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Chapter 12
Intermolecular Forces:
Liquids, Solids, and Phase Changes
12-1
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Intermolecular Forces:
Liquids, Solids, and Phase Changes
12.1 An Overview of Physical States and Phase Changes
12.2 Quantitative Aspects of Phase Changes
12.3 Types of Intermolecular Forces
12.4 Properties of the Liquid State
12.5 The Uniqueness of Water
12.6 The Solid State: Structure, Properties, and Bonding
12-2
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ATTRACTIVE FORCES
electrostatic in nature
Intramolecular forces
bonding forces
These forces exist within each molecule.
They influence the chemical properties of the substance.
Intermolecular forces
nonbonding forces
These forces exist between molecules.
They influence the physical properties of the substance.
12-3
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Phase Changes
exothermic
sublimination
vaporizing
melting
solid
liquid
freezing
endothermic
12-4
gas
condensing
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Table 12.1
A Macroscopic Comparison of Gases, Liquids, and Solids
State
Shape and Volume
Compressibility Ability to Flow
Gas
Conforms to shape and volume
of container
high
high
Liquid
Conforms to shape of container;
volume limited by surface
very low
moderate
Solid
Maintains its own shape and
volume
almost none
almost none
12-5
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Figure 12.1
Heats of vaporization and fusion for several common substances.
12-6
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Figure 12.2
12-7
Phase changes and their enthalpy changes.
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Figure 12.3
A cooling curve for the conversion of gaseous water to ice.
12-8
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Quantitative Aspects of Phase Changes
Within a phase, a change in heat is accompanied by a change in
temperature which is associated with a change in average Ek as
the most probable speed of the molecules changes.
q = (amount)(molar heat capacity)(T)
During a phase change, a change in heat occurs at a constant
temperature, which is associated with a change in Ep, as the
average distance between molecules changes.
q = (amount)(enthalpy of phase change)
12-9
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Figure 12.4
12-10
Liquid-gas equilibrium.
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Figure 12.5
12-11
The effect of temperature on the distribution of
molecular speed in a liquid.
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Figure 12.6
Vapor pressure as a function of
temperature and intermolecular forces.
12-12
Figure 12.7
A linear plot of vapor
pressure- temperature
relationship.
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The Clausius-Clapeyron Equation
ln P =
-Hvap 
1 
   C
R
T 
P2
-Hvap  1
1 
ln
=
  
R
P1
T2 T1
12-13
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SAMPLE PROBLEM 12.1
The vapor pressure of ethanol is 115 torr at 34.90C. If Hvap of
ethanol is 40.5 kJ/mol, calculate the temperature (in 0C) when
the vapor pressure is 760 torr.
PROBLEM:
PLAN:
Using the Clausius-Clapeyron Equation
We are given 4 of the 5 variables in the Clausius-Clapeyron
equation. Substitute and solve for T2.
SOLUTION:
P2
-Hvap  1
1 
ln
=
  
P1
R T2 T1
760 torr
ln
115 torr
=
-40.5 x103 J/mol
8.314 J/mol*K
T2 = 350K = 770C
12-14
34.90C = 308.0K
1
1
T2
308K
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Figure 12.8
Phase diagrams for CO2 and H2O.
CO2
12-15
H 2O
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Figure 12.9
Covalent and van der Waals radii.
van der Waal’s distance
bond length
covalent radius
van der Waal’s radius
12-16
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Figure 12.10
Periodic trends in covalent and van der Waals radii (in pm).
12-17
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12-18
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12-19
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Figure 12.11
Polar molecules and dipole-dipole forces.
solid
liquid
12-20
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THE HYDROGEN BOND
a dipole-dipole intermolecular force
A hydrogen bond may occur when an H atom in a molecule,
bound to small highly electronegative atom with lone pairs of
electrons, is attracted to the lone pairs in another molecule.
The elements which are so electronegative are N, O, and F.
H
hydrogen bond
acceptor
..
O
..
O
..
..
..
..
F
..
hydrogen bond
donor
hydrogen bond
acceptor
hydrogen bond
acceptor
12-21
H
..
..
N
..
F
..
hydrogen bond
donor
H
..
N
hydrogen bond
donor
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Figure 12.12
12-22
Dipole moment and boiling point.
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SAMPLE PROBLEM 12.2
PROBLEM:
Which of the following substances exhibits H bonding? For
those that do, draw two molecules of the substance with the H
bonds between them.
O
(a)
PLAN:
12-23
C2H6
(b) CH3OH
(c)
CH3C NH2
Find molecules in which H is bonded to N, O or F. Draw H
bonds in the format -B:
H-A-.
SOLUTION:
(b)
Drawing Hydrogen Bonds Between Molecules
of a Substance
(a) C2H6 has no H bonding sites.
H
H C O H
H
H
H O C H
H
H
(c)
H
O
H N CH3C CH C
3
N H
O
CH3C N H
H N
CH3C
O
H
H
O
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Figure 12.13
12-24
Hydrogen bonding and boiling point.
