1. Chemistry: Atoms First
Julia Burdge & Jason Overby
Chapter 12
Intermolecular Forces and the Physical Properties of Liquids and Solids
Kent L. McCorkle
Cosumnes River College
Sacramento, CA

2. Intermolecular Forces and the Physical properties of Liquids and Solids12
12.1 Intermolecular Forces
Dipole-Dipole Interactions
Hydrogen Bonding
Dispersion Forces
Ion-Dipole Interactions
12.2 Properties of Liquids
Surface Tension
Viscosity
Vapor Pressure
12.3 Crystal Structure
Unit Cells
Packing Spheres
Closest Packing
12.4 Types of Crystals
Ionic Crystals
Covalent Crystals
Molecular Crystals
Metallic Crystals
12.5 Amorphous Solids
12.6 Phase Changes
Liquid-Vapor Phase Transition
Solid-Liquid Phase Transition
Solid-Vapor Phase Transition
12.7 Phase Diagrams

3. 12.1 Intermolecular Forces
Intermolecular forces are attractive forces that hold particles together in the condensed phases.
The magnitude (and type) of intermolecular forces is what determines whether the particles that make up a substance are a gas, liquid, or solid.
Gas
Liquid
Solid

4. Intermolecular Forces
Attractive forces that act between atoms or molecules in a pure substance are collectively called van der Waals forces.
Dipole-dipole interactions are attractive forces that act between polar molecules.
The magnitude of the attractive forces depends on the magnitude of the dipole.

5. Intermolecular Forces

6. Hydrogen Bonding
Hydrogen bonding is a special type of dipole-dipole interaction.
Hydrogen bonding only occurs in molecules that contain H bonded to a small, highly electronegative atom such as N, O, or F. F H F

7. Hydrogen Bonding

8. Dispersion Forces
Dispersion forces or London dispersion forces result from the Coulombic attractions between instantaneous dipoles of non-polar molecules.

9. Intermolecular Forces

10. Worked Example 12.1
What kind(s) of intermolecular forces exist in (a) CCl4(l), (b) CH3COOH(l), (c) CH3COCH3(l), and (d) H2S(l).
Strategy Draw Lewis dot structures and apply VSEPR theory to determine whether each molecule is polar or nonpolar. Nonpolar molecules exhibit dispersion forces only. Polar molecules exhibit dipole-dipole interactions and dispersion forces. Polar molecules with N–H , F–H, or O–H bonds exhibit dipole-dipole interactions (including hydrogen bonding) and dispersion forces.
(a) (b) (c) (d)

11. Worked Example 12.1 (cont.) (a) (b) (c) (d)
Solution (a) CCl4 is nonpolar, so the only intermolecular forces are dispersion forces.
(b) CH3COOH is polar and contains an O–H bond, so it exhibits dipole-dipole interactions (including hydrogen bonding) and dispersion forces.
(c) CH3COCH3 is polar but does not contain N–H , F–H, or O–H bonds, so it exhibits dipole-dipole interactions and dispersion forces.
(d) H2S is polar but does not contain N–H , F–H, or O–H bonds, so it exhibits dipole-dipole interactions and dispersion forces.
Think About It Being able to draw correct Lewis structures is, once again, vitally important. Review, if you need to, the procedure for drawing them.

12. Ion-Dipole Interactions
Ion-dipole interactions are Coulombic attractions between ions (either positive or negative) and polar molecules.

13. Properties of Liquids
12.2
Surface tension is the amount of energy required to stretch or increase the surface of a liquid by a unit area.
The stronger the intermolecular forces, the higher the surface tension.

14. Capillary action is the movement of a liquid up a narrow tube.
Properties of Liquids
Capillary action is the movement of a liquid up a narrow tube.
Two types of forces bring about capillary action:
cohesion is the attraction between like molecules
adhesion is the attraction between unlike molecules
Adhesive forces are greater than cohesive forces
Cohesive forces are
greater than adhesive forces

15. Properties of Liquids
Viscosity is a measure of a fluid’s resistance to flow.
The higher the viscosity the more slowly a liquid flows.
Liquids that have higher intermolecular forces have higher viscosities.

