1
11

2
INTERMOLECULAR FORCES
Hydrogen bonds
Van der
Waals’ forces
Dipole-dipole forces
London
Dispersion forces

3
Johannes van der Waals
(1837 − 1923).
Fritz London
(1900 − 1954).

4
Permanent dipole
3 types of dipoles
Instantaneous dipole
Induced dipole

5
A permanent dipole exists in all polar molecules as a result of the difference in the electronegativity of bonded atoms.

6
An instantaneous dipole is a temporary dipole that exists as a result of fluctuation in the electron cloud.

7
An instantaneous dipole is a temporary dipole that exists as a result of fluctuation in the electron cloud.

8
An induced dipole is a temporary dipole that is created due to the influence of neighbouring dipole (which may be a permanent or an instantaneous dipole).
Permanent dipole

9
11.2

Van der Waals’ forces
10
Dipole-
Dipole
Interaction
Dipole-
Induced
Dipole
Interaction
Instantaneous
Dipole-
Induced Dipole
Interaction
London dispersion forces

• Electrostatic interactions between polar molecules
11

• In a sample containing many polar molecules
A balance of attraction and repulsion holding the molecules together
12

• When a non-polar molecule approaches a polar molecule (with a permanent dipole), a dipole will be induced in the non-polar molecule.
13
Dispersion forces exist among all molecules and contribute most to the overall van der Waals’ forces.

Polarization
Polarizability : - A measure of how easily the electron cloud of an atom/molecule can be distorted to induce a dipole
14

15
In general, size of electron cloud 
 electron cloud is less controlled by positive nuclei
 extent of electron cloud distortion 
 polarizability 
 stronger dispersion forces

11.2 Van der Waals’ forces (SB p.277)
• The instantaneous dipole arises from constant movement of electrons.
• Induces dipoles in neighbouring atoms or molecules
16

17

18

19

20
Evidence for the presence of London dispersion forces
1. Condensation of noble gases at low temperatures to form liquids and solids
 presence of attractive forces between non-polar atoms
E.g. Xe(g)  Xe(s)  H sub
= -14.9 kJ mol  1

21
Evidence for the presence of London dispersion forces
2. The non-ideal behaviour of gases



P
 a n
V
2




V
 bn

 nRT van der Waals’ equation

11.2 Van der Waals’ forces (SB p.279)
Much weaker than covalent bonds
Less than 10% the strength of covalent bonds van der Waals’ radius > covalent radius
I
2
22

23
Q.59
The electron clouds of adjacent iodine molecules would repel each other strongly until the equilibrium van der Waals’ distance is restored.

24
The strength of van der Waals’ forces can be estimated by melting point, boiling point, enthalpy change of fusion or enthalpy change of vapourization.
Higher m.p./b.p./  H fusion
/  H vap
 stronger van der Waals’ forces

25
Depends on three factors (in decreasing order of importance) : -
1. Size of molecule
2. Surface area of molecule
3. Polarity of molecule

Molecule Boiling point
( o C)
Helium
Neon
Argon
-269
-246
-186
Fluorine
Chlorine
Bromine
-188
-34.7
58.8
Methane
Ethane
Propane
26
-162
-88.6
-42.2
Sometimes !
Size of molecule  
Size of electron cloud 
Polarizability 
Dispersion forces 

27
The van der Waals’ forces also increase with the surface area of the molecule.
∵ van der Waals' forces are short-ranged forces
Atoms or molecules must come close together for significant induction of dipoles.

28
Pentane (C
5
H
12
)
2,2-dimethylpropane
(C
5
H
12
)
Both are non-polar
Same no. of electrons
Boiling point: 36.1
°
C Boiling point: 9.5
°
C

rod-shaped spherical in shape pentane molecules larger contact area
2,2-dimethylpropane molecules smaller contact area
29

Pentane (C
5
H
12
)
Boiling point
= 36.1
 C
Larger contact surface area
 Higher chance of forming induced dipoles
 stronger dispersion forces
30

2,2-dimethylpropane
(C
5
H
12
)
Boiling point
= 9.5
 C
Smaller contact surface area
 lower chance of forming induced dipoles
 weaker dispersion forces
31

For molecules with comparable molecular sizes and shapes, dispersion forces are approximately equal.
Then, strength of van der Waals’ forces depends on the polarity of molecules involved
Polar/polar > polar/non-polar > non-polar/non-polar
32

33
H
3
C
 + 
C O
Dipole-dipole forces
H
3
C
+
Dispersion forces
H
3
C H
3
C
RMM = 58.0, b.p. = 50  C
 + 
C O
H
2
C
H
3
C C
H
2
RMM = 58.0, b.p. = 0  C
CH
3
H
3
C
Dispersion forces only
H
2
C
C
H
2
CH
3

