Intermolecular Forces:
Liquids and Solids
● Phases and Phase Diagrams
● Liquids and Liquid Properties
● Intermolecular Forces
● Heating Curves
● Introduction to Solids
● Cubic Packing Arrangements
● Closest-Packed Structures
● Density of a Crystalline Solid
● Ionic Solids and Interstitial Sites
● The Born-Haber Cycle
Intermolecular Forces:
Liquids and Solids
● Phases and Phase Diagrams
● Liquids and Liquid Properties
● Intermolecular Forces
● Heating Curves
● Introduction to Solids
● Cubic Packing Arrangements
● Closest-Packed Structures
● Density of a Crystalline Solid
● Ionic Solids and Interstitial Sites
● The Born-Haber Cycle
Phases and Phase Diagrams
very compressible
d ≈ 1 – 10 g L−1 at SATP
condensation
vaporization
deposition
sublimation
fusion
incompressible
d ≈ 1 – 10 g mL−1
freezing
Phases and Phase Diagrams
A few definitions:
STP vs SATP
STP = Standard Temperature and Pressure (0 oC, 100 kPa)
SATP = Standard Ambient Temperature and Pressure (25 oC, 100 kPa)
100 kPa = 1 bar
Phases and Phase Diagrams
Differences between the different states of Matter:
Gas: Molecules move randomly and the intermolecular separations are
large (i.e. most of a gas is empty space).
Liquids and Solids: the molecular motions are quite restricted and the
intermolecular separations are small.
Solids: The molecules are often, but not always, arranged in regular,
repeating patterns.
Substances exist in different phases and phase changes occur because
molecules exert forces on each other. (Without intermolecular forces, all
substances would behave as ideal gases!!) It takes energy to overcome the
attractive intermolecular forces that cause molecules to aggregate.
Therefore, sublimation, fusion and evaporation are all endothermic
processes
A given substance will exist as a solid, liquid or gas depending on the
temperature and pressure of the sample. A phase diagram shows the stable
phases at each temperature and pressure.
Phases and Phase Diagrams
Phase diagram of I2
Phases and Phase Diagrams
Take note of the following points:
1. Solid is the most stable phase at low T and high P. Gas is the stable phase at high T
and low P.
Phases and Phase Diagrams
Take note of the following points:
2. The S-L line shows the T’s and P’s at which both solid and liquid are stable and can coexist. It
also shows us how the melting temperature changes with pressure.
Phases and Phase Diagrams
Take note of the following points:
3. The L-G line curve shows the T’s and P’s at which both liquid and gas are stable and can
coexist. It also shows us how the boiling temperature changes with pressure.
Phases and Phase Diagrams
Take note of the following points:
4. For most substances, the S-L line has a positive slope, but for a few substances (most notably,
water but also bismuth and antimony), it has a negative slope!
For most
substances
Phases and Phase Diagrams
Take note of the following points:
4. For most substances, the S-L line has a positive slope, but for a few substances (most notably,
water but also bismuth and antimony), it has a negative slope!
For water
Phases and Phase Diagrams
Take note of the following points:
5. At the triple point, all three phases are stable and coexist.
Phases and Phase Diagrams
Take note of the following points:
6. The G-L line ends abruptly at the critical point (Tc, Pc)
Phases and Phase Diagrams
Phase diagram of I2
What is the phase of I2 at 25 oC and 1 atm? We are dealing with a solid.
25 oC and 1 atm
25 oC
Phases and Phase Diagrams
Phase diagram of CO2
What is the phase of CO2 at 25 oC and 1 atm? We are dealing with a gas.
1 atm
25 oC and 1 atm
25 oC
Phases and Phase Diagrams
Phase diagram of H2O
What is the phase of H2O at 25 oC and 1 atm? We are dealing with a liquid.
25 oC and 1 atm
1 atm
25 oC
Phases and Phase Diagrams
Take note of the following points:
4. For most substances, the S-L line has a positive slope, but for a few substances (most notably,
water but also bismuth and antimony), it has a negative slope!
The slope of the S-L line is negative!
Phases and Phase Diagrams
Take note of the following points:
1m
1m
A column of water 1 m ×1 m × 10 m
occupies a volume of 10 m3 or
10,000 L. 1 L of water weighs 1 kg.
10,000 L of water weigh 10,000 kg.
The pressure exerted by 10,000 kg
of water equals:
9.8 m2/s×10,000 kg/(1 m×1 m)
 105 Pa  1 atm
10 m
10 m of water generates a 1 atm
additional pressure.
Phases and Phase Diagrams
Take note of the following points:
We find liquid water at the bottom of the ocean.
The slope of the S-L line is negative!
Phases and Phase Diagrams
Take note of the following points:
Polymorphism: The existence of a solid substance in more than one form.
Other forms of ice
obtained at several
thousands of
atmospheres
Phases and Phase Diagrams
To summarize, a typical phase diagram looks like this:
supercritical
solid-liquid
coexistence line
fluid
Solid
P
critical point
B
Pc
Liquid
A
1 atm
liquid-vapour coexistence line
“normal”
melting point
Gas
Tfus
“normal” boiling point
Tvap
triple point
T
Tc
Phases and Phase Diagrams
Supercritical fluid:
P
Pc
solid-liquid
coexistence
line
supercritical
fluid
(S)
B
(L)
(G)
critical
point
As one moves from A to B, the pressure
increases and the density of the gas
increases until it equals the density of
the liquid. At this point, gas and liquid
are indistinguishable, the interface
between liquid and gas vanishes and
we have a supercritical fluid.
A
liquid-vapour
coexistence line
Tc
T
If a gas is at T > TC (Point A in diagram),
increasing the pressure of the gas does
not yield a liquid but rather a
supercritical fluid (Point B).
To take a gas at T > TC and transform it into a liquid, the temperature must first be
reduced below TC. Then the pressure is increased to pass the liquid-vapor coexistence
curve.
Phases and Phase Diagrams
P
Pc
solid-liquid
coexistence
line
supercritical
fluid
(S)
B
(L)
(G)
critical
point
A
liquid-vapour
coexistence line
Tc
T
The phase boundary between liquid benzene
and its vapour disappears at Tc.
T increases from below TC to above TC
Supercritical fluid:
Below Tc, the
phase boundary is
clearly visible.
Just below Tc, the
phase boundary is
barely visible.
At Tc, the phase
boundary
disappears.
Phases and Phase Diagrams
Supercritical fluid:
P
solid-liquid
coexistence line
supercritical
fluid
critical
point
(S)
Pc
(L)
liquid-vapour
coexistence line
(G)
Tc
T
Did you know? Supercritical CO2 is used to extract caffeine from coffee beans. The
extracted caffeine can be sold to pharmaceutical or beverage companies. The critical
point for CO2 is fairly low (Tc = 31 oC) and so, supercritical CO2 can be used at ambient
temperatures without causing decomposition or “denaturing” of other compounds.
Because it has low toxicity, a low critical temperature and is nonflammable, supercritical
CO2 is becoming an increasingly important industrial and commercial solvent.
Phases and Phase Diagrams
Examples: For a particular substance, the S-L coexistence curve has a negative slope.
P
solid-liquid
coexistence line
supercritical
fluid
critical
point
(S)
Pc
(L)
PT
liquid-vapour
coexistence line
(G)
TT
Tc
T
a) What phase changes are possible if the pressure is increased at constant temperature T?
Assume that T is less than Ttp, where Ttp is the triple point temperature.
Gas  deposition  solid  melting  liquid
Phases and Phase Diagrams
Examples: For a particular substance, the S-L coexistence curve has a negative slope.
P
solid-liquid
coexistence line
supercritical
fluid
critical
point
(S)
Pc
(L)
PT
liquid-vapour
coexistence line
(G)
TT
Tc
T
b) What phase changes are possible if the pressure is increased at constant temperature T,
assuming Ttp < T < Tc, where Ttp and Tc are the triple point and critical point temperatures,
respectively.
Gas  condensation  liquid
Phases and Phase Diagrams
Examples: For a particular substance, the S-L coexistence curve has a negative slope.
P
solid-liquid
coexistence line
supercritical
fluid
critical
point
(S)
Pc
(L)
PT
liquid-vapour
coexistence line
(G)
TT
Tc
T
c) True or False? The melting temperature increases as the pressure increases.
False
Phases and Phase Diagrams
Examples: For a particular substance, the S-L coexistence curve has a negative slope.
P
solid-liquid
coexistence line
supercritical
fluid
critical
point
(S)
Pc
(L)
PT
liquid-vapour
coexistence line
(G)
TT
Tc
T
d) True or False? The solid is more dense than the liquid.
At a given temperature, when we increase the pressure, the density increases and
the solid becomes a liquid. False
Phases and Phase Diagrams
Examples: For a particular substance, the triple point is at 57 ºC and 5.1 atm, and the critical
point is at 31oC and 73 atm.
73 atm
25 oC and 73 atm
25 oC
a)
What is the phase of this substance at 25oC and 73 atm?
We are dealing with a liquid.
Phases and Phase Diagrams
Examples: For a particular substance, the triple point is at 57 ºC and 5.1 atm, and the critical
point is at 31oC and 73 atm.
-60 oC and
75 atm
-60 oC and
0.001 atm
-60 oC
b) What phase changes occur if the pressure is decreased from 75 atm to 0.001 atm at −60 oC?
Assume that the solid-liquid line has a positive slope.
Solid  sublimation  gas
REVIEW
Phases and Phase Diagrams
very compressible
d ≈ 1 – 10 g L−1 at SATP
condensation
vaporization
deposition
sublimation
fusion
incompressible
d ≈ 1 – 10 g mL−1
freezing
REVIEW
Phases and Phase Diagrams
To summarize, a typical phase diagram looks like this:
P
supercritical
fluid
Solid
Liquid
Gas
T
REVIEW
Phases and Phase Diagrams
To summarize, a typical phase diagram looks like this:
supercritical
solid-liquid
coexistence line
fluid
Solid
P
critical point
B
Pc
Liquid
A
1 atm
liquid-vapour coexistence line
“normal”
melting point
Gas
Tfus
“normal” boiling point
Tvap
triple point
T
Tc
Phases and Phase Diagrams
Examples 12-45: Which substances listed in the table can exist as liquids at room
temperature (~ 20 oC)?
Substance
Tc, K
Pc, atm
H2
33.3
12.8
N2
126.2
33.5
O2
154.8
50.1
CH4
191.1
45.8
CO2
304.2
72.9
HCl
324.6
82.1
NH3
405.7
112.5
SO2
431.0
77.7
H2O
647.3
218.3
P
solid-liquid
coexistence
line
supercritical
fluid
critical
point
(S)
Pc
(L)
(G)
T = 20 oC Tc
liquid-vapour
coexistence line
T
Phases and Phase Diagrams
Examples 12-45: Which substances listed in the table can exist as liquids at room
temperature (~ 20 oC)?
Substance
Tc, K
Pc, atm
H2
33.3
12.8
N2
126.2
33.5
O2
154.8
50.1
CH4
191.1
45.8
CO2
304.2
72.9
HCl
324.6
82.1
NH3
405.7
112.5
SO2
431.0
77.7
H2O
647.3
218.3
P
solid-liquid
coexistence
line
supercritical
fluid
critical
point
(S)
Pc
(L)
(G)
liquid-vapour
coexistence line
T = 20 oCTc = 240 oC
Phases and Phase Diagrams
Examples 12-45: Which substances listed in the table can exist as liquids at room
temperature (~ 20 oC)?
Substance
Tc, K
Pc, atm
H2
33.3
12.8
N2
126.2
33.5
O2
154.8
50.1
CH4
191.1
45.8
CO2
304.2
72.9
HCl
324.6
82.1
NH3
405.7
112.5
SO2
431.0
77.7
H2O
647.3
218.3
P
solid-liquid
coexistence
line
supercritical
fluid
critical
point
(S)
Pc
(L)
(G)
T = 20 oC Tc
liquid-vapour
coexistence line
T
20 oC < TC or TC > 293.15 K
Gases that can be liquified at room temperature are said to be “non-permanent gases”.
Gases that cannot be liquified at room temperature are said to be “permanent gases”.
Phases and Phase Diagrams
Examples 12-51: Phase diagram of phosphorous
a) Indicate the phases present in the regions labeled with a question mark?
P
(L)
?
(S)
43 atm
(G)
?
590 oC
T
Phases and Phase Diagrams
Examples 12-51: Phase diagram of phosphorous
b) A sample of solid red phosphorous cannot be melted by heating in a container open to the
atmosphere. Explain why this is so?
P
(L)
(S)
43 atm
(G)
1 atm
590 oC
T
Solid phosphorous can only be sublimed (S  G) if it is heated at P = 1 atm.
Phases and Phase Diagrams
Examples 12-51: Phase diagram of phosphorous
c) Trace the phase changes that occur when the pressure on a sample is reduced from Point A
to B, at constant temperature.
P
A
(L)
(S)
43 atm
B
(G)
590 oC
Solid  condensation  Liquid  vaporization  Gas
T
Intermolecular Forces:
Liquids and Solids
● Phases and Phase Diagrams
● Liquids and Liquid Properties
● Intermolecular Forces
● Heating Curves
● Introduction to Solids
● Cubic Packing Arrangements
● Closest-Packed Structures
● Density of a Crystalline Solid
● Ionic Solids and Interstitial Sites
● The Born-Haber Cycle
Intermolecular Forces:
Liquids and Solids
● Phases and Phase Diagrams
● Liquids and Liquid Properties
● Intermolecular Forces
● Heating Curves
● Introduction to Solids
● Cubic Packing Arrangements
● Closest-Packed Structures
● Density of a Crystalline Solid
● Ionic Solids and Interstitial Sites
● The Born-Haber Cycle
Liquids and Liquid Properties
Condensation
P
solid-liquid
coexistence line
supercritical
fluid
critical
point
(S)
(L)
(G)
liquid-vapour
coexistence line
T
We know that if the temperature of a gas is lowered sufficiently, the gas will condense to
a liquid. Why is this? As T is lowered, the average kinetic energy of the molecules
decreases. At some point, the molecules will no longer have enough kinetic energy to
overcome the attractive forces that draw the molecules together. Consequently, the
molecules cluster together to form a liquid.
Liquids and Liquid Properties
Freezing
P
solid-liquid
coexistence line
supercritical
fluid
critical
point
(S)
(L)
(G)
liquid-vapour
coexistence line
T
The freezing of a liquid can be explained in the same way: If the temperature of a liquid
is lowered sufficiently, the molecules will not have enough kinetic energy to overcome
attractive forces that draw the molecules closer together
 the liquid freezes.
Liquids and Liquid Properties
The physical properties of a liquid depend on the strength and nature of the
intermolecular forces. We shall examine why the following physical properties are
different from substance to substance.
vapour pressure
= equilibrium pressure of vapour that forms above a
liquid in a closed container
normal boiling point (Tvap)
= temperature at which the vapour pressure of the liquid
equals 1 atm
surface tension (g)
= energy required to increase the surface area of a liquid
viscosity (η)
 provides a measure of a fluid’s resistance to flow; the
speed of flow through a tube is inversely proportional
to the viscosity
In general, the stronger the intermolecular attractions, the higher
the boiling point, the greater the surface tension, the higher the
viscosity and the lower the vapour pressure.
Liquids and Liquid Properties
vapour pressure
= equilibrium pressure of vapour that forms above a
liquid in a closed container
to vacuum
air
liquid
Liquids and Liquid Properties
vapour pressure
= equilibrium pressure of vapour that forms above a
liquid in a closed container
to vacuum
air
liquid
Liquid N2
Liquids and Liquid Properties
vapour pressure
= equilibrium pressure of vapour that forms above a
liquid in a closed container
to vacuum
air
solid
Liquid N2
Liquids and Liquid Properties
vapour pressure
= equilibrium pressure of vapour that forms above a
liquid in a closed container
to vacuum
vacuum
solid
Liquid N2
Liquids and Liquid Properties
vapour pressure
= equilibrium pressure of vapour that forms above a
liquid in a closed container
to vacuum
vacuum
solid
Liquid N2
Liquids and Liquid Properties
vapour pressure
= equilibrium pressure of vapour that forms above a
liquid in a closed container
to vacuum
vacuum
solid
Liquids and Liquid Properties
vapour pressure
= equilibrium pressure of vapour that forms above a
liquid in a closed container
to vacuum
vacuum
liquid
Liquids and Liquid Properties
vapour pressure
= equilibrium pressure of vapour that forms above a
liquid in a closed container
to vacuum
liquid
Liquids and Liquid Properties
The physical properties of a liquid depend on the strength and nature of the
intermolecular forces. We shall examine why the following physical properties are
different from substance to substance.
vapour pressure
= equilibrium pressure of vapour that forms above a
liquid in a closed container
normal boiling point (Tvap)
= temperature at which the vapour pressure of the liquid
equals 1 atm
surface tension (g)
= energy required to increase the surface area of a liquid
viscosity (η)
 provides a measure of a fluid’s resistance to flow; the
speed of flow through a tube is inversely proportional
to the viscosity
In general, the stronger the intermolecular attractions, the higher
the boiling point, the greater the surface tension, the higher the
viscosity and the lower the vapour pressure.
Liquids and Liquid Properties
Clausius-Clapeyron equation
P
solid-liquid
coexistence line
supercritical
fluid
critical
point
(S)
(L)
(G)
liquid-vapour
coexistence line
T
The L-G line shows us how
1. the vapour pressure of a liquid changes with temperature
2. the boiling temperature of a liquid changes with pressure
Liquids and Liquid Properties
Clausius-Clapeyron equation
P
solid-liquid
coexistence line
supercritical
fluid
critical
point
(S)
(L)
(G)
liquid-vapour
coexistence line
T
Along the L-G line, both liquid and gas co-exist.
At equilibrium, the rate of evaporation equals
the rate of condensation
Vapour X(g)
Liquid X(l)
Liquids and Liquid Properties
Clausius-Clapeyron equation
The variation of vapour pressure with temperature is modeled reasonably well by the
Clausius-Clapeyron equation:
P 
ln  2 
 P1 
H vap  1
o
 
