Melting and boiling points
Methyl chloride, methyl bromide, ethyl chloride and some chlorofluoromethanes are gases at room temperature. Higher members are liquids or solids. Molecules of organic halogen compounds are generally polar. Due to greater polarity as well as higher molecular mass as compared to the parent hydrocarbon, the intermolecular forces of attraction (dipole-dipole and van der Waals) are stronger in the halogen derivatives. That is why the boiling points of chlorides, bromides and iodides are considerably higher than those of the hydrocarbons of comparable molecular mass. The attractions get stronger as the molecules get bigger in size and have more electrons.
Density
Bromo, iodo and polychloro derivatives of hydrocarbons are heavier than water. The density increases with increase in number of carbon atoms, halogen atoms and atomic mass of the halogen atoms.
If chloromethane has a boiling point of 250 K, what will be the probable boiling point of methane?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 110 K
The passage states that, because of greater polarity and higher molecular mass than the parent hydrocarbon, the intermolecular forces (dipole-dipole and van der Waals) are stronger in the halogen derivatives. So a haloalkane boils at a higher temperature than the hydrocarbon it comes from.
Chloromethane (\( \ce{CH3Cl} \), molar mass 50.5 g mol−1) is a polar molecule. Methane (\( \ce{CH4} \), 16 g mol−1) is non-polar and much lighter, so only weak van der Waals forces hold its molecules together. Its boiling point must therefore lie well below 250 K.