Question

Amongst \(\mathrm{H}_{2} \mathrm{O}, \mathrm{H}_{2} \mathrm{~S}, \mathrm{H}_{2} \mathrm{Se}\) and \(\mathrm{H}_{2}\) Te the one with the highest boiling point is: (a) \(\mathrm{H}_{2} \mathrm{O}\) because of hydrogen bonding (b) \(\mathrm{H}_{2}\) Te because of higher molecular weight (c) \(\mathrm{H}_{2} \mathrm{~S}\) because of hydrogen bonding (d) \(\mathrm{H}_{2}\) Se because of lower molecular weight.

Short answer

(a) \(\mathrm{H}_{2} \mathrm{O}\) because of hydrogen bonding.

Step-by-step solution

Analyze Boiling Point Determinants

Boiling point is determined by intermolecular forces such as hydrogen bonding, van der Waals forces, dipole-dipole interactions, and the molecular weight of the compound.

Consider Hydrogen Bonding

Hydrogen bonding is a strong type of dipole-dipole interaction that significantly raises the boiling point. Among the compounds \(\mathrm{H}_2\mathrm{O}, \mathrm{H}_2\mathrm{~S}, \mathrm{H}_2\mathrm{Se},\) and \(\mathrm{H}_2\mathrm{Te}\), only \(\mathrm{H}_2\mathrm{O}\) can form hydrogen bonds because of the high electronegativity of oxygen compared to sulfur, selenium, and tellurium.

Evaluate Effect of Molecular Weight

The boiling point generally increases with molecular weight due to stronger van der Waals forces. Among \(\mathrm{H}_2\mathrm{~S}, \mathrm{H}_2\mathrm{Se}, \mathrm{H}_2\mathrm{Te}\), the molecular weight increases in the order \(\mathrm{H}_2\mathrm{~S} < \mathrm{H}_2\mathrm{Se} < \mathrm{H}_2\mathrm{Te}\). However, this increase is not as significant as the effect of hydrogen bonding present in \(\mathrm{H}_2\mathrm{O}\).

Identify Compound with Highest Boiling Point

Given the strong hydrogen bonding present in \(\mathrm{H}_2\mathrm{O}\), it has a much higher boiling point compared to the others, even though \(\mathrm{H}_2\mathrm{Te}\) has the highest molecular weight. Therefore, \(\mathrm{H}_2\mathrm{O}\) has the highest boiling point among the given compounds.

Key concepts

Hydrogen Bonding

Hydrogen bonding is a compelling feature of water (\(\mathrm{H}_2\mathrm{O}\)) that significantly affects its physical properties, particularly the boiling point. Hydrogen bonds are a type of strong dipole-dipole interaction, occurring between a hydrogen atom and a more electronegative atom like oxygen, nitrogen, or fluorine.
In water, each molecule can potentially form four hydrogen bonds due to the presence of two hydrogen atoms and two lone pairs on the oxygen atom. This extensive hydrogen bonding network imparts a higher boiling point to water than you would expect based on its molecular weight alone.

• Water molecules are tightly held together due to these bonds, requiring more energy (in the form of heat) to break apart.
• Hydrogen bonds are more powerful than other van der Waals forces, although not as strong as covalent bonds.
• Among \(\mathrm{H}_2\mathrm{S}, \mathrm{H}_2\mathrm{Se},\) and \(\mathrm{H}_2\mathrm{Te}\), only \(\mathrm{H}_2\mathrm{O}\) forms hydrogen bonds prominently, explaining its higher boiling point.

Molecular Weight

Molecular weight is a significant factor influencing a substance's boiling point but is often secondary to hydrogen bonding. In the periodic table, moving from oxygen to tellurium, elements increase in atomic weight. The sequence is such that \(\mathrm{H}_2\mathrm{S} < \mathrm{H}_2\mathrm{Se} < \mathrm{H}_2\mathrm{Te}\). This increased molecular weight generally leads to a higher boiling point because the increased mass contributes to stronger intermolecular van der Waals forces.

• However, higher molecular weight does not always result in higher boiling points when competing with hydrogen bonding.
• In the case of \(\mathrm{H}_2\mathrm{O}\), the effect of hydrogen bonding is so significant that it outstrips the influence of molecular weight, thus giving it the highest boiling point among the given molecules.
• Molecular weight becomes more critical when comparing compounds that do not exhibit hydrogen bonding.

van der Waals Forces

Van der Waals forces are weak intermolecular forces arising from temporary dipoles in molecules. These forces include attractions known as London dispersion forces and dipole-dipole interactions. As molecular weight increases, these forces usually become more prominent, thus increasing the boiling point.

• Heavier molecules such as \(\mathrm{H}_2\mathrm{Te}\) experience stronger van der Waals forces than lighter ones like \(\mathrm{H}_2\mathrm{S}\).
• In the hydrogen chalcogenide series, these forces contribute to the increasing trend in boiling points from \(\mathrm{H}_2\mathrm{S}\) to \(\mathrm{H}_2\mathrm{Te}\).
• Despite the stronger van der Waals forces in heavier molecules, they are subservient to the robust hydrogen bonding in \(\mathrm{H}_2\mathrm{O}\), making it the exception to the general trend.

Intermolecular Forces

Intermolecular forces are the attractions that occur between molecules, and they directly influence physical properties like boiling and melting points. Understanding these forces helps explain why some substances are liquids at room temperature, while others are gases.
There are several types of intermolecular forces:

• Hydrogen Bonding: This is the strongest of the common intermolecular forces (among non-ionic interactions) and explains the anomalously high boiling point of \(\mathrm{H}_2\mathrm{O}\).
• Van der Waals Forces: This umbrella term includes both London dispersion forces, which are weak and present in all molecules, and stronger dipole-dipole interactions present in polar substances.
• Without strong hydrogen bonds, the boiling points of \(\mathrm{H}_2\mathrm{S}, \mathrm{H}_2\mathrm{Se},\) and \(\mathrm{H}_2\mathrm{Te}\) primarily reflect the influence of these van der Waals forces.
• Understanding the balance of these forces is critical for predicting and explaining boiling points, as highlighted in the exercise.