R

 T2

1

T1 
The quantities appearing in this equation are described below.
P2
P1
o
H vap
R
=
=
vapour pressure at temperature T2
vapour pressure at temperature T1
=
=
standard enthalpy of vaporization
8.3145 J K−1 mol−1
Extremely important: Pay attention to the units!
Liquids and Liquid Properties
Clausius-Clapeyron equation
P
supercritical
solid-liquid
coexistence line
fluid
critical
point
(S)
(L)
liquid-vapour
coexistence line
(P2, T2)
(P1, T1)
(G)
T
P 
ln  2 
 P1 
H vap  1
o
 
R

 T2

1

T1 
Liquids and Liquid Properties
Clausius-Clapeyron equation
P
supercritical
solid-liquid
coexistence line
fluid
critical
point
(S)
(L)
(P2, T2)
(P1, T1)
liquid-vapour
coexistence line
(G)
T
Example:
a) If the vapour pressure of P4(l) is 10 Torr at 128 oC and 400 Torr at 251 oC, then what
is vapHo?
 vap H o  1 1 
 P2 
  
Ln   
P
R
 1
 T2 T1 
P 
 10 
 RLn  2 
 8.3145Ln

P
400


 1   Ho 
 52.4 kJ / mol
 vap H o 
vap
1
1
1 1


128  273.15 251  273.15
T2 T1
Liquids and Liquid Properties
P1 = 10 Torr
T1 = 128 oC
P2 = 400 Torr
T2 = 251 oC
vapHo = 52.4 kJ/mol
Clausius-Clapeyron equation
Example:
b) What is the normal boiling point of P4(l)?
 vap H o  1 1 
 P2 
  
Ln   
R  T2 T1 
 P1 

 P2  1 1
R
   
Ln
o
 vap H
 P1  T2 T1
T2 
T2 
 P2 
1 1
R
 
 
Ln
T2 T1  vap H o  P1 
1
 P2 
1
R
 

Ln
o
T1  vap H
 P1 
1
 553.8 K
1
8.3145  760 

Ln

128  273.15 52,400  10 
Liquids and Liquid Properties
Clausius-Clapeyron equation
Example:
c) What is the vapour pressure at 200 oC?
 vap H o  1 1 
 P2 
  
