Arrange the following compounds in increasing order of their boiling…

The correct answer is Option (3) → A, C, D, B

Haloalkanes are polar molecules held together by dipole-dipole and van der Waals forces. Two factors decide their boiling points:

• Size of the halogen: for the same alkyl group, the boiling point rises as the halogen gets bigger and heavier: R–F < R–Cl < R–Br < R–I.

• Number of halogen atoms: each extra halogen adds mass and surface area, so \( \ce{CH3Cl} \) < \( \ce{CH2Cl2} \) < \( \ce{CHCl3} \) < \( \ce{CCl4} \).

Both effects follow the molar mass, which gives the order here:

• A. Methylfluoride: \( \ce{CH3F} \), 34 g mol⁻¹, b.p. −78 °C

• C. Trichloromethane: \( \ce{CHCl3} \), 119.5 g mol⁻¹, b.p. 61 °C

• D. Tetrachloromethane: \( \ce{CCl4} \), 154 g mol⁻¹, b.p. 77 °C

• B. Dibromomethane: \( \ce{CH2Br2} \), 174 g mol⁻¹, b.p. 97 °C

Shortcut: \( \ce{CH3F} \) (A), the only gas, is the lowest and \( \ce{CH2Br2} \) (B), the heaviest, is the highest. Only Option (3) starts with A and ends with B.