11.1 Intermolecular Forces

A Collective Work

Chapter 11: Liquids and Solids

11.1 Intermolecular Forces

Learning Outcomes

Describe the types of intermolecular forces possible between atoms or molecules in condensed phases (dispersion forces, dipole-dipole attractions, and hydrogen bonding)
Compare the relative strengths of intermolecular forces
Identify the types of intermolecular forces experienced by specific molecules based on their structures
Explain the relation between the intermolecular forces present within a substance and the temperatures associated with changes in its physical state

As was the case for gaseous substances, the kinetic molecular theory may be used to explain the behavior of solids and liquids. In the following description, the term particle will be used to refer to an atom, molecule, or ion. Note that we will use the popular phrase “intermolecular attraction” to refer to attractive forces between the particles of a substance, regardless of whether these particles are molecules, atoms, or ions.

Consider these two aspects of the molecular-level environments in solid, liquid, and gaseous matter:

Particles in a solid are tightly packed together and often arranged in a regular pattern; in a liquid, they are close together with no regular arrangement; in a gas, they are far apart with no regular arrangement.
Particles in a solid vibrate about fixed positions and do not generally move in relation to one another; in a liquid, they move past each other but remain in essentially constant contact; in a gas, they move independently of one another except when they collide.

A drawing of three glass containers is shown, comparing the energy and movement of three states of matter: crystalline solid, liquid, and gas. — Figure 11.1.1. Transitions between solid, liquid, and gaseous states of a substance occur when conditions of temperature or pressure favor the associated changes in intermolecular forces. (Note: The space between particles in the gas phase is much greater than shown.)

The differences in the properties of a solid, liquid, or gas reflect the strengths of the attractive forces between the atoms, molecules, or ions that make up each phase. The phase in which a substance exists depends on the relative extents of its intermolecular force (IMFs) and the kinetic energies (KE) of its molecules. IMFs are the various forces of attraction that may exist between the atoms and molecules of a substance due to electrostatic phenomena, as will be detailed in this chapter. These forces serve to hold particles close together, whereas the particles’ KE provides the energy required to overcome the attractive forces and thus increase the distance between particles. Figure 11.1.1 illustrates how changes in physical state may be induced by changing the temperature, hence, the average KE, of a given substance.

Image a shows a brown colored beverage in a glass with condensation on the outside. Image b shows a body of water with fog hovering above the surface of the water. — Figure 11.1.2. Condensation forms when water vapor in the air is cooled enough to form liquid water, such as (a) on the outside of a cold beverage glass or (b) in the form of fog. (credit a: modification of work by Jenny Downing; credit b: modification of work by Cory Zanker)

A butane lighter is shown. — Figure 11.1.3. Gaseous butane is compressed within the storage compartment of a disposable lighter, resulting in its condensation to the liquid state. (credit: modification of work by “Sam-Cat”/Flickr)

As an example of the processes depicted in this figure, consider a sample of water. When gaseous water is cooled sufficiently, the attractions between $H_{2} O$ molecules will be capable of holding them together when they come into contact with each other; the gas condenses, forming liquid $H_{2} O$ . For example, liquid water forms on the outside of a cold glass as the water vapor in the air is cooled by the cold glass, as seen in Figure 11.1.2.

We can also liquefy many gases by compressing them, if the temperature is not too high. The increased pressure brings the molecules of a gas closer together, such that the attractions between the molecules become strong relative to their KE. Consequently, they form liquids. Butane, $C_{4} H_{10}$ , is the fuel used in disposable lighters and is a gas at standard temperature and pressure. Inside the lighter’s fuel compartment, the butane is compressed to a pressure that results in its condensation to the liquid state, as shown in Figure 11.1.3

Gaseous butane is compressed within the storage compartment of a disposable lighter, resulting in its condensation to the liquid state.

Finally, if the temperature of a liquid becomes sufficiently low, or the pressure on the liquid becomes sufficiently high, the molecules of the liquid no longer have enough KE to overcome the IMF between them, and a solid forms. A more thorough discussion of these and other changes of state, or phase transitions, is provided in a later section of this chapter.

Forces between Molecules

Under appropriate conditions, the attractions between all gas molecules will cause them to form liquids or solids. This is due to intermolecular forces, not intramolecular forces. Intramolecular forces are those within the molecule that keep the molecule together, for example, the bonds between the atoms. Intermolecular forces are the attractions between molecules, which determine many of the physical properties of a substance. Figure 11.1.4 illustrates these different molecular forces. The strengths of these attractive forces vary widely, though usually the IMFs between small molecules are weak compared to the intramolecular forces that bond atoms together within a molecule. For example, to overcome the IMFs in one mole of liquid $HCl$ and convert it into gaseous $HCl$ requires only about 17 kilojoules. However, to break the covalent bonds between the hydrogen and chlorine atoms in one mole of $HCl$ requires about 25 times more energy—430 kilojoules.

