrm{I}^{-} \longrightarrow \mathrm{I}_3^{-}$The triiodide ion $\mathrm{I}_3^{-}$is soluble in water. Option C: Solid $\mathrm{I}_2$ is freely soluble in $\mathrm{CCl}_4$This statement is correct. lodine $(\mathrm{I}_2)$ is non-polar and thus dissolves well in non-polar solvents such as carbon tetrachloride $(\mathrm{CCl}_4)$. The similar polarity allows $\mathrm{I}_2$ to dissolve freely in $\mathrm{CCl}_4$. Option D: Solid $\mathrm{I}_2$ is freely soluble in hot waterThis statement is incorrect. While solubility of substances generally increases with temperature, iodine remains only slightly more soluble in hot water than in cold water. It does not become "freely soluble" in hot water due to its inherent non-polar nature.Therefore, the correct statements are:Option B and Option C.