1102. Enthalpy of Vaporization -
Vaporization is an endothermic process. The cooling effect can be evident when you leave a swimming pool or
a shower. When the water on your skin evaporates, it removes heat from your skin and causes you to feel cold. The energy change associated with the vaporization process is the enthalpy of vaporization, $\Delta H_{\text {vap }}$. For example, the vaporization of water at standard temperature is represented by:

$$
\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{H}_{2} \mathrm{O}(g) \quad \Delta H_{\mathrm{vap}}=44.01 \mathrm{~kJ} / \mathrm{mol}
$$

As described in the chapter on thermochemistry, the reverse of an endothermic process is exothermic. And so, the condensation of a gas releases heat:

$$
\mathrm{H}_{2} \mathrm{O}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l) \quad \Delta H_{\mathrm{con}}=-\Delta H_{\mathrm{vap}}=-44.01 \mathrm{~kJ} / \mathrm{mol}
$$


1104. Using Enthalpy of Vaporization -
One way our body is cooled is by evaporation of the water in sweat (Figure 10.25). In very hot climates, we can lose as much as 1.5 L of sweat per day. Although sweat is not pure water, we can get an approximate value of the amount of heat removed by evaporation by assuming that it is. How much heat is required to evaporate 1.5 L of water ( 1.5 kg ) at $T=37{ }^{\circ} \mathrm{C}$ (normal body temperature); $\Delta H_{\text {vap }}=43.46 \mathrm{~kJ} / \mathrm{mol}$ at $37^{\circ} \mathrm{C}$.

FIGURE 10.25 Evaporation of sweat helps cool the body. (credit: "Kullez"/Flickr)


1105. Solution -
We start with the known volume of sweat (approximated as just water) and use the given information to convert to the amount of heat needed:

$$
1.5 \mathrm{~L} \times \frac{1000 \mathrm{~g}}{1 \mathrm{~g}} \times \frac{1 \mathrm{mot}}{18 \frac{\mathrm{~g}}{\mathrm{o}}} \times \frac{43.46 \mathrm{~kJ}}{1 \mathrm{mot}}=3.6 \times 10^{3} \mathrm{~kJ}
$$

Thus, 3600 kJ of heat are removed by the evaporation of 1.5 L of water.


1106. Check Your Learning -
How much heat is required to evaporate 100.0 g of liquid ammonia, $\mathrm{NH}_{3}$, at its boiling point if its enthalpy of vaporization is $4.8 \mathrm{~kJ} / \mathrm{mol}$ ?


1107. Answer: -
28 kJ


1108. Melting and Freezing -
When we heat a crystalline solid, we increase the average energy of its atoms, molecules, or ions and the solid gets hotter. At some point, the added energy becomes large enough to partially overcome the forces holding the molecules or ions of the solid in their fixed positions, and the solid begins the process of transitioning to the liquid state, or melting. At this point, the temperature of the solid stops rising, despite the continual input of heat, and it remains constant until all of the solid is melted. Only after all of the solid has melted will continued
heating increase the temperature of the liquid (Figure 10.26).

FIGURE 10.26 (a) This beaker of ice has a temperature of $-12.0^{\circ} \mathrm{C}$. (b) After 10 minutes the ice has absorbed enough heat from the air to warm to $0^{\circ} \mathrm{C}$. A small amount has melted. (c) Thirty minutes later, the ice has absorbed more heat, but its temperature is still $0^{\circ} \mathrm{C}$. The ice melts without changing its temperature. (d) Only after all the ice has melted does the heat absorbed cause the temperature to increase to $22.2^{\circ} \mathrm{C}$. (credit: modification of work by Mark Ott)

If we stop heating during melting and place the mixture of solid and liquid in a perfectly insulated container so no heat can enter or escape, the solid and liquid phases remain in equilibrium. This is almost the situation with a mixture of ice and water in a very good thermos bottle; almost no heat gets in or out, and the mixture of solid ice and liquid water remains for hours. In a mixture of solid and liquid at equilibrium, the reciprocal processes of melting and freezing occur at equal rates, and the quantities of solid and liquid therefore remain constant. The temperature at which the solid and liquid phases of a given substance are in equilibrium is called the melting point of the solid or the freezing point of the liquid. Use of one term or the other is normally dictated by the direction of the phase transition being considered, for example, solid to liquid (melting) or liquid to solid (freezing).


1108. Melting and Freezing -
The enthalpy of fusion and the melting point of a crystalline solid depend on the strength of the attractive forces between the units present in the crystal. Molecules with weak attractive forces form crystals with low melting points. Crystals consisting of particles with stronger attractive forces melt at higher temperatures.

The amount of heat required to change one mole of a substance from the solid state to the liquid state is the enthalpy of fusion, $\Delta \mathrm{H}_{\text {fus }}$ of the substance. The enthalpy of fusion of ice is $6.0 \mathrm{~kJ} / \mathrm{mol}$ at $0^{\circ} \mathrm{C}$. Fusion (melting) is an endothermic process:

$$
\mathrm{H}_{2} \mathrm{O}(s) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l) \quad \Delta H_{\text {fus }}=6.01 \mathrm{~kJ} / \mathrm{mol}
$$

The reciprocal process, freezing, is an exothermic process whose enthalpy change is $-6.0 \mathrm{~kJ} / \mathrm{mol}$ at $0^{\circ} \mathrm{C}$ :

$$
\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{H}_{2} \mathrm{O}(s)
$$

$$
\Delta H_{\mathrm{frz}}=-\Delta H_{\mathrm{fus}}=-6.01 \mathrm{~kJ} / \mathrm{mol}
$$


1109. Sublimation and Deposition -
Some solids can transition directly into the gaseous state, bypassing the liquid state, via a process known as sublimation. At room temperature and standard pressure, a piece of dry ice (solid $\mathrm{CO}_{2}$ ) sublimes, appearing to gradually disappear without ever forming any liquid. Snow and ice sublime at temperatures below the melting point of water, a slow process that may be accelerated by winds and the reduced atmospheric pressures at high altitudes. When solid iodine is warmed, the solid sublimes and a vivid purple vapor forms (Figure 10.27). The reverse of sublimation is called deposition, a process in which gaseous substances condense directly into the solid state, bypassing the liquid state. The formation of frost is an example of deposition.

FIGURE 10.27 Sublimation of solid iodine in the bottom of the tube produces a purple gas that subsequently deposits as solid iodine on the colder part of the tube above. (credit: modification of work by Mark Ott)

Like vaporization, the process of sublimation requires an input of energy to overcome intermolecular attractions. The enthalpy of sublimation, $\Delta \mathrm{H}_{\text {sub }}$, is the energy required to convert one mole of a substance from the solid to the gaseous state. For example, the sublimation of carbon dioxide is represented by:

$$
\mathrm{CO}_{2}(s) \longrightarrow \mathrm{CO}_{2}(g) \quad \Delta H_{\text {sub }}=26.1 \mathrm{~kJ} / \mathrm{mol}
$$

Likewise, the enthalpy change for the reverse process of deposition is equal in magnitude but opposite in sign to that for sublimation:

$$
\mathrm{CO}_{2}(g) \longrightarrow \mathrm{CO}_{2}(s) \quad \Delta H_{\mathrm{dep}}=-\Delta H_{\mathrm{sub}}=-26.1 \mathrm{~kJ} / \mathrm{mol}
$$


1109. Sublimation and Deposition -
Consider the extent to which intermolecular attractions must be overcome to achieve a given phase transition. Converting a solid into a liquid requires that these attractions be only partially overcome; transition to the gaseous state requires that they be completely overcome. As a result, the enthalpy of fusion for a substance is less than its enthalpy of vaporization. This same logic can be used to derive an approximate relation between the enthalpies of all phase changes for a given substance. Though not an entirely accurate description, sublimation may be conveniently modeled as a sequential two-step process of melting followed by vaporization in order to apply Hess's Law. Viewed in this manner, the enthalpy of sublimation for a substance may be estimated as the sum of its enthalpies of fusion and vaporization, as illustrated in Figure 10.28. For example:

FIGURE 10.28 For a given substance, the sum of its enthalpy of fusion and enthalpy of vaporization is approximately equal to its enthalpy of sublimation.


1110. Heating and Cooling Curves -
In the chapter on thermochemistry, the relation between the amount of heat absorbed or released by a substance, $q$, and its accompanying temperature change, $\Delta T$, was introduced:

$$
q=m c \Delta T
$$

where $m$ is the mass of the substance and $c$ is its specific heat. The relation applies to matter being heated or cooled, but not undergoing a change in state. When a substance being heated or cooled reaches a temperature corresponding to one of its phase transitions, further gain or loss of heat is a result of diminishing or enhancing intermolecular attractions, instead of increasing or decreasing molecular kinetic energies. While a substance is undergoing a change in state, its temperature remains constant. Figure 10.29 shows a typical heating curve.

Consider the example of heating a pot of water to boiling. A stove burner will supply heat at a roughly constant rate; initially, this heat serves to increase the water's temperature. When the water reaches its boiling point, the temperature remains constant despite the continued input of heat from the stove burner. This same temperature is maintained by the water as long as it is boiling. If the burner setting is increased to provide heat at a greater rate, the water temperature does not rise, but instead the boiling becomes more vigorous (rapid). This behavior is observed for other phase transitions as well: For example, temperature remains constant while the change of state is in progress.

FIGURE 10.29 A typical heating curve for a substance depicts changes in temperature that result as the substance absorbs increasing amounts of heat. Plateaus in the curve (regions of constant temperature) are exhibited when the substance undergoes phase transitions.


1111. EXAMPLE 10.10 -
Total Heat Needed to Change Temperature and Phase for a Substance
How much heat is required to convert 135 g of ice at $-15^{\circ} \mathrm{C}$ into water vapor at $120^{\circ} \mathrm{C}$ ?


1112. Solution -
The transition described involves the following steps:

1. Heat ice from $-15^{\circ} \mathrm{C}$ to $0^{\circ} \mathrm{C}$
2. Melt ice
3. Heat water from $0^{\circ} \mathrm{C}$ to $100^{\circ} \mathrm{C}$
4. Boil water
5. Heat steam from $100^{\circ} \mathrm{C}$ to $120^{\circ} \mathrm{C}$

The heat needed to change the temperature of a given substance (with no change in phase) is: $q=m \times c \times \Delta T$ (see previous chapter on thermochemistry). The heat needed to induce a given change in phase is given by $q=$ $n \times \Delta H$.

Using these equations with the appropriate values for specific heat of ice, water, and steam, and enthalpies of fusion and vaporization, we have:

$$
q_{\text {total }}=(m \cdot c \cdot \Delta T)_{\text {ice }}+n \cdot \Delta H_{\text {fus }}+(m \cdot c \cdot \Delta T)_{\text {water }}+n \cdot \Delta H_{\mathrm{vap}}+(m \cdot c \cdot \Delta T)_{\text {steam }}
$$


1112. Solution -
$$
\begin{aligned}
& =\left(135 \mathrm{~g} \cdot 2.09 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C} \cdot 15^{\circ} \mathrm{C}\right)+\left(135 \cdot \frac{1 \mathrm{~mol}}{18.02 \mathrm{~g}} \cdot 6.01 \mathrm{~kJ} / \mathrm{mol}\right) \\
& +\left(135 \mathrm{~g} \cdot 4.18 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C} \cdot 100^{\circ} \mathrm{C}\right)+\left(135 \mathrm{~g} \cdot \frac{1 \mathrm{~mol}}{18.02 \mathrm{~g}} \cdot 40.67 \mathrm{~kJ} / \mathrm{mol}\right) \\
& +\left(135 \mathrm{~g} \cdot 1.84 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C} \cdot 20^{\circ} \mathrm{C}\right) \\
& =4230 \mathrm{~J}+45.0 \mathrm{~kJ}+56,500 \mathrm{~J}+305 \mathrm{~kJ}+4970 \mathrm{~J}
\end{aligned}
$$

Converting the quantities in J to kJ permits them to be summed, yielding the total heat required:

$$
=4.23 \mathrm{~kJ}+45.0 \mathrm{~kJ}+56.5 \mathrm{~kJ}+305 \mathrm{~kJ}+4.97 \mathrm{~kJ}=416 \mathrm{~kJ}
$$


1113. Check Your Learning -
What is the total amount of heat released when 94.0 g water at $80.0^{\circ} \mathrm{C}$ cools to form ice at $-30.0^{\circ} \mathrm{C}$ ?