A diagram showing a strong internal intramolecular force and a weak external intermolecular force. — Figure 11.1.4. Intramolecular forces keep a molecule intact. Intermolecular forces hold multiple molecules together and determine many of a substance’s properties.

All of the attractive forces between neutral atoms and molecules are known as van der Waals force, although they are usually referred to more informally as intermolecular attraction. We will consider the various types of IMFs in the next three sections of this chapter.

Dispersion Forces

One of the three van der Waals forces is present in all condensed phases, regardless of the nature of the atoms or molecules composing the substance. This attractive force is called the London dispersion force in honor of German-born American physicist Fritz London who, in 1928, first explained it. This force is often referred to as simply the dispersion force. Because the electrons of an atom or molecule are in constant motion (or, alternatively, the electron’s location is subject to quantum-mechanical variability), at any moment in time, an atom or molecule can develop a temporary, instantaneous dipole if its electrons are distributed asymmetrically. The presence of this dipole can, in turn, distort the electrons of a neighboring atom or molecule, producing an induced dipole. These two rapidly fluctuating, temporary dipoles thus result in a relatively weak electrostatic attraction between the species—a so-called dispersion force like that illustrated in Figure 11.1.5.

Two pairs of molecules are shown where each molecule has one larger blue side labeled “delta sign, negative sign” and a smaller red side labeled “delta sign, positive sign.” Toward the middle of the both molecules, but still on each distinct side, is a black dot. Between the two images is a dotted line labeled, “Attractive force.” In the first image, the red and blue sides are labeled, “Unequal distribution of electrons.” Below both images are brackets. The brackets are labeled, “Temporary dipoles.” — Figure 11.1.5. Dispersion forces result from the formation of temporary dipoles, as illustrated here for two nonpolar diatomic molecules.

Dispersion forces that develop between atoms in different molecules can attract the two molecules to each other. The forces are relatively weak, however, and become significant only when the molecules are very close. Larger and heavier atoms and molecules exhibit stronger dispersion forces than do smaller and lighter atoms and molecules. $F_{2}$ and $Cl_{2}$ are gases at room temperature (reflecting weaker attractive forces); $Br_{2}$ is a liquid, and $I_{2}$ is a solid (reflecting stronger attractive forces). Trends in observed melting and boiling points for the halogens clearly demonstrate this effect, as seen in Table 11.1.1.

Table 11.1.1. Melting and Boiling Points of the Halogens

Halogen

Molar Mass

Atomic Radius

Melting Point

Boiling Point

fluorine,

F_{2}

38 g/mol

72 pm

53 K

85 K

chlorine,

Cl_{2}

71 g/mol

99 pm

172 K

238 K

bromine,

Br_{2}

160 g/mol

114 pm

266 K

332 K

iodine,

I_{2}

254 g/mol

133 pm

387 K

457 K

astatine,

At_{2}

420 g/mol

150 pm

575 K

610 K

The increase in melting and boiling points with increasing atomic/molecular size may be rationalized by considering how the strength of dispersion forces is affected by the electronic structure of the atoms or molecules in the substance. In a larger atom, the valence electrons are, on average, farther from the nuclei than in a smaller atom. Thus, they are less tightly held and can more easily form the temporary dipoles that produce the attraction. The measure of how easy or difficult it is for another electrostatic charge (for example, a nearby ion or polar molecule) to distort a molecule’s charge distribution (its electron cloud) is known as polarizability. A molecule that has a charge cloud that is easily distorted is said to be very polarizable and will have large dispersion forces; one with a charge cloud that is difficult to distort is not very polarizable and will have small dispersion forces.

Example 11.1.1: London Forces and Their Effects

Order the following compounds of a group 14 element and hydrogen from lowest to highest boiling point: $CH_{4}$ , $SiH_{4}$ , $GeH_{4}$ , and $SnH_{4}$ . Explain your reasoning.

