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

Short Answer

Expert verified
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 by step solution

01

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.
02

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.
03

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 \(\mathrm{CCl}_{4}\).
04

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.

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