# Question #3b50b

May 31, 2015

Sulfur dichloride, $S C {l}_{2}$, will have the highest boiling point.

A substance's boiling point depends on the strength of the intermolecular forces its molecules exhibit.

The stronger the intermolecular forces, the higher the boiling point. Likewise, the weaker the intermolecular bonds, the lower the boiling point.

More often than not, polar molecules will form stronger intermolecular bonds when compared with nonpolar molecules, so polarity is usually a good indicator of relatively high boiling points.

Right from the get-go, krypton, $K r$, which exists as isolated atoms rather than molecules, is eliminated. Actually, krypton will have the lowest boiling point of the group because it can only form weak London dispersion forces and has a relatively small size.

Chlorine, $C {l}_{2}$, methane, $C {H}_{4}$, and boron trichloride, $B C {l}_{3}$, are also eliminated because they are nonpolar molecules. These molecules can only exhibit London dispersion forces, so their respective boling points will still be relatively low.

This leaves you with sulfur dichloride, $S C {l}_{2}$.

This time, you're dealing with a polar molecule. The two lone pairs of electrons present on the sulfur atom (not visible in the above image) and the polar $\text{S-Cl}$ bonds imply that a permanent dipole moment is present.

This implies that the molecule can exhibit dipole-dipole interactions, which are considerably stronger than London dispersion forces.

As a result, sulfur dichloride will have the highest boiling point of the group.