L.S.T. Leung Chik Wai Memorial School
F.6 Chemistry
Chapter 10: Intermolecular Forces
Chapt.10: p.1
INTERMOLECULAR FORCES
I.
II.
Introduction
Intermolecu1ar force refer to the weak short-ranged electrostatic interactions between uncharged
molecules or neutral atoms (such as noble gases).
Intermolecular forces are independent of any intramolecular bonding forces.
Usually, the term “van der Waals’ forces” is used to include all types of weak intermolecular
forces
Intermolecular forces. arise from the polarity of molecules or neutral atoms.
Polarity of Molecules
Permanent, instantaneous and induced dipoles lead to van der Waals’ forces.
A) Permanent Dipole
<1> In a diatomic molecule which contains atoms of different electronegativities, the
bonded electrons will be unsymmetrically distributed, i.e., the electron cloud is displaced.
• The more electronegative atom (e.g. Cl) tends to attract the bonded electron pair to
itself. It becomes slightly negatively charged (-)
• The less electronegative atom (e.g. H) will become electron deficient and slight
positively charged (+)
<2> As the electron cloud is distorted, the distribution of charge is non-symmetrical, the
separation of charge produces a permanent dipole in the molecule.
<3> In polyatomic molecule which have overall dipole moments e.g. trichioromethane,
propanone, permanent dipoles also exist.
(B)
Instantaneous Dipole
<1> In non-polar molecules (e.g the halogens, noble gases. benzene, tetrachloromethane,etc.)
there are no permanent dipoles because the molecules are electrically symmetrical, i.e.
the negative charge of electrons is evenly distributed around the nucleus.
<2> However, the electrons in a molecules are in continual motion. At any particular moment,
the electron cloud may not be evenly spread over the molecule. Hence there may be
more negative charge on the side of the molecule than on the other at a given time.
L.S.T. Leung Chik Wai Memorial School
F.6 Chemistry
Chapter 10: Intermolecular Forces
Chapt.10: p.2
<3> The separation of positive and negative charge produces a dipole.
Fluctuation of electron cloud in a non-polar molecules (or neutral atoms) causes
displacements between nuclei and electrons, giving rise to instantaneous (or temporary)
dipole
(C) Induced Dipole
<1> When a particle having a dipole (either permanent. instantaneous or induced) approaches
another particle (with or without a dipole), the dipole on the first dipole induces a new
dipole on the second dipole.
III. Van der Waals’ Forces
There are three types of van der Waals forces:
1.
dipole-dipole interactions(includes hydrogen bonding)
2.
3.
dipole-induced interactions
induced dipole—induced dipole interactions
They arise from the permanent, instantaneous or induced dipoles described above.
L.S.T. Leung Chik Wai Memorial School
F.6 Chemistry
Chapter 10: Intermolecular Forces
(A)
Chapt.10: p.3
Dipole-dipole interactions
Dipole-dipole interaction occurs only between unsymmetrical molecules with permanent dipole
moment (i.e polar molecules).
Permanent dipole-dipole forces are the attractions between the negative end of one polar
molecule and the positive end of another
Note:
<1> Factors affecting the strength or dipole-dipole interaction
(a) Difference in electronegativties between the bonded atoms.
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(b) The shape of molecule (contact surface area)
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<2> Polar molecules are attracted to each other more strongly than non-polar molecules of
similar mass.
A polar molecule will normally have higher melting and
boiling points than a non-polar mo1ecule similar molecular mass,
as a result of dipole-dipole interaction.
Example :
Propanone has a higher boiling point than butane even though they both
have the same molecular mass.
b.p. : 57 0C
b.p. : -1 0C
L.S.T. Leung Chik Wai Memorial School
F.6 Chemistry
Chapter 10: Intermolecular Forces
Chapt.10: p.4
(B) Dipole—induced dipole Interactions
Molecules without permanent dipole moment(non-polar molecules) may
interact through instantaneous dipole.
<1> When a polar molecule approaches another molecule, this molecule will be polarized, i.e. a
new dipole is induced on the second molecule:
<2> There are dipole-induced dipole interaction between the permanent and induced dipole. Such
interactions are weaker than dipole-dipole interaction.
(C)
Induced dipole—induced dipole interactions
Instantaneous dipole can be formed due to the continual motion of the electron cloud in a molecule.
Such instantaneous dipole can then induce a new dipole in neighbouring molecules/atoms. This
results in an attractive force between the two dipoles.
<1> Induced dipole—induced dipole interactions are even weaker than dipole—induced dipole
interactions.
<2> Induced dipole—induced dipole interactions cause liquefaction of non—polar
molecules/atoms, such as noble gases, nitrogen at low temperatures.