1114. Answer: -
68.7 kJ


1115. LEARNING OBJECTIVES -
By the end of this section, you will be able to:

- Explain the construction and use of a typical phase diagram
- Use phase diagrams to identify stable phases at given temperatures and pressures, and to describe phase transitions resulting from changes in these properties
- Describe the supercritical fluid phase of matter

In the previous module, the variation of a liquid's equilibrium vapor pressure with temperature was described. Considering the definition of boiling point, plots of vapor pressure versus temperature represent how the boiling point of the liquid varies with pressure. Also described was the use of heating and cooling curves to determine a substance's melting (or freezing) point. Making such measurements over a wide range of pressures yields data that may be presented graphically as a phase diagram. A phase diagram combines plots of pressure versus temperature for the liquid-gas, solid-liquid, and solid-gas phase-transition equilibria of a substance. These diagrams indicate the physical states that exist under specific conditions of pressure and temperature, and also provide the pressure dependence of the phase-transition temperatures (melting points, sublimation points, boiling points). A typical phase diagram for a pure substance is shown in Figure 10.30.

FIGURE 10.30 The physical state of a substance and its phase-transition temperatures are represented graphically in a phase diagram.

To illustrate the utility of these plots, consider the phase diagram for water shown in Figure 10.31.

FIGURE 10.31 The pressure and temperature axes on this phase diagram of water are not drawn to constant scale in order to illustrate several important properties.


1115. LEARNING OBJECTIVES -
FIGURE 10.31 The pressure and temperature axes on this phase diagram of water are not drawn to constant scale in order to illustrate several important properties.

We can use the phase diagram to identify the physical state of a sample of water under specified conditions of pressure and temperature. For example, a pressure of 50 kPa and a temperature of $-10^{\circ} \mathrm{C}$ correspond to the region of the diagram labeled "ice." Under these conditions, water exists only as a solid (ice). A pressure of 50 kPa and a temperature of $50^{\circ} \mathrm{C}$ correspond to the "water" region-here, water exists only as a liquid. At 25 kPa and $200^{\circ} \mathrm{C}$, water exists only in the gaseous state. Note that on the $\mathrm{H}_{2} \mathrm{O}$ phase diagram, the pressure and temperature axes are not drawn to a constant scale in order to permit the illustration of several important features as described here.

The curve BC in Figure 10.31 is the plot of vapor pressure versus temperature as described in the previous module of this chapter. This "liquid-vapor" curve separates the liquid and gaseous regions of the phase diagram and provides the boiling point for water at any pressure. For example, at 1 atm , the boiling point is $100^{\circ} \mathrm{C}$. Notice that the liquid-vapor curve terminates at a temperature of $374^{\circ} \mathrm{C}$ and a pressure of 218 atm ,
indicating that water cannot exist as a liquid above this temperature, regardless of the pressure. The physical properties of water under these conditions are intermediate between those of its liquid and gaseous phases. This unique state of matter is called a supercritical fluid, a topic that will be described in the next section of this module.


1115. LEARNING OBJECTIVES -
The solid-vapor curve, labeled AB in Figure 10.31, indicates the temperatures and pressures at which ice and water vapor are in equilibrium. These temperature-pressure data pairs correspond to the sublimation, or deposition, points for water. If we could zoom in on the solid-gas line in Figure 10.31, we would see that ice has a vapor pressure of about 0.20 kPa at $-10^{\circ} \mathrm{C}$. Thus, if we place a frozen sample in a vacuum with a pressure less than 0.20 kPa , ice will sublime. This is the basis for the "freeze-drying" process often used to preserve foods, such as the ice cream shown in Figure 10.32.

FIGURE 10.32 Freeze-dried foods, like this ice cream, are dehydrated by sublimation at pressures below the triple point for water. (credit: "lwao"/Flickr)

The solid-liquid curve labeled BD shows the temperatures and pressures at which ice and liquid water are in equilibrium, representing the melting/freezing points for water. Note that this curve exhibits a slight negative slope (greatly exaggerated for clarity), indicating that the melting point for water decreases slightly as pressure increases. Water is an unusual substance in this regard, as most substances exhibit an increase in melting point with increasing pressure. This behavior is partly responsible for the movement of glaciers, like the one shown in Figure 10.33. The bottom of a glacier experiences an immense pressure due to its weight that can melt some of the ice, forming a layer of liquid water on which the glacier may more easily slide.

FIGURE 10.33 The immense pressures beneath glaciers result in partial melting to produce a layer of water that provides lubrication to assist glacial movement. This satellite photograph shows the advancing edge of the Perito Moreno glacier in Argentina. (credit: NASA)

The point of intersection of all three curves is labeled B in Figure 10.31. At the pressure and temperature represented by this point, all three phases of water coexist in equilibrium. This temperature-pressure data
pair is called the triple point. At pressures lower than the triple point, water cannot exist as a liquid, regardless of the temperature.


1117. Determining the State of Water -
Using the phase diagram for water given in Figure 10.31, determine the state of water at the following temperatures and pressures:
(a) $-10^{\circ} \mathrm{C}$ and 50 kPa
(b) $25^{\circ} \mathrm{C}$ and 90 kPa
(c) $50^{\circ} \mathrm{C}$ and 40 kPa
(d) $80^{\circ} \mathrm{C}$ and 5 kPa
(e) $-10^{\circ} \mathrm{C}$ and 0.3 kPa
(f) $50^{\circ} \mathrm{C}$ and 0.3 kPa


1118. Solution -
Using the phase diagram for water, we can determine that the state of water at each temperature and pressure given are as follows: (a) solid; (b) liquid; (c) liquid; (d) gas; (e) solid; (f) gas.


1119. Check Your Learning -
What phase changes can water undergo as the temperature changes if the pressure is held at 0.3 kPa ? If the pressure is held at 50 kPa ?


1120. Answer: -
At $0.3 \mathrm{kPa}: \mathrm{s} \longrightarrow \mathrm{g}$ at $-58^{\circ} \mathrm{C}$. At $50 \mathrm{kPa}: \mathrm{s} \longrightarrow 1$ at $0^{\circ} \mathrm{C}, \mathrm{l} \longrightarrow \mathrm{g}$ at $78^{\circ} \mathrm{C}$

Consider the phase diagram for carbon dioxide shown in Figure 10.34 as another example. The solid-liquid curve exhibits a positive slope, indicating that the melting point for $\mathrm{CO}_{2}$ increases with pressure as it does for most substances (water being a notable exception as described previously). Notice that the triple point is well above 1 atm, indicating that carbon dioxide cannot exist as a liquid under ambient pressure conditions. Instead, cooling gaseous carbon dioxide at 1 atm results in its deposition into the solid state. Likewise, solid carbon dioxide does not melt at 1 atm pressure but instead sublimes to yield gaseous $\mathrm{CO}_{2}$. Finally, notice that the critical point for carbon dioxide is observed at a relatively modest temperature and pressure in comparison to water.

FIGURE 10.34 A phase diagram for carbon dioxide is shown. The pressure axis is plotted on a logarithmic scale to accommodate the large range of values.


1121. Determining the State of Carbon Dioxide -
Using the phase diagram for carbon dioxide shown in Figure 10.34, determine the state of $\mathrm{CO}_{2}$ at the following temperatures and pressures:
(a) $-30^{\circ} \mathrm{C}$ and 2000 kPa
(b) $-90^{\circ} \mathrm{C}$ and 1000 kPa
(c) $-60^{\circ} \mathrm{C}$ and 100 kPa
(d) $-40^{\circ} \mathrm{C}$ and 1500 kPa
(e) $0^{\circ} \mathrm{C}$ and 100 kPa
(f) $20^{\circ} \mathrm{C}$ and 100 kPa


1122. Solution -
Using the phase diagram for carbon dioxide provided, we can determine that the state of $\mathrm{CO}_{2}$ at each temperature and pressure given are as follows: (a) liquid; (b) solid; (c) gas; (d) liquid; (e) gas; (f) gas.


1123. Check Your Learning -
Identify the phase changes that carbon dioxide will undergo as its temperature is increased from $-100^{\circ} \mathrm{C}$ while holding its pressure constant at 1500 kPa . At 50 kPa . At what approximate temperatures do these phase changes occur?


1124. Answer: -
at $1500 \mathrm{kPa}: \mathrm{s} \longrightarrow 1$ at $-55^{\circ} \mathrm{C}, 1 \longrightarrow \mathrm{~g}$ at $-10^{\circ} \mathrm{C}$;
at $50 \mathrm{kPa}: \mathrm{s} \longrightarrow \mathrm{g}$ at $-60^{\circ} \mathrm{C}$


1125. Supercritical Fluids -
If we place a sample of water in a sealed container at $25^{\circ} \mathrm{C}$, remove the air, and let the vaporizationcondensation equilibrium establish itself, we are left with a mixture of liquid water and water vapor at a pressure of 0.03 atm . A distinct boundary between the more dense liquid and the less dense gas is clearly
observed. As we increase the temperature, the pressure of the water vapor increases, as described by the liquid-gas curve in the phase diagram for water (Figure 10.31), and a two-phase equilibrium of liquid and gaseous phases remains. At a temperature of $374^{\circ} \mathrm{C}$, the vapor pressure has risen to 218 atm , and any further increase in temperature results in the disappearance of the boundary between liquid and vapor phases. All of the water in the container is now present in a single phase whose physical properties are intermediate between those of the gaseous and liquid states. This phase of matter is called a supercritical fluid, and the temperature and pressure above which this phase exists is the critical point (Figure 10.35). Above its critical temperature, a gas cannot be liquefied no matter how much pressure is applied. The pressure required to liquefy a gas at its critical temperature is called the critical pressure. The critical temperatures and critical pressures of some common substances are given in the following table.

| Substance | Critical Temperature $\left({ }^{\circ} \mathrm{C}\right)$ | Critical Pressure (kPa) |
| :--- | :--- | :--- |
| hydrogen | -240.0 | 1300 |
| nitrogen | -147.2 | 3400 |
| oxygen | -118.9 | 5000 |
| carbon dioxide | 31.1 | 7400 |
| ammonia | 132.4 | 11,300 |
| sulfur dioxide | 157.2 | 7800 |
| water | 374.0 | 22,000 |


1125. Supercritical Fluids -
FIGURE 10.35 (a) A sealed container of liquid carbon dioxide slightly below its critical point is heated, resulting in (b) the formation of the supercritical fluid phase. Cooling the supercritical fluid lowers its temperature and pressure below the critical point, resulting in the reestablishment of separate liquid and gaseous phases (c and d). Colored floats illustrate differences in density between the liquid, gaseous, and supercritical fluid states. (credit: modification of work by "mrmrobin"/YouTube)


1126. LINK TO LEARNING -
Observe the liquid-to-supercritical fluid transition (http://openstax.org/l/16supercrit) for carbon dioxide.

