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Which type of intermolecular force accounts for each of these differences? (a) \(\mathrm{CH}_{3} \mathrm{OH}\) boils at \(65^{\circ} \mathrm{C} ; \mathrm{CH}_{3} \mathrm{SH}\) boils at \(6^{\circ} \mathrm{C} .(\mathbf{b}) \mathrm{Xe}\) is a liquid at atmospheric pressure and \(120 \mathrm{K},\)whereas Ar is a gas under the same conditions. (c) Kr, atomic weight 84 amu, boils at \(120.9 \mathrm{K},\) whereas \(\mathrm{Cl}_{2},\) molecular weight about 71 amu, boils at 238 \(\mathrm{K}\) . (d) Acetone boils at \(56^{\circ} \mathrm{C},\) whereas 2 -methylpropane boils at \(-12^{\circ} \mathrm{C}\) .

Short Answer

Expert verified
(a) CH3OH exhibits hydrogen bonding, leading to a higher boiling point (65°C) than CH3SH (6°C), which has weaker dipole-dipole forces. (b) Xe has stronger dispersion forces due to more electrons, making it a liquid at 120K, while Ar remains a gas. (c) Cl2 has a higher boiling point (238K) due to the presence of dipole-dipole interactions, while Kr (120.9K) only has weak dispersion forces. (d) Acetone has stronger dipole-dipole forces due to its polar carbonyl group, leading to its higher boiling point (56°C) compared to 2-methylpropane (-12°C), which has weak dispersion forces.

Step by step solution

01

Intermolecular Forces for CH3OH and CH3SH

Both compounds have dispersion forces present; however, CH3OH exhibits hydrogen bonding, whereas CH3SH does not. Hydrogen bonding, present in CH3OH, is stronger than the dipole-dipole forces present in CH3SH, leading to a higher boiling point for CH3OH (65°C) as compared to CH3SH (6°C).
02

Intermolecular Forces for Xe and Ar

Both noble gases, Xe and Ar, generally have only weak dispersion forces between their atoms; however, as Xe has more electrons than Ar, it presents stronger dispersion forces. This leads to Xe being a liquid at atmospheric pressure and 120K, whereas Ar remains a gas under similar conditions.
03

Intermolecular Forces for Kr and Cl2

Both Kr (a noble gas) and Cl2 (a diatomic molecule) present dispersion forces among their particles. Despite having a lower molecular weight, Cl2 has an additional dipole-dipole interaction between its molecules due to its polar nature. Thus, the boiling point of Cl2 (238K) is higher than that of Kr (120.9K), which only has weak dispersion forces.
04

Intermolecular Forces for Acetone and 2-Methylpropane

Acetone has a polar carbonyl group which results in the presence of dipole-dipole forces between its molecules. On the other hand, 2-methylpropane has only weak dispersion forces since it is a nonpolar molecule. The stronger dipole-dipole forces present in acetone lead to its higher boiling point (56°C) as compared to 2-methylpropane (-12°C).

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Key Concepts

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

Hydrogen Bonding
Hydrogen bonding is a special type of intermolecular force that occurs when hydrogen atoms are directly bonded to small, highly electronegative atoms like nitrogen, oxygen, or fluorine. In the context of the exercise, CH3OH (methanol) exhibits hydrogen bonding due to the presence of an -OH group. The hydrogen in the -OH group forms a bond with the oxygen atom of another methanol molecule.
This bonding significantly increases its intermolecular attractions, leading to a higher boiling point.
Hydrogen bonds are important because:
  • They are relatively strong compared to other intermolecular forces.
  • They significantly influence the physical properties of substances, such as boiling and melting points.
  • They are responsible for the unique properties of water, such as its high surface tension and specific heat capacity.
These factors explain why methanol, with hydrogen bonds, boils at a much higher temperature than CH3SH (methyl mercaptan), which lacks hydrogen bonding.
Dispersion Forces
Dispersion forces, also known as London dispersion forces, are the weakest of all intermolecular forces. They arise from temporary fluctuations in electron distribution in atoms and molecules, leading to transient dipoles and resulting attractions. Despite their weaknesses, dispersion forces are present in all molecular interactions.
For the noble gases Xe (xenon) and Ar (argon), dispersion forces are the primary type of intermolecular attraction present.
  • Xenon, with its greater number of electrons, experiences stronger dispersion forces than argon.
  • These stronger forces in xenon are due to the larger electron cloud, allowing for more significant temporary dipoles.
  • This results in xenon being able to liquefy at higher temperatures than argon.
Thus, even though both noble gases are chemically inert, their physical states at similar conditions differ, largely due to the strength of dispersion forces.
Dipole-Dipole Interactions
Dipole-dipole interactions occur in molecules that have permanent molecular dipoles, where there is a distinct imbalance in charge distribution. This leads to an electrostatic attraction between the positive end of one dipole and the negative end of another.
A great example from the exercise is Acetone, which features a carbonyl group (C=O) that creates a dipole moment.
This interaction is compared to 2-methylpropane, which has no dipole-dipole interactions due to its symmetrical, non-polar structure.
  • Dipole-dipole interactions are generally stronger than dispersion forces but weaker than hydrogen bonding.
  • They play a crucial role in determining the boiling points of polar compounds.
Understanding these interactions is essential in explaining why polar molecules typically have higher boiling points than their non-polar counterparts.
Boiling Point
The boiling point of a substance is the temperature at which its vapor pressure equals the atmospheric pressure. It is a critical property that helps us understand the strength of intermolecular forces within a liquid.
  • Higher boiling points usually indicate stronger intermolecular forces.
  • As exemplified in the exercise, CH3OH and CH3SH have different boiling points due to hydrogen bonding in the former.
  • Similarly, the boiling point differences between Xe and Ar, and Kr and Cl2, are results of varying dispersion and dipole-dipole forces present.
Boiling point is affected by a multitude of factors such as molecular weight, type of intermolecular interactions, and molecular structure. Understanding these concepts gives insight into the physical behavior of substances under different conditions.

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Most popular questions from this chapter

(a) List the following molecules in order of increasing polar-izability: GeCl_ \(_{4}, \mathrm{CH}_{4}, \mathrm{SiCl}_{4}, \mathrm{SiH}_{4},\) and \(\mathrm{GeBr}_{4}\) . (b) Predict the order of boiling points of the substances in part (a).

True or false: (a) For molecules with similar molecular weights, the dispersion forces become stronger as the molecules become more polarizable. (b) For the noble gases the dispersion forces decrease while the boiling points increase as you go down the column in the periodic table. (c) In terms of the total attractive forces for a given substance, dipole- dipole interactions, when present, are always greater than dispersion forces.( \(\mathbf{d}\) ) All other factors being the same, dispersion forces between linear molecules are greater than those between molecules whose shapes are nearly spherical. (e) The larger the atom, the more polarizable it is.

(a) What is the significance of the critical point in a phase diagram? (b) Why does the line that separates the gas and liquid phases end at the critical point?

(a) What is the relationship between surface tension and temperature? (b) What is the relationship between viscosity and temperature? (c) Why do substances with high surface tension also tend to have high viscosities?

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 \(\left(-0.5^{\circ} \mathrm{C}\) compared to \(-11.7^{\circ} \mathrm{C}\right) .\) Explain.

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