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Chapter 10: Problem 132

You and a friend each synthesize a compound with the formula \(\mathrm{XeCl}_{2} \mathrm{~F}_{2}\). Your compound is a liquid and your friend's compound is a gas (at the same conditions of temperature and pressure). Explain how the two compounds with the same formulas can exist in different phases at the same conditions of pressure and temperature.

Short Answer

Expert verified
The two compounds with the same chemical formula, \(\mathrm{XeCl}_{2} \mathrm{F}_{2}\), can exist in different phases (liquid and gas) at the same temperature and pressure due to differences in their molecular structures, which lead to differing intermolecular forces. The liquid compound has stronger intermolecular forces than the gaseous compound, resulting in a higher boiling point and allowing it to remain in the liquid phase under the given conditions.

Step by step solution

01

(Step 1: Understand the Intermolecular Forces)

Intermolecular forces are interactions that occur between neighboring molecules. These forces can include dispersion forces (London dispersion forces), dipole-dipole forces, and hydrogen bonding. The type and strength of intermolecular forces dictate the physical properties of matter, such as boiling points, melting points, and phase states.
02

(Step 2: Identify Molecular Structures)

The two different compounds have the same formula, \(\mathrm{XeCl}_{2} \mathrm{F}_{2}\), but might have different molecular structures. The atoms may occupy different positions in the two compounds, leading to differing bond polarities and unique intermolecular forces. For example, one compound might have a linear structure, while the other is bent or has a different three-dimensional configuration.
03

(Step 3: Investigate Intermolecular Forces)

After identifying the different molecular structures, we need to investigate the specific intermolecular forces in each compound. Compounds with stronger intermolecular forces require more energy to break the forces, resulting in a higher boiling point. This implies that the compound which exists as a liquid would have stronger intermolecular forces than the compound in gas form at the same temperature and pressure.
04

(Step 4: Compare Boiling Points)

Boiling points are a representation of intermolecular forces. A compound with a higher boiling point has stronger intermolecular forces holding its molecules together, making them harder to separate. In this case, the liquid compound has greater intermolecular forces than the gaseous one because it's held together more tightly.
05

(Step 5: Conclusion)

Both compounds with the chemical formula \(\mathrm{XeCl}_{2} \mathrm{F}_{2}\) can exist in different phases, one as a liquid and one as a gas, at the same conditions of temperature and pressure because they have different molecular structures. These distinct structures lead to differing intermolecular forces, which in turn cause variations in boiling and melting points. As a result, the liquid compound has stronger intermolecular forces than the gaseous one, allowing it to exist as a liquid rather than a gas under the same temperature and pressure conditions.

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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: \(\mathrm{D}\) is a symbol often used for \({ }^{2} \mathrm{H}\), the deuterium isotope of hydrogen.)

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.

In each of the following groups of substances, pick the one that has the given property. Justify your answer. a. highest boiling point: \(\mathrm{HBr}, \mathrm{Kr}\), or \(\mathrm{Cl}_{2}\) b. highest freezing point: \(\mathrm{H}_{2} \mathrm{O}, \mathrm{NaCl}\), or \(\mathrm{HF}\) c. lowest vapor pressure at \(25^{\circ} \mathrm{C}: \mathrm{Cl}_{2}, \mathrm{Br}_{2}\), or \(\mathrm{I}_{2}\) d. lowest freezing point: \(\mathrm{N}_{2}, \mathrm{CO}\), or \(\mathrm{CO}_{2}\) e. lowest boiling point: \(\mathrm{CH}_{4}, \mathrm{CH}_{3} \mathrm{CH}_{3}\), or \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{3}\) f. highest boiling point: \(\mathrm{HF}, \mathrm{HCl}\), or \(\mathrm{HBr}\) g. lowest vapor pressure at \(25^{\circ} \mathrm{C}: \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{3}, \mathrm{CH}_{3} \mathrm{CCH}_{3}\), or \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}\)

What type of solid (network, metallic, Group \(8 \mathrm{~A}\), ionic, or molecular) will each of the following substances form? a. \(\mathrm{Kr}\) d. \(S i \mathrm{O}_{2}\) b. \(\mathrm{SO}_{2}\) e. \(\mathrm{NH}_{3}\) c. \(\mathrm{N} i\) f. \(\mathrm{Pt}\)

Predict which substance in each of the following pairs would have the greater intermolecular forces. a. \(\mathrm{CO}_{2}\) or \(\mathrm{OCS}\) b. \(\mathrm{SeO}_{2}\) or \(\mathrm{SO}_{2}\) c. \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}\) or \(\mathrm{H}_{2} \mathrm{NCH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}\) d. \(\mathrm{CH}_{3} \mathrm{CH}_{3}\) or \(\mathrm{H}_{2} \mathrm{CO}\) e. \(\mathrm{CH}_{3} \mathrm{OH}\) or \(\mathrm{H}_{2} \mathrm{CO}\)
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