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Polarizability and Charged-Induced Dipole Forces
distortion of an electron cloud
•Polarizability increases down a group
size increases and the larger electron clouds are further
from the nucleus
•Polarizability decreases left to right across a period
increasing Zeff shrinks atomic size and holds the electrons
more tightly
•Cations are less polarizable than their parent atom
because they are smaller.
•Anions are more polarizable than their parent atom
because they are larger.
12-25
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Figure 12.14
Dispersion forces among nonpolar molecules.
separated
Cl2
molecules
12-26
instantaneous
dipoles
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Figure 12.15
Molar mass and boiling point.
12-27
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Figure 12.16
Molecular shape and boiling point.
fewer points for
dispersion
forces to act
more points for
dispersion
forces to act
12-28
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SAMPLE PROBLEM 12.3
PROBLEM:
Predicting the Type and Relative Strength of
Intermolecular Forces
For each pair of substances, identify the dominant
intermolecular forces in each substance, and select the
substance with the higher boiling point.
(a) MgCl2 or PCl3
(b) CH3NH2 or CH3F
(c) CH3OH or CH3CH2OH
PLAN:
CH3
(d) Hexane (CH3CH2CH2CH2CH2CH3)
CH3CCH2CH3
or 2,2-dimethylbutane
CH3
Use the formula, structure and Table 2.2 (button).
•Bonding forces are stronger than nonbonding(intermolecular) forces.
•Hydrogen bonding is a strong type of dipole-dipole force.
•Dispersion forces are decisive when the difference is molar mass or
molecular shape.
12-29
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SAMPLE PROBLEM 12.3
Predicting the Type and Relative Strength of
Intermolecular Forces
continued
SOLUTION:
(a) Mg2+ and Cl- are held together by ionic bonds while PCl3 is covalently
bonded and the molecules are held together by dipole-dipole interactions. Ionic
bonds are stronger than dipole interactions and so MgCl2 has the higher boiling
point.
(b) CH3NH2 and CH3F are both covalent compounds and have bonds which are
polar. The dipole in CH3NH2 can H bond while that in CH3F cannot. Therefore
CH3NH2 has the stronger interactions and the higher boiling point.
(c) Both CH3OH and CH3CH2OH can H bond but CH3CH2OH has more CH for
more dispersion force interaction. Therefore CH3CH2OH has the higher boiling
point.
(d) Hexane and 2,2-dimethylbutane are both nonpolar with only dispersion
forces to hold the molecules together. Hexane has the larger surface area,
thereby the greater dispersion forces and the higher boiling point.
12-30
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Figure 12.17
Summary diagram for analyzing the intermolecular forces in a sample.
INTERACTING PARTICLES
(atoms, molecules, ions)
ions present
ions only
IONIC BONDING
(Section 9.2)
ions not present
polar molecules only
DIPOLE-DIPOLE
FORCES
ion + polar molecule
ION-DIPOLE FORCES
nonpolar
molecules only
DISPERSION
FORCES only
H bonded to
N, O, or F
HYDROGEN
BONDING
polar + nonpolar
molecules
DIPOLEINDUCED DIPOLE
FORCES
DISPERSION FORCES ALSO PRESENT
12-31
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Figure 12.18
The molecular basis of surface tension.
hydrogen bonding
occurs across the surface
and below the surface
the net vector
for attractive
forces is downward
hydrogen bonding
occurs in three
dimensions
12-32
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Table 12.3
Surface Tension and Forces Between Particles
Surface Tension
Substance
Formula
(J/m2) at 200C
diethyl ether
CH3CH2OCH2CH3
1.7x10-2
dipole-dipole; dispersion
ethanol
CH3CH2OH
2.3x10-2
H bonding
butanol
CH3CH2CH2CH2OH
2.5x10-2
H bonding; dispersion
H2O
7.3x10-2
H bonding
Hg
48x10-2
metallic bonding
water
mercury
12-33
Major Force(s)
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Figure 12.19
Shape of water or mercury meniscus in glass.
capillarity
stronger
cohesive forces
adhesive forces
H 2O
12-34
Hg
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Table 12.4 Viscosity of Water at Several Temperatures
viscosity - resistance to flow
Temperature(0C)
Viscosity
(N*s/m2)*
20
1.00x10-3
40
0.65x10-3
60
0.47x10-3
80
0.35x10-3
*The units of viscosity are newton-seconds per square meter.
12-35
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Figure 12.20
The H-bonding ability of the water molecule.
hydrogen bond donor
hydrogen bond acceptor
12-36
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The Unique Nature of Water
•great solvent properties due to polarity and
hydrogen bonding ability
•exceptional high specific heat capacity
•high surface tension and capillarity
•density differences of liquid and solid states
12-37
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Figure 12.21
12-38
The hexagonal structure of ice.
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Figure 12.22
12-39
The striking beauty of crystalline solids.