16. The number of molecules with enough kinetic energy to escape.
Properties of Liquids
Vapor pressure is also dependent on intermolecular forces.
If a molecule at the surface of a liquid has enough kinetic energy, it can escape to the gas phase in a process called vaporization.
T1 < T2
The number of molecules with enough kinetic energy to escape.

17. Evaporation: H2O(l) → H2O(g) Condensation: H2O(l) ← H2O(g)
Properties of Liquids
The vapor pressure increases until the rate of evaporation equals the rate of condensation.
H2O(l) ⇌ H2O(g)
Evaporation: H2O(l) → H2O(g)
Condensation: H2O(l) ← H2O(g)
When the forward process and reverse process are occurring at the same rate, the system is in dynamic equilibrium.

18. Properties of Liquids
The vapor pressure increases until the rate of evaporation equals the rate of condensation.
H2O(l) ⇌ H2O(g)

19. Properties of Liquids
The vapor pressure increases with temperature.

20. Properties of Liquids
The Clausius-Clapeyron equation relates the natural log of vapor pressure and the reciprocal of absolute temperature.
ln P = natural log of vapor pressure
ΔHvap = the molar heat of vaporization
R = the gas constant (8.314 J/K•mol)
T = the kelvin temperature
C is an experimentally determined constant

21. Properties of Liquids
The Clausius-Clapeyron equation:
Plotting ln P versus 1/T is a line with a slope of −ΔH/R.
ΔH is assumed to be independent of temperature.

22. Properties of Liquids
The Clausius-Clapeyron equation can be rearranged into a two point form:

23. Worked Example 12.2
Diethyl ether is a volatile, highly flammable organic liquid that today is used mainly as a solvent. (It was used as an anesthetic during the nineteenth century and as a recreational intoxicant early in the twentieth century during prohibition, when ethanol was difficult to obtain.) The vapor pressure of diethyl ether is 401 mmHg at 18°C, and its molar heat of vaporization is 26 kJ/mol. Calculate its vapor pressure at 32°C.
Strategy Given the vapor pressure at one temperature, P1, use the equation below to calculate the vapor pressure at a second temperature, P2.
Temperature must be expressed in kelvins, so T1 = K and T2 = K. Because the molar heat of vaporization is given in kJ/mol, we will have to convert it to J/mol for the units of R to cancel properly: ΔHvap = 2.6×104 J/mol. The inverse function of ln x is ex.
ln = − P1 P2
ΔHvap R 1 T2 T1

24. Worked Example 12.2 (cont.) Solution ln = P1 P2 2.6×104 J/mol
8.314 J/K∙mol 1 K − K
= −0.4928 P1 P2
= e− = P1
0.6109
= P2
401 mmHg
0.6109
P2 =
= 6.6×102 mmHg
Think About It It is easy to switch P1 and P2 or T1 and T2 accidentally and get the wrong answer to a problem such as this. One way to help safeguard against this common error is to verify that the vapor pressure is higher at the higher temperature.

25. Crystal Structure
12.3
A crystalline solid possess rigid and long-range order; its atoms, molecules, or ions occupy specific positions.
A unit cell is the basic repeating structural unit of a crystalline solid.

26. Crystal Structure
There are seven types of unit cells.

27. Crystal Structure
The coordination number is defined as the number of atoms surrounding an atom in a crystal lattice.
The value of the coordination number indicates how tightly the atoms are packed together.
The basic repeating unit in the array of atoms is called a simple cubic cell.

28. Crystal Structure
There are three types of cubic cells.

29. Crystal Structure
In a body-centered cubic cell (bcc) the spheres in each layer rest in the depressions between spheres in the previous layer.
The coordination number is 8.

30. Crystal Structure
In a face-centered cubic cell (fcc) the coordination number is 12.

31. Most of a cell’s atoms are shared by neighboring cells.
Crystal Structure
Most of a cell’s atoms are shared by neighboring cells.
A corner atom is shared by eight unit cells.
An edge atom is shared by four unit cells.
A face-centered atom is shared by two unit cells.