Like a gas, a supercritical fluid will expand and fill a container, but its density is much greater than typical gas densities, typically being close to those for liquids. Similar to liquids, these fluids are capable of dissolving nonvolatile solutes. They exhibit essentially no surface tension and very low viscosities, however, so they can more effectively penetrate very small openings in a solid mixture and remove soluble components. These properties make supercritical fluids extremely useful solvents for a wide range of applications. For example, supercritical carbon dioxide has become a very popular solvent in the food industry, being used to decaffeinate coffee, remove fats from potato chips, and extract flavor and fragrance compounds from citrus
oils. It is nontoxic, relatively inexpensive, and not considered to be a pollutant. After use, the $\mathrm{CO}_{2}$ can be easily recovered by reducing the pressure and collecting the resulting gas.


1128. The Critical Temperature of Carbon Dioxide -
If we shake a carbon dioxide fire extinguisher on a cool day ( $18{ }^{\circ} \mathrm{C}$ ), we can hear liquid $\mathrm{CO}_{2}$ sloshing around inside the cylinder. However, the same cylinder appears to contain no liquid on a hot summer day ( $35^{\circ} \mathrm{C}$ ). Explain these observations.


1129. Solution -
On the cool day, the temperature of the $\mathrm{CO}_{2}$ is below the critical temperature of $\mathrm{CO}_{2}, 304 \mathrm{~K}$ or $31^{\circ} \mathrm{C}$, so liquid $\mathrm{CO}_{2}$ is present in the cylinder. On the hot day, the temperature of the $\mathrm{CO}_{2}$ is greater than its critical temperature of $31^{\circ} \mathrm{C}$. Above this temperature no amount of pressure can liquefy $\mathrm{CO}_{2}$ so no liquid $\mathrm{CO}_{2}$ exists in the fire extinguisher.


1130. Check Your Learning -
Ammonia can be liquefied by compression at room temperature; oxygen cannot be liquefied under these conditions. Why do the two gases exhibit different behavior?


1131. Answer: -
The critical temperature of ammonia is 405.5 K , which is higher than room temperature. The critical temperature of oxygen is below room temperature; thus oxygen cannot be liquefied at room temperature.


1133. Decaffeinating Coffee Using Supercritical $\mathbf{C O}_{\mathbf{2}}$ -
Coffee is the world's second most widely traded commodity, following only petroleum. Across the globe, people love coffee's aroma and taste. Many of us also depend on one component of coffee-caffeine-to help us get going in the morning or stay alert in the afternoon. But late in the day, coffee's stimulant effect can keep you from sleeping, so you may choose to drink decaffeinated coffee in the evening.

Since the early 1900s, many methods have been used to decaffeinate coffee. All have advantages and disadvantages, and all depend on the physical and chemical properties of caffeine. Because caffeine is a somewhat polar molecule, it dissolves well in water, a polar liquid. However, since many of the other 400-plus compounds that contribute to coffee's taste and aroma also dissolve in $\mathrm{H}_{2} \mathrm{O}$, hot water decaffeination processes can also remove some of these compounds, adversely affecting the smell and taste of the decaffeinated coffee. Dichloromethane $\left(\mathrm{CH}_{2} \mathrm{Cl}_{2}\right)$ and ethyl acetate $\left(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{C}_{2} \mathrm{H}_{5}\right)$ have similar polarity to caffeine, and are therefore very effective solvents for caffeine extraction, but both also remove some flavor and aroma components, and their use requires long extraction and cleanup times. Because both of these solvents are toxic, health concerns have been raised regarding the effect of residual solvent remaining in the decaffeinated coffee.


1133. Decaffeinating Coffee Using Supercritical $\mathbf{C O}_{\mathbf{2}}$ -
Supercritical fluid extraction using carbon dioxide is now being widely used as a more effective and environmentally friendly decaffeination method (Figure 10.36). At temperatures above 304.2 K and pressures above $7376 \mathrm{kPa}, \mathrm{CO}_{2}$ is a supercritical fluid, with properties of both gas and liquid. Like a gas, it penetrates deep into the coffee beans; like a liquid, it effectively dissolves certain substances. Supercritical carbon dioxide extraction of steamed coffee beans removes $97-99 \%$ of the caffeine, leaving coffee's flavor and aroma compounds intact. Because $\mathrm{CO}_{2}$ is a gas under standard conditions, its removal from the extracted coffee beans is easily accomplished, as is the recovery of the caffeine from the extract. The caffeine recovered from coffee beans via this process is a valuable product that can be used subsequently as an additive to other foods or drugs.

FIGURE 10.36 (a) Caffeine molecules have both polar and nonpolar regions, making it soluble in solvents of varying polarities. (b) The schematic shows a typical decaffeination process involving supercritical carbon dioxide.


1134. LEARNING OBJECTIVES -
By the end of this section, you will be able to:

- Define and describe the bonding and properties of ionic, molecular, metallic, and covalent network crystalline solids
- Describe the main types of crystalline solids: ionic solids, metallic solids, covalent network solids, and molecular solids
- Explain the ways in which crystal defects can occur in a solid

When most liquids are cooled, they eventually freeze and form crystalline solids, solids in which the atoms, ions, or molecules are arranged in a definite repeating pattern. It is also possible for a liquid to freeze before its molecules become arranged in an orderly pattern. The resulting materials are called amorphous solids or noncrystalline solids (or, sometimes, glasses). The particles of such solids lack an ordered internal structure and are randomly arranged (Figure 10.37).

FIGURE 10.37 The entities of a solid phase may be arranged in a regular, repeating pattern (crystalline solids) or
randomly (amorphous).
Metals and ionic compounds typically form ordered, crystalline solids. Substances that consist of large molecules, or a mixture of molecules whose movements are more restricted, often form amorphous solids. For examples, candle waxes are amorphous solids composed of large hydrocarbon molecules. Some substances, such as silicon dioxide (shown in Figure 10.38), can form either crystalline or amorphous solids, depending on the conditions under which it is produced. Also, amorphous solids may undergo a transition to the crystalline state under appropriate conditions.

(a)

(b)

FIGURE 10.38 (a) Silicon dioxide, $\mathrm{SiO}_{2}$, is abundant in nature as one of several crystalline forms of the mineral quartz. (b) Rapid cooling of molten $\mathrm{SiO}_{2}$ yields an amorphous solid known as "fused silica".

Crystalline solids are generally classified according to the nature of the forces that hold its particles together. These forces are primarily responsible for the physical properties exhibited by the bulk solids. The following sections provide descriptions of the major types of crystalline solids: ionic, metallic, covalent network, and molecular.


1135. Ionic Solids -
Ionic solids, such as sodium chloride and nickel oxide, are composed of positive and negative ions that are held together by electrostatic attractions, which can be quite strong (Figure 10.39). Many ionic crystals also have high melting points. This is due to the very strong attractions between the ions-in ionic compounds, the attractions between full charges are (much) larger than those between the partial charges in polar molecular compounds. This will be looked at in more detail in a later discussion of lattice energies. Although they are hard, they also tend to be brittle, and they shatter rather than bend. Ionic solids do not conduct electricity; however, they do conduct when molten or dissolved because their ions are free to move. Many simple compounds formed by the reaction of a metallic element with a nonmetallic element are ionic.

FIGURE 10.39 Sodium chloride is an ionic solid.


1136. Metallic Solids -
Metallic solids such as crystals of copper, aluminum, and iron are formed by metal atoms Figure 10.40. The structure of metallic crystals is often described as a uniform distribution of atomic nuclei within a "sea" of delocalized electrons. The atoms within such a metallic solid are held together by a unique force known as metallic bonding that gives rise to many useful and varied bulk properties. All exhibit high thermal and electrical conductivity, metallic luster, and malleability. Many are very hard and quite strong. Because of their
malleability (the ability to deform under pressure or hammering), they do not shatter and, therefore, make useful construction materials. The melting points of the metals vary widely. Mercury is a liquid at room temperature, and the alkali metals melt below $200^{\circ} \mathrm{C}$. Several post-transition metals also have low melting points, whereas the transition metals melt at temperatures above $1000^{\circ} \mathrm{C}$. These differences reflect differences in strengths of metallic bonding among the metals.

FIGURE 10.40 Copper is a metallic solid.


1137. Covalent Network Solid -
Covalent network solids include crystals of diamond, silicon, some other nonmetals, and some covalent compounds such as silicon dioxide (sand) and silicon carbide (carborundum, the abrasive on sandpaper). Many minerals have networks of covalent bonds. The atoms in these solids are held together by a network of covalent bonds, as shown in Figure 10.41. To break or to melt a covalent network solid, covalent bonds must be broken. Because covalent bonds are relatively strong, covalent network solids are typically characterized by hardness, strength, and high melting points. For example, diamond is one of the hardest substances known and melts above $3500^{\circ} \mathrm{C}$.

FIGURE 10.41 A covalent crystal contains a three-dimensional network of covalent bonds, as illustrated by the structures of diamond, silicon dioxide, silicon carbide, and graphite. Graphite is an exceptional example, composed of planar sheets of covalent crystals that are held together in layers by noncovalent forces. Unlike typical covalent solids, graphite is very soft and electrically conductive.


1138. Molecular Solid -
Molecular solids, such as ice, sucrose (table sugar), and iodine, as shown in Figure 10.42, are composed of neutral molecules. The strengths of the attractive forces between the units present in different crystals vary widely, as indicated by the melting points of the crystals. Small symmetrical molecules (nonpolar molecules), such as $\mathrm{H}_{2}, \mathrm{~N}_{2}, \mathrm{O}_{2}$, and $\mathrm{F}_{2}$, have weak attractive forces and form molecular solids with very low melting points (below $-200^{\circ} \mathrm{C}$ ). Substances consisting of larger, nonpolar molecules have larger attractive forces and melt at higher temperatures. Molecular solids composed of molecules with permanent dipole moments (polar
molecules) melt at still higher temperatures. Examples include ice (melting point, $0^{\circ} \mathrm{C}$ ) and table sugar (melting point, $185^{\circ} \mathrm{C}$ ).

carbon dioxide

iodine

FIGURE 10.42 Carbon dioxide $\left(\mathrm{CO}_{2}\right)$ consists of small, nonpolar molecules and forms a molecular solid with a melting point of $-78^{\circ} \mathrm{C}$. Iodine $\left(\mathrm{I}_{2}\right)$ consists of larger, nonpolar molecules and forms a molecular solid that melts at $114^{\circ} \mathrm{C}$.


1139. Properties of Solids -
A crystalline solid, like those listed in Table 10.4, has a precise melting temperature because each atom or molecule of the same type is held in place with the same forces or energy. Thus, the attractions between the units that make up the crystal all have the same strength and all require the same amount of energy to be broken. The gradual softening of an amorphous material differs dramatically from the distinct melting of a crystalline solid. This results from the structural nonequivalence of the molecules in the amorphous solid. Some forces are weaker than others, and when an amorphous material is heated, the weakest intermolecular attractions break first. As the temperature is increased further, the stronger attractions are broken. Thus amorphous materials soften over a range of temperatures.

