Question

Why is the boiling point of ethyl iodide (b.p. 72 °C) higher than the boiling point of ethyl fluoride (b.p. −37 °C) even though the dipole–dipole interaction should be better in C₂H₅F? Is it because of higher molecular mass of iodine?

Answer 1

The higher boiling point of ethyl iodide compared to ethyl fluoride is primarily due to the larger molecular mass of iodine, which leads to stronger London dispersion forces despite the stronger dipole-dipole interactions present in ethyl fluoride.

Step-by-step explanation:

The boiling point of ethyl iodide is higher than that of ethyl fluoride primarily due to the difference in molecular mass and the resultant strength of London dispersion forces.

While ethyl fluoride may have stronger dipole-dipole interactions because of its polar bonds, this effect is outweighed by the much larger number of electrons in ethyl iodide which amplify its London dispersion forces.

Dispersion forces are the weakest type of intermolecular forces, but their strength increases significantly with the number of electrons in a molecule, leading to higher boiling points.

Therefore, despite ethyl fluoride having stronger dipole-dipole interactions, the heavier iodine atom in ethyl iodide results in stronger dispersion forces that more substantially influence its higher boiling point.

This is similar to the case of other molecules like ICI and Br₂, where despite similar masses, the polar ICI has both dispersion forces and dipole-dipole interactions, leading to higher boiling points compared to the nonpolar Br₂ which only exhibits London dispersion forces.