Rationalize the difference in boiling points in each pair: (a) HF (20 °C) and HCl (-85 °C) (b) CHCl3 (61 °C) and CHBr3 (150 °C) (c) Br2 (59 °C) and ICl (97 °C)

Verified step by step guidance

1. Identify the types of intermolecular forces present in each compound. HF has hydrogen bonding, while HCl has dipole-dipole interactions and London dispersion forces. Hydrogen bonding is stronger than dipole-dipole interactions, leading to a higher boiling point for HF.
2. For CHCl3 and CHBr3, consider the molecular weight and the type of intermolecular forces. Both have dipole-dipole interactions and London dispersion forces, but CHBr3 has a higher molecular weight, leading to stronger London dispersion forces and a higher boiling point.
3. Examine the molecular structure and polarity of Br2 and ICl. Br2 is nonpolar with only London dispersion forces, while ICl is polar with dipole-dipole interactions in addition to London dispersion forces, resulting in a higher boiling point for ICl.
4. Discuss how molecular weight affects boiling points. Generally, as molecular weight increases, the strength of London dispersion forces increases, leading to higher boiling points.
5. Summarize the impact of intermolecular forces on boiling points: stronger intermolecular forces (hydrogen bonding, dipole-dipole interactions) result in higher boiling points compared to weaker forces (London dispersion forces).