Types of Crystalline Solids and Their Properties


1139. Properties of Solids -
Types of Crystalline Solids and Their Properties

| Type of Solid | Type of Particles | Type of Attractions | Properties | Examples |
| :---: | :---: | :---: | :---: | :---: |
| ionic | ions | ionic
 bonds | hard, brittle, conducts electricity as a liquid but not as a solid, high to very high melting points | $\begin{aligned} & \mathrm{NaCl}, \\ & \mathrm{Al}_{2} \mathrm{O}_{3} \end{aligned}$ |
| metallic | atoms of electropositive elements | metallic
 bonds | shiny, malleable, ductile, conducts heat and electricity well, variable hardness and melting temperature | $\begin{aligned} & \mathrm{Cu}, \mathrm{Fe}, \mathrm{Ti}, \\ & \mathrm{~Pb}, \mathrm{U} \end{aligned}$ |
| covalent network | atoms of electronegative elements | covalent bonds | very hard, not conductive, very high melting points | C
 (diamond), $\mathrm{SiO}_{2}, \mathrm{SiC}$ |
| molecular | molecules (or atoms) | IMFs | variable hardness, variable brittleness, not conductive, low melting points | $\begin{aligned} & \mathrm{H}_{2} \mathrm{O}, \mathrm{CO}_{2}, \\ & \mathrm{I}_{2}, \\ & \mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11} \end{aligned}$ |

TABLE 10.4


1141. Graphene: Material of the Future -
Carbon is an essential element in our world. The unique properties of carbon atoms allow the existence of carbon-based life forms such as ourselves. Carbon forms a huge variety of substances that we use on a daily basis, including those shown in Figure 10.43. You may be familiar with diamond and graphite, the two most common allotropes of carbon. (Allotropes are different structural forms of the same element.) Diamond is one of the hardest-known substances, whereas graphite is soft enough to be used as pencil lead. These very different properties stem from the different arrangements of the carbon atoms in the different allotropes.

FIGURE 10.43 Diamond is extremely hard because of the strong bonding between carbon atoms in all directions. Graphite (in pencil lead) rubs off onto paper due to the weak attractions between the carbon layers. An image of a graphite surface shows the distance between the centers of adjacent carbon atoms. (credit left photo: modification of work by Steve Jurvetson; credit middle photo: modification of work by United States Geological Survey)

You may be less familiar with a recently discovered form of carbon: graphene. Graphene was first isolated in 2004 by using tape to peel off thinner and thinner layers from graphite. It is essentially a single sheet (one atom thick) of graphite. Graphene, illustrated in Figure 10.44, is not only strong and lightweight, but it is also an excellent conductor of electricity and heat. These properties may prove very useful in a wide range of applications, such as vastly improved computer chips and circuits, better batteries and solar cells, and stronger and lighter structural materials. The 2010 Nobel Prize in Physics was awarded to Andre Geim and Konstantin Novoselov for their pioneering work with graphene.
Graphene sheet

Buckyball

Nanotube

Stacked sheets

FIGURE 10.44 Graphene sheets can be formed into buckyballs, nanotubes, and stacked layers.


1142. Crystal Defects -
In a crystalline solid, the atoms, ions, or molecules are arranged in a definite repeating pattern, but occasional defects may occur in the pattern. Several types of defects are known, as illustrated in Figure 10.45. Vacancies are defects that occur when positions that should contain atoms or ions are vacant. Less commonly, some atoms or ions in a crystal may occupy positions, called interstitial sites, located between the regular positions for atoms. Other distortions are found in impure crystals, as, for example, when the cations, anions, or molecules of the impurity are too large to fit into the regular positions without distorting the structure. Trace amounts of impurities are sometimes added to a crystal (a process known as doping) in order to create defects in the structure that yield desirable changes in its properties. For example, silicon crystals are doped with varying amounts of different elements to yield suitable electrical properties for their use in the manufacture of semiconductors and computer chips.

FIGURE 10.45 Types of crystal defects include vacancies, interstitial atoms, and substitutions impurities.


1143. LEARNING OBJECTIVES -
By the end of this section, you will be able to:

- Describe the arrangement of atoms and ions in crystalline structures
- Compute ionic radii using unit cell dimensions
- Explain the use of X-ray diffraction measurements in determining crystalline structures

Over $90 \%$ of naturally occurring and man-made solids are crystalline. Most solids form with a regular arrangement of their particles because the overall attractive interactions between particles are maximized, and the total intermolecular energy is minimized, when the particles pack in the most efficient manner. The regular arrangement at an atomic level is often reflected at a macroscopic level. In this module, we will explore some of the details about the structures of metallic and ionic crystalline solids, and learn how these structures are determined experimentally.


1144. The Structures of Metals -
We will begin our discussion of crystalline solids by considering elemental metals, which are relatively simple because each contains only one type of atom. A pure metal is a crystalline solid with metal atoms packed closely together in a repeating pattern. Some of the properties of metals in general, such as their malleability and ductility, are largely due to having identical atoms arranged in a regular pattern. The different properties of one metal compared to another partially depend on the sizes of their atoms and the specifics of their spatial arrangements. We will explore the similarities and differences of four of the most common metal crystal geometries in the sections that follow.


1145. Unit Cells of Metals -
The structure of a crystalline solid, whether a metal or not, is best described by considering its simplest repeating unit, which is referred to as its unit cell. The unit cell consists of lattice points that represent the locations of atoms or ions. The entire structure then consists of this unit cell repeating in three dimensions, as illustrated in Figure 10.46.

FIGURE 10.46 A unit cell shows the locations of lattice points repeating in all directions.
Let us begin our investigation of crystal lattice structure and unit cells with the most straightforward structure and the most basic unit cell. To visualize this, imagine taking a large number of identical spheres, such as tennis balls, and arranging them uniformly in a container. The simplest way to do this would be to make layers in which the spheres in one layer are directly above those in the layer below, as illustrated in Figure 10.47. This arrangement is called simple cubic structure, and the unit cell is called the simple cubic unit cell or primitive cubic unit cell.

FIGURE 10.47 When metal atoms are arranged with spheres in one layer directly above or below spheres in another layer, the lattice structure is called simple cubic. Note that the spheres are in contact.

In a simple cubic structure, the spheres are not packed as closely as they could be, and they only "fill" about $52 \%$ of the volume of the container. This is a relatively inefficient arrangement, and only one metal (polonium, Po) crystallizes in a simple cubic structure. As shown in Figure 10.48, a solid with this type of arrangement consists of planes (or layers) in which each atom contacts only the four nearest neighbors in its layer; one atom directly above it in the layer above; and one atom directly below it in the layer below. The number of other particles that each particle in a crystalline solid contacts is known as its coordination number. For a polonium atom in a simple cubic array, the coordination number is, therefore, six.

FIGURE 10.48 An atom in a simple cubic lattice structure contacts six other atoms, so it has a coordination number of six.


1145. Unit Cells of Metals -
FIGURE 10.48 An atom in a simple cubic lattice structure contacts six other atoms, so it has a coordination number of six.

In a simple cubic lattice, the unit cell that repeats in all directions is a cube defined by the centers of eight atoms, as shown in Figure 10.49. Atoms at adjacent corners of this unit cell contact each other, so the edge length of this cell is equal to two atomic radii, or one atomic diameter. A cubic unit cell contains only the parts of these atoms that are within it. Since an atom at a corner of a simple cubic unit cell is contained by a total of eight unit cells, only one-eighth of that atom is within a specific unit cell. And since each simple cubic unit cell has one atom at each of its eight "corners," there is $8 \times \frac{1}{8}=1$ atom within one simple cubic unit cell.

8 corners

Simple cubic lattice cell
FIGURE 10.49 A simple cubic lattice unit cell contains one-eighth of an atom at each of its eight corners, so it contains one atom total.


1147. Calculation of Atomic Radius and Density for Metals, Part 1 -
The edge length of the unit cell of alpha polonium is 336 pm .
(a) Determine the radius of a polonium atom.
(b) Determine the density of alpha polonium.


1148. Solution -
Alpha polonium crystallizes in a simple cubic unit cell:

(a) Two adjacent Po atoms contact each other, so the edge length of this cell is equal to two Po atomic radii: $1=$ $2 r$. Therefore, the radius of Po is $r=\frac{1}{2}=\frac{336 \mathrm{pm}}{2}=168 \mathrm{pm}$.
(b) Density is given by density $=\frac{\text { mass }}{\text { volume }}$. The density of polonium can be found by determining the density of its unit cell (the mass contained within a unit cell divided by the volume of the unit cell). Since a Po unit cell contains one-eighth of a Po atom at each of its eight corners, a unit cell contains one Po atom.

The mass of a Po unit cell can be found by:

$$
1 \text { Po unit cell } \times \frac{1 \text { Po atom }}{1 \text { Po unit cell }} \times \frac{1 \mathrm{~mol} \mathrm{Po}}{6.022 \times 10^{23} \text { Po atoms }} \times \frac{208.998 \mathrm{~g}}{1 \mathrm{~mol} \mathrm{Po}}=3.47 \times 10^{-22} \mathrm{~g}
$$

The volume of a Po unit cell can be found by:

$$
V=l^{3}=\left(336 \times 10^{-10} \mathrm{~cm}\right)^{3}=3.79 \times 10^{-23} \mathrm{~cm}^{3}
$$

(Note that the edge length was converted from pm to cm to get the usual volume units for density.)
Therefore, the density of $\mathrm{Po}=\frac{3.471 \times 10^{-22} \mathrm{~g}}{3.79 \times 10^{-23} \mathrm{~cm}^{3}}=9.16 \mathrm{~g} / \mathrm{cm}^{3}$


1149. Check Your Learning -
The edge length of the unit cell for nickel is 0.3524 nm . The density of Ni is $8.90 \mathrm{~g} / \mathrm{cm}^{3}$. Does nickel crystallize in a simple cubic structure? Explain.


1150. Answer: -
No. If Ni was simple cubic, its density would be given by:
1 Ni atom $\times \frac{1 \mathrm{~mol} \mathrm{Ni}}{6.022 \times 10^{23} \mathrm{Ni} \text { atoms }} \times \frac{58.693 \mathrm{~g}}{1 \mathrm{~mol} \mathrm{Ni}}=9.746 \times 10^{-23} \mathrm{~g}$
$V=l^{3}=\left(3.524 \times 10^{-8} \mathrm{~cm}\right)^{3}=4.376 \times 10^{-23} \mathrm{~cm}^{3}$
Then the density of Ni would be $=\frac{9.746 \times 10^{-23} \mathrm{~g}}{4.376 \times 10^{-23} \mathrm{~cm}^{3}}=2.23 \mathrm{~g} / \mathrm{cm}^{3}$
Since the actual density of Ni is not close to this, Ni does not form a simple cubic structure.

Most metal crystals are one of the four major types of unit cells. For now, we will focus on the three cubic unit cells: simple cubic (which we have already seen), body-centered cubic unit cell, and face-centered cubic unit cell-all of which are illustrated in Figure 10.50. (Note that there are actually seven different lattice systems, some of which have more than one type of lattice, for a total of 14 different types of unit cells. We leave the more complicated geometries for later in this module.)


1151. Lattice point locations -
FIGURE 10.50 Cubic unit cells of metals show (in the upper figures) the locations of lattice points and (in the lower figures) metal atoms located in the unit cell.

