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Which member in each pair has the stronger intermolecular dispersion forces? (a) \(\mathrm{H}_{2} \mathrm{O}\) or \(\mathrm{CH}_{3} \mathrm{OH},\) (b) \(\mathrm{CBr}_{3} \mathrm{CBr}_{3}\) or \(\mathrm{CCl}_{3} \mathrm{CCl}_{3}\) (c) \(\mathrm{C}\left(\mathrm{CH}_{3}\right)_{4}\) or \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\).

Short Answer

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(a) CH3OH, (b) CBr3CBr3, (c) CH3CH2CH2CH2CH3

Step by step solution

01

Analyze (a) H2O vs CH3OH

The dispersion forces depend on the molar mass and size of the molecule. Although both molecules can hydrogen bond, for this question we need to consider only dispersion forces. Both have similar sizes and similar environments for dispersion forces because of their hydrogen bonding, but CH3OH has a slightly higher molar mass. This generally leads to stronger dispersion forces.
02

Analyze (b) CBr3CBr3 vs CCl3CCl3

Dispersion forces increase with molecular weight and the total number of electrons. CBr3CBr3 has a higher molar mass than CCl3CCl3 due to the bromine atoms being heavier than chlorine atoms. Thus, CBr3CBr3 experiences stronger dispersion forces, as there are more electrons to contribute to temporary dipoles.
03

Analyze (c) C(CH3)4 vs CH3CH2CH2CH2CH3

Comparing the two hydrocarbons' shapes and molar masses, both have similar molar mass. However, CH3CH2CH2CH2CH3 (n-pentane) is a straight chain, whereas C(CH3)4 (tetramethylmethane) is more compact. Straight-chain molecules like n-pentane typically have stronger dispersion forces than their branched counterparts due to a greater surface area for intermolecular interactions.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Dispersion Forces
Dispersion forces, also known as London dispersion forces, are a type of weak intermolecular force. These forces arise from the movement of electrons in atoms and molecules, creating temporary dipoles. These temporary dipoles can influence nearby molecules, causing an attraction between them. Dispersion forces are present in all molecules, whether polar or nonpolar, but they are the only forces acting among nonpolar molecules.

The strength of dispersion forces depends on several factors, including the total number of electrons in the molecule and its overall shape. Molecules with a larger number of electrons typically exhibit stronger dispersion forces. That's because a larger electron cloud is more easily distorted, leading to stronger charge fluctuations. Additionally, the molecular shape affects the surface area available for interactions.

Even small or light molecules can have significant dispersion forces if they have a high number of electrons or a shape that promotes surface stability. Understanding these forces is essential when predicting physical properties like boiling and melting points.
Molecular Weight
Molecular weight is crucial in understanding dispersion forces as it greatly influences their strength. It refers to the sum of the atomic weights of all the atoms in a molecule. As the molecular weight increases, the capacity for molecules to exhibit stronger dispersion forces also increases.

This is because larger molecules have more electrons, which increases the potential for temporary dipole formation. Take, for instance, the comparison between \(\mathrm{CBr}_{3} \mathrm{CBr}_{3}\) and \(\mathrm{CCl}_{3} \mathrm{CCl}_{3}\). The bromine atoms in \(\mathrm{CBr}_{3} \mathrm{CBr}_{3}\) are heavier than chlorine atoms. Therefore, \(\mathrm{CBr}_{3} \mathrm{CBr}_{3}\) has a higher molecular weight, leading to stronger dispersion forces.

Knowledge of molecular weight allows chemists to estimate the strength of dispersion forces and predict how substances will behave under various conditions, such as how easily they might condense or evaporate.
Molar Mass
Closely related to molecular weight is the concept of molar mass. While molecular weight is an absolute measure of mass in atomic mass units (amu), molar mass represents the mass of a molecule per mole of a substance, typically in grams per mole (g/mol).

