Hydrogen Bonding
The boiling point of HF is anomalously higher than HCl despite HF having a lower molar mass; this is primarily due to which type of intermolecular interaction?
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Solution
Intermolecular hydrogen bonding
Hydrogen bonding is a special type of dipole-dipole interaction that occurs when a hydrogen atom is covalently bonded to a highly electronegative, small atom (F, O, or N) and is attracted to a lone pair on another electronegative atom in a neighbouring molecule. In HF, the large electronegativity difference between H (2.1) and F (4.0) creates a highly polar H-F bond. The fluorine atom has three lone pairs and acts as a hydrogen bond acceptor, while the H atom acts as a hydrogen bond donor. This intermolecular hydrogen bonding creates an associated liquid state in HF, requiring extra energy to break these H-bonds upon boiling, hence the anomalously high boiling point (~19.5 degrees C) compared to HCl (-85 degrees C). London dispersion forces exist in all molecules but are generally weaker and increase with molecular size/surface area; they cannot explain why the lighter HF boils higher than heavier HCl. Dipole-dipole interactions are present in HCl as well, but HCl cannot form hydrogen bonds since Cl is not small and electronegative enough to support strong H-bonding. Ionic bonding is absent since HF is a covalent molecule. This is a direct application of NCERT Class 11 Chapter 4 (Chemical Bonding) on intermolecular forces and hydrogen bonding. Plausibility: the trend is reversed for HF compared to HCl, HBr, HI, which clearly signals a special interaction, confirmed as hydrogen bonding.
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About This Question
- Subject
- chemistry
- Chapter
- chemical bonding and molecular structure
- Topic
- hydrogen bonding
- Difficulty
- Easy
- Year
- 2025
Solution
Correct Answer:
Intermolecular hydrogen bonding
Hydrogen bonding is a special type of dipole-dipole interaction that occurs when a hydrogen atom is covalently bonded to a highly electronegative, small atom (F, O, or N) and is attracted to a lone pair on another electronegative atom in a neighbouring molecule. In HF, the large electronegativity difference between H (2.1) and F (4.0) creates a highly polar H-F bond. The fluorine atom has three lone pairs and acts as a hydrogen bond acceptor, while the H atom acts as a hydrogen bond donor. This intermolecular hydrogen bonding creates an associated liquid state in HF, requiring extra energy to break these H-bonds upon boiling, hence the anomalously high boiling point (~19.5 degrees C) compared to HCl (-85 degrees C). London dispersion forces exist in all molecules but are generally weaker and increase with molecular size/surface area; they cannot explain why the lighter HF boils higher than heavier HCl. Dipole-dipole interactions are present in HCl as well, but HCl cannot form hydrogen bonds since Cl is not small and electronegative enough to support strong H-bonding. Ionic bonding is absent since HF is a covalent molecule. This is a direct application of NCERT Class 11 Chapter 4 (Chemical Bonding) on intermolecular forces and hydrogen bonding. Plausibility: the trend is reversed for HF compared to HCl, HBr, HI, which clearly signals a special interaction, confirmed as hydrogen bonding.
This easy difficulty chemistry question is from the chapter chemical bonding and molecular structure, covering the topic of hydrogen bonding. It appeared in the 2025 exam.
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