Some metals crystallize in an arrangement that has a cubic unit cell with atoms at all of the corners and an atom in the center, as shown in Figure 10.51. This is called a body-centered cubic (BCC) solid. Atoms in the corners of a BCC unit cell do not contact each other but contact the atom in the center. A BCC unit cell contains two atoms: one-eighth of an atom at each of the eight corners ( $8 \times \frac{1}{8}=1$ atom from the corners) plus one atom from the center. Any atom in this structure touches four atoms in the layer above it and four atoms in the layer below it. Thus, an atom in a BCC structure has a coordination number of eight.

Body-centered cubic structure
FIGURE 10.51 In a body-centered cubic structure, atoms in a specific layer do not touch each other. Each atom touches four atoms in the layer above it and four atoms in the layer below it.

Atoms in BCC arrangements are much more efficiently packed than in a simple cubic structure, occupying about $68 \%$ of the total volume. Isomorphous metals with a BCC structure include $\mathrm{K}, \mathrm{Ba}, \mathrm{Cr}, \mathrm{Mo}, \mathrm{W}$, and Fe at room temperature. (Elements or compounds that crystallize with the same structure are said to be isomorphous.)


1151. Lattice point locations -
Many other metals, such as aluminum, copper, and lead, crystallize in an arrangement that has a cubic unit cell with atoms at all of the corners and at the centers of each face, as illustrated in Figure 10.52. This arrangement is called a face-centered cubic (FCC) solid. A FCC unit cell contains four atoms: one-eighth of an atom at each of the eight corners ( $8 \times \frac{1}{8}=1$ atom from the corners) and one-half of an atom on each of the six faces ( $6 \times \frac{1}{2}=3$ atoms from the faces). The atoms at the corners touch the atoms in the centers of the adjacent faces along the face diagonals of the cube. Because the atoms are on identical lattice points, they have identical environments.

Face-centered cubic structure
FIGURE 10.52 A face-centered cubic solid has atoms at the corners and, as the name implies, at the centers of the faces of its unit cells.

Atoms in an FCC arrangement are packed as closely together as possible, with atoms occupying $74 \%$ of the volume. This structure is also called cubic closest packing (CCP). In CCP, there are three repeating layers of hexagonally arranged atoms. Each atom contacts six atoms in its own layer, three in the layer above, and three in the layer below. In this arrangement, each atom touches 12 near neighbors, and therefore has a coordination number of 12 . The fact that FCC and CCP arrangements are equivalent may not be immediately obvious, but why they are actually the same structure is illustrated in Figure 10.53.

Side view

Top view

Rotated view

Cubic closest packed structure
FIGURE 10.53 A CCP arrangement consists of three repeating layers (ABCABC...) of hexagonally arranged atoms. Atoms in a CCP structure have a coordination number of 12 because they contact six atoms in their layer, plus three atoms in the layer above and three atoms in the layer below. By rotating our perspective, we can see that a CCP structure has a unit cell with a face containing an atom from layer A at one corner, atoms from layer B across a diagonal (at two corners and in the middle of the face), and an atom from layer C at the remaining corner. This is the same as a face-centered cubic arrangement.


1151. Lattice point locations -
Because closer packing maximizes the overall attractions between atoms and minimizes the total intermolecular energy, the atoms in most metals pack in this manner. We find two types of closest packing in simple metallic crystalline structures: CCP, which we have already encountered, and hexagonal closest packing (HCP) shown in Figure 10.54. Both consist of repeating layers of hexagonally arranged atoms. In both types, a second layer (B) is placed on the first layer (A) so that each atom in the second layer is in contact with three atoms in the first layer. The third layer is positioned in one of two ways. In HCP, atoms in the third layer are directly above atoms in the first layer (i.e., the third layer is also type A), and the stacking consists of alternating type A and type B close-packed layers (i.e., ABABAB $\cdot \cdot$ ). In CCP, atoms in the third layer are not above atoms in either of the first two layers (i.e., the third layer is type C), and the stacking consists of alternating type $A$, type $B$, and type $C$ close-packed layers (i.e., ABCABCABC…). About two-thirds of all metals crystallize in closest-packed arrays with coordination numbers of 12. Metals that crystallize in an HCP structure include $\mathrm{Cd}, \mathrm{Co}, \mathrm{Li}, \mathrm{Mg}, \mathrm{Na}$, and Zn , and metals that crystallize in a CCP structure include $\mathrm{Ag}, \mathrm{Al}, \mathrm{Ca}$, $\mathrm{Cu}, \mathrm{Ni}, \mathrm{Pb}$, and Pt .

FIGURE 10.54 In both types of closest packing, atoms are packed as compactly as possible. Hexagonal closest packing consists of two alternating layers (ABABAB...). Cubic closest packing consists of three alternating layers (ABCABCABC...).


1153. Calculation of Atomic Radius and Density for Metals, Part 2 -
Calcium crystallizes in a face-centered cubic structure. The edge length of its unit cell is 558.8 pm .
(a) What is the atomic radius of Ca in this structure?
(b) Calculate the density of Ca.


1154. Solution -
(a) In an FCC structure, Ca atoms contact each other across the diagonal of the face, so the length of the diagonal is equal to four Ca atomic radii $(\mathrm{d}=4 r)$. Two adjacent edges and the diagonal of the face form a right triangle, with the length of each side equal to 558.8 pm and the length of the hypotenuse equal to four Ca atomic radii:

$$
a^{2}+a^{2}=d^{2} \longrightarrow(558.8 \mathrm{pm})^{2}+(558.5 \mathrm{pm})^{2}=(4 r)^{2}
$$

Solving this gives $r=\sqrt{\frac{(558.8 \mathrm{pm})^{2}+(558.5 \mathrm{pm})^{2}}{16}}=197.6 \mathrm{pm}$ for a Ca radius.
(b) Density is given by density $=\frac{\text { mass }}{\text { volume }}$. The density of calcium can be found by determining the density of its unit cell: for example, the mass contained within a unit cell divided by the volume of the unit cell. A facecentered Ca unit cell has one-eighth of an atom at each of the eight corners ( $8 \times \frac{1}{8}=1$ atom $)$ and one-half of an atom on each of the six faces $6 \times \frac{1}{2}=3$ atoms), for a total of four atoms in the unit cell.

The mass of the unit cell can be found by:

$$
1 \mathrm{Ca} \text { unit cell } \times \frac{4 \mathrm{Ca} \text { atoms }}{1 \mathrm{Ca} \text { unit cell }} \times \frac{1 \mathrm{~mol} \mathrm{Ca}}{6.022 \times 10^{23} \mathrm{Ca} \text { atoms }} \times \frac{40.078 \mathrm{~g}}{1 \mathrm{~mol} \mathrm{Ca}}=2.662 \times 10^{-22} \mathrm{~g}
$$

The volume of a Ca unit cell can be found by:


1154. Solution -
The volume of a Ca unit cell can be found by:

$$
V=a^{3}=\left(558.8 \times 10^{-10} \mathrm{~cm}\right)^{3}=1.745 \times 10^{-22} \mathrm{~cm}^{3}
$$

(Note that the edge length was converted from pm to cm to get the usual volume units for density.)
Then, the density of $\mathrm{Ca}=\frac{2.662 \times 10^{-22} \mathrm{~g}}{1.745 \times 10^{-22} \mathrm{~cm}^{3}}=1.53 \mathrm{~g} / \mathrm{cm}^{3}$


1155. Check Your Learning -
Silver crystallizes in an FCC structure. The edge length of its unit cell is 409 pm .
(a) What is the atomic radius of Ag in this structure?
(b) Calculate the density of Ag.


1156. Answer: -
(a) 144 pm ; (b) $10.5 \mathrm{~g} / \mathrm{cm}^{3}$

In general, a unit cell is defined by the lengths of three axes ( $a, b$, and $c$ ) and the angles ( $\alpha, \beta$, and $\gamma$ ) between them, as illustrated in Figure 10.55. The axes are defined as being the lengths between points in the space lattice. Consequently, unit cell axes join points with identical environments.

FIGURE 10.55 A unit cell is defined by the lengths of its three axes ( $a, b$, and $c$ ) and the angles ( $\alpha, \beta$, and $\gamma$ ) between the axes.

There are seven different lattice systems, some of which have more than one type of lattice, for a total of fourteen different unit cells, which have the shapes shown in Figure 10.56.


1156. Answer: -
There are seven different lattice systems, some of which have more than one type of lattice, for a total of fourteen different unit cells, which have the shapes shown in Figure 10.56.

| System/Axes/Angles | Unit Cells |
| :---: | :---: |
| Cubic $\begin{gathered} a=b=c \\ \alpha=\beta=\gamma=90^{\circ} \end{gathered}$ | Simple
 Face-centered
 Body-centered |
| Tetragonal $\begin{gathered} a=b \neq c \\ \alpha=\beta=\gamma=90^{\circ} \end{gathered}$ | Simple
 Body-centered |
| Orthorhombic $\begin{gathered} a \neq b \neq c \\ \alpha=\beta=\gamma=90^{\circ} \end{gathered}$ | Simple
 Body-centered
 Base-centered
 Face-centered |
| Monoclinic $\begin{gathered} a \neq b \neq c \\ \alpha=\gamma=90^{\circ} ; \beta \neq 90^{\circ} \end{gathered}$ | Simple
 Base-centered |
| Triclinic $\begin{gathered} a \neq b \neq c \\ \alpha \neq \beta \neq \gamma \neq 90^{\circ} \end{gathered}$ |  |
| Hexagonal $\begin{gathered} a=b \neq c \\ \alpha=\beta=90^{\circ} ; \gamma=120^{\circ} \end{gathered}$ |  |
| Rhombohedral $\begin{gathered} a=b=c \\ \alpha=\beta=\gamma \neq 90^{\circ} \end{gathered}$ |  |

FIGURE 10.56 There are seven different lattice systems and 14 different unit cells.


1157. The Structures of Ionic Crystals -
Ionic crystals consist of two or more different kinds of ions that usually have different sizes. The packing of these ions into a crystal structure is more complex than the packing of metal atoms that are the same size.

Most monatomic ions behave as charged spheres, and their attraction for ions of opposite charge is the same in every direction. Consequently, stable structures for ionic compounds result (1) when ions of one charge are
surrounded by as many ions as possible of the opposite charge and (2) when the cations and anions are in contact with each other. Structures are determined by two principal factors: the relative sizes of the ions and the ratio of the numbers of positive and negative ions in the compound.

In simple ionic structures, we usually find the anions, which are normally larger than the cations, arranged in a closest-packed array. (As seen previously, additional electrons attracted to the same nucleus make anions larger and fewer electrons attracted to the same nucleus make cations smaller when compared to the atoms from which they are formed.) The smaller cations commonly occupy one of two types of holes (or interstices) remaining between the anions. The smaller of the holes is found between three anions in one plane and one anion in an adjacent plane. The four anions surrounding this hole are arranged at the corners of a tetrahedron, so the hole is called a tetrahedral hole. The larger type of hole is found at the center of six anions (three in one layer and three in an adjacent layer) located at the corners of an octahedron; this is called an octahedral hole. Figure 10.57 illustrates both of these types of holes.

FIGURE 10.57 Cations may occupy two types of holes between anions: octahedral holes or tetrahedral holes.
Depending on the relative sizes of the cations and anions, the cations of an ionic compound may occupy tetrahedral or octahedral holes, as illustrated in Figure 10.58. Relatively small cations occupy tetrahedral holes, and larger cations occupy octahedral holes. If the cations are too large to fit into the octahedral holes, the anions may adopt a more open structure, such as a simple cubic array. The larger cations can then occupy the larger cubic holes made possible by the more open spacing.


