Problem 16 - Chemistry: The Central Science

Question

What type of intermolecular force accounts for the following differences in each case? (a) \(\mathrm{CH}_{3} \mathrm{OH}\) boils at \(65^{\circ} \mathrm{C}, \mathrm{CH}_{3} \mathrm{SH}\) boils at \(6^{\circ} \mathrm{C}\). (b) Xe is liquid at atmos- pheric pressure and \(120 \mathrm{~K}\), whereas \(\mathrm{Ar}\) is a gas. (c) \(\mathrm{Kr}\), atomic weight 84 , boils at \(120.9 \mathrm{~K}\), whereas \(\mathrm{Cl}_{2}\), molecular weight about 71, boils at \(238 \mathrm{~K} .\) (d) Acetone boils at \(56{ }^{\circ} \mathrm{C}\), whereas 2 -methylpropane boils at \(-12{ }^{\circ} \mathrm{C}\).

Step-by-Step Solution

Answer

\( \) In summary: (a) Methanol (\(CH_3OH\)) has stronger intermolecular forces due to hydrogen bonding, compared to methanethiol (\(CH_3SH\)) which lacks hydrogen bonding. (b) Xenon (Xe) experiences greater London dispersion forces compared to Argon (Ar) due to its larger size and more electrons, making it liquid at 120 K while Ar is a gas. (c) Chlorine (Cl₂) has stronger dipole-dipole forces compared to the London dispersion forces in Krypton (Kr), leading to a higher boiling point for Cl₂. (d) Acetone has stronger dipole-dipole interactions due to its polar C=O bond, resulting in a higher boiling point compared to 2-methylpropane which only has weaker London dispersion forces.

Step 1: (Case a): Methanol vs Methanethiol

We need to determine the intermolecular force responsible for the difference in boiling points between \(CH_3OH\) (methanol) and \(CH_3SH\) (methanethiol). Methanol has hydrogen bonding due to the presence of a highly polar O-H bond which results in a strong type of dipole-dipole interaction. Methanethiol lacks hydrogen bonding as it only has a S-H bond, which makes it a less polar molecule. The intermolecular forces in \(CH_3OH\) are stronger due to hydrogen bonding, which leads to higher boiling points compared to \(CH_3SH\).

Step 2: (Case b): Xe vs Ar

We need to determine the intermolecular force responsible for Xe being liquid at atmospheric pressure and 120 K, while Ar is a gas. Both Xenon (Xe) and Argon (Ar) are noble gases, meaning they have a filled valence electron shell. As a result, these elements exhibit weak intermolecular forces called London dispersion forces. Since Xenon is larger and has more electrons than Argon, it has greater London dispersion forces, which increase with the size and number of electrons present. Thus, the stronger London dispersion forces in Xe are responsible for its liquid state at 120 K compared to Ar, which is in the gaseous state.

Step 3: (Case c): Kr vs Cl₂

Here we need to compare Krypton (Kr) with a molecular weight of 84 and boiling point of 120.9 K and chlorine (Cl₂) with a molecular weight of about 71 and boiling point of 238 K. Kr is a noble gas similar to Xe and Ar, so it also experiences London dispersion forces. Chlorine (Cl₂) is a diatomic molecule formed by halogens, which experience dipole-dipole interactions because of the difference in electronegativity between the two atoms in each molecule. In addition, Cl₂ will also experience London dispersion forces as they are present in all molecules regardless of their polarity. Comparatively, the dipole-dipole forces in Cl₂ are stronger than the London dispersion forces in Kr, which leads to a higher boiling point for Cl₂.

Step 4: (Case d): Acetone vs 2-methylpropane

We need to compare the boiling points of acetone (\(C_3H_6O\), boiling point: 56 °C) and 2-methylpropane (\(C_4H_{10}\), boiling point: -12 °C). Acetone is a polar molecule due to the presence of the C=O bond, which leads to dipole-dipole interactions between the molecules. On the other hand, 2-methylpropane is a nonpolar molecule as its shape is symmetrical, and it only experiences London dispersion forces. The stronger dipole-dipole interactions in acetone result in a higher boiling point compared to 2-methylpropane, which relies solely on weaker London dispersion forces.

Key Concepts

Hydrogen Bonding

Hydrogen bonding is a special type of intermolecular force that explains why some molecules have higher boiling points compared to others. It occurs when hydrogen is covalently bonded to highly electronegative atoms like oxygen, nitrogen, or fluorine. This bond creates a significant polarity, leading to especially strong types of dipole-dipole interactions. For example, in methanol (\(CH_3OH\)), hydrogen atoms are linked to oxygen, resulting in hydrogen bonding.

This strong bond is why methanol has a comparatively high boiling point of \(65^{\circ} C\). In contrast, methanethiol (\(CH_3SH\)) lacks hydrogen bonding because sulfur is not as electronegative as oxygen. Thus, it boils at a much lower temperature (\(6^{\circ} C\)). The strength of hydrogen bonding affects many physical properties, including boiling points, making it a crucial concept in understanding molecular interactions.

• Water is another classic example of hydrogen bonding. It has a high boiling point because hydrogen and oxygen form these bonds.
• Hydrogen bonds are directional, meaning they form specific angles and can strongly influence the structure of the molecule.

Understanding hydrogen bonding helps in explaining the unique properties of molecules like water and methanol.

London Dispersion Forces

London dispersion forces are a universal intermolecular force that can be quite weak but crucial in certain contexts. These forces arise due to temporary fluctuations in electron distribution within atoms or molecules, creating temporary dipoles. Although they are the weakest type of intermolecular force, London dispersion forces increase with more electrons present.

Noble gases, like xenon (\(Xe\)) and argon (\(Ar\)), exhibit only London dispersion forces because they lack permanent dipoles. Despite this, xenon, having more electrons than argon, experiences stronger dispersion forces. This difference is why xenon can be a liquid at 120 K, while argon remains a gas.

• Larger atoms and molecules typically exhibit stronger London dispersion forces due to more significant fluctuations in electron density.
• All atoms and molecules experience London dispersion forces, but they are most noticeable in noble gases and nonpolar molecules.

Recognizing the role of London dispersion forces is vital, especially when comparing elements and their states or looking at nonpolar substances.

Dipole-Dipole Interactions

Dipole-dipole interactions are another crucial type of intermolecular force, occurring between molecules with permanent dipoles. These interactions happen when partially positive ends of a molecule attract the partially negative ends of another molecule. They are stronger than London dispersion forces but not as strong as hydrogen bonds.

To illustrate, acetone (\(C_3H_6O\)), with its polar C=O bond, exhibits dipole-dipole interactions, leading to its higher boiling point compared to 2-methylpropane (\(C_4H_{10}\)), which is nonpolar and relies on weaker London dispersion forces. Acetone's molecules align themselves in a manner that their opposite charges attract.

On the flip side, chlorine (\(Cl_2\)) experiences dipole-dipole interactions due to its polar bonds, albeit in a lesser degree than molecules capable of hydrogen bonding. These forces contribute to the higher boiling point of chlorine compared to krypton (\(Kr\)), which primarily exhibits London dispersion forces.

• Dipole-dipole interactions are directional, meaning the orientation of molecules can enhance or diminish the force strength.
• These interactions are more significant in polar molecules and should be considered along with other intermolecular forces.

Understanding dipole-dipole interactions aids in explaining the boiling points and physical behavior of various substances in different environments.