32. Crystal Structure
A simple cubic cell has the equivalent of only one complete atom contained within the cell.

33. Crystal Structure
A body-centered cubic cell has two equivalent atoms:
A face-centered cubic cell contains four complete atoms:

34. Hexagonal close-packed (hcp) structure:
Crystal Structure
Hexagonal close-packed (hcp) structure:
Site directly over an atom in layer A
Close packing starts with a layer of atoms (A)
Atoms in the second layer (B) fit into the depressions of the first layer
Hexagonal close-packed structure.

35. Cubic close-packed (ccp) structure:
Crystal Structure
Cubic close-packed (ccp) structure:
Site directly over an atom in layer A
(hcp)
Site NOT directly over an atom in layer A (ccp)
Cubic close-packed structure 35

36. Crystal Structure Closest packing: Hexagonal close-packing (hcp)
Cubic close-packing (ccp) corresponds to a face-centered cubit cell. 36

37. Edge length (a) and radius (r) are related:
Crystal Structure
Edge length (a) and radius (r) are related:
Simple cubic
Body-centered cubic
Face-centered cubic 37

38. Worked Example 12.3
Gold crystallizes in a cubic close-packed structure (face-centered cubic unit cell) and has a density of 19.3 g/cm3. Calculate the atomic radius of an Au atom in angstroms (Å).
Strategy Using the given density and the mass of gold continued within a face-centered cubic unit cell, determine the volumes of the unit cell. Then, use the volume to determine the value of a, and use the equation a = √8r to find r. Be sure to use consistent units for mass, length, and volume. The face-centered cubic unit cell contains a total of four atoms of gold [six faces, each shared by two unit cells, and eight corners, each shared by eight unit cells]. d = m/V and V = a3.
Solution First, we determine the mass of gold (in grams) contained within a unit cell:
m = × × = 1.31×10-21 g/unit cell
4 atoms
unit cell
1 mol
6.022×1023 atoms
197.0 g Au
1 mol Au

39. Worked Example 12.3 (cont.) Solution
Then we calculate the volume of the unit cell in cm3:
V = = = 6.78×10-23 cm3
Using the calculated volume and the relationship V = a3 (rearranged to solve for a), we determine the length of a side of a unit cell:
a = = √6.78×10-23 cm3 = 4.08×10-8 cm
Using the relationship provided a = √8r (rearranged to solve for r), we determine the radius of a gold atom in centimeters.
r = = = 1.44×10-8 cm
Finally, we convert centimeters to angstroms:
1.44×10-8 cm × × = 1.44 Å m d
1.31×10-21 g
19.3 g/cm3
Think About It Atomic radii tend to be on the order of 1 Å, so this answer is reasonable. 3 a √8
4.08×10-8 cm √8
1×10-2 m
1 cm
1 Å
1×10-10 m

40. Types of Crystals
12.4
Ionic crystals are composed of charged ions that are held together by Coulombic attraction.
The unit cell of an ionic compound can be defined be either the positions of the anions or the positions of the cations.

41. Crystal structures of three ionic compounds:
Types of Crystals
Crystal structures of three ionic compounds:
CsCl
Simple cubic lattice
ZnS
Zincblende structure
(based on FCC)
CaF2
fluorite structure
(based on FCC) 41

42. Worked Example 12.4
How many of each ion are contained within a unit cell of ZnS?
Strategy Determine the contribution of each ion in the unit cell based on its position. Referring to the figure, the unit cell has four Zn2+ ions completely contained within the unit cell, and S2- ions at each of the eight corners and at each of the six faces. Interior ions (those completely contained within the unit cell) contribute one, those at the corners each contribute one-eighth, and those on the faces contribute one-half.
Think About It Make sure that the ratio of cations to anions that you determine for a unit cell matches the ratio expressed in the compound’s empirical formula.
Solution The ZnS unit cell contains four Zn2+ ions (interior) and four S2- ions [8 × (corners) and 6 × (faces)] 1 8 1 2

