Chapter 11
States of
Matter: Liquids
and Solids
Contents and Concepts
1. Comparison of Gases, Liquids, and Solids
Changes of State
2. Phase Transitions
3. Phase Diagrams
11 | 2
Liquid State
1. Properties of Liquids: Surface Tension and
Viscosity
2. Intermolecular Forces; Explaining Liquid
Properties
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Solid State
1. Classification of Solids by Type of Attraction of
Units
2. Crystalline Solids; Crystal Lattices and Unit
Cells
3. Structures of Some Crystalline Solids
4. Calculations Involving Unit-Cell Dimensions
5. Determining Crystal Structure by X-Ray
Diffraction
11 | 4
Learning Objectives
1. Comparison of Gases, Liquids, and Solids
a. Recall the definitions of gas, liquid, and
solid given in Section 1.4.
b. Compare a gas, a liquid, and a solid using a
kinetic molecular theory description.
c. Recall the ideal gas law and the van der
Waals equation for gases (there are no
similar simple equations for liquids and
solids).
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Changes of State
2. Phase Transitions
a. Define change of state (phase transition).
b. Define melting, freezing, vaporization,
sublimation, and condensation.
c. Define vapor pressure.
d. Describe the process of reaching a dynamic
equilibrium that involves the vaporization of
a liquid and condensation of its vapor.
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2. Phase Transitions (con’t)
e. Describe the process of boiling.
f. Define freezing point and melting point.
g. Define heat (enthalpy) of fusion and heat
(enthalpy) of vaporization.
h. Calculate the heat required for a phase
change of a given mass of substance.
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2. Phase Transitions (con’t)
i. Describe the general dependence of the
vapor pressure (ln P) on the temperature
(T).
j. State the Clausius–Clapeyron equation (the
two-point form).
k. Calculate the vapor pressure at a given
temperature.
l. Calculate the heat of vaporization from
vapor pressure.
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3. Phase Diagrams
a. Define phase diagram.
b. Describe the melting-point curve and the
vapor-pressure curves (for the liquid and
the solid) in a phase diagram.
c. Define triple point.
d. Define critical temperature and critical
pressure.
e. Relate the conditions for the liquefaction of
a gas to its critical temperature.
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Liquid State
4. Properties of Liquids; Surface Tension and
Viscosity
a. Define surface tension.
b. Describe the phenomenon of capillary rise.
c. Define viscosity.
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5. Intermolecular Forces; Explaining Liquid
Properties
a. Define intermolecular forces.
b. Define dipole–dipole force.
c. Describe the alignment of polar molecules
in a substance.
d. Define London (dispersion) forces.
e. Note that London forces tend to increase
with molecular mass.
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5. Intermolecular Forces; Explaining Liquid
Properties (con’t)
f. Relate the properties of liquids to the
intermolecular forces involved.
g. Define hydrogen bonding.
h. Identify the intermolecular forces in a
substance.
i. Determine relative vapor pressures on the
basis of intermolecular attractions.
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Solid State
6. Classification of Solids by Type of Attraction
of Units
a. Define molecular solid, metallic solid, ionic
solid, and covalent network solid.
b. Identify types of solids.
c. Relate the melting point of a solid to its
structure.
d. Determine relative melting points based on
types of solids.
e. Relate the hardness and electrical
conductivity of a solid to its structure.
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7. Crystalline Solids; Crystal Lattices and Unit
Cells
a. Define crystalline solid and amorphous
solid.
b. Define crystal lattice and unit cell of a
crystal lattice.
c. Define simple cubic unit cell, body-centered
cubic unit cell, and face-centered cubic unit
cell.
d. Determine the number of atoms in a unit
cell.
e. Describe the two kinds of crystal defects.
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8. Structures of Some Crystalline Solids
a. Define hexagonal close-packed structure
and cubic close-packed structure.
b. Define coordination number.
c. Note the common structures (face-centered
cubic and body-centered cubic) of metallic
solids.
d. Describe the three types of cubic structures
of ionic solids.
e. Describe the covalent network structure of
diamond and graphite.
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9. Calculations Involving Unit-Cell Dimensions
a. Calculate atomic mass from unit-cell
dimension and density.
b. Calculate unit-cell dimension from unit-cell
type and density.
10. Determining Crystal Structure by X-Ray
Diffraction
a. Describe how constructive and destructive
interference give rise to a diffraction
pattern.
b. Note that diffraction of x rays from a crystal
gives information about the positions of
