# Question #94828

Aug 20, 2015

The answer is (b) $\text{Ar"<"HCl"<"HI}$

#### Explanation:

This question wants you to recognize the relationship between boiling point and the strength of the intermolecular forces of attraction.

More specifically, you must be able to link stronger intermolecular forces of attraction with higher boiling points and weaker intermolecular forces of attraction with lower boiling points.

Right from the start, the correct answer must be (b) because argon exhibits the weakest inermolecular forces of attraction between its atoms.

Argon atoms are very small and only exhibit weak London dispersion forces, so argon's boiling point will be very, very low.

The more interesting analysis can be done for hydrogen chloride and hydrogen iodide.

The fact that hydrogen chloride has a lower boiling point than hydrogen iodide can be attributed to the magnitude of the London dispersion forces these two molecules exhibit.

Hydrogen chloride molecules exhibit dipole-dipole interactions and London dispersion forces. On the other hand, the bond between iodine and hydrogen can be consideed nonpolar because of the smalle difference in electronegativity between the two atoms.

This means that the hydrogen iodide molecules do not exhibit dipole-dipole interactions, at least not to any significant amount.

However, they exhibit significant London dispersion forces due to the size and polarizability of the iodine atom. In fact, these London dispersion forces become so important as you go down group 17, that hydrogen iodide ends up with a higher boiling point than hydrogen chloride.