1157. The Structures of Ionic Crystals -
FIGURE 10.58 A cation's size and the shape of the hole occupied by the compound are directly related.
There are two tetrahedral holes for each anion in either an HCP or CCP array of anions. A compound that crystallizes in a closest-packed array of anions with cations in the tetrahedral holes can have a maximum cation:anion ratio of $2: 1$; all of the tetrahedral holes are filled at this ratio. Examples include $\mathrm{Li}_{2} \mathrm{O}, \mathrm{Na}_{2} \mathrm{O}, \mathrm{Li}_{2} \mathrm{~S}$,
and $\mathrm{Na}_{2} \mathrm{~S}$. Compounds with a ratio of less than $2: 1$ may also crystallize in a closest-packed array of anions with cations in the tetrahedral holes, if the ionic sizes fit. In these compounds, however, some of the tetrahedral holes remain vacant.


1159. Occupancy of Tetrahedral Holes -
Zinc sulfide is an important industrial source of zinc and is also used as a white pigment in paint. Zinc sulfide crystallizes with zinc ions occupying one-half of the tetrahedral holes in a closest-packed array of sulfide ions. What is the formula of zinc sulfide?


1160. Solution -
Because there are two tetrahedral holes per anion (sulfide ion) and one-half of these holes are occupied by zinc ions, there must be $\frac{1}{2} \times 2$, or 1 , zinc ion per sulfide ion. Thus, the formula is ZnS .


1161. Check Your Learning -
Lithium selenide can be described as a closest-packed array of selenide ions with lithium ions in all of the tetrahedral holes. What it the formula of lithium selenide?


1162. Answer: -
$\mathrm{Li}_{2} \mathrm{Se}$

The ratio of octahedral holes to anions in either an HCP or CCP structure is 1:1. Thus, compounds with cations in octahedral holes in a closest-packed array of anions can have a maximum cation:anion ratio of 1:1. In NiO , $\mathrm{MnS}, \mathrm{NaCl}$, and KH, for example, all of the octahedral holes are filled. Ratios of less than 1:1 are observed when some of the octahedral holes remain empty.


1164. Stoichiometry of Ionic Compounds -
Sapphire is aluminum oxide. Aluminum oxide crystallizes with aluminum ions in two-thirds of the octahedral holes in a closest-packed array of oxide ions. What is the formula of aluminum oxide?


1165. Solution -
Because there is one octahedral hole per anion (oxide ion) and only two-thirds of these holes are occupied, the ratio of aluminum to oxygen must be $\frac{2}{3}: 1$, which would give $\mathrm{Al}_{2 / 3} \mathrm{O}$. The simplest whole number ratio is $2: 3$, so the formula is $\mathrm{Al}_{2} \mathrm{O}_{3}$.


1166. Check Your Learning -
The white pigment titanium oxide crystallizes with titanium ions in one-half of the octahedral holes in a closest-packed array of oxide ions. What is the formula of titanium oxide?


1167. Answer: -
$\mathrm{TiO}_{2}$

In a simple cubic array of anions, there is one cubic hole that can be occupied by a cation for each anion in the array. In CsCl , and in other compounds with the same structure, all of the cubic holes are occupied. Half of the cubic holes are occupied in $\mathrm{SrH}_{2}, \mathrm{UO}_{2}, \mathrm{SrCl}_{2}$, and $\mathrm{CaF}_{2}$.

Different types of ionic compounds often crystallize in the same structure when the relative sizes of their ions and their stoichiometries (the two principal features that determine structure) are similar.


1168. Unit Cells of Ionic Compounds -
Many ionic compounds crystallize with cubic unit cells, and we will use these compounds to describe the general features of ionic structures.

When an ionic compound is composed of cations and anions of similar size in a 1:1 ratio, it typically forms a simple cubic structure. Cesium chloride, CsCl , (illustrated in Figure 10.59) is an example of this, with $\mathrm{Cs}^{+}$and $\mathrm{Cl}^{-}$having radii of 174 pm and 181 pm , respectively. We can think of this as chloride ions forming a simple cubic unit cell, with a cesium ion in the center; or as cesium ions forming a unit cell with a chloride ion in the center; or as simple cubic unit cells formed by $\mathrm{Cs}^{+}$ions overlapping unit cells formed by $\mathrm{Cl}^{-}$ions. Cesium ions and chloride ions touch along the body diagonals of the unit cells. One cesium ion and one chloride ion are present per unit cell, giving the l:l stoichiometry required by the formula for cesium chloride. Note that there is no lattice point in the center of the cell, and CsCl is not a BCC structure because a cesium ion is not identical to a chloride ion.

Simple cubic structure
FIGURE 10.59 Ionic compounds with similar-sized cations and anions, such as CsCl , usually form a simple cubic structure. They can be described by unit cells with either cations at the corners or anions at the corners.
We have said that the location of lattice points is arbitrary. This is illustrated by an alternate description of the CsCl structure in which the lattice points are located in the centers of the cesium ions. In this description, the cesium ions are located on the lattice points at the corners of the cell, and the chloride ion is located at the center of the cell. The two unit cells are different, but they describe identical structures.


1168. Unit Cells of Ionic Compounds -
When an ionic compound is composed of a 1:1 ratio of cations and anions that differ significantly in size, it typically crystallizes with an FCC unit cell, like that shown in Figure 10.60. Sodium chloride, NaCl , is an example of this, with $\mathrm{Na}^{+}$and $\mathrm{Cl}^{-}$having radii of 102 pm and 181 pm , respectively. We can think of this as chloride ions forming an FCC cell, with sodium ions located in the octahedral holes in the middle of the cell edges and in the center of the cell. The sodium and chloride ions touch each other along the cell edges. The unit cell contains four sodium ions and four chloride ions, giving the 1:1 stoichiometry required by the formula, NaCl .

Face-centered simple cubic structure
FIGURE 10.60 Ionic compounds with anions that are much larger than cations, such as NaCl , usually form an FCC structure. They can be described by FCC unit cells with cations in the octahedral holes.

The cubic form of zinc sulfide, zinc blende, also crystallizes in an FCC unit cell, as illustrated in Figure 10.61. This structure contains sulfide ions on the lattice points of an FCC lattice. (The arrangement of sulfide ions is identical to the arrangement of chloride ions in sodium chloride.) The radius of a zinc ion is only about $40 \%$ of the radius of a sulfide ion, so these small $\mathrm{Zn}^{2+}$ ions are located in alternating tetrahedral holes, that is, in one half of the tetrahedral holes. There are four zinc ions and four sulfide ions in the unit cell, giving the empirical formula ZnS .

ZnS face-centered unit cell
FIGURE 10.61 ZnS , zinc sulfide (or zinc blende) forms an FCC unit cell with sulfide ions at the lattice points and much smaller zinc ions occupying half of the tetrahedral holes in the structure.


1168. Unit Cells of Ionic Compounds -
A calcium fluoride unit cell, like that shown in Figure 10.62, is also an FCC unit cell, but in this case, the cations are located on the lattice points; equivalent calcium ions are located on the lattice points of an FCC lattice. All of the tetrahedral sites in the FCC array of calcium ions are occupied by fluoride ions. There are four calcium ions and eight fluoride ions in a unit cell, giving a calcium:fluorine ratio of 1:2, as required by the chemical formula, $\mathrm{CaF}_{2}$. Close examination of Figure 10.62 will reveal a simple cubic array of fluoride ions with calcium ions in one half of the cubic holes. The structure cannot be described in terms of a space lattice of points on the fluoride ions because the fluoride ions do not all have identical environments. The orientation of the four calcium ions about the fluoride ions differs.

$\mathrm{CaF}_{2}$ face-centered unit cell
FIGURE 10.62 Calcium fluoride, $\mathrm{CaF}_{2}$, forms an FCC unit cell with calcium ions (green) at the lattice points and fluoride ions (red) occupying all of the tetrahedral sites between them.


1169. Calculation of Ionic Radii -
If we know the edge length of a unit cell of an ionic compound and the position of the ions in the cell, we can calculate ionic radii for the ions in the compound if we make assumptions about individual ionic shapes and contacts.


1171. Calculation of Ionic Radii -
The edge length of the unit cell of LiCl ( NaCl -like structure, FCC ) is 0.514 nm or $5.14 \AA$. Assuming that the lithium ion is small enough so that the chloride ions are in contact, as in Figure 10.60, calculate the ionic radius for the chloride ion.

Note: The length unit angstrom, $\AA$, is often used to represent atomic-scale dimensions and is equivalent to $10^{-10} \mathrm{~m}$.


1172. Solution -
On the face of a LiCl unit cell, chloride ions contact each other across the diagonal of the face:

Drawing a right triangle on the face of the unit cell, we see that the length of the diagonal is equal to four chloride radii (one radius from each corner chloride and one diameter-which equals two radii-from the chloride ion in the center of the face), so $d=4 r$. From the Pythagorean theorem, we have:

$$
a^{2}+a^{2}=d^{2}
$$

which yields:

$$
(0.514 \mathrm{~nm})^{2}+(0.514 \mathrm{~nm})^{2}=(4 r)^{2}=16 r^{2}
$$

Solving this gives:

$$
r=\sqrt{\frac{(0.514 \mathrm{~nm})^{2}+(0.514 \mathrm{~nm})^{2}}{16}}=0.182 \mathrm{~nm}(1.82 \AA) \text { for a } \mathrm{Cl}^{-} \text {radius. }
$$


1173. Check Your Learning -
The edge length of the unit cell of KCl ( NaCl -like structure, FCC ) is 6.28 Å. Assuming anion-cation contact along the cell edge, calculate the radius of the potassium ion. The radius of the chloride ion is $1.82 \AA$.


1174. Answer: -
The radius of the potassium ion is $1.33 \AA$.

It is important to realize that values for ionic radii calculated from the edge lengths of unit cells depend on numerous assumptions, such as a perfect spherical shape for ions, which are approximations at best. Hence, such calculated values are themselves approximate and comparisons cannot be pushed too far. Nevertheless, this method has proved useful for calculating ionic radii from experimental measurements such as X-ray crystallographic determinations.


1175. X-Ray Crystallography -
The size of the unit cell and the arrangement of atoms in a crystal may be determined from measurements of the diffraction of X-rays by the crystal, termed X-ray crystallography. Diffraction is the change in the direction of travel experienced by an electromagnetic wave when it encounters a physical barrier whose dimensions are comparable to those of the wavelength of the light. X-rays are electromagnetic radiation with wavelengths about as long as the distance between neighboring atoms in crystals (on the order of a few $\AA$ ).

When a beam of monochromatic X-rays strikes a crystal, its rays are scattered in all directions by the atoms within the crystal. When scattered waves traveling in the same direction encounter one another, they undergo interference, a process by which the waves combine to yield either an increase or a decrease in amplitude (intensity) depending upon the extent to which the combining waves' maxima are separated (see Figure 10.63).

Constructive interface
(a)

Destructive interface
(b)

FIGURE 10.63 Light waves occupying the same space experience interference, combining to yield waves of greater (a) or lesser (b) intensity, depending upon the separation of their maxima and minima.