Reference: http://web.umr.edu/~gbert/INTERACT/intermolecular.HTM
L.S.T. Leung Chik Wai Memorial School
F.6 Chemistry
Chapter 10: Intermolecular Forces
IV.
Chapt.10: p.5
Strength of Van der Waals’ Forces
The relative strength of the three types of van der Waals’ forces is in the order :
Dipole-dipole > dipole-induced dipole > induced dipole- induced dipole
The strength of van der Waals’ forces depends on several factors such as polarizability of
molecules, molecular size and & shapes, temperature and pressure.
(A)
Polarizability
A molecule with a dipole moment will, polarize a large atom (or molecule) much more easily
than a small one.
If a molecule/an atom is more polarizable, its electron cloud will be more easily displaced.
Formation of dipole is more likely to occur. This causes larger intermolecular forces.
Note : Larger the atom  greater the polarizability greater the Van der Waal’s forces.
(B)
Size of electron cloud
As the number of electrons in a molecule(or an atom) increases the electron cloud will be more
diffuse and polarizable. There is more displacement in the electron cloud and unequal
distribution of charge. This results in stronger van der Waal’s forces.
Evidence :
Alkane
Larger alkane members have higher boiling points and densities
melting
point /°C
Boiling
point /°C
State at
298 K
Density/
g cm3
CH4
- 182
-161
gas
0.424(L)
C2H6
- 183
- 88
gas
0.546 (L)
C3H8
- 187
-42
gas
0.582 (L)
C5H10
- 138
0
gas
0.579 (L)
C6H12
- 130
36
liquid
0.626
C7H14
- 95
69
liquid
0.659
L.S.T. Leung Chik Wai Memorial School
F.6 Chemistry
Chapter 10: Intermolecular Forces
Chapt.10: p.6
Interpretation :
<1> The boiling point increases as the molecular size and electron cloud of the alkane
molecules increases. As the electron clouds are more easily polarized, the van der Waals’
forces (induced dipole-induced dipole ) between thehydrocarbon molecules increases.
More energy is required to separate neighbouring molecules.
<2> Besides, there are more attractive interactions between the molecules as the molecular
size increases, so that
(a) the density of alkanes increases
(b) there is a change of physical state from gas to liquid.
Evidence: The boiling points of noble cases increase down the group
Interpretation
<1> In noble gases, as the molecular mass increases, the molecular size (and hence the size of
electron cloud) of the noble gas atoms increases. The electron cloud is more easily
polarized and instantaneous dipole is more easily induced.
<2> Van der Waals’ forces between molecules become stronger down the group, resulting in
progressively higher boiling points.
Exercise 1:
Explain the following facts
(a) The boiling points of the halogens increase down the group.
F2(-187°C) < C12(-35°C) < Br2(59°C) < I2(183°C)
(b) The bond dissociation energy of fluorine (159 kJ mol-1) is less than that of chlorine.
(239°C)
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L.S.T. Leung Chik Wai Memorial School
F.6 Chemistry
Chapter 10: Intermolecular Forces
Chapt.10: p.7
Exercise 2
Comment on the following statement :
“Covalent bonding in oxygen is stronger than that in nitrogen because the boiling points of N2
and O2 are -196°C and -183°C respectively” I
Exercise 3
Arrange the following molecules in order of increasing boiling points:
C2H5Cl, CH4, C2H6
Explain your order in terms of intermolecular forces.
(C)
Contact between molecules
When molecules can get closer to one another, they have more contact surfaces. This allows
more interactions between the molecules, resulting in greater van der Waals’ fores.
Evidence : Branched alkanes have lower boiling points than straight chain isomers.
Consider the two C5H12 isomers
Melting point / 0C
Boiling point / 0C
-130
36
-16
10
L.S.T. Leung Chik Wai Memorial School
F.6 Chemistry
Chapter 10: Intermolecular Forces
Chapt.10: p.8
Interpretation
<1> The boiling point of pentane is higher than that of 2,2-dimethylpropane.
Reason
Branching prevents close contact of neighbouring molecules. As the branching decreases
(like pentane) the electron cloud gets less spherical. The molecules can get close together,
leading to stronger van der Waals’ forces.. The boiling points there fore increases.
<2> The melting point of pentane. is lower than that of 2,2-dimethyipropane.
Reason
2,2-dimethyipropane is more compact and symmetrical, it packs more efficiently in a
crystal lattice. It has a melting point higher than pentane.
(D) Temperature
At low temperature, the random movement of molecules slows down. The molecules come
closer together and interact more strongly. resulting in an increase in van der Waals’ forces
Evidence : Noble gases have low freezing points which increase down the group.