Show Solution

Applying the skills acquired in the chapter on chemical bonding and molecular geometry, all of these compounds are predicted to be nonpolar, so they may experience only dispersion forces: the smaller the molecule, the less polarizable and the weaker the dispersion forces; the larger the molecule, the larger the dispersion forces. The molar masses of $CH_{4}$ , $SiH_{4}$ , $GeH_{4}$ , and $SnH_{4}$ are approximately 16 g/mol, 32 g/mol, 77 g/mol, and 123 g/mol, respectively. Therefore, $CH_{4}$ is expected to have the lowest boiling point and $SnH_{4}$ the highest boiling point. The ordering from lowest to highest boiling point is expected to be $CH_{4}$ < $SiH_{4}$ < $GeH_{4}$ < $SnH_{4}$ .

A graph of the actual boiling points of these compounds versus the period of the Group 14 element shows this prediction to be correct:

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The shapes of molecules also affect the magnitudes of the dispersion forces between them. For example, boiling points for the isomers n-pentane, isopentane, and neopentane (shown in Figure 11.1.6) are 36 °C, 27 °C, and 9.5 °C, respectively. Even though these compounds are composed of molecules with the same chemical formula, $C_{5} H_{12}$ , the difference in boiling points suggests that dispersion forces in the liquid phase are different, being greatest for n-pentane and least for neopentane. The elongated shape of n-pentane provides a greater surface area available for contact between molecules, resulting in correspondingly stronger dispersion forces. The more compact shape of isopentane offers a smaller surface area available for intermolecular contact and, therefore, weaker dispersion forces. Neopentane molecules are the most compact of the three, offering the least available surface area for intermolecular contact and, hence, the weakest dispersion forces. This behavior is analogous to the connections that may be formed between strips of VELCRO brand fasteners: the greater the area of the strip’s contact, the stronger the connection.

Three images of molecules are shown. The first shows a cluster of large, gray spheres each bonded together and to several smaller, white spheres. There is a gray, jagged line and then the mirror image of the first cluster of spheres is shown. Above these two clusters is the label, “Small contact area, weakest attraction,” and below is the label, “neopentane boiling point: 9.5 degrees C.” The second shows a chain of three gray spheres bonded by the middle sphere to a fourth gray sphere. Each gray sphere is bonded to several smaller, white spheres. There is a jagged, gray line and then the mirror image of the first chain appears. Above these two chains is the label, “Less surface area, less attraction,” and below is the label, “isopentane boiling point: 27 degrees C.” The third image shows a chain of five gray spheres bonded together and to several smaller, white spheres. There is a jagged gray line and then the mirror image of the first chain appears. Above these chains is the label, “Large contact area, strong attraction,” and below is the label, “n-pentane boiling point 36 degrees C.” — Figure 11.1.6. The strength of the dispersion forces increases with the contact area between molecules, as demonstrated by the boiling points of these pentane isomers.

Dipole-Dipole Attractions

Recall from the chapter on chemical bonding and molecular geometry that polar molecules have a partial positive charge on one side and a partial negative charge on the other side of the molecule—a separation of charge called a dipole. Consider a polar molecule such as hydrogen chloride, $HCl$ . In the $HCl$ molecule, the more electronegative $Cl$ atom bears the partial negative charge, whereas the less electronegative $H$ atom bears the partial positive charge. An attractive force between $HCl$ molecules results from the attraction between the positive end of one $HCl$ molecule and the negative end of another. This attractive force is called a dipole-dipole attraction—the electrostatic force between the partially positive end of one polar molecule and the partially negative end of another, as illustrated in Figure 11.1.7.

Two diagrams illustrating dipole-dipole attraction — Figure 11.1.7. This image shows two arrangements of polar molecules, such as $HCl$ , that allow an attraction between the partial negative end of one molecule and the partial positive end of another.

The effect of a dipole-dipole attraction is apparent when we compare the properties of $HCl$ molecules to nonpolar $F_{2}$ molecules. Both $HCl$ and $F_{2}$ consist of the same number of atoms and have approximately the same molecular mass. At a temperature of 150 K, molecules of both substances would have the same average KE. However, the dipole-dipole attractions between $HCl$ molecules are sufficient to cause them to “stick together” to form a liquid, whereas the relatively weaker dispersion forces between nonpolar $F_{2}$ molecules are not, and so this substance is gaseous at this temperature. The higher normal boiling point of $HCl$ (188 K) compared to $F_{2}$ (85 K) is a reflection of the greater strength of dipole-dipole attractions between $HCl$ molecules, compared to the attractions between nonpolar $F_{2}$ molecules. We will often use values such as boiling or freezing points, or enthalpies of vaporization or fusion, as indicators of the relative strengths of IMFs of attraction present within different substances.