Other examples : -
1. Graphite layers of large surface area
 strong van der Waals’ forces
2. Polyethene vs ethene
(m.p. > 100  C) (m.p. =  169  C)
34

% contribution to the overall van der Waals' forces
Molecule Dipoledipole interaction
Dipoleinduced dipole interaction
Instantaneous dipoleinduced dipole interaction
C
4
H
10
HCl
0
15
0
4
100
81
35

Q.60(a)
CH
3
Cl < CH
3
Br < CH
3
I b.p./  C -24.2 3.56
42.4
The strength of dispersion forces increases with molecular size/mass .
Thus, b.p. increases with molecular size/mass
Although chloromethane is more polar, the effect of dispersion forces outweights that of dipole-dipole forces.
36

Q.60(b)
CH
3
H
3
C
C
CH
3
CH
3
9.5
 C
<
H
3
C
H
2
C
CH
CH
3
CH
3
27.7
 C
<
Less spherical
Greater surface area
H
3
C
H
2
C
C
H
2
H
2
C
CH
3
36.1
 C
37

Q.60(c)
F
2
F
2
Cl
2
ClF
< ClF < Cl
2
CH
< CH
2
2
Cl
Cl
2
2
-188  C -100  C -34.0
 C 39.6
 C
ClF > F
2
. It is because
1. ClF has a greater molecular size than F
2 and thus has stronger dispersion forces than F
2
2. ClF is polar and its molecules are held by both dipole-dipole forces and dispersion forces.
38

Q.60(c)
F
2
< ClF < Cl
2
> CH
2
Cl
2
-188  C -100  C -34.0
 C 39.6
 C
Cl
2
> ClF. It is because
1. Cl
2 has a greater molecular size than ClF and thus has stronger dispersion forces than ClF.
2.Although ClF is polar, the effect of dispersion forces outweights that of dipole-dipole forces.
39

Q.60(c)
F
2
< ClF < Cl
2
> CH
2
Cl
2
-188  C -100  C -34.0
 C 39.6
 C
CH
2
Cl
2
> Cl
2
. It is because
1. CH
2
Cl
2 has a greater molecular size than Cl thus has stronger dispersion forces than Cl
2
2
.
and
2.CH
2
Cl
2 is polar and its molecules are held by both dipole-dipole forces and dispersion forces.
40

Q.60(d)
NO < C
2
H
6 b.p./  C -151 -89
RMM 28.0 28.0
41

1 pm = 0.001 nm
1 nm = 10  9 m
42

43
NO < C
2
H
6 b.p./  C -151 -89
RMM 28.0 28.0
C
2
H
6
> NO. It is because
1. C
2
H
6 has a greater molecular size and contact surface area than NO and thus has stronger dispersion forces than NO.
2.Although NO is polar, the effect of dispersion forces outweights that of dipole-dipole forces.

44
The melting of a solid involves the separation of molecules from a regularly packed molecular crystal.
Thus, m.p. of a solid depends on
1. The strength of van der Waals’ forces
2.
Packing efficiency of molecules in the crystal lattice

45






Q.61
m.p.
H
3
C
H
2
C
CH
CH
3
CH
3
-160  C
<
H
3
C
H
2
C
C
H
2
H
2
C
CH
3
-136  C
CH
3
<
H
3
C
C
CH
3
CH
3
-20  C
Increasing symmetry
Increasing packing efficiency
46

47
Q.61
H
3
C
H
2
C
CH
CH
3
CH
3
<
H
3
C
H
2
C
C
H
2
H
2
C
CH
3
Greater surface area
Stronger van der Waals’ forces

48
11.4

49
A molecular crystal is a structure which consists of individual molecules packed together in a regular arrangement by weak intermolecular forces.

f.c.c. structure
50
A unit cell of iodine crystal showing the orientation of I
2 molecules

f.c.c. structure
51
A unit cell of dry ice (CO
2
)

Fullerenes are molecules composed entirely of carbon atoms , in the form of hollow spheres or hollow tubes .
52

The first fullerene discovered was buckminsterfullerene (C
60
) .
53
Buckminsterfullerene.
A soccer ball.

54
R.F. Curl
Discovered C
60
H.W. Kroto in 1985
R.E. Smalley
Awarded Nobel prize for Chemistry in 1996

C
60
Cutting at
12 vertices
55 icosahedron
正二十面體 truncated icosahedron

56
12 pentagons by cutting at 12 vertices.

57
20 hexagons by cutting 20 triangular faces.

58
Named after the architect
Richard Buckminster Fuller
A geodesic dome

59
Each carbon atom is connected to three other carbon atoms by one double covalent bond and two single covalent bonds .
Buckminsterfuller ene

60
Each pentagon is connected to five hexagons
Each hexagon is connected to three pentagons and three hexagons alternately .