Ln   
P
R
 1
 T2 T1 
  vap H o  1 1 
P2
  
 exp 
P1
R  T2 T1 

P1 = 10 Torr
T1 = 128 oC
P2 = 400 Torr
T2 = 251 oC
vapHo = 52.4 kJ/mol
  vap H o
P2  P1 exp 
R

 52,400 
1
1

P2 10 exp 


 109 torr
8
.
3145
200

273
.
15
128

273
.
15



 1 1 
  
 T2 T1 
REVIEW
Liquids and Liquid Properties
The physical properties of a liquid depend on the strength and nature of the
intermolecular forces. We shall examine why the following physical properties are
different from substance to substance.
vapour pressure
= equilibrium pressure of vapour that forms above a
liquid in a closed container
normal boiling point (Tvap)
= temperature at which the vapour pressure of the liquid
equals 1 atm
surface tension (g)
= energy required to increase the surface area of a liquid
viscosity (η)
 provides a measure of a fluid’s resistance to flow; the
speed of flow through a tube is inversely proportional
to the viscosity
In general, the stronger the intermolecular attractions, the higher
the boiling point, the greater the surface tension, the higher the
viscosity and the lower the vapour pressure.
Intermolecular Forces:
Liquids and Solids
● Phases and Phase Diagrams
● Liquids and Liquid Properties
● Intermolecular Forces
● Heating Curves
● Introduction to Solids
● Cubic Packing Arrangements
● Closest-Packed Structures
● Density of a Crystalline Solid
● Ionic Solids and Interstitial Sites
● The Born-Haber Cycle
Intermolecular Forces
Dipole moments and polarizabilities of a
few molecules. (Based on data from
Physical Chemistry, 6th Edition, by P.
Atkins, published by Freeman, 1998).
The purpose of this section is to understand how
molecules interact with each other and how these
interactions help us to understand trends in the
physical properties of compounds (e.g. boiling points,
vapour pressures, viscosity and surface tensions of
liquids; densities and melting points of solids;
deviations from ideal gas behaviour, etc.)
Generally speaking, differences in the physical
properties of substances can often be rationalized by
considering how molecules interact with each other at
the molecular level. We’ll look at some of the types of
intermolecular forces that act between pairs of
molecules.
μ (in
α/αHe
debye)
H2
0
4.1
HF
1.91
2.6
HCl
1.08
13
HBr
0.80
18
CO
0.12
10
CO2
0
13
H2O
1.85
7.5
NH3
1.47
11
He
0
1
Ar
0
8.4
CH4
0
13
CCl4
0
53
Intermolecular Forces
A. Dipole-dipole forces
Some molecules possess a permanent dipole moment, μ, because the bond dipoles do
not cancel out. Such molecules are said to be “polar” because one end of the molecule
is slightly positive and the other end is slightly negative. The charge distribution of a
polar molecule can be represented by an arrow that points from the positive end to the
negative end. The dipole moments of a few molecules were given in the previous table.
Polar molecules (i.e. dipoles) tend to orient themselves in a “head-to-tail” manner, as
shown below:
The molecules are not organized into perfect straight line because the molecules are in
motion (i.e. each one possesses kinetic energy) and they “jiggle” out of perfect
alignment.
Digging Deeper
For a pair of interacting polar molecules with
dipole moments of μA and μB:
Potential Energy  
μA
μB
d
 A2   B2
d6
Intermolecular Forces
A. Dipole-dipole forces
Example: Iodine chloride, ICl, and bromine, Br2, have exactly the same number of electrons, and
it is reasonable to assume that these molecules are essentially the same size. Yet the boiling
points of ICl(l) and Br2(l) are quite different, 97 oC and 59 oC, respectively. Use your knowledge
of dipole-dipole interactions to explain why ICl(l) has a higher boiling point than Br2(l).
Intermolecular Forces
A. Dipole-dipole forces
Example: Iodine chloride, ICl, and bromine, Br2, have exactly the same number of electrons, and
it is reasonable to assume that these molecules are essentially the same size. Yet the boiling
points of ICl(l) and Br2(l) are quite different, 97 oC and 59 oC, respectively. Use your knowledge
of dipole-dipole interactions to explain why ICl(l) has a higher boiling point than Br2(l).
Chlorine is more electronegative than bromine. Consequently, I-Cl shows a dipole moment:
(d)ICl(d)
Br2 on the other hand cannot generate a dipole moment because the molecule is made of two
identical bromine atoms.
Intermolecular Forces
B. London dispersion forces
The electrons in a molecule are in constant motion and at any particular instant, there may be
an asymmetric distribution of electrons in the molecule (i.e. with a greater number of electrons
at one end than at the other end). The asymmetric distribution of electrons gives rise to an
instantaneous and temporary dipole moment (inst ). The formation of inst in a molecule causes
(or induces) the formation of a dipole in neighbouring molecules. The induced dipole moment
is ind . (Notice the head-to-tail arrangement of the instantaneous and induced dipole
moments.)
Molecule A
---- ---
- - --
inst
Molecule B
---- ---
- - --
ind
There is an attraction between inst and ind . The strength of the interaction increases
as the “polarizabilities” of the molecules increase.
Intermolecular Forces
B. London dispersion forces
provides a measure of the extent to which the charge cloud
of a molecule can be distorted (i.e. polarized) by another
molecule.
Polarizability
(α)
The charge cloud of a large molecule is diffuse and easily polarized. The charge cloud of a small,
compact molecule is not easily polarized. The polarizabilities of a few molecules were given
earlier. Note that the larger the molecule, the larger the polarizability.
Digging Deeper
For a pair of molecules with polarizabilities of αA and αB:
αA
αB
d
Potential Energy  
 A  B
d6
Intermolecular Forces
B. London dispersion forces
Remarks:
London dispersion forces are most attractive when the molecules are large because large
molecules have larger, more diffuse (i.e. more polarizable) charge clouds.
London dispersion forces always contribute to the molecular interactions because all molecules
have charge clouds and are therefore polarizable to some extent.
Molecule A
---- ---
- - --
inst
Molecule B
---- --ind
- - --
Intermolecular Forces
B. London dispersion forces
μ (in
Example: Methane (CH4) and carbon
tetrachloride (CCl4) are both nonpolar molecules.
Use your knowledge of London dispersion forces
to explain why the boiling point of CCl4(l) is much
higher than that of CH4(l).
The C-Cl bonds are very
polar, but the bond dipoles
cancel. The CCl4 molecule is
nonpolar.
Carbon tetrachloride (CCl4) is a molecule that is much
larger than methane (CH4). Consequently, CCl4
contains many more electrons which can generate an
instantaneous dipole moment more readily than CH4.
These dipole moments generate strong intermolecular
forces between CCl4 molecules which are more difficult
to break than in CH4. Consequently, Tb of CCl4 (78oC)
is larger than Tb of CH4 (162 oC).
α/αHe
debye)
H2
0
4.1
HF
1.91
2.6
HCl
1.08
13
HBr
0.80
18
CO
0.12
10
CO2
0
13
H2O
1.85
7.5
NH3
1.47
11
He
0
1
Ar
0
8.4
CH4
0
13
CCl4
0
53
Intermolecular Forces
C. Hydrogen bonding forces
A special type of bond forms between molecules when the molecules contain a
hydrogen atom bonded to N, O, or F. When H is bonded to N, O or F, the H atom
carries a significant positive charge and it is strongly attracted to a lone pair on another
molecule! When a hydrogen atom which is covalently bonded to one atom is
simultaneously attracted to the lone pair on another atom, it is “bridging” two molecules,
as shown below. Such a bond is called a hydrogen bond.
H
X
Y
An intermolecular hydrogen bond “bridges” two
molecules.
An intramolecular hydrogen bond bridges two
parts of the same molecule.
intermolecular hydrogen bond
Did you know? “inter” means “between” and
“intra” means “within”.
Intermolecular Forces
C. Hydrogen bonding forces
Note carefully:
H is covalently bonded to X but is simultaneously attracted to a lone pair of electrons on Y.
Both X and Y must be N, O or F
Hydrogen bonds are the strongest type of intermolecular force (but they are still weak in
comparison to covalent and ionic bonding forces)
intermolecular forces
H
X
chemical bonding
forces
Y
dipole-dipole
& LDFs
H bonds
covalent & ionic bonds
0.10-10
kJ mol−1
10−40
kJ mol−1
100’s or 1000’s
kJ mol−1
intermolecular hydrogen bond
Intermolecular Forces
C. Hydrogen bonding forces
Hydrogen bonds are important!
H bonds between H2O’s in ice give the solid an open structure
H bonds between H2O’s in water give the liquid a high BP, high surface tension and a large heat
capacity.
Hydrogen bonding in water. This is Figure 12-7 of
Petrucci 10e. Used with permission.
Intermolecular Forces
C. Hydrogen bonding forces
Hydrogen bonds are important!
H bonds are especially important in
biology (e.g. H bonds keep the two helices
of DNA together; the structures and
functions of proteins and enzymes are
determined by H bonds)
The helical structures of proteins (above) and DNA (on the right) are stabilized by hydrogen
bonds. These are Figures 28-12 and 28-26 of Petrucci 10e. Used with permission.
Intermolecular Forces
C. Hydrogen bonding forces
Hydrogen bonds are important!
H bonds are especially important in biology (e.g. H bonds keep the two helices of DNA together;
the structures and functions of proteins and enzymes are determined by H bonds)
REVIEW
Intermolecular Forces
A. Dipole-dipole forces
Polar molecules with a dipole moment .
B. London dispersion forces
The formation of inst in a molecule causes (or induces) the formation of a dipole in
neighbouring molecules.
Molecule A
---- ---
- - --
inst
Molecule B
---- ---
- - --
ind
C. Hydrogen bonding forces
A special type of bond forms between molecules when the molecules contain a
hydrogen atom bonded to N, O, or F.
intermolecular hydrogen bond
H
X
Y
Intermolecular Forces
C. Hydrogen bonding forces
A special type of bond forms between molecules when the molecules contain a
hydrogen atom bonded to N, O, or F. When H is bonded to N, O or F, the H atom
carries a significant positive charge and it is strongly attracted to a lone pair on another
molecule!
intermolecular hydrogen bond
H
X
Y
Electronegativity Scale
Intermolecular Forces
C. Hydrogen bonding forces
Example: Dichloroethene, C2H2Cl2, has several isomeric forms. Use your knowledge of
intermolecular forces to predict whether (Z)-1,2-dichloroethene or (E)-1,2-dichloroethene
has the higher boiling point. Lewis structures are given below.
H
H
C
Tb = 60 oC
Cl
Cl
vs
C
Cl
(Z)-1,2-dichloroethene, also
called cis-1,2-dichloroethene
H
C
C
Tb = 48 oC
H
Cl
(E)-1,2-dichloroethene, also called
trans-1,2-dichloroethene
The chlorine atom being more electronegative than carbon induces a dipole moment in
the C-Cl bond. However, those dipole moments are opposite in (E)-1,2-dichloroethane
and they cancel each other. The dipole moments of the C-Cl bond do not cancel each
other in (Z)-1,2-dichloroethane. Thus the Z-isomer has a permanent dipole moment
which induces strong intermolecular forces. The Z-isomer has a higher boiling point.
A word of warning!
Don’t over generalize the results of this example. You might end up making the wrong
prediction!
Intermolecular Forces
C. Hydrogen bonding forces
Example: Consider the trans and cis isomers of C4H4O4. Which one has the higher melting
point?
O
O
OH
HO
Tm = 300 oC
OH
OH
O
Tm = 140 oC
O
Fumaric acid (trans)
Maleic acid (cis)
Fumaric acid generates intermolecular H-bonds which leads to the formation of a
network where all the molecules are associated with another. Maleic acid forms
intramolecular H-bonds and the intermolecular forces between molecules are weaker.
Consequently, maleic acid has a lower melting point than fumaric acid.
Intermolecular Forces
C. Hydrogen bonding forces
Example: Consider the trans and cis isomers of C4H4O4. Which one has the higher melting
point?
O
O
OH
HO
Tm = 300 oC
OH
OH
O
Tm = 140 oC
O
Fumaric acid (trans)
Maleic acid (cis)
Fumaric acid generates intermolecular H-bonds which leads to the formation of a
network where all the molecules are associated with another. Maleic acid forms
intramolecular H-bonds and the intermolecular forces between molecules are weaker.
Consequently, maleic acid has a lower melting point than fumaric acid.
Intermolecular Forces
C. Hydrogen bonding forces
Example: Use your knowledge of intermolecular forces to predict whether CH3COCH3(l) or
CH3CH2CH2OH(l) has the higher boiling point.
O
Tb = 56 oC
O
H 3C
CH3
acetone
H
Tb = 97 oC
1-propanol
Acetone and 1-propanol have a similar number of C- and O-atoms. They both induce a
dipole moment since O is more electronegative than C. However, 1-propanol can form
H-bonds and acetone cannot. Consequently, acetone boils at a lower temperature than
1-propanol.
Intermolecular Forces
C. Hydrogen bonding forces
Example: Which one of the liquids, HO-CH2CH2-OH or CH3CH2OH, has the highest vapour
pressure at room temperature?
Tb = 196 oC
H
O
O
O
Ethylene glycol
H
H
Tb = 78 oC
Ethanol
A single ethylene glycol (EG) can form more H-bonds with other EG molecules than
ethanol can form with other ethanol molecules. Consequently, EG boils at a higher
temperature. EG has a lower vapor pressure than ethanol.
Intermolecular Forces
C. Hydrogen bonding forces
Example: Use your knowledge of intermolecular forces to predict the order of boiling points for
H2O, H2S (hydrogen sulfide), H2Se (hydrogene selenide), and H2Te (hydrogen telluride).
All these molecules are bent, and as such, generate a dipole moment.
X
H
H

