Arrange the following compounds in decreasing order of their boiling points: $NH_3$, HF, $SiH_4$, $CH_4$

Analyze the question: We need to rank four hydrides from highest to lowest boiling point. Boiling point depends on the strength of the intermolecular forces.

Concept:

• Hydrogen bonding is the strongest of these forces. It occurs when H is bonded to a very electronegative atom (F, O, N).
• Without hydrogen bonding, only weak van der Waals (London dispersion) forces act. These get stronger as molecular mass increases.

Solution:

1. HF and NH₃ both form hydrogen bonds, so they boil at much higher temperatures than SiH₄ and CH₄. • HF boiling point is about +19.5 °C. F is the most electronegative element, so the H–F bond is very polar and the hydrogen bonds are very strong. • NH₃ boiling point is about -33 °C. N is less electronegative than F, and each NH₃ has only one lone pair to accept hydrogen bonds, so the bonding is weaker than in HF. • So HF > NH₃.

2. SiH₄ and CH₄ are non-polar and have only London dispersion forces. • SiH₄ has the larger molar mass (32 g/mol) and a bigger electron cloud. Its boiling point is about -112 °C. • CH₄ has a molar mass of 16 g/mol. Its boiling point is about -161 °C. • So SiH₄ > CH₄.

Decreasing order: HF > NH₃ > SiH₄ > CH₄

This matches option A.

Why the others are wrong:

• B, C and D place CH₄ or SiH₄ above the hydrogen-bonded compounds, or put NH₃ above HF. Both orderings contradict the intermolecular force strengths.

Tip: Remember the order of strength: hydrogen bonding > dipole-dipole > London dispersion. Within a group of non-hydrogen-bonded molecules, the heavier one boils higher. A quick way to remember the hydrogen-bonded atoms is "FON" (F, O, N).