The bar chart compares the boiling points of some alkyl halides.What…

The correct answer is Option (3) → About 310 K; bromine is a bigger and heavier atom than chlorine, so the van der Waals forces are stronger.

For the same alkyl group, a bigger and heavier halogen atom gives stronger attractions between the molecules, so more energy is needed to boil the compound.

• Boiling point of bromoethane — about 310 K. In the CH₃CH₂X group, the blue bar (bromides) ends a little above the 300 K line. The red bar of the same group, about 285 K, is chloroethane.

• Reason — stronger van der Waals forces. Bromine is bigger and heavier than chlorine and has more electrons, so the van der Waals forces between bromoethane molecules are stronger than those between chloroethane molecules.

Why the other options are incorrect

Option (1) About 285 K; the C–Br bond is more polar than the C–Cl bond. — Incorrect. 285 K is the red bar, chloroethane; and chlorine is more electronegative than bromine, so the C–Cl bond is the more polar one.

Option (2) About 310 K; the C–Br bond is more polar than the C–Cl bond. — Incorrect. The value is right, but the C–Br bond is less polar than the C–Cl bond; the higher boiling point comes from the bigger, heavier bromine atom.

Option (4) About 285 K; bromine is a bigger and heavier atom than chlorine, so the van der Waals forces are stronger. — Incorrect. The reason is right, but 285 K is the boiling point of chloroethane (red bar), not of bromoethane.

NCERT (2026-27 edition):
6.6 Physical Properties – “The attractions get stronger as the molecules get bigger in size and have more electrons. The pattern of variation of boiling points of different halides is depicted in Fig. 6.1. For the same alkyl group, the boiling points of alkyl halides decrease in the order: RI> RBr> RCl> RF. This is because with the increase in size and mass of halogen atom, the magnitude of van der Waal forces increases.”

Fig. 6.1: Comparison of boiling points of some alkyl halides