Intermolecular Forces
C. Hydrogen bonding forces
Example: Use your knowledge of intermolecular forces to predict the order of boiling points for
H2O, H2S, H2Se and H2Te.
All these molecules are bent, and as such, generate a dipole moment.
X
H
H

Tb
120
100
-59.6 oC
-41.3 oC
-2 oC
80
T b , oC
60
40
Tb (oC) = 0.6055xM (g/mol) - 83.785
20
0
-20
-40
HO
2
Tb =-6060 oC
H2S
Tb = 60 oC
-80
0
20
40
H2Te
Tb = 2 oC
H2Se
Tb = 41 oC
60
80
100
Mass, g/mol
120
140
160
Intermolecular Forces
C. Hydrogen bonding forces
Example: Use your knowledge of intermolecular forces to predict the order of boiling points for
H2O, H2S, H2Se and H2Te.
All these molecules are bent, and as such, generate a dipole moment.
X
H
H

Tb
+100 oC
-59.6 oC
-41.3 oC
-2 oC
120
H2O
Tb = 100 oC
100
80
T b , oC
60
40
Tb (oC) = 0.6055xM (g/mol) - 83.785
20
0
-20
H2S
Tb = 60 oC
-40
-60
-80
0
20
40
H2Te
Tb = 2 oC
H2Se
Tb = 41 oC
60
80
100
Mass, g/mol
120
140
160
Intermolecular Forces
C. Hydrogen bonding forces
Example: Use your knowledge of intermolecular forces to predict the order of boiling points for
H2O, H2S, H2Se and H2Te.
By increasing the mass and the size of the molecule H2X, one increases the polarizability, and
thus the strength of the interactions between the H2X molecules. The exception is H2O because
water forms H-bonds:
H2S < H2Se < H2Te < H2O
Tb
+100 oC
-59.6 oC
-41.3 oC
-2 oC
120
H2O
Tb = 100 oC
100
80
T b , oC
60
40
Tb (oC) = 0.6055xM (g/mol) - 83.785
20
0
-20
H2S
Tb = 60 oC
-40
-60
-80
0
20
40
H2Te
Tb = 2 oC
H2Se
Tb = 41 oC
60
80
100
Mass, g/mol
120
140
160
Intermolecular Forces
C. Hydrogen bonding forces
Example: Use your knowledge of intermolecular forces to predict the order of boiling points for
H2O, H2S, H2Se and H2Te.
It should be noted that similar arguments can be applied to the hydrides of the group 15
and group 17 elements. Thus, in order of increasing BP, we have:
PH3 < AsH3 < SbH3 < NH3 and HCl < HBr < HI < HF
Intermolecular Forces:
Liquids and Solids
● Phases and Phase Diagrams
● Liquids and Liquid Properties
● Intermolecular Forces
● Heating Curves
● Introduction to Solids
● Cubic Packing Arrangements
● Closest-Packed Structures
● Density of a Crystalline Solid
● Ionic Solids and Interstitial Sites
● The Born-Haber Cycle
Review this section
on your own.
Heating Curves
A heating curve shows us how the temperature varies with the amount of heat added. Consider
heating a sample of ice from ti = –10 oC to t = 150 oC at constant pressure.
T
warming
vapour
(slope3)
tf
boiling liquid
tvap
warming
liquid
(slope2)
warming
solid
(slope1)
melting solid
tfus
ti
qgas
qliq
qsol
Hfus
Hvap
Q
Intermolecular Forces:
Liquids and Solids
● Phases and Phase Diagrams
● Liquids and Liquid Properties
● Intermolecular Forces
● Heating Curves
● Introduction to Solids
● Cubic Packing Arrangements
● Closest-Packed Structures
● Density of a Crystalline Solid
● Ionic Solids and Interstitial Sites
● The Born-Haber Cycle
Introduction to Solids
Solids can be classified as crystalline or as amorphous, depending on whether we have regular
or irregular packing of the atoms, molecules or ions that make up the solid. Crystalline and
amorphous solids have rather different physical characteristics.
Crystalline solids
• regular repeating patterns
• “sharp” melting points
Amorphous solids
• irregular packing
• melt over a temperature range
Examples of amorphous solids include rubber, polystyrene, window glass, candle wax,
and cotton candy. We can get irregular packing (amorphous solids) if there are impurities
in the sample when the liquid freezes, or if the molecules are large or have flexible
structures. For molecules that are large or very flexible (i.e. polymers!), it is statistically
most probable that the molecules will be irregularly packed when the liquid freezes.
We shall focus exclusively on crystalline solids.
Introduction to Solids
For crystalline solids, we use the following concepts to characterize the pattern of the
packing arrangement.
Crystal lattice
three dimensional array of points that
shows how the structural units are
arranged in space.
Each structural unit could be a single atom, a molecule or an ion. In C(s), the structural
unit is a carbon atom. In I2(s), the structural unit is an I2 molecule. In NH4NO3(s), the
structural units are NH4+ and NO3− ions.
Unit cell
the smallest building block that possesses
the symmetry of the crystal lattice; a solid
sample of any size can be “built” by
stacking together unit cells.
Introduction to Solids
Crystalline solids are classified according to the nature of the bonding, or according to
the geometry and symmetry of the packing arrangement. Let’s focus first on the nature
of the bonding.
Crystalline solids
ionic
Network
covalent
molecular
In an ionic solid,
In a network
positive and
covalent solid,
negative ions are
atoms are held in
held in their lattice their lattice positions
positions by (strong) by (strong) covalent
ionic bonding
bonds.
forces.
In a molecular solid,
molecules are held
in their lattice
positions by (weak)
intermolecular
forces.
oC
H2O Tm = 0
NaCl Tm = 800
SiO2 Tm= 1600
LiF
Tm = 848 oC SiC
Tm =2830 oC S
Tm = 113 oC
8
o
ZnO Tm = 1974 C Cdiamond Tm = 4440 oC
I2
Tm = 114 oC
(at P = 12.4 GPa)
oC
oC
metallic
In a metallic solid,
metal cations are
held in their lattice
positions by
(strong) metallic
bonding forces.
Cu
Tm = 1083 oC
Ag
Tm = 962 oC
W
Tm = 3422 oC
Introduction to Solids
Crystalline solids are classified according to the nature of the bonding, or according to
the geometry and symmetry of the packing arrangement. Let’s focus first on the nature
of the bonding.
Crystalline solids
ionic
Network
covalent
molecular
In an ionic solid,
In a network
positive and
covalent solid,
negative ions are
atoms are held in
held in their lattice their lattice positions
positions by (strong) by (strong) covalent
ionic bonding
bonds.
forces.
In a molecular solid,
molecules are held
in their lattice
positions by (weak)
intermolecular
forces.
oC
H2O Tm = 0
NaCl Tm = 800
SiO2 Tm= 1600
LiF
Tm = 848 oC SiC
Tm =2830 oC S
Tm = 113 oC
8
o
ZnO Tm = 1974 C Cdiamond Tm = 4440 oC
I2
Tm = 114 oC
(at P = 12.4 GPa)
oC
oC
metallic
In a metallic solid,
metal cations are
held in their lattice
positions by
(strong) metallic
bonding forces.
Cu
Tm = 1083 oC
Ag
Tm = 962 oC
W
Tm = 3422 oC
Introduction to Solids
The Periodic Table can help understand the differences in Tm between NaCl, LiF, and
ZnO.
NaCl Tm = 800 oC
1 q A qC
Electrostatic Force 
LiF
Tm = 848 oC
2
4