61
Graphite is planar because it is made of hexagonal rings linked together.

62
In C
60
, pentagonal rings prevent the sheet from being planar, making it spherical.

0.1434 nm
0.1386 nm
Why are there two types of bond in C
60
?
63

The surface of the sphere is NOT planar
 2p z orbitals are NOT parallel to one another
 Delocalization of  e  s is NOT favourable
64

Family of fullerene s
C
28
C
32
C
50
Some of the more stable members of the fullerene family. (a) C
28
(b) C
32
(c) C
50
(d) C
70
C
70
65

66
Molecular structure
C
60 molecules held by dispersion forces

Fullerene molecules are held together by weak van der Waals’ forces .
Substance
Graphite
Diamond
Buckminsterfullerene
Melting point ( ° C)
3730
3550
1070
67

Giant covalent structure
68
Molecular structure

Buckminsterfullerenes are relatively strong and hard compared with most other molecular solids.
The C
60 molecules are packed closely together in solid state.
buckminsterfullerene molecule (C
60
)
69

Pure buckminsterfullerene (C
60
) is an electrical insulator .(no delocalized electrons) buckminsterfullerene
The buckminsterfullerene with potassium atoms filling the spaces between its molecules is a superconductor . Its formula is K
3
C
60
.
potassium atom
70

Carbon nanotube (CNT) or buckytube
71
First discovered by Dr. Sumio Iijima in 1991

Carbon nanotube (CNT) or buckytube
72
It is formed by carbon atoms arranged in a long cylindrical hollow tube .

Carbon nanotube (CNT) or buckytube
73
The diameter of a nanotube is in the order of a few nanometres (10  9 m) .

Carbon nanotube (CNT) or buckytube
74

75
Graphite sheet

Carbon nanotube (CNT) or buckytube
76

Properties of nanotubes
The tensile strength of carbon nanotubes is exceptionally high due to the strong covalent bonds holding the atoms together
The strongest materials on earth.
~100 times stronger than steel
Applications : clothes, sports equipments, space elevators…
77

Properties of nanotubes
Carbon nanotube is an electrical conductor because of the movement of delocalized electrons along the graphite sheets.
Depending on their structures, carbon nanotubes can be semi-conducting or as electrically conductive as metals .
78

79
11.5

Look at the boiling points of some simple hydrides of Group IV to VII elements
(p.87).
80

B.p.  as molecular size 
Group 4 hydrides are non-polar, only dispersion forces exist
Dispersion forces  as molecular size  .
81

All are polar
B.p.  as molecular size  (dispersion > dipole-dipole)
However, H
2
O, HF and NH
3 have abnormally high b.p.
There exist unusually strong dipole-dipole forces (H-bond)
82

When a hydrogen atom is directly bonded to a highly electronegative atom
(e.g. fluorine, oxygen and nitrogen), a highly polar bond is formed.
83
2.1
4.0
2.1
3.5
2.1
3.0

84
Electrostatic attractions exist between this partial positive charge and the lone pair electrons on a highly electronegative atom (i.e. fluorine, oxygen or nitrogen) of another molecule.
These attractions are called hydrogen bonds

85 hydrogen bond

hydrogen bond
86
Formation of hydrogen bonds between H
2
O molecules.

Reasons for abnormal strength of H-bond
1. the polarity of H–X bond is great when X is F ,
O , or N.
2. H atom does not have inner electrons.
 its nucleus (proton) is partially exposed due to unequal sharing of electron.
 The partial positive charge on H is so concentrated that it can come very close to the lone pair of a small & highly electronegative atom (F, O or N)
 Abnormally strong dipole-dipole forces
87

88
Two essential requirements for the formation of a hydrogen bond:
• One molecule must contain at least one
H atom attached to a highly electronegative atom (i.e. F, O or N).
• The other molecule must contain an F,
O or N atom that provides the lone pair of electrons.

Identify the hydrogen atoms of the following species that are capable of forming hydrogen bonding with water molecules.
89 adenine
Soluble in water glucose

An exceptional case : -

Cl
 Cl C
 +
H
H-bond
O C
CH
3
Cl

CH
3
Due to the combined effect of the three electronegative Cl atoms, the H atom becomes sufficiently positive to form hydrogen bond
90

91
Relative strength of van der Waals’ forces, hydrogen bond and covalent bond
Phenomenon
He(s)  He(g)
Energy absorbed
(kJ mol -1 )
0.11
Forces overcome
Van der Waals’ forces
H
2
O(s)  H
2
O(g) 46.90
Hydrogen bonds
O
2
(g)  2O(g) 494.00
Covalent bonds

Q.65
Tendency of H-bond formation : -
 EN 0.4 0.4 0.9 0.9 1.4 1.9
C – H < S – H < Cl – H < N – H < O – H < F – H
No lone pair on C
N is smaller than Cl
H can come closer
92