Example 11.1.2: Dipole-Dipole Forces and Their Effects

Predict which will have the higher boiling point: $N_{2}$ or $CO$ . Explain your reasoning.

Show Solution

$CO$ and $N_{2}$ are both diatomic molecules with masses of about 28 amu, so they experience similar London dispersion forces. Because $CO$ is a polar molecule, it experiences dipole-dipole attractions. Because $N_{2}$ is nonpolar, its molecules cannot exhibit dipole-dipole attractions. The dipole-dipole attractions between $CO$ molecules are comparably stronger than the dispersion forces between nonpolar $N_{2}$ molecules, so $CO$ is expected to have the higher boiling point.

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Hydrogen Bonding

Nitrosyl fluoride ( $ONF$ , molecular mass 49 amu) is a gas at room temperature. Water ( $H_{2} O$ , molecular mass 18 amu) is a liquid, even though it has a lower molecular mass. We clearly cannot attribute this difference between the two compounds to dispersion forces. Both molecules have about the same shape and $ONF$ is the heavier and larger molecule. It is, therefore, expected to experience more significant dispersion forces. Additionally, we cannot attribute this difference in boiling points to differences in the dipole moments of the molecules. Both molecules are polar and exhibit comparable dipole moments. The large difference between the boiling points is due to a particularly strong dipole-dipole attraction that may occur when a molecule contains a hydrogen atom bonded to a fluorine, oxygen, or nitrogen atom (the three most electronegative elements). The very large difference in electronegativity between the $H$ atom (2.1) and the atom to which it is bonded (4.0 for an $F$ atom, 3.5 for an $O$ atom, or 3.0 for a N atom), combined with the very small size of a $H$ atom and the relatively small sizes of $F$ , $O$ , or $N$ atoms, leads to highly concentrated partial charges with these atoms. Molecules with $F - H$ , $O - H$ , or $N - H$ moieties are very strongly attracted to similar moieties in nearby molecules, a particularly strong type of dipole-dipole attraction called hydrogen bonding. Examples of hydrogen bonds include $HF \dots HF$ , $H_{2} O \dots HOH$ , and $H_{3} N \dots HNH_{2}$ , in which the hydrogen bonds are denoted by dots. Figure 11.1.8 illustrates hydrogen bonding between water molecules.

Five water molecules are shown near one another, but not touching. A dotted line lies between many of the hydrogen atoms on one molecule and the oxygen atom on another molecule. — Figure 11.1.8. Water molecules participate in multiple hydrogen-bonding interactions with nearby water molecules.

Despite use of the word “bond,” keep in mind that hydrogen bonds are intermolecular attractive forces, not intramolecular attractive forces (covalent bonds). Hydrogen bonds are much weaker than covalent bonds, only about 5 to 10% as strong, but are generally much stronger than other dipole-dipole attractions and dispersion forces.

Hydrogen bonds have a pronounced effect on the properties of condensed phases (liquids and solids). For example, consider the trends in boiling points for the binary hydrides of group 15 ( $NH_{3}$ , $PH_{3}$ , $AsH_{3}$ , and $SbH_{3}$ ), group 16 hydrides ( $H_{2} O$ , $H_{2} S$ , $H_{2} Se$ , and $H_{2} Te$ ), and group 17 hydrides ( $HF$ , $HCl$ , $HBr$ , and $HI$ ). The boiling points of the heaviest three hydrides for each group are plotted in Figure 11.1.9. As we progress down any of these groups, the polarities of the molecules decrease slightly, whereas the sizes of the molecules increase substantially. The effect of increasingly stronger dispersion forces dominates that of increasingly weaker dipole-dipole attractions, and the boiling points are observed to increase steadily.

A line graph is shown where the y-axis is labeled “Boiling point (, degree sign, C )” and has values of “ negative 150” to “150” from bottom to top in increments of 50. The x-axis is labeled “Period” and has values of “0” to “5” in increments of 1. Three lines are shown on the graph and are labeled in the legend. The red line is labeled as “halogen family,” the blue is “oxygen family” and the green is “nitrogen family.” The first point on the red line is labeled “question mark” and is at point “2, negative 120”. The second point on the line is labeled “H C l” and is at point “3, negative 80” while the third point on the line is labeled “H B r” and is at point “4, negative 60”. The fourth point on the line is labeled “H I” and is at point “5, negative 40.” The first point on the green line is labeled “question mark” and is at point “2, negative 125.” The second point on the line is labeled “P H, subscript 3” and is at point “3, negative 80” while the third point on the line is labeled “A s H, subscript 3” and is at point “4, negative 55.” The fourth point on the line is labeled “S b H, subscript 3” and is at point “5, negative 10.” The first point on the blue line is labeled “question mark” and is at point “2, negative 80.” The second point on the line is labeled “H, subscript 2, S” and is at point “3, negative 55” while the third point on the line is labeled “H, subscript 2, S e” and is at point “4, negative 45.” The fourth point on the line is labeled “H, subscript 2, T e” and is at point “5, negative 3.” — Figure 11.1.9. For the group 15, 16, and 17 hydrides, the boiling points for each class of compounds increase with increasing molecular mass for elements in periods 3, 4, and 5.