r
o
A
C
ZnO Tm = 1974 oC
Introduction to Solids
Crystalline solids are classified according to the nature of the bonding, or according to
the geometry and symmetry of the packing arrangement. Let’s focus first on the nature
of the bonding.
Crystalline solids
ionic
Network
covalent
molecular
In an ionic solid,
In a network
positive and
covalent solid,
negative ions are
atoms are held in
held in their lattice their lattice positions
positions by (strong) by (strong) covalent
ionic bonding
bonds.
forces.
In a molecular solid,
molecules are held
in their lattice
positions by (weak)
intermolecular
forces.
oC
H2O Tm = 0
NaCl Tm = 800
SiO2 Tm= 1600
LiF
Tm = 848 oC SiC
Tm =2830 oC S
Tm = 113 oC
8
o
ZnO Tm = 1974 C Cdiamond Tm = 4440 oC
I2
Tm = 114 oC
(at P = 12.4 GPa)
oC
oC
metallic
In a metallic solid,
metal cations are
held in their lattice
positions by
(strong) metallic
bonding forces.
Cu
Tm = 1083 oC
Ag
Tm = 962 oC
W
Tm = 3422 oC
Introduction to Solids
Crystalline solids are classified according to the nature of the bonding, or according to
the geometry and symmetry of the packing arrangement. Let’s focus first on the nature
of the bonding.
Crystalline solids
ionic
Network
covalent
molecular
In an ionic solid,
In a network
positive and
covalent solid,
negative ions are
atoms are held in
held in their lattice their lattice positions
positions by (strong) by (strong) covalent
ionic bonding
bonds.
forces.
In a molecular solid,
molecules are held
in their lattice
positions by (weak)
intermolecular
forces.
oC
H2O Tm = 0
NaCl Tm = 800
SiO2 Tm= 1600
LiF
Tm = 848 oC SiC
Tm =2830 oC S
Tm = 113 oC
8
o
ZnO Tm = 1974 C Cdiamond Tm = 4440 oC
I2
Tm = 114 oC
(at P = 12.4 GPa)
oC
oC
metallic
In a metallic solid,
metal cations are
held in their lattice
positions by
(strong) metallic
bonding forces.
Cu
Tm = 1083 oC
Ag
Tm = 962 oC
W
Tm = 3422 oC
Introduction to Solids
Crystalline solids are classified according to the nature of the bonding, or according to
the geometry and symmetry of the packing arrangement. Let’s focus first on the nature
of the bonding.
Crystalline solids
ionic
Network
covalent
molecular
In an ionic solid,
In a network
positive and
covalent solid,
negative ions are
atoms are held in
held in their lattice their lattice positions
positions by (strong) by (strong) covalent
ionic bonding
bonds.
forces.
In a molecular solid,
molecules are held
in their lattice
positions by (weak)
intermolecular
forces.
oC
H2O Tm = 0
NaCl Tm = 800
SiO2 Tm= 1600
LiF
Tm = 848 oC SiC
Tm =2830 oC S
Tm = 113 oC
8
o
ZnO Tm = 1974 C Cdiamond Tm = 4440 oC
I2
Tm = 114 oC
(at P = 12.4 GPa)
oC
oC
metallic
In a metallic solid,
metal cations are
held in their lattice
positions by
(strong) metallic
bonding forces.
Cu
Tm = 1083 oC
Ag
Tm = 962 oC
W
Tm = 3422 oC
Introduction to Solids
Metallic solids can be viewed as an array of metal cations bathing in a sea of valence
electrons.
Electron sea
model of metallic
bonding
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
A “sea” of
delocalized
valence
electrons
Introduction to Solids
When we focus on the geometry of the packing arrangements, we find that there are
seven basic shapes for unit cells. The shape of each unit cell is described in terms of
three lengths (a, b and c) and three angles (, , and g). We will focus primarily on
cubic unit cells. The others are mentioned only to emphasize that there are other
types/shapes of unit cells besides cubic unit cells.
cubic
a=b=c
 =  = g = 90o
trigonal
a=b=c
tetragonal
 =  = g ≠ 90o
a=b≠c
hexagonal
 =  = g = 90o
a=b≠c

o;

=

=
90
monoclinic
and g = 120o
a≠b≠c
= g = 90o
triclinic
and  ≠ 90o
a≠b≠c
orthorhombic  ≠  ≠ g ≠ 90o
g
a≠b≠c

 =  = g = 90o
a
b
c
Introduction to Solids
When we focus on the geometry of the packing arrangements, we find that there are
seven basic shapes for unit cells. The shape of each unit cell is described in terms of
three lengths (a, b and c) and three angles (, , and g). We will focus primarily on
cubic unit cells. The others are mentioned only to emphasize that there are other
types/shapes of unit cells besides cubic unit cells.
cubic
a=b=c
 =  = g = 90o
trigonal
a=b=c
tetragonal
 =  = g ≠ 90o
a=b≠c
hexagonal
 =  = g = 90o
a=b≠c
o
monoclinic  =  = 90 ;o
and g = 120
a≠b≠c
= g = 90o
triclinic
and  ≠ 90o
a≠b≠c
orthorhombic  ≠  ≠ g ≠ 90o
a≠b≠c
 =  = g = 90o
a
a
a
Intermolecular Forces:
Liquids and Solids
● Phases and Phase Diagrams
● Liquids and Liquid Properties
● Intermolecular Forces
● Heating Curves
● Introduction to Solids
● Cubic Packing Arrangements
● Closest-Packed Structures
● Density of a Crystalline Solid
● Ionic Solids and Interstitial Sites
● The Born-Haber Cycle
REVIEW
Introduction to Solids
For crystalline solids, we use the following concepts to characterize the pattern of the
packing arrangement.
Crystal lattice
three dimensional array of points that
shows how the structural units are
arranged in space.
Each structural unit could be a single atom, a molecule or an ion. In C(s), the structural
unit is a carbon atom. In I2(s), the structural unit is an I2 molecule. In NH4NO3(s), the
structural units are NH4+ and NO3− ions.
Unit cell
the smallest building block that possesses
the symmetry of the crystal lattice; a solid
sample of any size can be “built” by
stacking together unit cells.
Cubic Packing Arrangements
To investigate the cubic packing arrangements, we are going to imagine packing
together identical, hard spheres each having a radius R.
A. Simple cubic packing
Let’s consider ”building” a simple cubic packing arrangement on top of a table. We
generate the first layer by arranging the spheres as shown (on the left) below.
Subsequent layers are added by lining up spheres with those in the previous layer as
shown (on the right).
Layer 3
Layer 1
Layer 2
Layer 1
Cubic Packing Arrangements
A. Simple cubic packing
(i) What is the relationship between the edge length of the unit cell (a) and the atom’s
radius (R)?
a
R
a
a
a = 2×R
Cubic Packing Arrangements
A. Simple cubic packing
(ii) What is the number of spheres contained in one unit cell?
a
R
a
a
Each cell contains eight one eighth of a sphere. Thus there are:
1
Number of spheres per unit cell  8  1
8
Cubic Packing Arrangements
A. Simple cubic packing
(iii)
What is the coordination number of a sphere?
a
R
a
a
Each sphere touches four spheres in the plane, one on top, and one underneath. Thus
each sphere touches 6 other spheres. The coordination number equals 6.
Cubic Packing Arrangements
A. Simple cubic packing
(iv)
What is the fraction of empty space in a simple cubic crystal?
a
R
a
a
The unit cell occupies a volume VUC = (a)3 = (2×R)3 = 8×R3.
4
3
The unit cell contains one sphere of volume: Vsphere  R 3
The fraction of empty volume = (VUC – Vsphere)/VUC = 1 – /6 = 0.476
Cubic Packing Arrangements
B. Body-centred cubic packing
(i) What is the relationship between the edge length of the unit cell (a) and the atom’s
radius (R)?
a
R
Since the atoms are touching along the cube’s diagonal (body diagonal), its length
equals:
R + 2×R + R = 4×R
Cubic Packing Arrangements
B. Body-centred cubic packing
(i) What is the relationship between the edge length of the unit cell (a) and the atom’s
radius (R)?
Body Diagonal = 4×R
a
What is the square diagonal?
a
a
Cubic Packing Arrangements
B. Body-centred cubic packing
(i) What is the relationship between the edge length of the unit cell (a) and the atom’s
radius (R)?
Body Diagonal = 4×R
a
(Square Diagonal)2 = a2 + a2 = 2×a2
Square Diagonal 
2a
a
2 a
a
a
Cubic Packing Arrangements
B. Body-centred cubic packing
(i) What is the relationship between the edge length of the unit cell (a) and the atom’s
radius (R)?
Body Diagonal = 4×R
a
(Square Diagonal)2 = a2 + a2 = 2×a2
Square Diagonal 
2a
2 a
(Body Diagonal)2 = a2 + (2×a)2 = 3×a2
Body Diagonal = 3×a = 4×R
R
3
a
4
or
a
4
R
3
a
Cubic Packing Arrangements
B. Body-centred cubic packing
(ii) What is the number of spheres contained in one unit cell?
a
R
Each corner is occupied by one eighth of an atom and there is a full atom at the center
of the unit cell. Thus the cell contains:
1
Number of atoms per unit cell 1  8  1  1  2
8
Cubic Packing Arrangements
B. Body-centred cubic packing
(iii)
What is the coordination number of a sphere?
a
R
The atom at the center of the unit cell touches 8 other atoms. The coordination number
equals 8.
Cubic Packing Arrangements
B. Body-centred cubic packing
(iv)
What is the fraction of empty space in a body centered cubic crystal?
a
R
3
64 3
 4 
3
R

