Choose from each pair the substance that, in the liquid state, would have the greater vapor pressure at a given temperature. Base your choice on predicted strengths of intermolecular forces. (a) \( \mathrm{BiBr}_{3} \) or \( \mathrm{BiCl}_{3} \); (b) \( \mathrm{CO}^{\text {a }} \) or \( \mathrm{CO}_{2} \); (c) \( \mathrm{N}_{2} \) or NO; (d) \( \mathrm{CH}_{3} \mathrm{COOH}^{\circ} \) or \( \mathrm{HCOOCH}_{3} \).

To determine which substance in each pair has a greater vapor pressure in the liquid state, we need to consider the strengths of their intermolecular forces. Substances with **weaker intermolecular forces** will have **higher vapor pressures** because their molecules can escape into the vapor phase more easily. Let's analyze each pair: ### (a) \( \mathrm{BiBr}_{3} \) vs. \( \mathrm{BiCl}_{3} \) - **Intermolecular Forces**: Both compounds are similar in structure, but \( \mathrm{BiBr}_{3} \) has larger bromine atoms compared to chlorine in \( \mathrm{BiCl}_{3} \). Larger atoms are more polarizable, leading to **stronger London dispersion forces** in \( \mathrm{BiBr}_{3} \). - **Vapor Pressure**: **\( \mathrm{BiCl}_{3} \)** has weaker intermolecular forces and therefore a **higher vapor pressure**. ### (b) \( \mathrm{CO} \) vs. \( \mathrm{CO}_{2} \) - **Intermolecular Forces**: \( \mathrm{CO} \) is a polar molecule with dipole-dipole interactions, whereas \( \mathrm{CO}_{2} \) is nonpolar and relies only on London dispersion forces. - **Vapor Pressure**: **\( \mathrm{CO}_{2} \)** has weaker intermolecular forces and thus a **higher vapor pressure**. ### (c) \( \mathrm{N}_{2} \) vs. NO - **Intermolecular Forces**: \( \mathrm{N}_{2} \) is nonpolar with only London dispersion forces. NO has a dipole moment, introducing dipole-dipole interactions which are stronger than the forces in \( \mathrm{N}_{2} \). - **Vapor Pressure**: **\( \mathrm{N}_{2} \)** has weaker intermolecular forces and therefore a **higher vapor pressure**. ### (d) \( \mathrm{CH}_{3}\mathrm{COOH} \) vs. \( \mathrm{HCOOCH}_{3} \) - **Intermolecular Forces**: \( \mathrm{CH}_{3}\mathrm{COOH} \) (acetic acid) can form strong hydrogen bonds due to the hydroxyl (-OH) group. \( \mathrm{HCOOCH}_{3} \) (methyl formate) cannot form hydrogen bonds as it lacks the hydroxyl group. - **Vapor Pressure**: **\( \mathrm{HCOOCH}_{3} \)** has weaker intermolecular forces and thus a **higher vapor pressure**. ### **Summary of Choices with Greater Vapor Pressure:** a) **\( \mathrm{BiCl}_{3} \)** b) **\( \mathrm{CO}_{2} \)** c) **\( \mathrm{N}_{2} \)** d) **\( \mathrm{HCOOCH}_{3} \)** Answer Summary: All are the second-listed compounds, which have higher vapor pressures: a) BiCl₃; b) CO₂; c) N₂; d) HCOOCH₃.