Q.66
Substance
NH
3
HF
H
2
O
Relative molecular mass
17
20
18
Boiling point
(
°
C)
-33.3
19.5
100
HF > NH
3 because
H – F bond is more polar than N – H bond
93

Q.66
Substance
NH
3
HF
H
2
O
Relative molecular mass
17
20
18
Boiling point
(
°
C)
-33.3
19.5
100
94
H
2
O > HF because
H
2
O can form H-bonds more extensively , regardless of the fact that H-F bond is more polar than H-O bond.

hydrogen bond
95
Each NH
3 molecule has only ONE
 On the average, each NH
3 only ONE hydrogen bond lone pair.
molecule can form

96
Each HF molecule has only ONE hydrogen atom.
 On the average, each HF molecule can form only ONE hydrogen bond

hydrogen bond
97
Each H
2
O molecule has TWO hydrogen atoms and
TWO lone pairs.
 On the average, each H
2
TWO hydrogen bonds
O molecule can form

a water molecule hydrogen bond
The lone pairs of oxygen atom of each water molecule forms hydrogen bonds with two hydrogen atoms of nearby water molecules hydrogen bond hydrogen atom oxygen atom
98

hydrogen bond
The two hydrogen atoms of each water molecule also form hydrogen bonds with the lone pairs of oxygen atoms of nearby water molecules.
hydrogen bond hydrogen atom oxygen atom
99

100
Each H
2
O molecule is bonded tetrahedrally to four H
2
O molecules
1
4
2
3

In solid ice, the tetrahedral arrangement repeats over and over again, resulting in an open and regular network structure of water molecules.
Open : the maximum number of hydrogen bonds can be formed
Regular : all molecules are held in positions by strong hydrogen bonds
101

102
The oxygen atoms in the structure of ice are arranged in a hexagonal shape.

103
The hexagonal symmetry of a snowflake reflects the structure of ice.

In liquid state, water molecules pack together more closely and randomly .
Hydrogen bonds are continuously formed and broken .
Liquid water takes the shapes of the containers
104

Most substances have higher densities in the solid state than in the liquid state.
liquid paraffin solid paraffin
Solid paraffin is denser than liquid paraffin.
ice water
105

Ice has a lower density than liquid water!
At 0
°
C, density of ice = 0.92 g cm −3 density of liquid water = 1.00 g cm −3
106 liquid paraffin solid paraffin ice water

107
In cold weather, ice forms a layer on the top of a pond.
Ice acts as an insulator for the water beneath.
This allows fish and other aquatic organisms to survive.

In ice, water molecules are arranged in an orderly manner in an open network structure because of extensive formation of hydrogen bonding .
108

109
In this open structure, water molecules are further apart than they are in liquid water.
ice liquid water melts open structure collapses water molecules tend to pack more closely together

Energy is absorbed to break some of the hydrogen bonds
Less H-bonds
 less stable
 Close & random
110
More H-bonds
 More stable
 Open & regular

The melting point (0
°
C) and boiling point
(100
°
C) of water are much higher than expected .
A lot of energy is required to overcome the hydrogen bonds between water molecules and separate them.
111

112
High surface tension of water allows water striders to ‘walk’ on it.

Surface tension of molecular liquids arises from intermolecular forces.
Stronger intermolecular forces leads to higher surface tension
Liquid
Hexane
Relative surface tension at 25  C
18.4
Methanol
Ethanol
Water
22.6
22.8
72.3
113

Water molecules at the surface are strongly attracted by neighboring molecules on the same surface.
The surface of water is like a tightly-stretched skin such that small insects can walk on it.
intermolecular forces
114

115
Water forms droplets rather than spreading out on leaf.

In a sample of water, each water molecule is attracted to neighboring water molecules in all directions and there is a balance of force .
116

There is an imbalance of force for the molecules at the surface.
The water molecules at the surface tend to be pulled inwards by other water molecules below the surface.
As a result, water forms droplets rather than spreading out on leaf.
In other words, water tends to reduce its surface area by taking the spherical shape.
117

118
The high surface tension of water allows water to be transported to the top of trees by capillary action .
The tallest tree on earth
115.56m

The resistance of a liquid to flow.
119
The higher the viscosity of a liquid, the more slowly it flows.
Viscosity arises from intermolecular forces

Strong hydrogen bonds hold water molecules together and do not allow them to move past one another easily.
Liquid Relative viscosity
Benzene 1
Water 15
Water has high melting and boiling points, high surface tension and high viscosity .
120
Surface tension

Consider an ethanol molecule.
hydroxyl group lone pairs of electrons
121

Ethanol molecules are held together by H-bonds .
 high boiling point hydrogen bond
H-bond strength
122

