Concept: Boiling points are influenced by the strength of intermolecular forces. Stronger intermolecular forces require more energy to overcome, leading to higher boiling points. The primary intermolecular forces are hydrogen bonding, dipole-dipole interactions, and London dispersion forces.

Why (A) is correct:

1. Hydrogen Bonding: Both HF and NH₃ exhibit hydrogen bonding. Hydrogen bonding is a particularly strong type of dipole-dipole interaction that occurs when hydrogen is bonded to a highly electronegative atom (F, O, or N).

2. Electronegativity and Bond Polarity: Fluorine is more electronegative than nitrogen. This means the H-F bond is more polar than the N-H bond, leading to stronger hydrogen bonds in HF compared to NH₃. Therefore, HF has a higher boiling point than NH₃.

3. H₂S: H₂S does not exhibit hydrogen bonding because sulfur is not sufficiently electronegative. Its primary intermolecular forces are dipole-dipole interactions (due to its bent structure and polar S-H bonds) and London dispersion forces. These forces are significantly weaker than hydrogen bonding.

4. Comparison: Consequently, HF has the strongest intermolecular forces (strongest hydrogen bonding), followed by NH₃ (weaker hydrogen bonding), and then H₂S (dipole-dipole and London dispersion forces). This leads to the order of boiling points: $\mathrm{HF}>{\mathrm{NH}}_3>{\mathrm{H}}_2\mathrm{S}$.

Option Analysis:

• A) $\mathrm{HF}>{\mathrm{NH}}_3>{\mathrm{H}}_2\mathrm{S}$: This order correctly reflects the relative strengths of intermolecular forces.
• B) $\mathrm{HF}>{\mathrm{H}}_2\mathrm{S}>{\mathrm{NH}}_3$: Incorrect. NH₃ has hydrogen bonding, which is stronger than the forces in H₂S.
• C) $\mathrm{HF}<{\mathrm{H}}_2\mathrm{S}<{\mathrm{NH}}_3$: Incorrect. This order is the reverse of the actual trend.
• D) $\mathrm{HF}<{\mathrm{NH}}_3<{\mathrm{H}}_2\mathrm{S}$: Incorrect. This order is also incorrect, as HF and NH₃ have stronger intermolecular forces than H₂S.

Correct Answer: (A)