Question

Without referring to Table \(6,\) predict which compound in each of the following pairs has the higher boiling point: (a) \(\mathrm{H}_{2} \mathrm{O}\) or \(\mathrm{H}_{2} \mathrm{Se}\) (b) \(\mathrm{H}_{2} \mathrm{~S}\) or \(\mathrm{H}_{2} \mathrm{Te}\)

Short answer

(a) \(\mathrm{H}_{2}\mathrm{O}\) has the higher boiling point; (b) \(\mathrm{H}_{2}\mathrm{Te}\) has the higher boiling point.

Step-by-step solution

Understand the Concept of Boiling Point

Boiling point is the temperature at which a liquid turns into a vapor. It is influenced by intermolecular forces, such as hydrogen bonding, dipole-dipole interactions, and London dispersion forces. Stronger intermolecular forces generally result in higher boiling points.

Identify the Types of Intermolecular Forces

For each pair, determine the types of intermolecular forces present. Water (\(\mathrm{H}_{2}\mathrm{O}\)) can form strong hydrogen bonds due to the presence of highly electronegative oxygen, whereas hydrogen selenide (\(\mathrm{H}_{2}\mathrm{Se}\)) forms weaker dipole-dipole interactions and dispersion forces. Similarly, both \(\mathrm{H}_{2}\mathrm{S}\) and \(\mathrm{H}_{2}\mathrm{Te}\) primarily exhibit dipole-dipole and dispersion forces, with \(\mathrm{H}_{2}\mathrm{Te}\) having stronger dispersion forces due to its larger size.

Compare Intermolecular Forces for Each Pair

(a) Compare \(\mathrm{H}_{2}\mathrm{O}\) and \(\mathrm{H}_{2}\mathrm{Se}\): \(\mathrm{H}_{2}\mathrm{O}\) has hydrogen bonding which is stronger than the dipole-dipole and dispersion forces in \(\mathrm{H}_{2}\mathrm{Se}\). Thus, \(\mathrm{H}_{2}\mathrm{O}\) has a higher boiling point.(b) Compare \(\mathrm{H}_{2}\mathrm{S}\) and \(\mathrm{H}_{2}\mathrm{Te}\): Although both have similar forces, \(\mathrm{H}_{2}\mathrm{Te}\) has stronger London dispersion forces due to its larger atomic mass, resulting in a higher boiling point than \(\mathrm{H}_{2}\mathrm{S}\).

Conclusion

Determine which compound in each pair has the higher boiling point. (a) \(\mathrm{H}_{2}\mathrm{O}\) has a higher boiling point than \(\mathrm{H}_{2}\mathrm{Se}\). (b) \(\mathrm{H}_{2}\mathrm{Te}\) has a higher boiling point than \(\mathrm{H}_{2}\mathrm{S}\).

Key concepts

Intermolecular Forces

Intermolecular forces are the forces of attraction and repulsion between interacting particles, such as molecules or atoms. These forces play a crucial role in determining the physical properties of a substance, including its boiling point.
This is because they influence how much energy is needed to separate the molecules in a liquid. The primary types of intermolecular forces include:

• Hydrogen bonding.
• Dipole-dipole interactions.
• London dispersion forces.

Each type of force varies in strength, impacting the observed boiling point of a substance.
Substances with stronger intermolecular forces will typically require more energy to vaporize, equating to a higher boiling point.

Hydrogen Bonding

Hydrogen bonding is a particularly strong type of dipole-dipole interaction. It occurs when hydrogen is bound to a highly electronegative atom such as oxygen, nitrogen, or fluorine. In this scenario, the hydrogen atom shares its electrons with the electronegative atom, resulting in a strong attraction between molecules.For example, water molecules (\(\mathrm{H}_{2}\mathrm{O}\)) demonstrate hydrogen bonding. Oxygen is highly electronegative, which allows it to attract the shared electrons from hydrogen more effectively.
This explains why water has a high boiling point compared to hydrogen selenide (\(\mathrm{H}_{2}\mathrm{Se}\)), where such strong bonding is absent.
In this sense, hydrogen bonds make a crucial contribution to the elevated boiling points in compounds containing them.

London Dispersion Forces

London dispersion forces are the weakest type of intermolecular force and are present in all molecules, regardless of polarity. These forces arise due to temporary dipoles. Temporary dipoles occur when electrons in a molecule or atom are distributed asymmetrically at any given time.Even though they are weak individually, London dispersion forces can become significant in larger atoms or molecules. This is because larger atoms have more electrons that can contribute to temporary dipoles.
This makes their cumulative effect more substantial. In the case of \(\mathrm{H}_{2}\mathrm{Te}\), the larger size compared to \(\mathrm{H}_{2}\mathrm{S}\) results in stronger London dispersion forces for \(\mathrm{H}_{2}\mathrm{Te}\).
Consequently, this makes it have a higher boiling point than \(\mathrm{H}_{2}\mathrm{S}\) even though the forces are weak at an individual level.