The molar mass can help predict and compare the impact of dispersion forces in different molecules. In example (a), comparing \(\mathrm{H}_{2} \mathrm{O}\) with \(\mathrm{CH}_{3} \mathrm{OH}\): Both have hydrogen bonding, but we focus on dispersion forces here. The molar mass of \(\mathrm{CH}_{3} \mathrm{OH}\) is slightly higher than \(\mathrm{H}_{2} \mathrm{O}\), leading to stronger dispersion forces.

Calculating and understanding molar mass is critical when determining the physical properties of substances, such as boiling points and solubility, which are influenced by the strength of intermolecular forces.
Molecular Shape
The shape of a molecule plays a significant role in determining the strength of dispersion forces. Specifically, how molecules are shaped affects their ability to pack closely and the overall surface area available for interaction.

In scenario (c), comparing \(\mathrm{C}(\mathrm{CH}_{3})_{4}\) with \(\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{CH}_{2}\mathrm{CH}_{2}\mathrm{CH}_{3}\), we observe that the latter, also known as n-pentane, is linear, while the former, tetramethylmethane, is more spherical. Linear molecules like n-pentane can pack together more closely than branched molecules, leading to larger surface areas and stronger dispersion forces.

This understanding of molecular shape and its impact on dispersion forces helps chemists design and synthesize molecules with desired physical and chemical properties, such as optimizing pharmaceuticals for better absorption and interactions within the body.

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Most popular questions from this chapter

Which member in each pair has the greater dispersion forces? (a) \(\mathrm{CH}_{3} \mathrm{OH}\) or \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH},(\mathbf{b}) \mathrm{NH}_{3}\) or \(\mathrm{N}\left(\mathrm{CH}_{3}\right)_{3},\) (c) \(\mathrm{CH}_{2} \mathrm{Cl}_{2}\) or \(\mathrm{CH}_{2} \mathrm{Br}_{2}\).

Butane and 2 -methylpropane, whose space-filling models are shown here, are both nonpolar and have the same molecular formula, \(\mathrm{C}_{4} \mathrm{H}_{10}\), yet butane has the higher boiling point \(\left(-0.5^{\circ} \mathrm{C}\right.\) compared to \(\left.-11.7{ }^{\circ} \mathrm{C}\right)\). Explain.

Ethylene glycol \(\left(\mathrm{HOCH}_{2} \mathrm{CH}_{2} \mathrm{OH}\right)\) and pentane \(\left(\mathrm{C}_{5} \mathrm{H}_{12}\right)\) are both liquids at room temperature and room pressure, and have about the same molecular weight. (a) One of these liquids is much more viscous than the other. Which one do you predict is more viscous? (b) One of these liquids has a much lower normal boiling point \(\left(36.1^{\circ} \mathrm{C}\right)\) compared to the other one \(\left(198^{\circ} \mathrm{C}\right) .\) Which liquid has the lower normal boiling point? (c) One of these liquids is the major component in antifreeze in automobile engines. Which liquid would you expect to be used as antifreeze? (d) One of these liquids is used as a "blowing agent" in the manufacture of polystyrene foam because it is so volatile. Which liquid would you expect to be used as a blowing agent?

Describe the intermolecular forces that must be overcome to convert these substances from a liquid to a gas: (a) \(\mathrm{CF}_{4}\) (b) \(\mathrm{NH}_{3}\), (c) \(\mathrm{BCl}_{3}\)

For many years drinking water has been cooled in hot climates by evaporating it from the surfaces of canvas bags of porous clay pots. How many grams of water can be cooled from 35 to \(20^{\circ} \mathrm{C}\) by the evaporation of \(60 \mathrm{~g}\) of water? (The heat of vaporization of water in this temperature range is \(2.4 \mathrm{~kJ} / \mathrm{g}\). The specific heat of water is \(4.18 \mathrm{~J} / \mathrm{g}-\mathrm{K}\).)

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