R


The unit cell occupies a volume VUC = (a) =
3
3
3


4
3
3
The unit cell contains two spheres, each sphere of volume: Vsphere  R
The fraction of empty volume = (VUC – 2×Vsphere)/VUC = 1 
3
  0.319
8
Cubic Packing Arrangements
C. Face-centred cubic packing
(i) What is the relationship between the edge length of the unit cell (a) and the atom’s
radius (R)?
a
R
Cubic Packing Arrangements
C. Face-centred cubic packing
(i) What is the relationship between the edge length of the unit cell (a) and the atom’s
radius (R)?
(Square Diagonal)2 = a2 + a2 = 2×a2
a
Square Diagonal = 2×a = 4×R
a2 2R
or
R
1
2 2
a
a
a
Cubic Packing Arrangements
C. Face-centred cubic packing
(ii) What is the number of spheres contained in one unit cell?
a
R
Each one of the eight corners is occupied by one eighth of an atom and each one of the
6 faces is occupied by one half of an atom.
1
1
Number of atoms per unit cell  6   8   3  1  4
2
8
Cubic Packing Arrangements
C. Face-centred cubic packing
(iii)
What is the coordination number of a sphere?
The atom at the center of the face touches 4 atoms in the plane shown on the figure.
Cubic Packing Arrangements
C. Face-centred cubic packing
(iii)
What is the coordination number of a sphere?
The atom at the center of the face touches 4 more atoms in the plane shown on the
figure.
Cubic Packing Arrangements
C. Face-centred cubic packing
(iii)
What is the coordination number of a sphere?
The atom at the center of the face touches 4 more atoms in the plane shown on the
figure.
Cubic Packing Arrangements
C. Face-centred cubic packing
(iii)
What is the coordination number of a sphere?
In total, the atom at the center of the face touches 12 atoms. The coordination number
equals 12.
Cubic Packing Arrangements
C. Face-centred cubic packing
(iv)
What is the fraction of empty space in a face-centered cubic crystal?
a
R
The unit cell occupies a volume VUC =
(a)3
= 2

3
2 R 16 2 R 3
4
3

 0.260
The fraction of empty volume = (VUC – 4×Vsphere)/VUC = 1 
3 2
3
The unit cell contains four spheres, each sphere of volume: Vsphere  R
Intermolecular Forces:
Liquids and Solids
● Phases and Phase Diagrams
● Liquids and Liquid Properties
● Intermolecular Forces
● Heating Curves
● Introduction to Solids
● Cubic Packing Arrangements
● Closest-Packed Structures
● Density of a Crystalline Solid
● Ionic Solids and Interstitial Sites
● The Born-Haber Cycle
Closest Packed Structures
Same layer as the layer
with the grey balls.
Different layer from the layer
with the grey balls.
Closest Packed Structures
“x” type dimple (vertically
aligned with a sphere in layer 1)
“y” type dimple (not
vertically aligned with a
sphere in layer 1 or layer 2)
There are two types of
“dimples” to choose from
when forming layer 3!!
Closest Packed Structures
“x” type dimple (vertically
aligned with a sphere in layer 1)
choose “x” only
“y” type dimple (not
vertically aligned with a
sphere in layer 1 or layer 2)
choose “y” only
Spheres in layer 3 are vertically
aligned with those in layer 1. That
is, layer 3 is a repeat of layer 1.
Spheres in layer 3 are not aligned
vertically with those in layer 1 or
layer 2. Layer 3 is a distinct layer.
ABAB ... closest-packing
ABCABC ... closest-packing
Closest Packed Structures
Spheres in layer 3 are vertically
aligned with those in layer 1. That
is, layer 3 is a repeat of layer 1.
Spheres in layer 3 are not aligned
vertically with those in layer 1 or
layer 2. Layer 3 is a distinct layer.
ABAB ... closest-packing
ABCABC ... closest-packing
The unit cell for ABAB ... closest-packing
is “hexagonal”. Thus, this type of
closest-packing is also called “hexagonal
closest-packing” or hcp for short.
The unit cell for ABCABC ... closestpacking is “face-centred cubic”.
Thus, this type of closest-packing is
also called “cubic closest-packing” or
ccp for short.
Closest Packed Structures
The unit cell for ABAB ... closest-packing
is “hexagonal”. Thus, this type of
closest-packing is also called “hexagonal
closest-packing” or hcp for short.
The unit cell for ABCABC ... closestpacking is “face-centred cubic”.
Thus, this type of closest-packing is
also called “cubic closest-packing” or
ccp for short.
You are expected to remember
that cubic closest-packing (ccp)
and face-centered cubic (fcc)
are the same!!
Intermolecular Forces:
Liquids and Solids
● Phases and Phase Diagrams
● Liquids and Liquid Properties
● Intermolecular Forces
● Heating Curves
● Introduction to Solids
● Cubic Packing Arrangements
● Closest-Packed Structures
● Density of a Crystalline Solid
● Ionic Solids and Interstitial Sites
● The Born-Haber Cycle
Density of a Crystalline Solid
The density of a solid depends on the microscopic details of the packing arrangement.
d
m mcell

V Vcell
M

natoms   atom 
NA 


a3
mass of one atom
Matom is molar mass, in g mol−1
NA = 6.022×1023 mol −1
Keep in mind:
The formula d = ncell Matom /(NA a3) should be used only for identical hard spheres in a
cubic lattice. If you have more than one type of sphere (e.g. of different sizes or masses)
or unit cell that is not cubic, then you should start from d = mcell / Vcell.
Density of a Crystalline Solid
Example: Tungsten, W, crystallizes in one of the three cubic arrangements. If the edge
length of the unit cell is 321 pm and the density is 18.5 g cm-3, then what type of crystal
lattice does tungsten have? What is the radius of a tungsten atom?
m mcell
d 
V Vcell
 M atom 

natoms  
NA 


a3
Matom = 184 g.mol1
Density of a Crystalline Solid
Example: Tungsten, W, crystallizes in one of the three cubic arrangements. If the edge
length of the unit cell is 321 pm and the density is 18.5 g cm-3, then what type of crystal
lattice does tungsten have? What is the radius of a tungsten atom?
m mcell
d 
V Vcell
 M atom 

natoms  
NA 


a3
Matom = 184 g.mol1
d  a3
natoms 
 M atom 


 NA 
18.5 ( g.cm 3 )  (32110 10 cm)3
natoms 
 2.003  2
1
 184 g.mol



23
1 
 6.022 10 mol 
Density of a Crystalline Solid
Example: Tungsten, W, crystallizes in one of the three cubic arrangements. If the edge
length of the unit cell is 321 pm and the density is 18.5 g cm-3, then what type of crystal
lattice does tungsten have? What is the radius of a tungsten atom?
Relationship
# of atoms/unit cell
between a and R
Simple cubic
Body-centered cubic
R = a/2
1
R
2
R
Face-centered cubic
4
3
a
4
3
 321  139 pm
4
R
1
2 2
a
Density of a Crystalline Solid
The edge length can be determined experimentally using x-ray diffraction. When x-rays
are passed through a crystalline solid, the x-rays are deflected from their paths by the
atoms of the solid and interfere with each other to produce an interference pattern – a
“diffraction pattern” – that can be analyzed to determine the geometry of the crystal
lattice.
Relationship
# of atoms/unit cell
between a and R
Simple cubic
1
Body-centered cubic
2
Face-centered cubic
4
R = a/2
R
R
3
a
4
1
2 2
a
REVIEW
Introduction to Solids
For crystalline solids, we use the following concepts to characterize the pattern of the
packing arrangement.
Crystal lattice
three dimensional array of points that
shows how the structural units are
arranged in space.
Each structural unit could be a single atom, a molecule or an ion. In C(s), the structural
unit is a carbon atom. In I2(s), the structural unit is an I2 molecule. In NH4NO3(s), the
structural units are NH4+ and NO3− ions.
Unit cell
the smallest building block that possesses
the symmetry of the crystal lattice; a solid
sample of any size can be “built” by
stacking together unit cells.
REVIEW
Introduction to Solids
When we focus on the geometry of the packing arrangements, we find that there are
seven basic shapes for unit cells. The shape of each unit cell is described in terms of
three lengths (a, b and c) and three angles (, , and g). We will focus primarily on
cubic unit cells. The others are mentioned only to emphasize that there are other
types/shapes of unit cells besides cubic unit cells.
cubic
a=b=c
 =  = g = 90o
trigonal
a=b=c
tetragonal
 =  = g ≠ 90o
a=b≠c
hexagonal
 =  = g = 90o
a=b≠c

o;

=

=
90
monoclinic
and g = 120o
a≠b≠c
= g = 90o
triclinic
and  ≠ 90o
a≠b≠c
orthorhombic  ≠  ≠ g ≠ 90o
g
a≠b≠c