Alcohols vs Thiols (p.90)
Alcohol CH
3
OH C
2
H
5
OH C
3
H
7
OH C
4
H
9
OH b.p.(  C) 64.5
78 97 117
Thiol CH
3
SH C
2
H
5
SH C
3
H
7
SH C
4
H
9
SH b.p.(  C) 5.8
37 67 97
Dispersion forces : Thiol > alcohol
Boiling point : Alcohol > thiol
123

hydrogen bonds
124 water ethanol
Ethanol and water are completely miscible

Ethanol is viscous because of the presence of extensive intermolecular hydrogen bonds.
Ethanol is viscous , completely miscible with water , and has a high boiling point .
125

126 propan-1-ol propane-1,2-diol propane-1,2,3-triol
Viscosity  as no. of OH groups per molecule 
Viscosity

Explain the following.
Water is easily absorbed by tissue paper rather than forming droplets on it.
Tissue paper is composed of cellulose which is a natural polymer made of glucose molecules.
Thus, tissue paper can form extensive hydrogen bonds with water molecules.
127

Carboxylic Acids
O
H
3
C
RMM = 60
C
O
H
H-bonds
H
O
O
C CH
3
Ethanoic acid exists as dimers , (CH
3
COOH)
2 in vapour phase or in non-polar solvents
,
RMM = 2  60 = 120
128

Q.67
Ethanoic acid molecules form H-bonds with polar solvent molecules rather than with other ethanoic molecules.
H
H
O
O
H
3
C C
H
O H
O
H
129

Q.67
Ethanoic acid molecules form H-bonds with polar solvent molecules rather than with other ethanoic molecules.
R
H
O
O
H
3
C C H
O H O
R
130

131
Intramolecular Hydrogen Bonding
O
H
O
H
O
N
O b.p. = 214  C N
O O b.p. = 279  C

132
Formation of intramolecular hydrogen bonds prevents the formation of intermolecular hydrogen bonds  lower boiling point
H
H
O
O O
N
O b.p. = 214  C N
O O b.p. = 279  C

Roles of Hydrogen Bonding in Biochemical Systems
Proteins : polymers of amino acids
Primary structure : sequence of amino acids
Peptide linkage
133

Secondary structures : -
1.
 -pleated sheet
2.
 -helix
134

1.
 -pleated sheet
Intermolecular H-bonds formed between peptide linkages of adjacent protein chains
135

C O C O
N C N C
H planar
C H C
Both N and C of the N – C bond are sp 2 hybridized to facilitate delocalization of  electrons
136

C O C O
N C N C
H C H planar
N – C bond has double bond character
C
 free rotation w.r.t. the bond axis is restricted
 N – H and C = O groups are held in opposite positions to facilitate the formation of inter-chain H-bonds
137

138

2.  -helical structure
Intramolecular hydrogen bond
C
O
H
N
139

2.  -helical structure
The molecular chains of protein can be held in position to give the
 -helical structure by forming intramolecular hydrogen bonds.
140

Both  - and  - structures were first suggested by Linus Pauling
.
141

Tertiary structure
3-D arrangements of secondary structures
Myoglobin
142

Quaternary structure
3-D arrangements of tertiary structures
Haemoglobin
143

Hydrogen bonding in DNA
DNA ( D eoxyribon N uclei A cid) carries genetic information hydrogen bonds
144

Effect of hydrogen bonding on DNA
The presence of intermolecular H-bonds helps maintain the double helical shape of DNA molecules. hydrogen bonds
145

Effect of hydrogen bonding on DNA
The double helical structure is maintained by intermolecular hydrogen bonds formed between specific base pairs
146

Effect of hydrogen bonding on DNA
147
C ytosine G uanine

Effect of hydrogen bonding on DNA
148
T hymine A denine

Effect of hydrogen bonding on DNA
Sequence of bases = genetic code
CAGACTTGCAAT…
GTCTGAACGTTA…
Or
149

150
Without hydrogen bond, life becomes impossible hydrogen bonds

151
Without hydrogen bond, life becomes impossible

152
Without hydrogen bond, life becomes impossible
H
 +
O O
O

H
 +
It allows oxygen to dissolve in water

153
• 3 different states: solid , liquid and gas
• Change of states involves breaking or forming of intermolecular forces of the molecular substances

Phase Diagram
A phase diagram is a graph summarizing the conditions of pressure and temperature under which the different phases of a substance are stable.
Phase  state
E.g. C in the same state may have different phases
Graphite, diamond, C
60
154

A phase is any homogeneous and physically distinct part of a system which is separated from other parts of the system by a definite physical boundary known as the phase boundary .
155

A system having two phases in the same liquid state oil
Phase boundary water
156

A system having three phases in two states oil
Phase boundaries water glass
157

A system having four phases in three states oil air glass
Phase boundaries water
158