atoms in the crystal.
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Comparing Gases, Liquids, and Solids
Gases are compressible fluids.
Liquids are relatively incompressible fluids.
Solids are nearly incompressible and rigid.
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A change of state or phase transition is a change
of a substance from one state (solid, liquid, gas) to
another. The specific name or names for each of
these transitions are given below.
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The vapor pressure of a liquid at a particular
temperature is the partial pressure of the vapor
over the liquid measured at equilibrium.
When a liquid is placed in a closed vessel, the
partial pressure of its vapor increases over time
until it reaches equilibrium. At equilibrium,
evaporation and condensation continue to occur,
but do so at the same rate. This situation, which is
called a dynamic equilibrium, is illustrated on the
next slide.
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The vapor pressure depends on the liquid and on
the temperature. This relationship is illustrated on
the next slide for four substances. Note that as
temperature increases, vapor pressure increases.
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The boiling point is the
temperature at which the
vapor pressure is equal to
the pressure on the liquid,
usually atmospheric
pressure. At this
temperature, bubbles of gas
form within the liquid, as
illustrated in the figure. The
normal boiling point is
measured at 1 atmosphere
pressure.
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When the pressure on the liquid increases, as is
the case with a pressure cooker, the boiling point
increases.
Conversely, when the pressure on the liquid
decreases, as is the case at high altitude, the
boiling point decreases.
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The freezing point is the temperature at which a
pure liquid changes to a crystalline solid (or
freezes).
The melting point is the temperature at which a
crystalline solid changes to a liquid (or melts).
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Any change of state requires that energy be added
to or removed from the system. As shown in the
graph, for water, there are two regions that remain
at the same temperature even as heat is added.
Those represent regions of phase change.
11 | 26
The heat of fusion, DHfus, is the heat needed to
melt a solid. It is given in kJ/mol.
For water, the phase change is represented by
H2O(s)  H2O(l); DHfus = 6.01 kJ/mol
The heat of vaporization, DHvap, is the heat
needed to vaporize a liquid. It is given in kJ/mol.
For water, the phase change is represented by
H2O(l)  H2O(g); DHvap = 40.7 kJ/mol
11 | 27
?
Shown to the right is a
representation of a closed
container in which you
have just placed 10 L of
H2O. This is the starting
point in time, t = 0.
Assume that all of the
water is in the liquid
phase at this point. The
dots represent a few of
the H2O molecules.
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Consider a time t = 1, when some time has passed
but the system has not reached equilibrium.
1. How will the level of the liquid H2O
compare to that at t =0?
2. How will the vapor pressure in the flask
compare to that at t = 0?
3. How will the number of H2O molecules in
the vapor state compare to that at t = 0?
4. How does the rate of evaporation in this
system compare to the rate of
condensation?
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Time t = 1:
1. The liquid level will be lower than at t = 0.
2. The vapor pressure will be higher than at
t = 0.
3. There will be more molecules of H2O in the
vapor state than at t = 0.
4. The rate of evaporation is greater than the
rate of condensation.
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Consider a time t = 2, when enough time has
passed for the system to reach equilibrium.
1. How will the level of the liquid H2O
compare to that at t = 1?
2. How will the vapor pressure in the flask
compare to that at t = 1?
3. How will the number of H2O molecules in
the vapor state compare to that at t = 1?
4. How does the rate of evaporation in this
system compare to the rate of
condensation?
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Time t = 2 (the system is at equilibrium):
1. The liquid level will be lower than at t = 1.
2. The vapor pressure will be higher than at
t = 1.
3. There will be more molecules of H2O in the
vapor state than at t = 1.
4. The rate of evaporation is equal to the rate
of condensation.
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When vapor pressure is plotted versus
temperature, a curve is obtained.
When the natural log (ln P) of the vapor pressure
is plotted against the reciprocal of temperature
(1/T), a straight-line graph is obtained.
The next slide shows this graph for four liquids.
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The Clausius–Clapeyron equation describes these
graphs:
 P2  ΔH v ap  1
1
 

