Let's analyze each option to determine its correctness: Option A: Solid I2 is freely soluble in water This statement is incorrect. lodine (I2) 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 I2 is freely soluble in water but only in presence of excess KI This statement is correct. When KI (potassium iodide) is present in excess, iodine forms a soluble complex, KI3 (potassium triiodide), which is much more soluble in water. The excess iodide ions from KI react with iodine: I2+I−⟶I3− The triiodide ion I3−is soluble in water. Option C: Solid I2 is freely soluble in CCl4 This statement is correct. lodine (I2) is non-polar and thus dissolves well in non-polar solvents such as carbon tetrachloride (CCl4). The similar polarity allows I2 to dissolve freely in CCl4. Option D: Solid I2 is freely soluble in hot water This 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.