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Chapter 11: Problem 17

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) \(\mathrm{BCl}_{3}\)

Short Answer

Expert verified
To convert (a) \(\mathrm{CF}_{4}\), (b) \(\mathrm{NH}_{3}\), and (c) \(\mathrm{BCl}_{3}\) from a liquid to a gas, the following intermolecular forces need to be overcome: (a) In \(\mathrm{CF}_{4}\), London dispersion forces (LDFs) must be overcome as it is a nonpolar molecule; (b) In \(\mathrm{NH}_{3}\), both hydrogen bonding and LDFs must be overcome due to the presence of highly electronegative nitrogen atoms; (c) In \(\mathrm{BCl}_{3}\), only LDFs need to be overcome as it is a nonpolar molecule. In all cases, increasing the temperature will increase the kinetic energy of the molecules, which must exceed the energy of the intermolecular forces to enable the substances to change from liquid to gas phase.

Step by step solution

01

Identify Intermolecular Forces in CF4

In \(\mathrm{CF}_{4}\), there are four fluorine atoms bonded to a carbon atom. The molecule has a tetrahedral molecular geometry. These molecules experience London dispersion forces (LDFs), which are weak intermolecular forces caused by temporary dipoles. Since \(\mathrm{CF}_{4}\) is a nonpolar molecule due to the symmetry in its structure, there are no other significant intermolecular forces present.
02

Identify Intermolecular Forces in NH3

In \(\mathrm{NH}_{3}\), nitrogen is bonded to three hydrogen atoms. The molecule has a trigonal pyramidal molecular geometry. Ammonia molecules have strong hydrogen bonding due to the highly electronegative nitrogen which attracts the hydrogen atoms creating strong dipoles. Apart from hydrogen bonding, \(\mathrm{NH}_{3}\) also experiences London dispersion forces due to the presence of temporary dipoles.
03

Identify Intermolecular Forces in BCl3

In \(\mathrm{BCl}_{3}\), boron is bonded to three chlorine atoms. The molecule has a trigonal planar geometry. This molecule is nonpolar due to the symmetry in its structure. Therefore, the only significant intermolecular force present in \(\mathrm{BCl}_{3}\) is London dispersion forces, caused by temporary dipoles.
04

Describe the Conversion of CF4 from Liquid to Gas

To convert \(\mathrm{CF}_{4}\) from a liquid to a gas, we must overcome the London dispersion forces present between the molecules. As the temperature is increased, the kinetic energy of the molecules increases, allowing them to move faster and farther apart from each other. When the kinetic energy becomes higher than the energy of the LDFs, the \(\mathrm{CF}_{4}\) molecules can escape from the liquid phase and enter the gas phase.
05

Describe the Conversion of NH3 from Liquid to Gas

To convert \(\mathrm{NH}_{3}\) from a liquid to a gas, we must overcome both hydrogen bonding and London dispersion forces. Similar to the case of \(\mathrm{CF}_{4}\), increasing the temperature will increase the kinetic energy of the molecules. In this case, we need to reach a temperature where the energy is high enough to break the strong hydrogen bonds and the weaker LDFs, allowing the \(\mathrm{NH}_{3}\) molecules to escape from the liquid phase and enter the gas phase.
06

Describe the Conversion of BCl3 from Liquid to Gas

To convert \(\mathrm{BCl}_{3}\) from a liquid to a gas, we must overcome the London dispersion forces present between the molecules. By increasing the temperature, we increase the kinetic energy, and once the kinetic energy of the molecules becomes sufficient to overcome the energy of the LDFs, the \(\mathrm{BCl}_{3}\) molecules can escape from the liquid phase and enter the gas phase.

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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},\( (b) \)\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.

(a) When you exercise vigorously, you sweat. How does this help your body cool? (b) A flask of water is connected to a vacuum pump. A few moments after the pump is turned on, the water begins to boil. After a few minutes, the water begins to freeze. Explain why these processes occur.

(a) Would you expect the viscosity of isopropanol, \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHOH},\) to be larger or smaller than the viscosity of ethanol, \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\) ? (b) Would you expect the viscosity of isopropanol to be smaller or larger than the viscosity of 1-propanol, $\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{2} \mathrm{OH}$ ?

The generic structural formula for a 1 -alkyl-3-methylimidazolium cation is where \(\mathrm{R}\) is \(\mathrm{a}-\mathrm{CH}_{2}\left(\mathrm{CH}_{2}\right)_{n} \mathrm{CH}_{3}\) alkyl group. The melt- ing points of the salts that form between the 1 -alkyl3-methylimidazolium cation and the \(\mathrm{PF}_{6}^{-}\) anion are as follows: $\mathrm{R}=\mathrm{CH}_{2} \mathrm{CH}_{3}\left(\mathrm{~m} \cdot \mathrm{p} .=60^{\circ} \mathrm{C}\right), \mathrm{R}=\mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}$ $\left(\mathrm{m} \cdot \mathrm{p} .=40^{\circ} \mathrm{C}\right), \mathrm{R}=\mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\left(\mathrm{~m} \cdot \mathrm{p} .=10^{\circ} \mathrm{C}\right),$ and $\mathrm{R}=\mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\left(\mathrm{~m} \cdot \mathrm{p} .=-61^{\circ} \mathrm{C}\right) .$ Why does the melting point decrease as the length of alkyl group increases?
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