Phase Diagram of Carbon Dioxide
Three regions in each of which only one phase is stable
P / atm
Solid
Liquid
Vapour
T /  C
159

Phase Diagram of Carbon Dioxide
The three regions meet at three lines , along which two phases coexist in equilibrium.
P / atm
Solid
Liquid
Vapour
T /  C
160

Phase Diagram of Carbon Dioxide
AT is the sublimation curve
P / atm
A
Solid
T
Liquid
Vapour
T /  C
161

Phase Diagram of Carbon Dioxide
CO
2 sublimation
(s) CO
2
(g)
P / atm
A
Solid
T
Liquid
Vapour
T /  C
162

Phase Diagram of Carbon Dioxide
AT shows the variation of sublimation temperature of carbon dioxide with external pressure
P / atm
A
Solid
T
Liquid
Vapour
T /  C
163

Phase Diagram of Carbon Dioxide
TB is the melting curve
P / atm
Solid
B positive slope
(most common)
Liquid
T
Vapour
A
T /  C
164

Phase Diagram of Carbon Dioxide
CO
2 melting
(s) CO
2
(l)
P / atm B
Liquid
Solid
T
Vapour
A
T /  C
165

Phase Diagram of Carbon Dioxide
TB shows the variation of melting temperature of carbon dioxide with external pressure
P / atm B
Liquid
Solid
T
Vapour
A
T /  C
166

Phase Diagram of Carbon Dioxide
TC is the boiling curve
P / atm B
C
Solid
Liquid
T
Vapour
A
T /  C
167

Phase Diagram of Carbon Dioxide
CO
2 boiling
(l) CO
2
(g)
P / atm B
C
Solid
Liquid
T
Vapour
A
T /  C
168

Phase Diagram of Carbon Dioxide
TC shows the variation of boiling temperature of carbon dioxide with external pressure
P / atm B
C
Solid
Liquid
T
Vapour
A
T /  C
169

Q.62

P / atm
A
Solid
T
Liquid
Vapour
170
T /  C

Q.62

P / atm
A
Solid
T
Liquid
Vapour
171
T /  C

Q.62

P / atm
A
Solid
T
Liquid
Vapour
172
T /  C

Q.62

P / atm
A
Solid
T
Liquid
Vapour
173
T /  C

Q.62
(c) Freezing by  T
P / atm
A
Solid
T
Liquid
Vapour
174
T /  C

Q.62
(c) Freezing by  P
P / atm
A
Solid
T
Liquid
Vapour
175
T /  C

Q.62
(d) Melting by  T
P / atm
A
Solid
T
Liquid
Vapour
176
T /  C

Q.62
(d) Melting by  P
P / atm
A
Solid
T
Liquid
Vapour
177
T /  C

Q.62

P / atm
A
Solid
T
Liquid
Vapour
178
T /  C

Q.62

P / atm
A
Solid
T
Liquid
Vapour
179
T /  C

180
T is the triple point where all three phases coexist in equilibrium.
P / atm B
C
Solid
Liquid
5.1 atm
T
Vapour
A
T /  C
 56.4
 C

Vapour
Triple point
Solid Liquid vapour pressure above solid = vapour pressure above liquid
181

Dry ice sublimes when heated at 1atm
182
P / atm
5.1 atm
1 atm
A
Solid
B
Liquid
C
T
Vapour
 56.4
 C
T /  C

Dry ice is so called because it never melts (goes wet ) at normal pressure
P / atm B
C
Solid
Liquid
5.1 atm
1 atm
A
T
Vapour
T /  C
 56.4
 C
183

184
At P > 5.1 atm , dry ice melts to give liquid CO
2 when heated
P / atm B
C
10 atm
Solid
Liquid
Triple point video
5.1 atm
T
Vapour
A
T /  C
 56.4
 C

TC curve terminates at C beyond which the boundary between liquid and vapour disappears
P / atm
P c
= 73atm
B
C
Liquid Critical point
Solid
T
Vapour
A
T /  C
T c
= 31  C
185

Above T c
, the vapour cannot be condensed no matter how high the external pressure is
P / atm
P c
= 73atm
B
C
Liquid
Solid
T
Gas
Vapour
A
T /  C
T c
= 31  C
186

P / atm
P c
= 73atm
A
As dense as a liquid
As mobile as a gas
B
C
Liquid
Solid
T
Vapour
T c
= 31  C
Supercritical fluid
Gas
T /  C
187

P / atm
P c
= 73atm
Decaffeination using supercritical CO
2
B
Liquid
Solid
T
Vapour
C
A
T c
= 31  C
Supercritical fluid
Gas
T /  C
188

