Question

In which of the following, the compounds are correctly arranged in the decreasing order of boiling points?

• A. \( \text{HF}>\text{H}_2\text{O}>\text{NH}_3>\text{PH}_3 \)
• B. \( \text{H}_2\text{O}>\text{HF}>\text{NH}_3>\text{PH}_3 \)
• C. \( \text{H}_2\text{O}>\text{HF}>\text{PH}_3>\text{NH}_3 \)
• D. \( \text{HF}>\text{NH}_3>\text{H}_2\text{O}>\text{PH}_3 \)

Hint:

When comparing boiling points of substances with hydrogen bonds (\(H_2O, HF, NH_3\)), remember the order is \(H_2O>HF>NH_3\). Water is highest due to the extensive 3D network of H-bonds it can form. Any comparable molecule without H-bonding (like \(PH_3\)) will have a significantly lower boiling point.

Answer 1

The Correct Option is B

Boiling point is determined by the strength of intermolecular forces. We need to compare the forces for the given compounds.
1. \(\text{H}_2\text{O}, \text{HF}, \text{NH}_3\): All three exhibit hydrogen bonding, which is a particularly strong type of dipole-dipole interaction. This leads to unusually high boiling points compared to other hydrides in their respective groups.
- \(\text{H}_2\text{O}\): Water has the highest boiling point (100 °C). Although the H-bond in HF is individually stronger due to fluorine's high electronegativity, each water molecule can form up to four hydrogen bonds with its neighbors (two as a donor, two as an acceptor), creating an extensive 3D network. This strong, extensive network requires the most energy to break.
- HF: Hydrogen fluoride has the next highest boiling point (19.5 °C). Each HF molecule can only form, on average, two hydrogen bonds (one as donor, one as acceptor), leading to zigzag chains rather than a 3D network.
- \(\text{NH}_3\): Ammonia has the lowest boiling point of the three (-33 °C). While it can potentially form multiple H-bonds, the N-H bond is less polar than the O-H or F-H bonds, making the individual hydrogen bonds weaker.
2. \(\text{PH}_3\) (Phosphine): Phosphorus is not very electronegative, so PH\(_3\) does not exhibit hydrogen bonding. Its intermolecular forces are weak van der Waals forces (dipole-dipole and London dispersion forces). As a result, its boiling point is much lower (-87.7 °C) than the other three.
Arranging them in decreasing order of boiling points:
H\(_2\)O (100 °C)>HF (19.5 °C)>NH\(_3\) (-33 °C)>PH\(_3\) (-87.7 °C).
This matches option (B).