Boiling Points of Group 16 Hydrides

The boiling point of a substance is the temperature at which it changes from a liquid to a gas. This transition requires overcoming the intermolecular forces holding the molecules together in the liquid state. The stronger the intermolecular forces, the more energy is required, and thus the higher the boiling point.

Let's examine the intermolecular forces present in the given compounds:

• \(\text{H}_2\text{O}\) (Water)
• \(\text{H}_2\text{S}\) (Hydrogen Sulfide)
• \(\text{H}_2\text{Se}\) (Hydrogen Selenide)

These are hydrides of elements in Group 16 of the periodic table (\(\text{O}\), \(\text{S}\), \(\text{Se}\)).

Intermolecular Forces Analysis

• \(\text{H}_2\text{O}\): Water is a polar molecule and exhibits strong hydrogen bonding in addition to dipole-dipole forces and London dispersion forces. Hydrogen bonding is a particularly strong type of intermolecular force.
• \(\text{H}_2\text{S}\) and \(\text{H}_2\text{Se}\): These are also polar molecules and exhibit dipole-dipole forces and London dispersion forces. Hydrogen bonding is not significant in \(\text{H}_2\text{S}\) and \(\text{H}_2\text{Se}\) to the extent it is in \(\text{H}_2\text{O}\), because sulfur and selenium are less electronegative than oxygen, making the \(\text{S-H}\) and \(\text{Se-H}\) bonds less polar and the hydrogen atoms less positive.

Comparing Boiling Points

When comparing substances with similar types of intermolecular forces, boiling point generally increases with increasing molecular weight due to stronger London dispersion forces. However, hydrogen bonding significantly increases the boiling point compared to substances with only dipole-dipole and London forces, even if they have higher molecular weights.

• \(\text{H}_2\text{O}\) (Molar mass ≈ 18 g/mol): Strong hydrogen bonding leads to an unusually high boiling point.
• \(\text{H}_2\text{S}\) (Molar mass ≈ 34 g/mol): Exhibits dipole-dipole and London forces.
• \(\text{H}_2\text{Se}\) (Molar mass ≈ 79 g/mol): Exhibits dipole-dipole and London forces. Being heavier than \(\text{H}_2\text{S}\), \(\text{H}_2\text{Se}\) has stronger London dispersion forces.

Considering the forces:

• The strong hydrogen bonding in \(\text{H}_2\text{O}\) makes its boiling point much higher than both \(\text{H}_2\text{S}\) and \(\text{H}_2\text{Se}\).
• Comparing \(\text{H}_2\text{S}\) and \(\text{H}_2\text{Se}\), \(\text{H}_2\text{Se}\) is heavier, so its London dispersion forces are stronger than those in \(\text{H}_2\text{S}\). Therefore, \(\text{H}_2\text{Se}\) has a higher boiling point than \(\text{H}_2\text{S}\).

The general trend for boiling points of these hydrides is: \(\text{H}_2\text{S}\) < \(\text{H}_2\text{Se}\) < \(\text{H}_2\text{O}\).

For example, approximate boiling points are:

Based on this comparison, \(\text{H}_2\text{S}\) has the lowest boiling point among \(\text{H}_2\text{O}\), \(\text{H}_2\text{S}\), and \(\text{H}_2\text{Se}\).