 =  = g = 90o
a
b
c
REVIEW
Introduction to Solids
When we focus on the geometry of the packing arrangements, we find that there are
seven basic shapes for unit cells. The shape of each unit cell is described in terms of
three lengths (a, b and c) and three angles (, , and g). We will focus primarily on
cubic unit cells. The others are mentioned only to emphasize that there are other
types/shapes of unit cells besides cubic unit cells.
cubic
a=b=c
 =  = g = 90o
trigonal
a=b=c
tetragonal
 =  = g ≠ 90o
a=b≠c
hexagonal
 =  = g = 90o
a=b≠c
o
monoclinic  =  = 90 ;o
and g = 120
a≠b≠c
= g = 90o
triclinic
and  ≠ 90o
a≠b≠c
orthorhombic  ≠  ≠ g ≠ 90o
a≠b≠c
 =  = g = 90o
a
a
a
REVIEW
Density of a Crystalline Solid
The edge length can be determined experimentally using x-ray diffraction. When x-rays
are passed through a crystalline solid, the x-rays are deflected from their paths by the
atoms of the solid and interfere with each other to produce an interference pattern – a
“diffraction pattern” – that can be analyzed to determine the geometry of the crystal
lattice.
Relationship
# of atoms/unit cell
between a and R
Simple cubic
1
Body-centered cubic
2
Face-centered cubic
4
R = a/2
R = 0.50×a
3
a
4
R = 0.43×a
R
R
1
2 2
a
R = 0.35×a
Intermolecular Forces:
Liquids and Solids
● Phases and Phase Diagrams
● Liquids and Liquid Properties
● Intermolecular Forces
● Heating Curves
● Introduction to Solids
● Cubic Packing Arrangements
● Closest-Packed Structures
● Density of a Crystalline Solid
● Ionic Solids and Interstitial Sites
● The Born-Haber Cycle
Ionic Solids and Interstitial Sites
Up to this point, we’ve focused on packing together identical hard spheres. Now, we’ll
extend that model so that we can describe the structures of binary ionic solids.
For many ionic solids, it is often the case that one of the ions forms a cubic lattice and
the other ion occupies “holes” in that lattice. In order to understand the structures of
ionic crystals, we must first examine the types of holes we can have.
“Holes” (or interstitial sites) are named according to the “coordination number” of a small
sphere that just fits into that hole. The types of holes that interest us the most are cubic,
tetrahedral and octahedral holes.
Ionic Solids and Interstitial Sites
trigonal hole
The small sphere has a
coordination number of three.
Ionic Solids and Interstitial Sites
tetrahedral hole
coordination # = 4
Ionic Solids and Interstitial Sites
octahedral hole
coordination # = 6
top view
side view
Ionic Solids and Interstitial Sites
cubic hole
coordination # = 8
Ionic Solids and Interstitial Sites
Example: Within the face-centered cubic (fcc) unit cell, there are both octahedral and
tetrahedral holes. Examine the fcc unit cell to identify the locations of the octahedral and
tetrahedral holes. How many holes of each type are there per unit cell?
One octahedral hole
¼ of an octahedral hole
Ionic Solids and Interstitial Sites
Example: Within the face-centered cubic (fcc) unit cell, there are both octahedral and
tetrahedral holes. Examine the fcc unit cell to identify the locations of the octahedral and
tetrahedral holes. How many holes of each type are there per unit cell?
One octahedral hole
¼ of an octahedral hole
There is ¼ of an octahedral hole (per edge) and an octahedral hole at the centre of the
cell.
# octahedral holes = ¼ hole/edge × 12 edges + 1 = 4 (per fcc unit cell)
For the fcc cell, there are 4 spheres per cell and 4 octahedral holes holes. The ratio of
spheres-to-holes is:
# spheres : # octahedral holes = 4 : 4 = 1 : 1
Ionic Solids and Interstitial Sites
Example: Within the face-centered cubic (fcc) unit cell, there are both octahedral and
tetrahedral holes. Examine the fcc unit cell to identify the locations of the octahedral and
tetrahedral holes. How many holes of each type are there per unit cell?
b
a
tetrahedral hole
c
d
There is one tetrahedral hole associated with each corner of the fcc cell (that is, the
sphere at each corner can be considered to be the “cap” of a tetrahedron). Thus,
# tetrahedral holes = 8 (per fcc cell)
For the fcc cell, there are 4 spheres per celland 8 tetrahedral holes. The ratios of
spheres-to-holes are:
# spheres : # tetrahedral holes = 4 : 8 = 1 : 2
Ionic Solids and Interstitial Sites
Example: Within the face-centered cubic (fcc) unit cell, there are both octahedral and
tetrahedral holes. Examine the fcc unit cell to identify the locations of the octahedral and
tetrahedral holes. How many holes of each type are there per unit cell?
b
a
tetrahedral hole
c
d
In general: In a closest-packed structure containing N spheres, where N is a very large
number, there are N octahedral holes and 2N tetrahedral holes.
Ionic Solids and Interstitial Sites
How big are tetrahedral, octahedral and cubic holes?
Octahedral holes. Let ro be the radius of a sphere that just fits into the octahedral hole.
Let ro be the radius of a sphere that just
fits into the octahedral hole.
(2 R) 2  (2 R) 2  ( R  2ro  R) 2
2ro
2R
8  R 2  4  ( R  ro ) 2
2  R 2  ( R  ro ) 2
2 R  R  ro
2R
ro  R (1  2 )  0.414  R
Ionic Solids and Interstitial Sites
How big are tetrahedral, octahedral and cubic holes?
Cubic holes: Let ro be the radius of a sphere that just fits into the cubic hole.
Let rc be the radius of a sphere that just
fits into the cubic hole. This sphere has
its centre at the midpoint of the bodydiagonal. Therefore, we need to find the
length of the body-diagonal.
2R
BD  R  2ro  R  2  ( R  ro )
Ionic Solids and Interstitial Sites
How big are tetrahedral, octahedral and cubic holes?
Cubic holes: Let ro be the radius of a sphere that just fits into the cubic hole.
BD  R  2ro  R  2  ( R  ro )
Diagonal2 =
a2  a2  2  a2
Diagonal =
a
2a
a
2 a
Ionic Solids and Interstitial Sites
How big are tetrahedral, octahedral and cubic holes?
Cubic holes: Let ro be the radius of a sphere that just fits into the cubic hole.
BD  R  2ro  R  2  ( R  ro )
BD 2  ( 2a) 2  a 2  2  a 2  a 2  3  a 2
BD  3  a  3  (2  R)  2 3R
BD  2  ( R  ro )  2 3R
2a
ro  ( 3 1)  R  0.732  R
a
Ionic Solids and Interstitial Sites
How big are tetrahedral, octahedral and cubic holes?
Tetrahedral holes: Let ro be the radius of a sphere that just fits into the tetrahedral hole.
b
a
b
a
c
2×ro
d
c
d
Ionic Solids and Interstitial Sites
How big are tetrahedral, octahedral and cubic holes?
Tetrahedral holes: Let ro be the radius of a sphere that just fits into the tetrahedral hole.
b
a
b
a
x?
c
2×ro
d
c
BD  R  2ro  R  2  ( R  ro )
d
Ionic Solids and Interstitial Sites
How big are tetrahedral, octahedral and cubic holes?
Tetrahedral holes: Let ro be the radius of a sphere that just fits into the tetrahedral hole.
b
a
b
2×R
a
x
d
c
d
x
c
x 2  x 2  2  x 2  ( 2 R) 2
x  2R
Ionic Solids and Interstitial Sites
How big are tetrahedral, octahedral and cubic holes?
Tetrahedral holes: Let ro be the radius of a sphere that just fits into the tetrahedral hole.
b
a
b
a
x=
c
2×ro
d
2R
d
c
BD  R  2ro  R  2  ( R  ro )
BD 2  4  ( R  ro )2  (2R)2  ( 2R) 2  6  R 2
BD  2  ( R  ro )  6  R
Ionic Solids and Interstitial Sites
How big are tetrahedral, octahedral and cubic holes?
Tetrahedral holes: Let ro be the radius of a sphere that just fits into the tetrahedral hole.
b
a
b
a
x=
c
2×ro
d
c
6
ro  (
 1)  R  0.225  R
2
2R
d
Ionic Solids and Interstitial Sites
Summary of hole types and the radius ratio rules
Trigonal
Tetrahedral
Octahedral
cubic
Holes
Trigonal
Tetrahedral
Octahedral
cubic
Relationship
Too small to worry about
0.225  R
0.414  R
0.732  R
Ionic Solids and Interstitial Sites
Radius Ratio Rules for Ionic Solids
Let R+ be the radius of the positive ion in a binary ionic solid and let R− be the radius of
the negative ion.
Positive ions occupy tetrahedral sites in the
R
lattice of negative ions. The coordination
0.225   0.414
R
number of each positive ion is 4.
Note: The positive ions are too big for the tetrahedral sites and thus the negative ions
are forced apart a little  this increases the attraction between the “+” and “−” ions and
decreases the repulsion between the “−” ions.
0.414 
R
 0.732
R
0.732 
R
R
Positive ions occupy octahedral sites in the
lattice of negative ions. The coordination
number of each positive ion is 6.
Positive ions occupy cubic sites in the lattice
of negative ions. The coordination number
of each positive ion is 8.
Note: The rules above are just guidelines. There are many exceptions.
Ionic Solids and Interstitial Sites
Sodium Chloride Structure (also called “rock salt structure”)
Cl− ions form a face-centered cubic (fcc)
lattice and Na+ ions occupy 100% of the
octahedral holes in the chloride lattice.
Na
Cl
Ionic Solids and Interstitial Sites
Cesium Chloride Structure
Cl− ions form a simple cubic lattice
and Cs+ ions occupy cubic holes in
the chloride lattice.
Cs
Cl
Ionic Solids and Interstitial Sites
Zinc Blende Structure
S2− ions form a face-centered cubic (fcc)
lattice and Zn2+ ions occupy half (50%)
of the tetrahedral hole.
Why are only half of the tetrahedral
holes occupied?
How many spheres:
4 S2
How many tetrahedral holes:
8 for only
4 Zn2+
Ionic Solids and Interstitial Sites
Zinc Blende Structure
S2− ions form a face-centered cubic (fcc)
lattice and Zn2+ ions occupy half (50%)
of the tetrahedral hole.
Why are only half of the tetrahedral
holes occupied? There are 4 S2− ions
per cell. There must be four Zn2+ ions
per cell because ZnS is a 1:1 salt.
Therefore, only 4 of 8 tetrahedral holes
are occupied by Zn2+ ions.
REVIEW
Introduction to Solids
For crystalline solids, we use the following concepts to characterize the pattern of the
packing arrangement.
Crystal lattice
three dimensional array of points that
shows how the structural units are
arranged in space.
Each structural unit could be a single atom, a molecule or an ion. In C(s), the structural
unit is a carbon atom. In I2(s), the structural unit is an I2 molecule. In NH4NO3(s), the
structural units are NH4+ and NO3− ions.
Unit cell
the smallest building block that possesses
the symmetry of the crystal lattice; a solid
sample of any size can be “built” by
stacking together unit cells.
REVIEW
Introduction to Solids
When we focus on the geometry of the packing arrangements, we find that there are
seven basic shapes for unit cells. The shape of each unit cell is described in terms of
three lengths (a, b and c) and three angles (, , and g). We will focus primarily on
cubic unit cells. The others are mentioned only to emphasize that there are other
types/shapes of unit cells besides cubic unit cells.
cubic
a=b=c
 =  = g = 90o
trigonal
a=b=c
tetragonal
 =  = g ≠ 90o
a=b≠c
hexagonal
 =  = g = 90o
a=b≠c

o;