Q.63
P / atm
P c
= 73atm
A
In winter, T < T c
CO
2 is in liquid phase
B
C
Liquid
Solid
T
Vapour
T c
= 31  C
Supercritical fluid
Gas
T /  C
189

Q.63
P / atm
P c
= 73atm
In summer, T > T c
CO
2 is in gas phase
B
Supercritical fluid
C
A
Solid
T
Liquid
Vapour
T c
= 31  C
Gas
T /  C
190

Q.64
P / atm
P c
= 73atm
As P  , CO
2
(l)  CO
2
(g)
So T  and CO
2
(g)  CO
2
(s)
B
C
Liquid
Solid
T
1 atm
Vapour
A
T c
= 31  C
Supercritical fluid
Gas
T /  C
191

Phase Diagram of Water
P / atm
B
Negative slope
(very rare)
Liquid
Solid
C
Supercritical fluid
Gas
T
Vapour
A
T /  C m.p.  when external P 
192

Phase Diagram of Water
P / atm
B
H
2
O(s) H
2
O(l)
Liquid
Solid
T
Vapour
A
C
Supercritical fluid
Gas
T /  C
193

Phase Diagram of Water
P / atm
B
H
2
O(s) H
2
O(l)
Liquid
Solid
T
Vapour
A
C
Supercritical fluid
Gas
T /  C
194

195
H
2
O(s) H
2
O(l)
Ice melts below 0  C when an extremely high pressure is applied to it.
This results in a decrease in friction between contact surfaces and makes possible ice-skating and the movement of tremendously massive glaciers .

Phase Diagram of Water
P / atm
B
C
Supercritical fluid
Liquid
1 atm
Solid Gas
0.006 atm T
Vapour
A
T /  C
0  C 0.01
 C 100  C
196

Phase Diagram of Water
P / atm
B
P c
=217.2atm
Liquid
Solid
T
Vapour
A
C
Supercritical fluid
Gas
T c
=374  C
T /  C
197

Critical temperature and the strength of intermolecular forces
Gas He H
2
Ne N
2
O
2
CO
2
NH
3
H
2
O
T c
(K) 4.2 33.3 44.5 126 154 304 405 647 non-polar
Higher T c
 stronger intermolecular forces
Polar with H bond
 greater deviation from ideal gas behaviour
198

Q.68
Cu(NH
3
)
4
SO
4
 NH
3
Reason : does not exist.
Unlike H
2
O, each NH three H atoms.
3 has one lone pair and
Thus, NH
3 cannot form hydrogen bonds in the same way as H
2
O in the crystal lattice.
199

200

11.2 Van der Waals’ forces (SB p.280)
Back
201
How is the enthalpy of vaporization related to intermolecular forces of a simple molecular substance like neon?
Answer
The enthalpy of vaporization of a substance is the energy needed to vaporize one mole of the substance at its boiling point. Consider a substance like neon, which consists of single atoms, Neon liquefies when the temperature is lowered to –246 o C at 1 atm.
The enthalpy of vaporization of the liquid at this temperature is
1.77 kJ mol -1 . Some of this energy is needed to push back the atmosphere when the vapour forms. The remaining energy must be supplied to overcome the intermolecular attractions. Because each molecule in a liquid is surrounded by several neighbouring molecules, this remaining energy is some multiple of a single molecule-molecule interaction. Typically, this multiple is about 5.

11.2 Van der Waals’ forces (SB p.280)
(a) Comment on the relative strength of van der Waals’ forces in solid, liquid and gaseous bromine.
Answer
(a) The relative strength of van der Waals’ forces decreases in the order:
Solid bromine > liquid bromine > gaseous bromine
The van der Waals’ forces are highly dependent on the distance between adjacent molecules. It decreases exponentially with the separation between the molecules. Going from solid to liquid and then to gaseous state, the separation between molecules increases, so the van der Waals’ forces become weaker and weaker.
202

11.2 Van der Waals’ forces (SB p.280)
(b) Plastics are substances which have very strong van der
Waals’ forces. Explain why the van der Waals’ forces are so strong in plastics.
Answer
(b) A large size of a molecule of plastics indicates that it has a large electron cloud which is more easily polarized. Therefore, the molecule of plastics is more likely induced to form an instantaneous dipole. Moreover, the molecule of plastics has an extensive surface area. These make plastics have very strong van der Waals’ forces between the molecules.
203

11.2 Van der Waals’ forces (SB p.280)
Back
(c) Arrange the following substances in an increasing order of boiling point:
(i) N
2
, O
2
, Cl
2
, Ne
(ii) H
2
, Br
2
, He
(c) (i) Ne < N
2
< O
2
< Cl
2
Answer
(ii) He < H
2
< Br
2
204