If we use this trend to predict the boiling points for the lightest hydride for each group, we would expect $NH_{3}$ to boil at about −120 °C, $H_{2} O$ to boil at about −80 °C, and HF to boil at about −110 °C. However, when we measure the boiling points for these compounds, we find that they are dramatically higher than the trends would predict, as shown in Figure 11.1.10. The stark contrast between our naïve predictions and reality provides compelling evidence for the strength of hydrogen bonding.

Example 11.1.3: Effect of Hydrogen Bonding on Boiling Points

Consider the compounds dimethylether ( $CH_{3} OCH_{3}$ ), ethanol ( $CH_{3} CH_{2} OH$ ), and propane ( $CH_{3} CH_{2} CH_{3}$ ). Their boiling points, not necessarily in order, are −42.1 °C, −24.8 °C, and 78.4 °C. Match each compound with its boiling point. Explain your reasoning.

Show Solution

The VSEPR-predicted shapes of $CH_{3} OCH_{3}$ , $CH_{3} CH_{2} OH$ , and $CH_{3} CH_{2} CH_{3}$ are similar, as are their molar masses (46 g/mol, 46 g/mol, and 44 g/mol, respectively), so they will exhibit similar dispersion forces. Since $CH_{3} CH_{2} CH_{3}$ is nonpolar, it may exhibit only dispersion forces. Because $CH_{3} OCH_{3}$ is polar, it will also experience dipole-dipole attractions. Finally, $CH_{3} CH_{2} OH$ has an $- OH$ group, and so it will experience the uniquely strong dipole-dipole attraction known as hydrogen bonding. So the ordering in terms of strength of IMFs, and thus boiling points, is $CH_{3} CH_{2} CH_{3}$ < $CH_{3} OCH_{3}$ < $CH_{3} CH_{2} OH$ . The boiling point of propane is −42.1 °C, the boiling point of dimethylether is −24.8 °C, and the boiling point of ethanol is 78.5 °C.

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Key Concepts and Summary

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 dipolar molecule for the partial positive end of another. The temporary dipole that results from the 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: $F$ , $O$ , or $N$ . Figure 11.1.11 summarizes the types of intermolecular forces and their relative strengths.

Figure 11.1.11. Summary of the types of intermolecular forces and their relative strengths.

Try It

Identify the intermolecular forces present in the following solids:
1. $CH_{3} CH_{2} OH$
2. $CH_{3} CH_{2} CH_{3}$
3. $CH_{3} CH_{2} Cl$
Arrange each of the following sets of compounds in order of increasing boiling point temperature:
1. $HCl$ , $H_{2} O$ , $SiH_{4}$
2. $F_{2}$ , $Cl_{2}$ , $Br_{2}$
3. $CH_{4}$ , $C_{2} H_{6}$ , $C_{3} H_{8}$
4. $O_{2}$ , $NO$ , $N_{2}$

Show Selected Solutions

1. (a) hydrogen bonding and dispersion forces;

(b) dispersion;

(c) dipole-dipole attraction and dispersion forces

2. (a) $SiH_{4}$ < $HCl$ < $H_{2} O$ ; (b) $F_{2}$ < $Cl_{2}$ < $Br_{2}$ ; (c) $CH_{4}$ < $C_{2} H_{6}$ < $C_{3} H_{8}$ ; (d) $N_{2}$ < $O_{2}$ < $NO$

Glossary

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

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

polarizability: measure of the ability of a charge to distort a molecule’s charge distribution (electron cloud)

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

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Learning Outcomes

Forces between Molecules

Dispersion Forces

Example 11.1.1: London Forces and Their Effects

Check Your Learning

Dipole-Dipole Attractions

Example 11.1.2: Dipole-Dipole Forces and Their Effects

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Hydrogen Bonding

Example 11.1.3: Effect of Hydrogen Bonding on Boiling Points

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Key Concepts and Summary

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Glossary

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