When X-rays of a certain wavelength, $\lambda$, are scattered by atoms in adjacent crystal planes separated by a distance, $d$, they may undergo constructive interference when the difference between the distances traveled by the two waves prior to their combination is an integer factor, $n$, of the wavelength. This condition is satisfied when the angle of the diffracted beam, $\theta$, is related to the wavelength and interatomic distance by the equation:

$$
n \lambda=2 d \sin \theta
$$

This relation is known as the Bragg equation in honor of W. H. Bragg, the English physicist who first explained this phenomenon. Figure 10.64 illustrates two examples of diffracted waves from the same two crystal planes. The figure on the left depicts waves diffracted at the Bragg angle, resulting in constructive interference, while that on the right shows diffraction and a different angle that does not satisfy the Bragg condition, resulting in destructive interference.


1175. X-Ray Crystallography -
FIGURE 10.64 The diffraction of X-rays scattered by the atoms within a crystal permits the determination of the distance between the atoms. The top image depicts constructive interference between two scattered waves and a resultant diffracted wave of high intensity. The bottom image depicts destructive interference and a low intensity diffracted wave.


1176. LINK TO LEARNING -
Visit this site (http://openstax.org/l/16bragg) for more details on the Bragg equation and a simulator that allows you to explore the effect of each variable on the intensity of the diffracted wave.

An X-ray diffractometer, such as the one illustrated in Figure 10.65, may be used to measure the angles at which X-rays are diffracted when interacting with a crystal as described earlier. From such measurements, the Bragg equation may be used to compute distances between atoms as demonstrated in the following example exercise.

FIGURE 10.65 (a) In a diffractometer, a beam of X-rays strikes a crystalline material, producing (b) an X-ray diffraction pattern that can be analyzed to determine the crystal structure.


1178. Using the Bragg Equation -
In a diffractometer, X-rays with a wavelength of 0.1315 nm were used to produce a diffraction pattern for copper. The first order diffraction $(n=1)$ occurred at an angle $\theta=25.25^{\circ}$. Determine the spacing between the diffracting planes in copper.


1179. Solution -
The distance between the planes is found by solving the Bragg equation, $n \lambda=2 d \sin \theta$, for $d$.
This gives: $d=\frac{n \lambda}{2 \sin \theta}=\frac{1(0.1315 \mathrm{~nm})}{2 \sin \left(25.25^{\circ}\right)}=0.154 \mathrm{~nm}$


1180. Check Your Learning -
A crystal with spacing between planes equal to 0.394 nm diffracts X-rays with a wavelength of 0.147 nm . What is the angle for the first order diffraction?


1181. Answer: -
$10.8^{\circ}$


1183. X-ray Crystallographer Rosalind Franklin -
The discovery of the structure of DNA in 1953 by Francis Crick and James Watson is one of the great achievements in the history of science. They were awarded the 1962 Nobel Prize in Physiology or Medicine, along with Maurice Wilkins, who provided experimental proof of DNA's structure. British chemist Rosalind Franklin made invaluable contributions to this monumental achievement through her work in measuring X-ray diffraction images of DNA. Early in her career, Franklin's research on the structure of coals proved helpful to the British war effort. After shifting her focus to biological systems in the early 1950s, Franklin and doctoral student Raymond Gosling discovered that DNA consists of two forms: a long, thin fiber formed when wet (type "B") and a short, wide fiber formed when dried (type "A"). Her X-ray diffraction images of DNA (Figure 10.66) provided the crucial information that allowed Watson and Crick to confirm that DNA forms a double helix, and to determine details of its size and structure. Franklin also conducted pioneering research on viruses and the RNA that contains their genetic information, uncovering new information that radically changed the body of knowledge in the field. After developing ovarian cancer, Franklin continued
to work until her death in 1958 at age 37. Among many posthumous recognitions of her work, the Chicago Medical School of Finch University of Health Sciences changed its name to the Rosalind Franklin University of Medicine and Science in 2004, and adopted an image of her famous X-ray diffraction image of DNA as its official university logo.

FIGURE 10.66 This illustration shows an X-ray diffraction image similar to the one Franklin found in her research. (credit: National Institutes of Health)


1184. Key Terms -
adhesive force force of attraction between molecules of different chemical identities
amorphous solid (also, noncrystalline solid) solid in which the particles lack an ordered internal structure
body-centered cubic (BCC) solid crystalline structure that has a cubic unit cell with lattice points at the corners and in the center of the cell
body-centered cubic unit cell simplest repeating unit of a body-centered cubic crystal; it is a cube containing lattice points at each corner and in the center of the cube
boiling point temperature at which the vapor pressure of a liquid equals the pressure of the gas above it
Bragg equation equation that relates the angles at which X-rays are diffracted by the atoms within a crystal
capillary action flow of liquid within a porous material due to the attraction of the liquid molecules to the surface of the material and to other liquid molecules
Clausius-Clapeyron equation mathematical relationship between the temperature, vapor pressure, and enthalpy of vaporization for a substance
cohesive force force of attraction between identical molecules
condensation change from a gaseous to a liquid state
coordination number number of atoms closest to any given atom in a crystal or to the central metal atom in a complex
covalent network solid solid whose particles are held together by covalent bonds
critical point temperature and pressure above which a gas cannot be condensed into a liquid
crystalline solid solid in which the particles are arranged in a definite repeating pattern
cubic closest packing (CCP) crystalline structure in which planes of closely packed atoms or ions are stacked as a series of three alternating layers of different relative orientations ( ABC )
deposition change from a gaseous state directly to a solid state
diffraction redirection of electromagnetic radiation that occurs when it encounters a physical barrier of appropriate dimensions
dipole-dipole attraction intermolecular attraction between two permanent dipoles
dispersion force (also, London dispersion force) attraction between two rapidly fluctuating,
temporary dipoles; significant only when particles are very close together
dynamic equilibrium state of a system in which reciprocal processes are occurring at equal rates
face-centered cubic (FCC) solid crystalline structure consisting of a cubic unit cell with lattice points on the corners and in the center of each face


1184. Key Terms -
dynamic equilibrium state of a system in which reciprocal processes are occurring at equal rates
face-centered cubic (FCC) solid crystalline structure consisting of a cubic unit cell with lattice points on the corners and in the center of each face
face-centered cubic unit cell simplest repeating unit of a face-centered cubic crystal; it is a cube containing lattice points at each corner and in the center of each face
freezing change from a liquid state to a solid state
freezing point temperature at which the solid and liquid phases of a substance are in equilibrium; see also melting point
hexagonal closest packing (HCP) crystalline structure in which close packed layers of atoms or ions are stacked as a series of two alternating layers of different relative orientations (AB)
hole (also, interstice) space between atoms within a crystal
hydrogen bonding occurs when exceptionally strong dipoles attract; bonding that exists when hydrogen is bonded to one of the three most electronegative elements: F, O, or N
induced dipole temporary dipole formed when the electrons of an atom or molecule are distorted by the instantaneous dipole of a neighboring atom or molecule
instantaneous dipole temporary dipole that occurs for a brief moment in time when the electrons of an atom or molecule are distributed asymmetrically
intermolecular force noncovalent attractive force between atoms, molecules, and/or ions
interstitial sites spaces between the regular particle positions in any array of atoms or ions
ionic solid solid composed of positive and negative ions held together by strong electrostatic attractions
isomorphous possessing the same crystalline structure
melting change from a solid state to a liquid state
melting point temperature at which the solid and liquid phases of a substance are in equilibrium; see also freezing point
metallic solid solid composed of metal atoms
molecular solid solid composed of neutral molecules held together by intermolecular forces of attraction
normal boiling point temperature at which a
liquid's vapor pressure equals 1 atm ( 760 torr)
octahedral hole open space in a crystal at the center of six particles located at the corners of an octahedron
phase diagram pressure-temperature graph summarizing conditions under which the phases of a substance can exist
polarizability measure of the ability of a charge to distort a molecule's charge distribution (electron cloud)
simple cubic structure crystalline structure with a cubic unit cell with lattice points only at the corners


1184. Key Terms -
polarizability measure of the ability of a charge to distort a molecule's charge distribution (electron cloud)
simple cubic structure crystalline structure with a cubic unit cell with lattice points only at the corners
simple cubic unit cell (also, primitive cubic unit cell) unit cell in the simple cubic structure
space lattice all points within a crystal that have identical environments
sublimation change from solid state directly to gaseous state
supercritical fluid substance at a temperature and pressure higher than its critical point; exhibits properties intermediate between those of gaseous and liquid states
surface tension energy required to increase the area, or length, of a liquid surface by a given amount


1185. Key Equations -
$h=\frac{2 T \cos \theta}{r \rho g}$
$P=A e^{-\Delta H_{\mathrm{vap}} / R T}$
$\ln P=-\frac{\Delta H_{\mathrm{vap}}}{R T}+\ln A$
$\ln \left(\frac{P_{2}}{P_{1}}\right)=\frac{\Delta H_{\mathrm{vap}}}{R}\left(\frac{1}{T_{1}}-\frac{1}{T_{2}}\right)$
$n \lambda=2 d \sin \theta$

Summary


1185. Key Equations - 1185.1. Intermolecular Forces
The physical properties of condensed matter (liquids and solids) can be explained in terms of the kinetic molecular theory. In a liquid, intermolecular attractive forces hold the molecules in contact, although they still have sufficient KE to move past each other.

Intermolecular attractive forces, collectively referred to as van der Waals forces, are responsible for the behavior of liquids and solids and are electrostatic in nature. Dipole-dipole attractions result from the electrostatic attraction of the partial negative end of one polar molecule for the partial positive end of another. The temporary dipole that results from the
tetrahedral hole tetrahedral space formed by four atoms or ions in a crystal
triple point temperature and pressure at which the vapor, liquid, and solid phases of a substance are in equilibrium
unit cell smallest portion of a space lattice that is repeated in three dimensions to form the entire lattice
vacancy defect that occurs when a position that should contain an atom or ion is vacant
van der Waals force attractive or repulsive force between molecules, including dipole-dipole, dipole-induced dipole, and London dispersion forces; does not include forces due to covalent or ionic bonding, or the attraction between ions and molecules
vapor pressure (also, equilibrium vapor pressure) pressure exerted by a vapor in equilibrium with a solid or a liquid at a given temperature
vaporization change from liquid state to gaseous state
viscosity measure of a liquid's resistance to flow
X-ray crystallography experimental technique for determining distances between atoms in a crystal by measuring the angles at which X-rays are diffracted when passing through the crystal
motion of the electrons in an atom can induce a dipole in an adjacent atom and give rise to the London dispersion force. London forces increase with increasing molecular size. Hydrogen bonds are a special type of dipole-dipole attraction that results when hydrogen is bonded to one of the three most electronegative elements: $\mathrm{F}, \mathrm{O}$, or N .


1185. Key Equations - 1185.2. Properties of Liquids
The intermolecular forces between molecules in the liquid state vary depending upon their chemical identities and result in corresponding variations in various physical properties. Cohesive forces between like molecules are responsible for a liquid's viscosity (resistance to flow) and surface tension
(elasticity of a liquid surface). Adhesive forces between the molecules of a liquid and different molecules composing a surface in contact with the liquid are responsible for phenomena such as surface wetting and capillary rise.


1185. Key Equations - 1185.3. Phase Transitions
Phase transitions are processes that convert matter from one physical state into another. There are six phase transitions between the three phases of matter. Melting, vaporization, and sublimation are all endothermic processes, requiring an input of heat to overcome intermolecular attractions. The reciprocal transitions of freezing, condensation, and deposition are all exothermic processes, involving heat as intermolecular attractive forces are established or strengthened. The temperatures at which phase transitions occur are determined by the relative strengths of intermolecular attractions and are, therefore, dependent on the chemical identity of the substance.


