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Chapter 12: Problem 13

Arrange the following compounds in order of increasing boiling point: \(\mathrm{RbF}, \mathrm{CO}_{2}, \mathrm{CH}_{3} \mathrm{OH}, \mathrm{CH}_{3} \mathrm{Br}\). Explain your arrangement.

Short Answer

Expert verified
The compounds arranged in order of increasing boiling point are: CO2, CH3Br, CH3OH, RbF.

Step by step solution

01

Understanding the types of forces in each compound

The first thing you should do is identify the type of intermolecular forces in each of these compounds. Rubidium Fluoride (RbF) is an ionic compound – it has ionic bonds. The other three compounds are molecular, which means their forces are related to covalent bonds exterior effect. Carbon Dioxide (CO2) is a non-polar molecule, so it only has London dispersion forces. Methanol (CH3OH) has a polar covalent bond, hence it has dipole-dipole forces and also hydrogen bonding due to the presence of a -OH group. Bromomethane (CH3Br) is also a polar molecule and has dipole-dipole forces, but it lacks a hydrogen bonding since there's no OH or NH groups.
02

Ranking the compounds by forces strength

Next you'll need to rank the types of intermolecular forces from weakest to strongest. The order (from weakest to strongest) is generally: London dispersion forces < dipole-dipole forces < Hydrogen bonding < Ionic bonding. Given that, it could be inferred that CO2 should have the lowest boiling point as it only possesses London dispersion forces, which are the weakest. Then comes CH3Br with dipole-dipole forces, followed by CH3OH with both dipole-dipole and hydrogen bonding (Stronger than just dipole-dipole). RbF would have the highest boiling point due to its ionic bonds, stronger than all the previous forces.
03

Final arrangement

After considering the strength of these intermolecular forces, you can then arrange the given compounds in order of increasing boiling points. Thus, the correct arrangement in order of increasing boiling point should be: CO2 < CH3Br < CH3OH < RbF

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