ln  
R  T1 T2 
 P1 
R = 8.314510 J/(mol K)
DHvap must be given in J/mol
This equation can be used to find the vapor
pressure, the heat of vaporization, or the
temperature.
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?
The fuel requirements of some homes
are supplied by propane gas, C3H8,
which is contained as a liquid in steel
cylinders. If a home uses 2.40 kg of
propane in an average day, how much
heat must be absorbed by the propane
cylinder each day to evaporate the
liquid propane, forming the gas that is
subsequently burned? The heat of
vaporization of propane is 16.9 kJ/mol.
11 | 36
We want to know how much heat it takes to evaporate
2.40 kg of propane, which has a heat of vaporization of
16.9 kJ/mol.
We will convert kilograms to grams, then grams to moles,
and finally moles to kilojoules.
103 g 1 mol 16.9 kJ
2.40 kg 



1kg 44.10 g
mol
9.20 × 102 kJ
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?
The vapor pressure of diethyl ether
(commonly known as simply ether) is
439.8 mmHg at 20.0°C. The heat of
vaporization of ether is 29.2 kJ/mol.
What is the vapor pressure of ether at
34.0°C?
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This problem uses the Clausius–Clapeyron
equation. When setting up your variables, it
simplifies your math if you let P2 be the unknown
vapor pressure.
P1 = 439.8 mmHg
T1 = 20.0°C
P2 = ?
T2 = 34°C
DHvap = 29.2 kJ/mol
 P2  ΔH v ap
ln  
R
 P1 
 1
1
 

 T1 T2 
Units are critical when working with this equation:
Temperatures must be in K, and DHvap in J/mol.
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Δ
H
P1 = 439.8 mmHg
v ap
 P2 
ln  
T1 = 293°C
R
 P1 
P2 = ?
T2 = 307°C
DHvap = 29.2 x 103 J/mol
 1
1
 

 T1 T2 
29.2  10 3 J


 1
P2
1 
mol
 


ln

8.315 J  293 K 307 K 
 439.8 mmHg 
mol  K
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Solve the right side first:
29.2  10 J