1185. Key Equations - 1185.4. Phase Diagrams
The temperature and pressure conditions at which a substance exists in solid, liquid, and gaseous states are summarized in a phase diagram for that substance. Phase diagrams are combined plots of three pressure-temperature equilibrium curves: solid-liquid, liquid-gas, and solid-gas. These curves represent the relationships between phasetransition temperatures and pressures. The point of intersection of all three curves represents the substance's triple point-the temperature and pressure at which all three phases are in equilibrium. At pressures below the triple point, a substance cannot exist in the liquid state, regardless of its temperature. The terminus of the liquid-gas curve represents the substance's critical point, the pressure and temperature above which a liquid phase cannot exist.


1185. Key Equations - 1185.5. The Solid State of Matter
Some substances form crystalline solids consisting
of particles in a very organized structure; others form amorphous (noncrystalline) solids with an internal structure that is not ordered. The main types of crystalline solids are ionic solids, metallic solids, covalent network solids, and molecular solids. The properties of the different kinds of crystalline solids are due to the types of particles of which they consist, the arrangements of the particles, and the strengths of the attractions between them. Because their particles experience identical attractions, crystalline solids have distinct melting temperatures; the particles in amorphous solids experience a range of interactions, so they soften gradually and melt over a range of temperatures. Some crystalline solids have defects in the definite repeating pattern of their particles. These defects (which include vacancies, atoms or ions not in the regular positions, and impurities) change physical properties such as electrical conductivity, which is exploited in the silicon crystals used to manufacture computer chips.


1185. Key Equations - 1185.6. Lattice Structures in Crystalline Solids
The structures of crystalline metals and simple ionic compounds can be described in terms of packing of spheres. Metal atoms can pack in hexagonal closestpacked structures, cubic closest-packed structures, body-centered structures, and simple cubic structures. The anions in simple ionic structures commonly adopt one of these structures, and the cations occupy the spaces remaining between the anions. Small cations usually occupy tetrahedral holes in a closest-packed array of anions. Larger cations usually occupy octahedral holes. Still larger cations can occupy cubic holes in a simple cubic array of anions. The structure of a solid can be described by indicating the size and shape of a unit cell and the contents of the cell. The type of structure and dimensions of the unit cell can be determined by X-ray diffraction measurements.


1186. Exercises - 1186.1. Intermolecular Forces
1. In terms of their bulk properties, how do liquids and solids differ? How are they similar?
2. In terms of the kinetic molecular theory, in what ways are liquids similar to solids? In what ways are liquids different from solids?
3. In terms of the kinetic molecular theory, in what ways are liquids similar to gases? In what ways are liquids different from gases?
4. Explain why liquids assume the shape of any container into which they are poured, whereas solids are rigid and retain their shape.
5. What is the evidence that all neutral atoms and molecules exert attractive forces on each other?
6. Open the PhET States of Matter Simulation (http://openstax.org/l/16phetvisual) to answer the following questions:
(a) Select the Solid, Liquid, Gas tab. Explore by selecting different substances, heating and cooling the systems, and changing the state. What similarities do you notice between the four substances for each phase (solid, liquid, gas)? What differences do you notice?
(b) For each substance, select each of the states and record the given temperatures. How do the given temperatures for each state correlate with the strengths of their intermolecular attractions? Explain.
(c) Select the Interaction Potential tab, and use the default neon atoms. Move the Ne atom on the right and observe how the potential energy changes. Select the Total Force button, and move the Ne atom as before. When is the total force on each atom attractive and large enough to matter? Then select the Component Forces button, and move the Ne atom. When do the attractive (van der Waals) and repulsive (electron overlap) forces balance? How does this relate to the potential energy versus the distance between atoms graph? Explain.
7. Define the following and give an example of each:
(a) dispersion force
(b) dipole-dipole attraction
(c) hydrogen bond
8. The types of intermolecular forces in a substance are identical whether it is a solid, a liquid, or a gas. Why then does a substance change phase from a gas to a liquid or to a solid?
9. Why do the boiling points of the noble gases increase in the order $\mathrm{He}<\mathrm{Ne}<\mathrm{Ar}<\mathrm{Kr}<\mathrm{Xe}$ ?
10. Neon and HF have approximately the same molecular masses.


1186. Exercises - 1186.1. Intermolecular Forces
10. Neon and HF have approximately the same molecular masses.
(a) Explain why the boiling points of Neon and HF differ.
(b) Compare the change in the boiling points of $\mathrm{Ne}, \mathrm{Ar}, \mathrm{Kr}$, and Xe with the change of the boiling points of $\mathrm{HF}, \mathrm{HCl}, \mathrm{HBr}$, and HI , and explain the difference between the changes with increasing atomic or molecular mass.
11. Arrange each of the following sets of compounds in order of increasing boiling point temperature:
(a) $\mathrm{HCl}, \mathrm{H}_{2} \mathrm{O}, \mathrm{SiH}_{4}$
(b) $\mathrm{F}_{2}, \mathrm{Cl}_{2}, \mathrm{Br}_{2}$
(c) $\mathrm{CH}_{4}, \mathrm{C}_{2} \mathrm{H}_{6}, \mathrm{C}_{3} \mathrm{H}_{8}$
(d) $\mathrm{O}_{2}, \mathrm{NO}, \mathrm{N}_{2}$
12. The molecular mass of butanol, $\mathrm{C}_{4} \mathrm{H}_{9} \mathrm{OH}$, is 74.14 ; that of ethylene glycol, $\mathrm{CH}_{2}(\mathrm{OH}) \mathrm{CH}_{2} \mathrm{OH}$, is 62.08 , yet their boiling points are $117.2^{\circ} \mathrm{C}$ and $174^{\circ} \mathrm{C}$, respectively. Explain the reason for the difference.
13. On the basis of intermolecular attractions, explain the differences in the boiling points of $n$-butane $\left(-1^{\circ} \mathrm{C}\right)$ and chloroethane $\left(12^{\circ} \mathrm{C}\right)$, which have similar molar masses.
14. On the basis of dipole moments and/or hydrogen bonding, explain in a qualitative way the differences in the boiling points of acetone $\left(56.2^{\circ} \mathrm{C}\right)$ and 1-propanol $\left(97.4^{\circ} \mathrm{C}\right)$, which have similar molar masses.


1186. Exercises - 1186.1. Intermolecular Forces
15. The melting point of $\mathrm{H}_{2} \mathrm{O}(s)$ is $0^{\circ} \mathrm{C}$. Would you expect the melting point of $\mathrm{H}_{2} \mathrm{~S}(s)$ to be $-85^{\circ} \mathrm{C}, 0^{\circ} \mathrm{C}$, or 185 ${ }^{\circ} \mathrm{C}$ ? Explain your answer.
16. Silane $\left(\mathrm{SiH}_{4}\right)$, phosphine $\left(\mathrm{PH}_{3}\right)$, and hydrogen sulfide $\left(\mathrm{H}_{2} \mathrm{~S}\right)$ melt at $-185^{\circ} \mathrm{C},-133^{\circ} \mathrm{C}$, and $-85^{\circ} \mathrm{C}$, respectively. What does this suggest about the polar character and intermolecular attractions of the three compounds?
17. Explain why a hydrogen bond between two water molecules is weaker than a hydrogen bond between two hydrogen fluoride molecules.
18. Under certain conditions, molecules of acetic acid, $\mathrm{CH}_{3} \mathrm{COOH}$, form "dimers," pairs of acetic acid molecules held together by strong intermolecular attractions:


1186. Exercises - 1186.1. Intermolecular Forces
Draw a dimer of acetic acid, showing how two $\mathrm{CH}_{3} \mathrm{COOH}$ molecules are held together, and stating the type of IMF that is responsible.
19. Proteins are chains of amino acids that can form in a variety of arrangements, one of which is a helix. What kind of IMF is responsible for holding the protein strand in this shape? On the protein image, show the locations of the IMFs that hold the protein together:

20. The density of liquid $\mathrm{NH}_{3}$ is $0.64 \mathrm{~g} / \mathrm{mL}$; the density of gaseous $\mathrm{NH}_{3}$ at STP is $0.0007 \mathrm{~g} / \mathrm{mL}$. Explain the difference between the densities of these two phases.
21. Identify the intermolecular forces present in the following solids:
(a) $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}$
(b) $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{3}$
(c) $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Cl}$


1186. Exercises - 1186.2. Properties of Liquids
22. The test tubes shown here contain equal amounts of the specified motor oils. Identical metal spheres were dropped at the same time into each of the tubes, and a brief moment later, the spheres had fallen to the heights indicated in the illustration.
Rank the motor oils in order of increasing viscosity, and explain your reasoning:

23. Although steel is denser than water, a steel needle or paper clip placed carefully lengthwise on the surface of still water can be made to float. Explain at a molecular level how this is possible.

FIGURE 10.67 (credit: Cory Zanker)
24. The surface tension and viscosity values for diethyl ether, acetone, ethanol, and ethylene glycol are shown here.

| Compound | Molecule | Surface Tension
 $(\mathbf{m N / m})$ | Viscosity
 $(\mathbf{m P a} \mathbf{s})$ |
| :---: | :---: | :---: | :---: |
| diethyl ether
 $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OC}_{2} \mathrm{H}_{5}$ |  | 17 | 0.22 |
| acetone
 $\mathrm{CH}_{3} \mathrm{COCH}_{3}$ |  | 23 | 0.31 |
| ethanol
 $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}$ |  | 22 | 1.07 |
| ethylene glycol
 $\mathrm{CH}_{2}(\mathrm{OH}) \mathrm{CH}(\mathrm{OH})$ |  | 48 | 16.1 |

(a) Explain their differences in viscosity in terms of the size and shape of their molecules and their IMFs.
(b) Explain their differences in surface tension in terms of the size and shape of their molecules and their


1186. Exercises - 1186.2. Properties of Liquids
(a) Explain their differences in viscosity in terms of the size and shape of their molecules and their IMFs.
(b) Explain their differences in surface tension in terms of the size and shape of their molecules and their

IMFs:
25. You may have heard someone use the figure of speech "slower than molasses in winter" to describe a process that occurs slowly. Explain why this is an apt idiom, using concepts of molecular size and shape, molecular interactions, and the effect of changing temperature.
26. It is often recommended that you let your car engine run idle to warm up before driving, especially on cold winter days. While the benefit of prolonged idling is dubious, it is certainly true that a warm engine is more fuel efficient than a cold one. Explain the reason for this.
27. The surface tension and viscosity of water at several different temperatures are given in this table.

| Water | Surface Tension (mN/m) | Viscosity (mPa s) |
| :---: | :---: | :---: |
| $0{ }^{\circ} \mathrm{C}$ | 75.6 | 1.79 |
| $20^{\circ} \mathrm{C}$ | 72.8 | 1.00 |
| $60^{\circ} \mathrm{C}$ | 66.2 | 0.47 |
| $100^{\circ} \mathrm{C}$ | 58.9 | 0.28 |

(a) As temperature increases, what happens to the surface tension of water? Explain why this occurs, in terms of molecular interactions and the effect of changing temperature.
(b) As temperature increases, what happens to the viscosity of water? Explain why this occurs, in terms of molecular interactions and the effect of changing temperature.
28. At $25^{\circ} \mathrm{C}$, how high will water rise in a glass capillary tube with an inner diameter of 0.63 mm ? Refer to Example 10.4 for the required information.
29. Water rises in a glass capillary tube to a height of 17 cm . What is the diameter of the capillary tube?