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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 \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{SH}_{4}\) (c) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Cl}\) or \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHCl}\).

Short Answer

Expert verified
In summary, the molecules with stronger intermolecular dispersion forces are: (a) Br2, (b) CH3CH2CH2CH2CH2SH, and (c) CH3CH2CH2Cl.

Step by step solution

01

Pair (a): Br2 and O2

To compare Br2 and O2, we will consider the size and polarizability of each molecule. Br2 has a larger atomic size than O2 because its atomic number is greater, which leads to a higher number of electrons and an increased electron cloud size. The polarizability of Br2 is also greater than O2 due to the larger electron cloud and the looser hold the nucleus has on the electrons. Therefore, Br2 has stronger intermolecular dispersion forces than O2.
02

Pair (b): CH3CH2CH2CH2SH and CH3CH2CH2CH2CH2SH4

There is a typo in the formula of the second compound. Assuming that it should be CH3CH2CH2CH2CH2SH, we can compare the two molecules. Both compounds have the same number of carbon and hydrogen atoms; the only difference is that the first compound contains a sulfur atom, while the second compound has an additional carbon and two hydrogen atoms. The number of electrons in the second compound is greater, making it larger and more polarizable. Therefore, CH3CH2CH2CH2CH2SH has stronger intermolecular dispersion forces than CH3CH2CH2CH2SH.
03

Pair (c): CH3CH2CH2Cl and (CH3)2CHCl

Comparing these two molecules, we can see that both contain the same number of carbon, hydrogen, and chlorine atoms. However, the arrangement of these atoms is different. In CH3CH2CH2Cl, the carbon atoms form a straight chain, while in (CH3)2CHCl, the carbon atoms form a branched chain. Molecules with more branching have a more compact shape, leading to a reduced surface area available for London dispersion forces. Hence, CH3CH2CH2Cl, with a more extended structure due to the straight chain, will have stronger intermolecular dispersion forces than (CH3)2CHCl.

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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 that occurs between molecules. They are the result of temporary fluctuations in the electron distribution within atoms or molecules, leading to temporary dipoles. These dipoles, in turn, induce dipoles in neighboring molecules, leading to an attraction between them.

These forces are universal and occur in all molecules, whether polar or nonpolar. Although dispersion forces are weak individually, they can become significant when summed up over large numbers of molecules, especially in molecules with large electron clouds.
  • Cumulative effect: More extensive molecules with larger electron clouds have stronger dispersion forces due to increased polarizability.
  • Universal presence: All molecules experience dispersion forces, but their strength can vary widely.
Understanding dispersion forces is crucial to predicting boiling points, solubilities, and other physical properties of substances.
Polarizability
Polarizability refers to the ability of an electron cloud around an atom or molecule to be distorted. This distortion occurs in the presence of an electric field, such as that from an external molecule or ion.

A larger, more diffuse electron cloud will be more susceptible to distortion, making the molecule more polarizable. Consequently, polarizability affects the strength of the dispersion forces: the higher the polarizability, the stronger the attractive forces. Examples of factors affecting polarizability include:
  • Size of the molecule: Larger atoms and molecules generally have higher polarizability.
  • Electron density: A denser electron cloud can be easily distorted, heightening polarizability.
Increasing polarizability enhances dispersion forces, influencing properties such as volatility and boiling points.
Molecular Structure
Molecular structure greatly influences the strength of dispersion forces. The arrangement of atoms in a molecule can lead to significant differences in surface area, thereby affecting how these forces are experienced.

For instance, in comparing linear and branched molecules, linear molecules typically have greater surface area contact and thus stronger dispersion forces.
  • Shape and Surface Area: Linear molecules have more surface in contact with others, enhancing dispersion forces compared to branched molecules which are more compact.
  • Compactness: Branched molecules reduce effective contact area, leading to weaker dispersion forces.
Hence, the shape and atomic connectivity within a molecule play vital roles in determining the strength of dispersion forces.
Chemical Comparisons
Chemical comparisons involve evaluating the properties of different molecules or compounds based on their chemical composition and structure. By studying characteristics such as molecular weight, shape, and polarizability, it's possible to predict which molecule will have stronger intermolecular forces, like dispersion forces.

Key metrics for comparisons include:
  • Molecular weight: Typically, the heavier the molecule, the stronger the dispersion forces due to a larger electron cloud.
  • Molecular shape: Straight-chained molecules often display stronger dispersion forces compared to their branched counterparts.
  • Electron cloud: A molecule with a larger electron cloud will usually have increased polarizability and thus stronger dispersion forces.
Understanding these principles assists in predicting and explaining behaviors such as boiling points and viscosities of various substances.

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Most popular questions from this chapter

As a metal such as lead melts, what happens to (a) the average kinetic energy of the atoms, (b) the average distance between the atoms?

Ethylene glycol \(\left(\mathrm{HOCH}_{2} \mathrm{CH}_{2} \mathrm{OH}\right)\) is the major component of antifreeze. It is a slightly viscous liquid, not very volatile at room temperature, with a boiling point of \(198^{\circ} \mathrm{C}\). Pentane \(\left(\mathrm{C}_{5} \mathrm{H}_{12}\right)\), which has about the same molecular weight, is a nonviscous liquid that is highly volatile at room temperature and whose boiling point is \(36.1^{\circ} \mathrm{C}\). Explain the differences in the physical properties of the two substances.

Benzoic acid, \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\), melts at \(122^{\circ} \mathrm{C}\). The density in the liquid state at \(130^{\circ} \mathrm{C}\) is \(1.08 \mathrm{~g} / \mathrm{cm}^{3}\). The density of solid benzoic acid at \(15^{\circ} \mathrm{C}\) is \(1.266 \mathrm{~g} / \mathrm{cm}^{3}\). (a) In which of these two states is the average distance between molecules greater? (b) Explain the difference in densities at the two temperatures in terms of the relative kinetic energies of the molecules.

Suppose the vapor pressure of a substance is measured at two different temperatures. (a) By using the ClausiusClapeyron equation (Equation 11.1) derive the following relationship between the vapor pressures, \(P_{1}\) and \(P_{2}\), and the absolute temperatures at which they were measured, \(T_{1}\) and \(T_{2}:\) $$ \ln \frac{P_{1}}{P_{2}}=-\frac{\Delta H_{\text {vap }}}{R}\left(\frac{1}{T_{1}}-\frac{1}{T_{2}}\right) $$ (b) Gasoline is a mixture of hydrocarbons, a major component of which is octane \(\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\right)\). Octane has a vapor pressure of \(13.95\) torr at \(25^{\circ} \mathrm{C}\) and a vapor pressure of \(144.78\) torr at \(75^{\circ} \mathrm{C}\). Use these data and the equation in part (a) to calculate the heat of vaporization of octane. (c) By using the equation in part (a) and the data given in part (b), calculate the normal boiling point of octane. Compare your answer to the one you obtained from Exercise 11.80. (d) Calculate the vapor pressure of octane at \(-30^{\circ} \mathrm{C}\).

Describe the intermolecular forces that must be overcome to convert these substances from a liquid to a gas: (a) \(\mathrm{SO}_{2}\), (b) \(\mathrm{CH}_{3} \mathrm{COOH}_{\text {, }}\) (c) \(\mathrm{H}_{2} \mathrm{~S}\).

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