 1
P
1 
mol
 


ln

8.315 J  293 K 307 K 
 439.8 mmHg 
mol  K
3
2
x or
Use
e


P2
  0.5466 inverse ln
ln
 439.8 mmHg 


P2

  1.727
 439.8 mmHg 
P2 = 760 mmHg
(3 significant figures)
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A phase diagram is a graphical way to summarize
the conditions under which the various states of a
substance are stable.
11 | 42
In a phase diagram, phases are separated by lines
that represent equilibrium between those phases.
The triple point is the point where all three phases
are in equilibrium.
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The phase diagram for
water illustrates the
components described on
the previous slide.
The slope of the AB line,
which shows the
equilibrium between solid
and liquid, slants to the
left. This is because the
density of the liquid is
greater than the density of
the solid.
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The phase diagram for
carbon dioxide is given
at right. Note that the
AB line slants to the
right.
At STP, carbon dioxide
is never a liquid. That is
why carbon dioxide
sublimes directly from
solid to gas.
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The phase diagram
for sulfur shows some
interesting
characteristics. The
existence of two
different solid
structures for sulfur,
rhombic and
monoclinic, gives rise
to more than one
triple point.
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The critical point gives the temperature (critical
temperature, TC) at which the liquid state can no longer
exist and the pressure at that temperature (critical
pressure). Above this temperature and pressure, there is
only one state, a supercritical fluid. This is illustrated below
for CO2, for which TC = 31°C.
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At 3200 m, the pressure on the liquid is less than
at sea level, which means the boiling point is
lower. The eggs will need to be cooked longer at
the higher altitude.
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Now we will look at some properties of liquids:
surface tension and viscosity.
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Surface tension is the energy required to increase
the surface area of a liquid by a unit amount. The
values are given in J/m2.
Surface tension arises
because the molecules at
the surface of a liquid
experience a net force
toward the center of the
liquid.
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Surface tension is the
property that allows
insects to walk on the
surface of water or a pin
to float.
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Surface tension is also related to capillary rise. In
this phenomenon, when a very-small-diameter
glass tube (a capillary) is placed in water, the
water rises.
The water is attracted to the glass, so a thin film
moves up the surface of the glass. This action
expands the surface area, so the water column
rises to counteract it. The process then repeats
itself.
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Surface tension also explains the development of
a meniscus, the curvature of the surface of a liquid
in a column.
When the liquid is attracted to the glass, the
curvature is concave, as is the case with water.
When the liquid is not attracted to the glass, the
curvature is convex, as is the case with mercury.
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Viscosity is the resistance
to flow exhibited by liquids
and gases.
This property differentiates
the flow of syrup from the
flow of water: Syrup is
more viscous than water.
In the figure, steel balls
were dropped into the
glycerol and the water at
the same time.
glycerol
water
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To explain the properties of liquids that we’ve just
studied, we need to learn about intermolecular
forces—that is, the forces of interaction between
molecules.
11 | 58
We can learn about intermolecular forces by
studying heats of vaporization.
Neon is normally a gas, but it liquefies at a
temperature of –246°C at 1 atm of pressure. The
heat of vaporization for liquid neon is 1.77 kJ/mol.
The DHvap for neon is in part the energy needed to
push back the atmosphere when the vapor forms
(0.23 kJ/mol) and in part the energy needed to
overcome intermolecular attractions (1.54 kJ/mol).
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Three types of intermolecular forces exist between
neutral molecules or atoms: dipole-dipole forces,
London forces, and hydrogen bonding forces.
11 | 60
The term van der Waals forces is a general term that
includes both dipole-dipole forces and London forces.
Polar molecules exhibit dipole-dipole forces that result in
alignment of the molecules. This phenomenon is illustrated
below for solid and liquid HCl.
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In nonpolar molecules, there is no dipole-dipole
force, yet there is still a force of attraction. In 1930,
Fritz London explained this relationship with what
we now call the London forces.
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In a nonpolar molecule, the
charge is uniformly distributed
over time. But in any one
instant, the charge is not
uniformly distributed. In that
instant, there is an
instantaneous dipole. To see
how this works, let’s look at a
neon atom.
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The instantaneous dipole of the neon atom
induces an instantaneous dipole in adjacent
atoms, resulting in an attractive force between
them.
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In the next moments, the electrons move, and the
instantaneous dipoles change. However, the
instantaneous dipoles tend to change together,
maintaining the attractive force between them.
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London forces are the weak attractive forces
between molecules resulting from the small,
instantaneous dipoles that occur because of the
varying positions of the electrons during their
motion about the nuclei.
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All molecules exhibit London forces.
London forces increase with increasing mass
(atomic number) because the presence of more
electrons causes a stronger instantaneous dipole.
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London forces are also larger for more polarizable
(more easily distorted) molecules.
On the next slide, each compound has the same
chemical formula, C5H12. However, in each
compound, the carbon atoms are bonded in a
different structure. Note how the heats of
vaporization differ for the molecules.
