Question

Which member in each pair has the stronger intermolecular dispersion forces: (a) \(\mathrm{Br}_{2}\) or \(\mathrm{O}_{2}\), (b) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{SH}\) or \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{SH},(\mathrm{c}) \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Cl}\) or \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHCl} ?\)

Short answer

In each pair, the molecule with the stronger intermolecular dispersion forces are: (a) \(\mathrm{Br}_{2}\), (b) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{SH}\), and (c) \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHCl}\). This is due to their larger size, electron cloud, and polarizability compared to their counterparts.

Step-by-step solution

Pair (a) comparison: \(\mathrm{Br}_{2}\) vs \(\mathrm{O}_{2}\)

Since both molecules are homonuclear diatomic molecules, the primary difference between them is their size and electron cloud. A larger electron cloud is more easily polarizable, which leads to stronger dispersion forces. Bromine is larger than oxygen and has more electrons, so its electron cloud is larger. Thus, \(\mathrm{Br}_{2}\) has stronger dispersion forces than \(\mathrm{O}_{2}\).

Pair (b) comparison: \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{SH}\) vs \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{SH}\)

In this case, both molecules have similar functional groups (Sulfur-containing). The major difference between them is the length of their carbon chains. A longer carbon chain will lead to a larger surface area and polarization, thus resulting in stronger dispersion forces. Therefore, \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{SH}\) has stronger dispersion forces than \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{SH}\).

Pair (c) comparison: \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Cl}\) vs \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHCl}\)

For this pair, both molecules have chlorine atoms, so the primary difference in their dispersion forces comes from the size and shape of the molecules. The molecule \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHCl}\) has a more compact shape and two methyl groups, which make it more polarizable. Therefore, \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHCl}\) has stronger dispersion forces than \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Cl}\).

Key concepts

Dispersion Forces

Dispersion forces, also known as London dispersion forces, are the weakest type of intermolecular forces. They play a crucial role in the interactions between molecules, especially nonpolar ones. These forces arise due to the temporary fluctuations in electron density within atoms and molecules, leading to temporary dipoles. When these dipoles form, they can induce dipoles in neighboring molecules, causing an attraction.

Dispersion forces are ubiquitous and can be found in all molecules, whether polar or nonpolar. They are particularly significant in larger molecules because the strength of these forces increases with more electrons. This is why molecules like \(\mathrm{Br}_{2}\) exhibit stronger dispersion forces compared to \(\mathrm{O}_{2}\), as bromine has a larger electron cloud.

Polarizability

Polarizability refers to the ease with which the electron cloud of a molecule can be distorted to form temporary dipoles. Larger atoms with more electrons tend to be more polarizable because their outer electrons are further from the nucleus and are less tightly held.

For instance, consider \(\mathrm{Br}_{2}\) and \(\mathrm{O}_{2}\). Bromine is more polarizable than oxygen due to its larger size and greater number of electrons. As a result, \(\mathrm{Br}_{2}\) has stronger dispersion forces than \(\mathrm{O}_{2}\).

In general, the more polarizable a molecule, the stronger the dispersion forces it will experience.

Molecular Size

Molecular size significantly influences the strength of intermolecular forces, particularly dispersion forces. Larger molecules have greater surface areas allowing for more interactions. Moreover, more electrons contribute to greater polarizability.

In comparing two molecules, the one with the longer carbon chain usually exhibits stronger dispersion forces. This is evident in molecules like \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{SH}\) versus \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{SH}\). The latter has an extended carbon chain, improving its ability to attract adjacent molecules through stronger dispersion forces.

This principle applies broadly: the larger the molecule, the stronger its dispersion forces.

Functional Groups

Functional groups are specific groups of atoms within molecules that determine the characteristics and chemical reactivity of the molecules. While the type of functional group can influence the overall dipole and reactivity, the presence of certain groups can change dispersion forces slightly.

In the context of dispersion forces, the specific functional group doesn't change the nature of these forces directly. However, the size and shape related to functional groups can affect molecule interactions. Consider \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Cl}\) and \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHCl}\); both contain a chlorine atom, but the shape and bulkiness of the groups influence the strength of dispersion forces.

The interplay between shape and type of functional group can thus slightly modify how these forces act between complex molecules.