Problem 27 - Introductory Chemistry Atoms First

Question

Would you expect \(\mathrm{CCl}_{4}\) or \(\mathrm{CBr}_{4}\) to have the higher boiling point? Explain your answer.

Step-by-Step Solution

Answer

We expect \(\mathrm{CBr}_{4}\) to have a higher boiling point than \(\mathrm{CCl}_{4}\) because it has a greater molecular weight, leading to stronger London dispersion forces, which are the dominant intermolecular forces in both compounds.

Step 1: Identify the types of intermolecular forces present in both compounds

In both \(\mathrm{CCl}_{4}\) and \(\mathrm{CBr}_{4}\), the central carbon atom is covalently bonded to four halogen atoms. Since both compounds are nonpolar due to their symmetrical shape, there are no dipole-dipole forces or hydrogen bonding present. The only type of intermolecular forces present in both compounds are London dispersion forces.

Step 2: Compare the London dispersion forces of both compounds

London dispersion forces increase with molecular weight and larger surface area. In \(\mathrm{CCl}_{4}\), the molecular weight is about 154 g/mol, while in \(\mathrm{CBr}_{4}\), the molecular weight is about 332 g/mol. Therefore, \(\mathrm{CBr}_{4}\) has a greater molecular weight.

Step 3: Relate London dispersion forces to boiling point

The stronger the intermolecular forces within a compound, the higher its boiling point will be. Since \(\mathrm{CBr}_{4}\) has a greater molecular weight, it will have stronger London dispersion forces than \(\mathrm{CCl}_{4}\).

Step 4: Conclude which compound has a higher boiling point

As a result of the stronger London dispersion forces present in \(\mathrm{CBr}_{4}\) compared to \(\mathrm{CCl}_{4}\), we expect \(\mathrm{CBr}_{4}\) to have a higher boiling point.

Key Concepts

Intermolecular Forces

Intermolecular forces are the attractions that occur between molecules. These forces play a critical role in determining the physical properties of substances, such as boiling points, melting points, and solubility. There are several different types of intermolecular forces, but in this context, we are focusing on London dispersion forces. These forces are generally weaker than other types of intermolecular forces such as hydrogen bonds and dipole-dipole interactions.
However, even these seemingly weak London dispersion forces can become quite significant, especially in larger and heavier molecules. The strength of the intermolecular forces directly influences how much energy is required to change the phase of a substance. In simpler terms, the stronger the forces between the molecules, the higher the temperature needed to boil the substance.

London Dispersion Forces

London dispersion forces are a type of van der Waals force and are the only type of intermolecular force present in nonpolar molecules like \({\mathrm{CCl}}_{4}\) and \({\mathrm{CBr}}_{4}\). They arise due to the temporary fluctuations in electron density which creates temporary dipoles. These forces depend on the size and the shape of the molecules.
- Larger molecules tend to have stronger London dispersion forces because they have more electrons available to create temporary dipoles.- The surface area of the molecule also affects these forces. The larger the surface area, the stronger the dispersion forces.Even though they are the weakest type of intermolecular force, London dispersion forces can have a significant impact on the boiling points of substances. This is especially true for heavier molecules, where dispersion forces are more pronounced.

Molecular Weight

Molecular weight, also known as molecular mass, is the sum of the atomic weights of all the atoms in a molecule. It is typically measured in grams per mole (g/mol). Molecular weight is a crucial factor in determining the strength of London dispersion forces. As a rule of thumb, the higher the molecular weight, the stronger the London dispersion forces.
This is because higher molecular weight compounds have more electrons, which increases the potential for forming temporary dipoles. In the case of \({\mathrm{CCl}}_{4}\) and \({\mathrm{CBr}}_{4}\), \({\mathrm{CBr}}_{4}\) has a significantly higher molecular weight than \({\mathrm{CCl}}_{4}\), resulting in stronger dispersion forces and thus a higher boiling point.

• \({\mathrm{CCl}}_{4}\) has a molecular weight of about 154 g/mol.
• \({\mathrm{CBr}}_{4}\) has a molecular weight of about 332 g/mol.

This substantial difference in molecular weight explains why \({\mathrm{CBr}}_{4}\) is expected to have a higher boiling point compared to \({\mathrm{CCl}}_{4}\).