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The straight-chain compound has the strongest
intermolecular forces and the highest heat of
vaporization because it is most flexible and,
therefore, most polarizable.
The most compact compound has the weakest
intermolecular forces and the lowest heat of
vaporization.
11 | 70
We have just related heat of vaporization to the
strength of intermolecular forces: The stronger the
intermolecular forces, the higher the heat of
vaporization.
We can do the same for vapor pressure, boiling
point, surface tension, and viscosity.
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The stronger the intermolecular forces, the lower
the vapor pressure at any given temperature.
The stronger the intermolecular forces, the higher
the boiling point.
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The stronger the intermolecular forces, the higher
the surface tension.
Viscosity depends in part on intermolecular forces:
The stronger the intermolecular forces, the more
viscous the liquid.
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Let’s look at Table 11.2 again.
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The compounds are listed in order of increasing
molar mass. We would expect that the vapor
pressures would decrease from top to bottom, but
this is not the case: The vapor pressures of water
and glycerol are much lower than anticipated.
How can this be explained?
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Let’s look at the structures of water, glycerol, and
methanol. These compounds have unusually low
vapor pressures and unusually high boiling points.
What do these molecules have in common?
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Each molecule has one or more –OH group.
Molecules with this group have an additional force
called the hydrogen bonding force.
11 | 77
Hydrogen bonding is a weak attractive force that
exists between hydrogen atoms bonded to a very
electronegative atom, X, and a lone pair of
electrons on another small, electronegative atom,
Y.
—X—H . . . Y—
Most often, X is F, O, or N, the smallest, most
electronegative elements.
11 | 78
Hydrogen bonding in water is illustrated here in
three different formats: ball-and-stick model,
electrostatic map showing electron density
distributions, and Lewis formula.
11 | 79
We can see the
impact of adding
hydrogen bonding
forces on the boiling
points of Group VI
hydrides.
11 | 80
The boiling points for
Group IVA, VA, and
VIIA hydrides are
graphed to the right.
Group IVA hydrides
follow the expected
pattern. In Group
VA, NH3 does not.
Likewise, in Group
VIIA, HF does not.
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11 | 82
a. There are several ways to determine this.
First, the products in the first reaction form an
explosive mix.
Second, the volume of gas from the first
reaction is three times that of the second
reaction.
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b. The first reaction requires breaking bonds. The
second reaction is a phase change that
involves disrupting intermolecular forces.
Because chemical bonds are much stronger
than intermolecular forces, the enthalpy of the
first reaction would be much higher than that of
the second reaction.
c. The enthalpy change for the first reaction could
be calculated using standard enthalpy of
formation values (Chapter 6) or bond energies
(Chapter 9).
11 | 84
?
Identify the intermolecular forces that
you would expect for each of the
following substances:
a. O2
b. H2O2
c. CHBr3
a. Nonpolar molecule: London forces
b. Polar molecule with O—H bond: London forces,
dipole-dipole forces, hydrogen bonding
c. Polar molecule: London forces, dipole-dipole
forces
11 | 85
A solid is a nearly incompressible state of matter
with a well-defined shape, because the units
(atoms, molecules, ions) making up the solid are in
close contact and reside in fixed positions or sites.
11 | 86
We will now look at solids in more detail. We will
first identify the basic unit that makes up the solid
and the forces between those units. We will then
use this information to classify solids into one of
four types: molecular solid, ionic solid, metallic
solid, or covalent network solid.
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11 | 88
A molecular solid consists of atoms or molecules.
It is held together by intermolecular forces.
A metallic solid consists of the positive cores of
metal atoms. It is held together by metallic
bonding, a “sea” of delocalized electrons.
11 | 89
An ionic solid is composed of cations and anions.
It is held together by ionic bonds, the electrical
attractions between oppositely charged particles.
A covalent network solid consists of atoms.
These atoms are held together in large chains or
networks by covalent bonds.
11 | 90
Carbon exists as a covalent network solid in both
the graphite and diamond forms.
11 | 91
Physical properties such as melting point,
hardness, and electrical conductivity are related to
solid structure.
Table 11.1 gives melting points and boiling points
for various solids, indicating the solid structure of
each.
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11 | 93
In general, molecular solids have lower melting
points, and covalent network solids and ionic
solids have high melting points.
The melting points of ionic solids vary owing to
differences in lattice energy that reflect the charge
and sizes of the ions involved.
The melting points of metals vary widely.
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Hardness depends on how easily the structural
units can be moved relative to one another.
Molecular solids tend to be soft. By contrast, threedimensional covalent network solids, such as
diamond, are very hard.
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Ionic compounds are brittle because they tend to
fracture easily along crystal planes.
Metals are malleable, so that they can be easily
shaped by hammering.
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Metals are good electrical conductors because of
their delocalized valence electrons.
Ionic solids do not conduct electricity. When
melted, however, they do conduct electricity. They
also conduct electricity when they are dissolved in
water.
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11 | 98
Solids can also be described as either crystalline
or amorphous.
A crystalline solid is composed of one or more
crystals with a well-defined ordered structure in
three dimensions.
11 | 99
An amorphous solid has a disordered structure
and lacks a well-defined arrangement of basic
units.