Elements
Freezing point / °C
Helium
-269.7
Neon
-248.6
Argon
- 189.4
Krypton
-157.3
Xenon
-111.9
Freezing points of noble gases
To freeze the noble gases, the temperature has to be lowered to decrease the kinetic energy of
the atoms. i.e. to slow down the their motion so that they come together.
Xenon, which has an atomic size and electron cloud larger than other noble gases, has greater
van der Waals forces. Therefore, its freezing point is higher than those of other noble gases.
(E)
Pressure
Molecules come closer together and interact more strongly under high pressure.This results in
an increase in van der Waals’ forces.
Exercise :
Butane (m.p. -138 C; b.p.0 0C) can be readily liquefied. Liquid butane is stored in steel
cylinders.
(a) Why is butane stored in steel cylinder?
(b) (1) What forces are acting between butane molecules in liquid
(2) Describe how butane molecules are arranged in the liquid state.
(c)
Explain why butane is liquefied under a pressure.
L.S.T. Leung Chik Wai Memorial School
F.6 Chemistry
Chapter 10: Intermolecular Forces
Chapt.10: p.9
(F) Comparison of the covalent and Van der Waal’s radii of non-metals to indicate the relative
Strength of covalent bonds and van der Waals’ forces
Covalent
radius:
It is the half distance between the nuclei of adjacent atoms of the same
element which are held together by single covalent bond.
Van der Waals’radius :
It is half the distance between the centres of 2 atoms in adjacent molecules
(which are not chemically bound together) in molecular crystal..
Example : Iodine molecules in its crystal structure
<1> Van der Waals’ radius (rv) > covalent radius (r)
Deduction :
<2> For iodine, the distance between iodine atoms in neighbouring iodine molecules is twice the
Van der Waals’ radius of iodine.
<3> The covalent radii and van der Waals’ radii of some molecules are given in the following table
Molecules
Covalent radius / nm
Van der Waals’ radius/ nm
112
0.037
0.12
N2
0.070
0.15
O2
0.066
0.14
F2
0.071
0.135
Cl2
0.099
0.181
Covalent radii and Van der Waal’s radii of some molecules
Conclusion:
The Van der Waals’ radius is significantly greater than the corresponding covalent radius. It can
be deduced that van der Waals’ forces are weaker than covalent bonds.
L.S.T. Leung Chik Wai Memorial School
F.6 Chemistry
Chapter 10: Intermolecular Forces
V.
Chapt.10: p.10
Hydrogen Bonding
(A)
Introduction
Hydrogen bond is a special type of dipole-dipole interaction in which a hydrogen atom is
situated between 2 small and electronegative atoms, one of which has a lone pair of
electrons.
<1> When a hydrogen atom is bonded to highly electronegative atom (e.g. N, O & F), the
bonded electron pair is attracted to the electronegative atom.
<2> The electron density of hydrogen atom is reduced as a result. The nucleus of a hydrogen atom
is exposed to attraction lone pair of electrons on another electronegative atom.
Where A- represent a electronegative atoms like N, O, or F
:B represent another electronegative atom possesses a lone pair electron
<3>
(i)
The essential elements requirements for a hydrogen bond to be formed
a hydrogen atom attached to a highly electronegative atom (like N, O or F)
(ii) an unshared pair electrons on another electronegative atom, to interact with the
hydrogen carrying a high partial positive charge.
Note:
(1) The small size of the electronegative atom would facilitate the close approach of
hydrogen atom and the formation of
hydrogen bond.
(2) Example : hydrogen bond in hydrogen fluoride (HF)
(a)
Fluorine atom in a HF molecule attracts the shared electron pair towards itself.
Fluorine becomes partially negative whereas hydrogen becomes partially
positive.
(b)
The nucleus of hydrogen atom, having a high positive charge density , attracts
the lone pair of electrons in F atom of another HF molecule.
Thus, in liquid state, HF molecules are linked by hydrogen bonding.
(c)
L.S.T. Leung Chik Wai Memorial School
F.6 Chemistry
Chapter 10: Intermolecular Forces
Chapt.10: p.11
Exercise :
Ammonia can act both as a donor and as an acceptor of hydrogen bonds. Indicate the interaction of two
ammonia molecules by a sketch.
(B)
Strength of hydrogen bonds
<1> Range of the H-bond energy 2 -50 kJ mol-1
<2> The more electronegative the atom bonded to hydrogen, the more electron deficient will
be the hydrogen atom. Thus, The proton of the hydrogen atom will be more exposed to
the attraction of lone pair electrons in other molecules. The stronger will be the hydrogen
bond.
(C)
Types of hydrogen bond
• intermolecular hydrogen bonds: occur between 2 separate molecules e.g. H2O, HF, NH3
• intramolecular hydrogen bonds: occur between the same molecule.
Example :
2-nitrophenol (intramolecular hydrogen bond) b.p. : 216°C
4-nitrophenol (intermoleular hydrogen bond)
b.p.