43. Worked Example 12.5
The edge length of the NaCl unit cell is 564 pm. Determine the density of NaCl in g/cm3.
Strategy Use the number of Na+ and Cl- ions in a unit cell (four of each) to determine the mass of a unit cell. Calculate volume using the edge length given in the problem statement. Density is mass divided by volume (d = m/V). Be careful to use units consistently.
The masses of Na+ and Cl- ions are amu and amu, respectively. The conversion factor from amu to grams is
so the masses of the Na+ and Cl- ions are 3.818×10-23 g and 5.887×10-23 g, respectively. The unit cell length is
564 pm × × = 5.64×10-8 cm
1 g
6.022×1023 amu
1×10-12 m
1 pm
1 cm
1×10-2 m

44. Worked Example 12.5 (cont.)
Solution The mass of the unit cell is 3.882×10-22 g (4 × 3.818×10-23 g + 4 × 5.887×10-23 g). The volume of a unit cell is 1.794×10-22 cm3 [(5.64×10-8 cm)3]. Therefore, the density is given by
d = = 2.16 g/cm3
3.882×10-22 g
1.794×10-22 cm3
Think About It If you were to hold a cubic centimeter (1 cm3) of salt in your hand, how heavy would you expect it to be? Common errors in this type of problem include errors of unit conversion–especially with regard to length and volume. Such errors can lead to results that are off by many orders of magnitude. Often you can use common sense to gauge whether or not a calculated answer is reasonable. For instance, simply getting the centimeter-meter conversion upside down would result in a calculated density of 2.16×1012 g/cm3! You know that a cubic centimeter of salt doesn’t have a mass that large. (That’s billions of kilograms!) If the magnitude of a result is not reasonable, go back and check your work.

45. Types of Crystals
In covalent crystals, atoms are held together in an extensive three-dimensional network entirely by covalent bonds. 45

46. Types of Crystals
In molecular crystals, the lattice points are occupied by molecules; the attractive forces between them are van der Waals forces and/or hydrogen bonding. 46

47. Worked Example 12.6
The metal iridium (Ir) crystallizes with a face-centered cubic unit cell. Given that the length of the edge of a unit cell is 383 pm, determine the density of iridium in g/cm3.
Strategy A face-centered metallic crystal contains four atoms per unit cell [8 × (corners) and 6 × (faces)]. Use the number of atoms per cell and the atomic mass to determine the mass of a unit cell. Calculate volume using the edge length given in the problem statement. Density is then mass divided by volume (d = m/V). Be sure to make all necessary unit conversions.
The mass of an Ir atom is amu. The conversion factor from amu to grams is
so the mass of an Ir atom is 3.192×10-22 g. The unit cell length is
383 pm × × = 3.83×10-8 cm 1 8 1 2
1 g
6.022×1023 amu
1×10-12 m
1 pm
1 cm
1×10-2 m

48. Worked Example 12.6 (cont.)
Solution The mass of the unit cell is 1.277×10-21 g (4 × 3.192×10-22 g). The volume of a unit cell is 5.618×10-23 cm3 [(3.83×10-8 cm)3]. Therefore, the density is given by
d = = 22.7 g/cm3
1.277×10-21 g
5.62×10-23 cm3
Think About It Metals typically have high densities, so common sense can help you decide whether or not your calculated answer is reasonable.

49. Electrons are delocalized over the entire crystal.
Types of Crystals
In metallic crystals, every lattice point is occupied by an atom of the same metal.
Electrons are delocalized over
the entire crystal.
Delocalized electrons make metals
good conductors.
Large cohesive force resulting from
delocalization makes metals strong. 49

50. Types of Crystals 50

51. Amorphous Solids
12.5
Amorphous solids lack a regular three-dimensional arrangement of atoms.
Glass is an amorphous solid.
Glass is a fusion product.
SiO2 is the chief component.
Na2O and B2O3 are typically fused with molten SiO2 and allowed to cool without crystallizing.