=

=
90
monoclinic
and g = 120o
a≠b≠c
= g = 90o
triclinic
and  ≠ 90o
a≠b≠c
orthorhombic  ≠  ≠ g ≠ 90o
g
a≠b≠c

 =  = g = 90o
a
b
c
REVIEW
Introduction to Solids
When we focus on the geometry of the packing arrangements, we find that there are
seven basic shapes for unit cells. The shape of each unit cell is described in terms of
three lengths (a, b and c) and three angles (, , and g). We will focus primarily on
cubic unit cells. The others are mentioned only to emphasize that there are other
types/shapes of unit cells besides cubic unit cells.
cubic
a=b=c
 =  = g = 90o
trigonal
a=b=c
tetragonal
 =  = g ≠ 90o
a=b≠c
hexagonal
 =  = g = 90o
a=b≠c
o
monoclinic  =  = 90 ;o
and g = 120
a≠b≠c
= g = 90o
triclinic
and  ≠ 90o
a≠b≠c
orthorhombic  ≠  ≠ g ≠ 90o
a≠b≠c
 =  = g = 90o
a
a
a
REVIEW
Density of a Crystalline Solid
The edge length can be determined experimentally using x-ray diffraction. When x-rays
are passed through a crystalline solid, the x-rays are deflected from their paths by the
atoms of the solid and interfere with each other to produce an interference pattern – a
“diffraction pattern” – that can be analyzed to determine the geometry of the crystal
lattice.
Relationship
# of atoms/unit cell
between a and R
Simple cubic
1
Body-centered cubic
2
Face-centered cubic
4
R = a/2
R = 0.50×a
3
a
4
R = 0.43×a
R
R
1
2 2
a
R = 0.35×a
REVIEW
Ionic Solids and Interstitial Sites
Summary of hole types and the radius ratio rules
Trigonal
Tetrahedral
Octahedral
cubic
Holes
Trigonal
Tetrahedral
Octahedral
cubic
Relationship
Too small to worry about
0.225  R
0.414  R
0.732  R
Ionic Solids and Interstitial Sites
Fluorite Structure
Ca2+ ions form a fcc lattice and F− ions
occupy all of the tetrahedral holes.
Thus, there are 4 Ca2+ ions per unit cell and
8 F− ions.
How many spheres:
4 Ca2+
How many tetrahedral holes:
8 F
Ionic Solids and Interstitial Sites
Antifluorite Structure
O2
Na
Sodium oxide, Na2O, adopts the so-called
antifluorite structure. The antifluorite
structure is the “reverse” of the fluorite
structure in the sense that the negative ions
form a fcc lattice and the positive ions
occupy all of the tetrahedral holes. In the
case of Na2O, the O2− ions form a fcc lattice
and the Na+ ions occupy 100% of the
tetrahedral holes.
Ionic Solids and Interstitial Sites
Example: In beryllium oxide, BeO, the oxide ions form a face-centred cubic lattice and
the beryllium ions occupy tetrahedral sites in the lattice of O2- ions. What fraction of the
holes is occupied by the Be2+ ions? (Ans: 50%)
How many spheres:
4 O2
How many tetrahedral holes:
8 tetrahedral holes for Be2
With 8 negative charges from O2 anions and 16 positive charges from Be2+ cations, the charge
balance would not work. Only 4 Be2+ generating 8 positive charges can be present.
50% of the tetrahedral holes can be occupied by Be2+ cations.
Ionic Solids and Interstitial Sites
Example: To “build” the sodium chloride structure, we would first arrange Cl ions in a
face-centred cubic structure and then insert Na+ ions into the octahedral sites of the Cl
lattice. Assume that Na+ and Cl are hard spheres with radii of 97 pm and 181 pm,
respectively. Calculate the density of sodium chloride, in g cm3. (Ans: 2.26 g cm−3)
a = 2×(97+181) = 556 pm
Na
4 spheres (Cl)
4 octahedral holes (Na+)
Cl
a
d
4  ( M Na   M Cl  )
4  (23  35.5 g / mol )
3


2
.
26
g
.
cm
a3  N A
(556 10 10 cm) 3  (6.022 10 23 mol 1 )
Ionic Solids and Interstitial Sites
Example: In cesium chloride, the Cl ions form a simple cubic lattice and the Cs+ ions
occupy cubic holes in the chloride lattice. However, the Cs+ ions force the Cl ions apart
so that none of the chloride ions are in direct contact with each other. If the density of
cesium chloride, CsCl, is 3.988 g cm3, then what is the distance (R+ + R) ? (Ans: 357
pm)
1 spheres (Cl)
1 cubic hole (Cs+)
d
Cs
a
Cl
a3
( M Cs   M Cl  )
a3  N A
a3 
( M Cs   M Cl  )
d  NA
where BD  2  ( R  R )  3  a
M Cs   M Cl 
132.9  35.5
8
3
4
.
12

10
cm  412 pm
3
23
d  NA
3.988 g.cm  6.022 10
Ionic Solids and Interstitial Sites
Example: In cesium chloride, the Cl ions form a simple cubic lattice and the Cs+ ions
occupy cubic holes in the chloride lattice. However, the Cs+ ions force the Cl ions apart
so that none of the chloride ions are in direct contact with each other. If the density of
cesium chloride, CsCl, is 3.988 g cm3, then what is the distance (R+ + R) ? (Ans: 357
pm)
a  412 pm
BD  3  a  3  412  714 pm
a
Cs
Cl
BD  714 pm  2  ( R  R )
R  R 
714 pm
 357 pm
2
Intermolecular Forces:
Liquids and Solids
● Phases and Phase Diagrams
● Liquids and Liquid Properties
● Intermolecular Forces
● Heating Curves
● Introduction to Solids
● Cubic Packing Arrangements
● Closest-Packed Structures
● Density of a Crystalline Solid
● Ionic Solids and Interstitial Sites
● The Born-Haber Cycle
The Born Haber Cycle
The stability of an ionic solid is quantified in terms of its lattice energy.
Lattice Energy = energy change when gas phase ions combine to form an ionic solid
Lattice energy (Hlattice) for NaCl(s)
Hlattice
Na+(g) + Cl−(g) → NaCl(s)
The lattice energy cannot be measured directly. However, we can obtain it indirectly
from other thermochemical data using the Born-Haber cycle.
The Born-Haber cycle involves the following steps:
• Write down the “formation reaction” for the solid from the elements
under standard conditions (T = 25 oC and P = 105 bar).
• Convert the elements into gas-phase atoms.
• Convert the atoms into gas-phase ions.
• Combine the ions to form the solid.
The Born Haber Cycle
Born-Haber cycle for NaCl(s)
• Write down the “formation reaction” for the solid.
• Convert the elements into gas-phase atoms.
• Convert the atoms into gas-phase ions.
• Combine the ions to form the solid.
Hfo
Na(s) + ½ Cl2(g)
NaCl(s)
Hsubo
(sublimation)
Na(g) +
IE(1)
(first ionization
energy)
Na+(g) +
½ DCl-Cl
(dissociation)
Cl(g)
EA(1)
(first electron
affinity)
Cl(g)
Hlatticeo
The Born Haber Cycle
Born-Haber cycle for NaCl(s)
State function does not depend on the path:
University of
d(home – UW) = constant
path(UW  home) varies
The Born Haber Cycle
Born-Haber cycle for NaCl(s)
• Write down the “formation reaction” for the solid.
• Convert the elements into gas-phase atoms.
• Convert the atoms into gas-phase ions.
• Combine the ions to form the solid.
Hfo
Na(s) + ½ Cl2(g)
NaCl(s)
Hsubo
(sublimation)
Na(g) +
IE(1)
(first ionization
energy)
Na+(g) +
½ DCl-Cl
(dissociation)
Cl(g)
EA(1)
(first electron
affinity)
Hlatticeo
Cl(g)
Apply Hess’ Law: Hfo = Hsubo + IE(1)Na(g) + ½ DCl-Cl + EA(1)Cl(g) + Hlattice
Thus:
Hlattice =  (Hsubo + IE(1)Na(g) + ½ DCl-Cl + EA(1)Cl(g))  Hfo
The Born Haber Cycle
Born-Haber cycle for MgF2(s)
• Write down the “formation reaction” for the solid.
• Convert the elements into gas-phase atoms.
• Convert the atoms into gas-phase ions.
• Combine the ions to form the solid.
Hfo
Mg(s) + F2(g)
MgF2(s)
Hsubo
(sublimation)
Mg(g) +
IE(1) + IE(2)
(first and second
ionization energy)
DF-F
(dissociation)
2 F(g)
EA(1)
(first electron
affinity)
Hlatticeo
Mg2+(g) + 2 F(g)
Apply Hess’ Law: Hfo = Hsubo + IE(1)Mg(g) +IE(2)Mg(g) + DF-F + 2×EA(1)F(g) + Hlattice
Thus:
Hlattice = (Hsubo + IE(1)Mg(g) +IE(2)Mg(g) + DF-F + 2×EA(1)F(g)) + Hfo
The Born Haber Cycle
Born-Haber cycle for NaCl(s)
Hess’ Law: Hfo = Hsubo + IE(1)Na(g) + ½ DCl-Cl + EA(1)Cl(g) + Hlattice
Born-Haber cycle for MgF2(s)
Hess’ Law: Hfo = Hsubo + IE(1)Mg(g) +IE(2)Mg(g) + DF-F + 2×EA(1)F(g) + Hlattice
By comparing the expressions we wrote for NaCl(s) and MgCl2(s), we can see that the
expression we obtain for ΔHfo will vary from salt to salt. Therefore, to obtain the correct
expression, you must first construct the Born-Haber cycle.
Intermolecular Forces:
Liquids and Solids
● Phases and Phase Diagrams
● Liquids and Liquid Properties
● Intermolecular Forces
● Heating Curves
● Introduction to Solids
● Cubic Packing Arrangements
● Closest-Packed Structures
● Density of a Crystalline Solid
● Ionic Solids and Interstitial Sites
● The Born-Haber Cycle