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Chapter 10: Problem 29

You have three covalent compounds with three very different boiling points. All of the compounds have similar molar mass and relative shape. Explain how these three compounds could have very different boiling points.

Short Answer

Expert verified
The different boiling points of the three covalent compounds with similar molar mass and relative shape can be explained by differences in their intermolecular forces. These forces include London dispersion forces, dipole-dipole interactions, and hydrogen bonding. Differences in the chemical composition and structure of the compounds result in varying strengths of these intermolecular forces, leading to the observed differences in boiling points.

Step by step solution

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1. Understanding Covalent Compounds

Covalent compounds are formed by the sharing of electrons between atoms, resulting in the formation of molecules. The boiling point of a compound depends on the strength of the forces holding the molecules together. The stronger these forces (called the intermolecular forces) are, the higher the boiling point will be. So, different intermolecular forces can result in different boiling points.
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2. Types of Intermolecular Forces

There are multiple types of intermolecular forces that could be affecting the boiling points of these three covalent compounds. These include: 1. London dispersion forces 2. Dipole-dipole interactions 3. Hydrogen bonding
03

3. London Dispersion Forces

London dispersion forces, also known as Van der Waals forces, are weak forces that occur between nonpolar molecules. These forces arise due to the temporary polarity created in molecules as a result of electron motion. The more electrons a molecule has, the greater the chance of temporary polarity forming, and thus, the stronger the London dispersion forces. However, as the three compounds have similar molar masses and shapes, the difference in London dispersion forces isn't enough to explain significant boiling point differences.
04

4. Dipole-Dipole Interactions

Dipole-dipole interactions are stronger than London dispersion forces and occur between polar molecules. Polar molecules have a net dipole moment due to the presence of polar covalent bonds. Polar covalent bonds form when the electrons in a bond are shared unequally between the two atoms. If the three covalent compounds are polar molecules to different extents, differing dipole-dipole interactions could result in differences in their boiling points.
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5. Hydrogen Bonding

Hydrogen bonding is an even stronger type of intermolecular force. It is a specific type of dipole-dipole interaction that occurs between molecules that have hydrogen atom(s) bonded to highly electronegative atoms such as nitrogen, oxygen, or fluorine. The strong hydrogen bond forces cause an increase in boiling point. The presence or absence of hydrogen bonding in one or more of these compounds could lead to the observed differences in their boiling points.
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6. Conclusion

The different boiling points of these three covalent compounds with similar molar mass and relative shape can be explained by differences in their intermolecular forces. These forces include London dispersion forces, dipole-dipole interactions, and hydrogen bonding. By understanding the chemical composition and structure of the compounds, we could identify which of these forces are more predominant in each compound and, therefore, explain the differences in boiling points.

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Most popular questions from this chapter

Some of the physical properties of \(\mathrm{H}_{2} \mathrm{O}\) and \(\mathrm{D}_{2} \mathrm{O}\) are as follows: Account for the differences. (Note: D is a symbol often used for \(^{2} \mathrm{H},\) the deuterium isotope of hydrogen.)

A 20.0 -g sample of ice at \(-10.0^{\circ} \mathrm{C}\) is mixed with 100.0 g water at \(80.0^{\circ} \mathrm{C}\) . Calculate the final temperature of the mixture assuming no heat loss to the surroundings. The heat capacities of \(\mathrm{H}_{2} \mathrm{O}(s)\) and \(\mathrm{H}_{2} \mathrm{O}(l)\) are 2.03 and \(4.18 \mathrm{J} / \mathrm{g} \cdot^{\circ} \mathrm{C},\) respectively, and the enthalpy of fusion for ice is 6.02 \(\mathrm{kJ} / \mathrm{mol} .\)

Aluminum has an atomic radius of 143 \(\mathrm{pm}\) and forms a solid with a cubic closest packed structure. Calculate the density of solid aluminum in \(\mathrm{g} / \mathrm{cm}^{3}\) .

Rationalize why chalk (calcium carbonate) has a higher melting point than motor oil (composed of large compounds containing only carbon and hydrogen), which has a higher melting point than water (a compound that exhibits hydrogen bonding).

In each of the following groups of substances, pick the one that has the given property. Justify each answer. a. highest boiling point: $\mathrm{CCl}_{4}, \mathrm{CF}_{4}, \mathrm{CBr}_{4}$ b. lowest freezing point: LiF, \(\mathrm{F}_{2}, \mathrm{HCl}\) c. smallest vapor pressure at $25^{\circ} \mathrm{C} : \mathrm{CH}_{3} \mathrm{OCH}_{3}, \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}$ , \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{3}\) d. greatest viscosity: $\mathrm{H}_{2} \mathrm{S}, \mathrm{HF}, \mathrm{H}_{2} \mathrm{O}_{2}$ e. greatest heat of vaporization: $\mathrm{H}_{2} \mathrm{CO}, \mathrm{CH}_{3} \mathrm{CH}_{3}, \mathrm{CH}_{4}$ f. smallest enthalpy of fusion: \(\mathrm{I}_{2}, \mathrm{CsBr}, \mathrm{CaO}\)
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