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Figure 12.23
The crystal lattice and the unit cell.
lattice point
unit
cell
unit
cell
portion of a 3-D lattice
12-40
portion of a 2-D lattice
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Figure 12.24 (1 of 3)
The three cubic unit cells.
Simple Cubic
1/8 atom at
8 corners
Atoms/unit cell = 1/8 * 8 = 1
coordination number = 6
12-41
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Figure 12.24 (2 of 3)
The three cubic unit cells.
Body-centered
Cubic
1/8 atom at
8 corners
1 atom at
center
Atoms/unit cell = (1/8*8) + 1 = 2
coordination number = 8
12-42
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Figure 12.24 (3 of 3)
The three cubic unit cells.
Face-centered
Cubic
1/8 atom at
8 corners
1/2 atom at
6 faces
coordination number = 12
12-43
Atoms/unit cell = (1/8*8)+(1/2*6) = 4
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Figure 12.26
Packing of spheres.
simple cubic
(52% packing efficiency)
body-centered cubic
(68% packing efficiency)
12-44
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Figure 12.26 (continued)
layer a
layer b
hexagonal
closest
packing
layer a
cubic closest
packing
layer c
closest packing of first
and second layers
abab… (74%)
hexagonal
unit cell
12-45
expanded
side views
abcabc… (74%)
face-centered
unit cell
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SAMPLE PROBLEM 12.4 Determining Atomic Radius from Crystal Structure
PROBLEM:
PLAN:
Barium is the largest nonradioactive alkaline earth metal. It has
a body-centered cubic unit cell and a density of 3.62 g/cm3.
What is the atomic radius of barium?
(Volume of a sphere: V = 4/3pr3)
We can use the density and molar mass to find the volume of 1 mol of
Ba. Since 68%(for a body-centered cubic) of the unit cell contains
atomic material, dividing by Avogadro’s number will give us the volume
of one atom of Ba. Using the volume of a sphere, the radius can be
calculated.
density of Ba (g/cm3)
radius of a Ba atom
reciprocal divided by M
volume of 1 mol Ba metal
V = 4/3pr3
volume of 1 Ba atom
multiply by packing efficiency
volume of 1 mol Ba atoms
12-46
divide by Avogadro’s number
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SAMPLE PROBLEM 12.4 Determining Atomic Radius from Crystal Structure
continued
SOLUTION:
Volume of Ba metal =
37.9 cm3/mol Ba
26 cm3
mol Ba atoms
r3
x
3.62 g
x 0.68
x
137.3 g Ba
mol Ba
= 37.9 cm3/mol Ba
= 26 cm3/mol Ba atoms
mol Ba atoms
6.022x1023 atoms
= 4.3x10-23 cm3/atom
-23
3
3V
3(4.3x10
cm
) = 2.2 x 10-8cm
r =3
3
4p
4 x 3.14
= 3V/4p

12-47
1 cm3
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Figure 12.27
Diffraction of x-rays by crystal planes.
12-48
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Figure 12.28
Formation of an x-ray diffraction pattern of the
protein hemoglobin.
12-49
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Figure 12.29
Cubic closest packing for
frozen argon.
12-50
Figure 12.30
Cubic closest packing
of frozen methane.
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Table 12.5
Particles
Atomic
Atoms
Molecular Molecules
Characteristics of the Major Types of Crystalline Solids
Interparticle
Forces
Dispersion
Dispersion,
dipole-dipole,
H bonds
Physical
Behavior
Examples (mp,0C)
Soft, very low mp, poor
thermal & electrical
conductors
Fairly soft, low to moderate
mp, poor thermal &
electrical conductors
Group 8A(18)
[Ne-249 to Rn-71]
Nonpolar - O2[-219],
C4H10[-138], Cl2
[-101], C6H14[-95]
Polar - SO2[-73],
CHCl3[-64], HNO3[42], H2O[0.0]
Ionic
Positive & Ion-ion
negative ions attraction
Hard & brittle, high mp,
good thermal & electrical
conductors when molten
NaCl [801]
CaF2 [1423]
MgO [2852]
Metallic
Atoms
Metallic bond
Soft to hard, low to very
high mp, excellent thermal
and electrical conductors,
malleable and ductile
Na [97.8]
Zn [420]
Fe [1535]
Network
Atoms
Covalent bond Very hard, very high mp,
usually poor thermal and
electrical conductors
12-51
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Figure 12.31
The sodium chloride structure.
expanded view
12-52
space-filling
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Figure 12.32
cubic closest packing
12-53
Crystal structures of metals.
hexagonal closest packing
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12-54
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Figure 12.33
12-55
Crystalline and amorphous silicon dioxide.
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Figure 12.35
12-56
The band of molecular orbitals in lithium metal.
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Figure 12.36
Electrical conductivity in a conductor, semiconductor, and insulator.
conductor
insulator
semiconductor
12-57
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Figure 12.37
The levitating power of a superconducting oxide.
rare earth magnet
superconducting ceramic disk
liquid nitrogen
12-58