Concept: Boiling point is influenced by intermolecular forces. Stronger intermolecular forces lead to higher boiling points. The main types of intermolecular forces are hydrogen bonding, dipole-dipole interactions, and London dispersion forces (van der Waals forces).

Why (A) is correct:

Amongst the hydrides of Group 16 elements (${\mathrm{H}}_2\mathrm{O},{\mathrm{H}}_2\mathrm{S},{\mathrm{H}}_2\mathrm{Se},{\mathrm{H}}_2\mathrm{Te}$), the boiling point generally increases down the group due to increasing molecular weight and thus stronger London dispersion forces. However, ${\mathrm{H}}_2\mathrm{O}$ has an exceptionally high boiling point compared to the other hydrides. This is because oxygen is highly electronegative and small, allowing it to form strong intermolecular hydrogen bonds. These hydrogen bonds require a significant amount of energy to break, leading to a much higher boiling point for water than would be predicted based on its molecular weight alone.

Option Analysis:

• A) ${\mathrm{H}}_2\mathrm{O}$ because of hydrogen bonding: This is correct. Water forms extensive hydrogen bonds, which are the strongest intermolecular forces among these compounds, leading to the highest boiling point.
• B) ${\mathrm{H}}_2\mathrm{Te}$ because of higher molecular weight: While ${\mathrm{H}}_2\mathrm{Te}$ has the highest molecular weight among the series (excluding ${\mathrm{H}}_2\mathrm{O}$), and therefore the strongest London dispersion forces, its boiling point is still lower than that of ${\mathrm{H}}_2\mathrm{O}$ due to the absence of strong hydrogen bonding.
• C) ${\mathrm{H}}_2\mathrm{S}$ because of hydrogen bonding: ${\mathrm{H}}_2\mathrm{S}$ does not exhibit significant hydrogen bonding because sulfur is less electronegative than oxygen and its larger size makes the hydrogen bond weaker. Its boiling point is much lower than water's.
• D) ${\mathrm{H}}_2\mathrm{Se}$ because of lower molecular weight: ${\mathrm{H}}_2\mathrm{Se}$ has a lower molecular weight than ${\mathrm{H}}_2\mathrm{Te}$ and thus weaker London dispersion forces, leading to a lower boiling point than ${\mathrm{H}}_2\mathrm{Te}$. It also does not form strong hydrogen bonds.

Correct Answer: (A)