A glass is an amorphous solid obtained by cooling
a liquid rapidly enough that its basic units are
“frozen” in random positions before they can
assume a crystalline arrangement.
11 | 100
The ordered structure of a crystal is described in
terms of its crystal lattice. A crystal lattice is the
geometric arrangement of lattice points of a crystal
in which we choose one lattice point at the same
location within each of the basic units of the
crystal.
11 | 101
The crystal
structure and
lattice for copper
is shown here.
The copper
atoms have
been shrunk to
make the
structure more
visible.
11 | 102
A unit cell is the smallest boxlike unit from which
you can imagine constructing a crystal by stacking
the units in three dimensions.
The shapes of the unit cells for the seven different
crystal systems are shown on the next slide. Each
unit cell features a different angle between the
faces and relationship between the dimensions of
the faces.
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11 | 104
11 | 105
We will now look in more detail at the cubic crystal
system.
There are three possible unit cells: simple cubic,
body-centered cubic, and face-centered cubic.
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11 | 107
We have assumed that crystals have perfect order.
In fact, real crystals have defects or imperfections.
Ruby is one example. It is primarily Al2O3, but
some Al3+ ions are replaced with Cr3+ ions, giving
the red color.
11 | 108
We now look in more detail at the structures of
several solids of different types: molecular,
metallic, ionic, and covalent network.
11 | 109
Molecular solids exhibit cubic and hexagonal
close-packed structures.
11 | 110
Closest packing of spheres begins with one layer,
A, followed by layer B over spaces in layer A. The
next layer can be either layer C or a repeat of layer
A.
11 | 111
When the pattern is ABABAB, a hexagonal closepacked (hcp) structure results.
When the pattern is ABCABC, a cubic closepacked structure results. This is identical to a
structure with a face-centered cubic (fcc) unit cell.
11 | 112
In any close-packed structure, each atom is
surrounded by other atoms. The number of
nearest neighbors is called the coordination
number.
11 | 113
Many metallic solids are close-packed crystals. In
a body-centered cubic (bcc) lattice, 68% of the
space is occupied by atoms. In a close-packed
(cp) structure, 74% of the space is occupied by
atoms.
The table on the next slide identifies the type of
solid each metal forms.
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11 | 115
The crystal structure of ionic solids is complicated
by the fact that we need to give both cations and
anions positions within the crystal structure.
We will examine the crystal structure of three
compounds with the general formula MX: CsCl,
NaCl, and ZnS.
11 | 116
In CsCl, the cation and the anion have
approximately the same size.
Two models of the unit cell are shown on the next
slide: a space-filling model (left) and a model with
the ions shrunk in size to emphasize the structure
(right).
11 | 117
The crystal structure has a Cs+ ion in the center
with a Cl- ion at each corner. We could also take
the unit cell to have a Cl- ion at the center and Cs+
ions at each corner. This gives a total of one Cs+
ion and one Cl- ion (8 corners times 1/8 each) per
unit cell.
11 | 118
In NaCl, the Cl- ions are much larger than the Na+
ions. As a result, the Cl- ions almost touch one
another.
11 | 119
The crystal structure involves two overlapping
lattices. Each has one ion at each corner and on
each face. This gives a total of four of each ion per
unit cell: 8 corners times 1/8 ion each + 6 faces
times ½ ion each per unit cell.
11 | 120
Zinc sulfide is polymorphic—that is, it crystallizes
into two different structures. The hexagonal
structure is called wurtzite; the cubic structure is
called zinc blende or sphalerite. We will discuss
the cubic structure.
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Each corner, each face,
and the center has a S2ion. There are also four
Zn2+ ions completely
within the unit cell. This
gives a total of four Zn2+
ions and four S2- ions (8
corners times 1/8 Cleach + 6 faces times ½
Cl- each) per unit cell.
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The structure of a covalent network solid is
determined primarily by the direction of the
covalent bonds.
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The structure of
diamond is similar to
that of ZnS, with a
carbon atom replacing
each Zn2+ position (dark
spheres) and each S2position (light spheres).
Each carbon is bonded
to four other carbons
and has a tetrahedral
geometry.
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The structure of graphite consists of large, flat
sheets of carbon atoms covalently bonded to form
hexagonal arrays. These sheets are stacked on
top of each other. They slide over one another
easily, making graphite a good lubricant.
The graphite structure has several possible
resonance formulas and, therefore, delocalized
electrons. That property makes graphite a good
conductor.
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The figure on the left shows one resonance
structure for graphite. The figure on the right
shows an electron micrograph of graphite in which
the layer structure is visible.
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The large atoms, A, occupy each corner and each
face. Eight corners times ⅛ atom each plus six
faces × ½ atom each gives a total of four A
atoms. The four small atoms, B, are completely
contained within the unit cell. With four A and
four B, the formula is AB.
Because the four B atoms are completely within
the unit cell, the unit cell is determined by the
arrangement of A atoms, which is facecentered cubic (fcc).
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Knowing the unit cell type and its dimensions allows
several calculations:
•
•
•
Using the cell’s dimensions, you can calculate its
volume.
From the density and volume of the crystal, you
can calculate its mass.
Using the mass of the crystal and the type of unit
cell, you can calculate the mass of an atom.
Alternately, from the density and the unit cell structure, you
can calculate the length of the unit cell edges.
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Determining Crystal Structure by X-Ray
Diffraction
The ordered structure of crystal planes can reflect x rays in
a diffraction pattern. This pattern can be analyzed to
determine the positions of the atoms in the unit cell.
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