52. Amorphous Solids

53. Noncrystalline (amorphous) quartz glass
Amorphous Solids
Noncrystalline (amorphous) quartz glass
Crystalline quartz

54. Phase Changes
12.6
A phase is a homogeneous part of a system that is separated from the rest of the system by a well defined boundary.
When a substance goes from one phase to another phase, it has undergone a phase change.
Example Phase Change
Freezing of water H2O(l) → H2O(s)
Evaporation (or vaporization) of water H2O(l) → H2O(g)
Melting (fusion) of ice H2O(s) → H2O(l)
Condensation of water vapor H2O(g) → H2O(l)
Sublimation of dry ice CO2(s) → CO2(g)
Deposition of iodine I2(g) → I2(s)

55. The six possible phase changes55

56. Phase Changes
The boiling point of a substance is defined as the temperature at which its vapor pressure equals the external atmospheric pressure.
The molar heat of vaporization (ΔHvap) is the amount of heat required to vaporize a mole of substance at its boiling point. 56

57. ice ⇌ water H2O(s) ⇌ H2O(l)
Phase Changes
The transformation of a liquid to a solid is called freezing.
The reverse process is called melting, or fusion.
The melting point (freezing point) of a solid (or liquid) is the temperature at which the solid and liquid phases coexist in equilibrium.
ice ⇌ water
H2O(s) ⇌ H2O(l)
In dynamic equilibrium, the forward and reverse process are occurring at the same rate. 57

58. Phase Changes
The molar heat of fusion (ΔHfus) is the energy required to melt 1 mol of a solid. 58

59. Phase Changes Heating curves: Liquid and vapor in equilibrium
Time
Liquid and vapor in equilibrium
Boiling point
Vapor
Melting point
Liquid
Solid and liquid in equilibrium
Solid 59

60. Solid I2 in equilibrium with its vapor
Phase Changes
Sublimation is the process by which molecules go directly from the solid phase to the vapor phase.
Deposition is reverse process of sublimation.
The molar enthalpy of sublimation (ΔHsub) of a substance is the energy required to sublime 1 mole of a solid.
ΔHsub = ΔHfus + ΔHvap
Solid I2 in equilibrium with its vapor 60

61. Worked Example 12.7
(a) Calculate the amount of heat deposited on the skin of a person burned by g of liquid water at 100.0°C and (b) the amount of heat deposited by 1.00 g of steam at 100.0°C. (c) Calculate the amount of energy necessary to warm g of water from 0.0°C to body temperature and (d) the amount of heat required to melt g of ice 0.0°C and then warm it to body temperature. (Assume that body temperature is 37.0°C.)

62. Worked Example 12.7 (cont.)
Strategy For the purpose of following the sign conventions, we can designate the water as the system and the body as the surroundings. (a) Heat is transferred from hot water to the skin in a single step: a temperature change. (b) The transfer of heat from steam to the skin takes place in two steps: a phase change and a temperature change. (c) Cold water is warmed to body temperature in a single step: a temperature change. (d) The melting of ice and the subsequent warming of the resulting liquid water takes place in two steps: a phase change and a temperature change. In each case, the heat transferred during a temperature change depends on the mass of the water, the specific heat of the water, and the change in temperature. For the phase changes, the heat transferred depends on the amount of water (in moles) and the molar heat of vaporization (ΔHvap) or molar heat of fusion (ΔHfus). In each case, the total energy transferred or required is the sum of the energy changes for the individual steps.
The specific heat is J/g∙°C for water and 1.99 J/g∙°C for steam. ΔHvap is kJ/mol and ΔHfus is 6.01 kJ/mol. Note: The ΔHvap of water is the amount of heat required to vaporize a mole of water. However, we want to know how much heat is deposited when water vapor condenses, so we use −40.79 kJ/mol.

