Iodine displacement and partitioning in immiscible layers

Steps:

• Chlorine gas reacts with aqueous NaI: Cl₂ + 2I⁻ → 2Cl⁻ + I₂, liberating iodine.
• Iodine initially colors the lower aqueous layer brown due to solubility in water.
• Excess iodine partitions into the immiscible upper hexane layer, turning it purple-brown.
• Observation: brown lower layer and colored upper layer.

Why A is correct:

• Iodine (I₂) is brown in aqueous media and more soluble in nonpolar hexane, coloring both layers per solubility laws.

Why the others are wrong:

• B: Upper hexane layer changes color as iodine dissolves into it.
• C: Initial aqueous NaI is colorless, not brown; it becomes brown, not colorless.
• D: Same error as C; initial solution is colorless and becomes brown.

Final answer: A