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At room temperature, \(\mathrm{Si}\) is a solid, \(\mathrm{CCl}_{4}\) is a liquid, and \(\mathrm{Ar}\) is a gas. List these substances in order of (a) increasing intermolecular energy of attraction, (b) increasing boiling point.

Short Answer

Expert verified
The substances arranged in order of increasing intermolecular energy of attraction are: \(Ar < CCl_{4} < Si\). The order of increasing boiling point is the same: \(Ar < CCl_{4} < Si\).

Step by step solution

01

Identify the intermolecular forces in each substance

To determine the intermolecular forces within a substance, we look at its molecular structure. Si is a covalent network solid, consisting of a 3D array of covalent bonds. CCl4 is a covalently bonded molecule with London dispersion forces (LDF) operating between the molecules. Ar is an noble gas that has weak London dispersion forces between its atoms, with no other forces like dipole-dipole or hydrogen bonding.
02

Arrange substances in order of increasing intermolecular energy of attraction

Now that we've identified the intermolecular forces, we can now arrange them based on their strength. The order of strength for intermolecular forces is: London Dispersion Forces < Dipole-Dipole Forces < Hydrogen Bonding < Covalent Network Solids In our case, we have: Ar (LDF) < CCl4 (LDF) < Si (Covalent Network Solid) Since Ar and CCl4 both exhibit LDF, notice the strength of LDF depends on the size of the molecule. Since CCl4 is much larger and heavier compared to Ar, it has stronger LDF. So, the order of substances in increasing intermolecular energy of attraction is: Ar < CCl4 < Si
03

Arrange substances in order of increasing boiling point

The boiling point of a substance is directly related to the strength of its intermolecular forces as substances with stronger intermolecular forces require more energy to overcome these forces. Thus, the order of increasing boiling point will follow the same order as of increasing intermolecular energy of attraction. Ar < CCl4 < Si

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Key Concepts

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

Boiling Point
Understanding the boiling point of substances is crucial in chemistry because it reflects the energy required to overcome intermolecular forces holding particles together. The boiling point is the temperature at which a substance changes from a liquid to a gas, a process that requires sufficient energy to break the intermolecular bonds. As we observed in CCl4, despite being a molecule with only London dispersion forces, its boiling point is higher than that of Ar because it has a larger mass, leading to stronger intermolecular London dispersion forces. On the other hand, Si, being a covalent network solid, requires a substantial amount of energy to disrupt its extensive 3D covalent bond network, resulting in a very high boiling point.

In practical terms, understanding the boiling point helps to predict and explain the physical state of a substance at a given temperature which is essential for various applications, from cooking to industrial processes. Lower boiling points indicate weaker intermolecular attractions as seen in Ar, and conversely, higher boiling points, as with Si, signify stronger intermolecular bonds.
Covalent Network Solid
Silicon (Si) is an archetypal covalent network solid characterized by each atom being covalently bonded to adjacent atoms in a continuous network that extends throughout the material. This intricate 3D arrangement gives covalent network solids exceptional stability and high melting and boiling points. Unlike other weaker forces, this strong bond formation between silicon atoms occurs over an immensely broad scale contributing to hardness and other unique properties. In terms of electrical conductivity, such materials generally behave as semiconductors.

Covalent network solids include diamonds (a form of carbon), silicon carbide (SiC), and boron nitride (BN), all of which have applications in areas that require materials with high durability and resistance to heat, such as in cutting tools or heat shields for spacecraft. By understanding the properties of covalent network solids, students can appreciate how and why these materials are integral to advanced technological applications.
London Dispersion Forces
London dispersion forces (LDF), are weak intermolecular forces found in all atoms and molecules. They arise from temporary shifts in electron density which produce a temporary polarity in the molecule, allowing it to attract other molecules momentarily. These forces are significantly weaker than other types of intermolecular forces such as ionic or covalent bonds but are present universally.

In the given exercise, LDF are responsible for the existence of liquid CCl4 and gaseous Ar at room temperature. While Ar is a noble gas with minimal LDF, CCl4 displays stronger LDF due to its larger electron cloud and greater polarizability, which in turn, increases the boiling point. Students can better understand the physical states of different substances by recognizing the impact of London dispersion forces, especially in substances that do not exhibit stronger types of intermolecular attractions.

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Most popular questions from this chapter

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 \(-30^{\circ} \mathrm{C}\).

Name the phase transition in each of the following situations and indicate whether it is exothermic or endothermic: (a) When ice is heated, it turns to water. (b) Wet clothes dry on a warm summer day. (c) Frost appears on a window on a cold winter day. (d) Droplets of water appear on a cold glass of beer.

The boiling points, surface tensions, and viscosities of water and several alcohols are as follows: (a) For ethanol, propanol, and \(n\)-butanol the boiling points, surface tensions, and viscosities all increase. What is the reason for this increase? (b) How do you explain the fact that propanol and ethylene glycol have similar molecular weights ( 60 versus \(62 \mathrm{amu}\) ), yet the viscosity of ethylene glycol is more than 10 times larger than propanol? (c) How do you explain the fact that water has the highest surface tension but the lowest viscosity?

The table below lists the density of \(\mathrm{O}_{2}\) at various temperatures and at \(1 \mathrm{~atm}\). Graph the data and predict the substance's normal boiling point.

Explain the following observations: (a) Water evaporates more quickly on a hot, dry day than on a hot, humid day. (b) It takes longer to cook an egg in boiling water at high altitudes than it does at lower altitudes.

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