63. Worked Example 12.7 (cont.)
Solution (a) ΔT = 37.0°C – 100.0°C = –63.0°C
q = msΔT = 1.00 g × ×–63.0°C
Thus, 1.00 g of water at 100.0°C deposits kJ of heat on the skin. (The negative sign indicates that heat is given off by the system and absorbed by the surroundings.)
(b)
q1 = nΔHvap = mol ×
q2 = msΔT = 1.00 g × ×–63.0°C
The overall energy deposited on the skin by 1.00 g of steam is the sum of q1 and q2:
–2.26 kJ + (–0.264 kJ) = –2.53 kJ
4.184 J
g∙°C
= –2.64×102 J = –0.264 kJ
1.00 g
18.02 g/mol
= mol water
−40.79 kJ
mol
= –2.26 kJ
4.184 J
g∙°C
= –2.64×102 J = –0.264 kJ

64. Worked Example 12.7 (cont.) Solution (c) ΔT = 37.0°C – 0.0°C = 37.0°C
q = msΔT = 1.00 g × ×37.0°C
The energy required to warm g of water from 0.0°C to 37.0°C is 15.5 kJ.
(d)
q1 = nΔHfus = 5.55 mol ×
q2 = msΔT = g × ×37.0°C
The energy rquired to melt g of ice at 0.0°C and warm it to 37.0°C is the sum of q1 and q2:
33.4 kJ kJ = 48.9 kJ
4.184 J
g∙°C
= 1.55×104 J = 15.5 kJ
Think About It In problems that include phase changes, the q values corresponding to the phase-change steps will be the largest contributions to the total. If you find that this is not the case in your solution, check to see if you have made the common error of neglecting to convert the q values corresponding to temperature changes from J to kJ.
100.0 g
18.02 g/mol
= 5.55 mol water
6.01 kJ
mol
= 33.4 kJ
4.184 J
g∙°C
= 1.55×104 J = 15.5 kJ

65. Phase Diagrams
12.6
A phase diagram summarizes the conditions at which a substance exists as a solid, liquid, or gas.
The triple point is the
only combination of
pressure and temperature
where three phases of a
substance exist in
equilibrium.
triple point

66. The phase diagram of water:
Phase Diagrams
The phase diagram of water: 66

67. Worked Example 12.8
Using the following phase diagram, (a) determine the normal boiling point and the normal melting point of the substance, (b) determine the physical state of the substance at 2 atm and 110°C, and (c) determine the pressure and temperature that correspond to the triple point of the substance.
Strategy Each point on the phase diagram corresponds to a pressure-temperature combination. The normal boiling and melting points are the temperatures at which the substance undergoes phase changes. These points fall on the phase boundary lines. The triple point is where the three phase boundaries meet.

68. Worked Example 12.8 (cont.)
Solution By drawing lines corresponding to a given pressure and/or temperature, we can determine the temperature at which a phase change occurs, or the physical state of the substance under specified conditions.
(a)
The normal boiling and melting points are ~140°C and ~205°C, respectively.

69. Worked Example 12.8 (cont.) Solution (b)
At 2 atm and 110°C the substance is a solid.

70. Worked Example 12.8 (cont.) Solution (c)
The triple point occurs at ~0.8 atm and ~115°C.
Think About It The triple point of this substance occurs at a pressure below atmospheric pressure. Therefore, it will melt rather than sublime when it is heated under ordinary conditions.

71. 12 Key Concepts Intermolecular Forces Dipole-Dipole Interactions
Hydrogen Bonding
Dispersion Forces
Ion-Dipole Interactions
Properties of Liquids
Surface Tension
Viscosity
Vapor Pressure
Crystal Structure
Unit Cells
Packing Spheres
Closest Packing
Types of Crystals
Ionic Crystals
Covalent Crystals
Molecular Crystals
Metallic Crystals
Amorphous Solids
Phase Changes
Liquid-Vapor Phase Transition
Solid-Liquid Phase Transition
Solid-Vapor Phase Transition
Phase Diagrams

72. Group Quiz #25
Which member of each pair has stronger intermolecular force (boiling point, heat of vaporization)? Explain your reasoning.
CH4 or CH3CH3
NH3 or NF3
CO2 or SO2
O2 or O3
I2 or Cl2 72 72