11.3 Van der Waals’ radii (SB p.284)
Back
What is the consequence of two molecules approaching each other at a distance less than the sum of their van der Waals’ radii?
Answer
The electron clouds of the two molecules will repel each other, and the distance between the two molecules will increase until the repulsion is just balanced by the attraction.
205

11.5 Hydrogen bonding (SB p.291)
The relative molecular masses and boiling points of five compounds are given below:
Compound Relative molecular mass
Boiling point
( o C)
Ammonia (NH
3
) 17 -33.4
Ethanol (C
2
H
5
OH)
Hydrogen fluoride (HF)
46
20
78
19.5
Methanol (CH
3
OH)
Water (H
2
O)
32
18
66
100
206

11.5 Hydrogen bonding (SB p.291)
(a) Ammonia, hydrogen fluoride and water have similar relative molecular masses, yet their boiling points are different. Explain why.
Answer
(a) H
2
O can form 2 hydrogen bonds per molecule while NH
3 and HF can only form 1 hydrogen bond per molecule.
Thus, the boiling point of water is higher than those of
NH
3 and HF. Besides, as F is more electronegative than N, the intermolecular hydrogen bond formed between HF molecules is stronger than that between NH
3 molecules.
207

11.5 Hydrogen bonding (SB p.291)
Back
(b) Ethanol and methanol have similar structures, yet their boiling points are different. Explain why.
Answer
(b) For molecules with similar structures, their boiling points depend on their relative molecular masses. As the relative molecular mass of ethanol is greater than that of methanol, the boiling point of ethanol is higher.
208

11.5 Hydrogen bonding (SB p.293)
Why it takes much longer time to boil an egg on a mountain peak?
Answer
The boiling point of water decreases with decreasing pressure.
Although water boils easily at mountain peak, the cooking of an egg takes longer time. It is because the amount of heat delivered to the egg is proportional to the temperature of water.
Back
209

11.5 Hydrogen bonding (SB p.296)
(a) The formation of a hydrogen bond between two molecules RAH and R’B may be represented as:
R  A  H · · · · · · · B  R’
(i) Suggest possible elements for A and B. What are their common features?
(ii)In which of the following ranges would you expect the strength of hydrogen bonds to lie?
0.1 – 10 kJ mol -1
10 – 50 kJ mol -1
100 – 400 kJ mol -1
Answer
210

11.5 Hydrogen bonding (SB p.296)
(a) (i) A and B can be nitrogen, oxygen or fluorine. All of them are highly electronegative atoms, thus they form highly polar molecules, resulting in the formation of hydrogen bonds.
(ii) 10 – 50 kJ mol -1
211

11.5 Hydrogen bonding (SB p.296)
(b) Benzoic acid has an apparent relative molecular mass of
244 in hexane, but only 122 in aqueous solution. With the aid of diagrams, explain this phenomenon.
Answer
212

11.5 Hydrogen bonding (SB p.296)
(b) The relative molecular mass of benzoic acid (C
6
H
5
COOH) is 122.
In hexane, benzoic acid molecules form dimers with hydrogen bondings between the molecules.
However, in water, the benzoic acid molecules form hydrogen bonds with the water molecules.
213

11.5 Hydrogen bonding (SB p.296)
(c) Cyclohexane (C
6
H
12
(C
6
H
12
O
6
) is insoluble in water whereas glucose
) is miscible with water in all proportions.
Answer
214

11.5 Hydrogen bonding (SB p.296)
Back
(c) Cyclohexane is non-
polar, and there are only weak van der Waals’
forces holding the molecules together. Thus, cyclohexane molecules do not form hydrogen bonds with water. On the other hand, glucose can form hydrogen bonds with water molecules via its

OH groups. Therefore, glucose is soluble in water but cyclohexane is not.
215
Cyclohexane Glucose

11.5 Hydrogen bonding (SB p.297)
Name the types of bonding or intermolecular forces that are broken and formed in the following processes.
• H
2
O(s)  H
2
O(g)
• 2Mg(s) + O
2
(g)  2MgO(s)
• H
2
(g) + F
2
(g)  2HF(g)
• 2Na(s) + 2H
2
O(l)  2NaOH(aq) + H
2
(g)
• CH
3
CH
2
OH(l) + 3O
2
(g)  2CO
2
(g) + 3H
2
O(l)
Answer
216

11.5 Hydrogen bonding (SB p.297)
Back
(a) Bond broken: hydrogen bond
(b) Bonds broken: metallic bond and covalent bond
Bond formed: ionic bond
(c) Bond broken: covalent bond
Bonds formed: covalent bond and hydrogen bond
(d) Bonds broken: covalent bond, metallic bond and hydrogen bond
Bonds formed: ionic bond and covalent bond
(e) Bonds broken: covalent bond and hydrogen bond
Bonds formed: covalent bond and hydrogen bond
217