: 259°C
(D) Effect of hydrogen bond
Hydrogen bonding is a strong dipole-dipole attraction. It influences physical properties such as
boiling points, melting points. viscosity and enthalpy of vaporization.
1
L.S.T. Leung Chik Wai Memorial School
F.6 Chemistry
Chapter 10: Intermolecular Forces
(1)
Chapt.10: p.12
Hydrogen bondings in the hydrides of Group IV, V. VI and VII
The following tables and graph show the boiling points of the hydrides of elements in Group IV,
V, VI and VII and the variations in the boiling points.
L.S.T. Leung Chik Wai Memorial School
F.6 Chemistry
Chapter 10: Intermolecular Forces
Chapt.10: p.13
Interpretation:
<1> The hydrides in any particular group have gradually increasing boiling points generally.
Reason:
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<2> Three hydrides HF, NH3, and H2O have much higher boiling points than expected.
Reason:
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Exercise 3
Account for the differences in boiling points between
(a) H2O and H2S
(b) CH4 and SiH4
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Exercise 4 :
The boiling points and standard enthalpies of formation of the hydrides of Group V elements
are as follows:
Comment on
(a) the variation of boiling points, and
(b) the relative stability (in relation to the constituent elements)
of the hydrides of Group V elements
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L.S.T. Leung Chik Wai Memorial School
F.6 Chemistry
Chapter 10: Intermolecular Forces
Chapt.10: p.14
Note
The presence of intermolecular hydrogen bonds in water molecules is also proved by the heat
enthalpy of vaporization of water, in relation to other Group VI hydrides.
More energy is needed to overcome the hydrogen bonds between water molecules before water
can vaporize. This results in enthalpy of vaporization.
(2)
Hydrogen bonding in alcohols
Hydrogen bonding occurs in any compound containing hydroxyl (—OH) groups.
<1> The molecules exist in pairs or in clusters. Energy is required to break the hydrogen bonds
before the molecules can escape as vapour. The boiling points of alcohols are therefore
higher than expected from the relative molecular mass.
<2> As the number of hydroxyl groups increases, there is a greater degree of hydrogen
bonding.
<3> Hydrogen bonding affects viscosity, as well as boiling point.
<4> Hydrogen bond is also formed between water molecules and alcohols (especially those of
low molecular mass.. Therefore, alcohols can dissolve in water.
Exercise 5:
Consider some physical data about butan-1-ol and 2-methylpropan-2-ol.
Density/ g cm-3
Boiling point /0C
Butan-1-ol
0.81
118
2-methylpropan-2-ol
0.79
82
Account for the differences in the densities and boiling points of the molecules.
L.S.T. Leung Chik Wai Memorial School
F.6 Chemistry
Chapter 10: Intermolecular Forces
(3)
Chapt.10: p.15
Hydrogen bonding in carboxylic acid
Evidence :
Explanation :
When ethanoic acid is dissolved in non-polar solvents or exists in vapour state,
its relative molecular mass is found to be 120, instead of the expected 60 Association of ethanoic acid molecules occurs:
Ethanoic acid molecules exists as dimers (i.e. 2 molecules bond together), as a
result of hydrogen bonding.
<1> In acids where the association is not complete, the relative molecular mass will be
between 60 and 120.
<2> Hydrogen bondings can form between water molecules and the carboxylic acid groups.
Therefore, carboxylic acids (of low molecular mass) can dissolve in water.
(E)
Relative strength of hydrogen bonding and van der Waals’ forces
covalent bonding > hydrogen bonding > van der Waals’ forces and dipole—dipole attraction.
The strength of hydrogen bond, covalent bond and van der Waals’ forces are compared in the
following table below:
Bond
Hydrogen bond
Covalent bond
Van der Waals’ forces
Bond strength (kJ mol )
2—50
100—900
less than 20
<1> Breaking or formation of intermolecular hydrogen bonds between molecules in liquids
would cause an enthalpy change when the liquids are mixed.
<2> From the enthalpy change, the strengths of hydrogen bonds formed between molecules
(e.g. trichloromethane and ethyl ethanoate) can be measured.
Exercise 6
What is the stronger bond in hydrogen fluoride molecule : the shorter. covalent bond. or the
longer hydrogen bond between different molecules?
L.S.T. Leung Chik Wai Memorial School
F.6 Chemistry
Chapter 10: Intermolecular Forces
Chapt.10: p.16
Determination of the strength of hydrogen bond formed between trichloromethane and ethyl ethanoate
The structural formulae of
<1>
trichloromethane and ethyl ethanoate are given below
Trichloroethane and ethyl ethanoate do not form hydrogen bonds with themselves. In
CHCl3, the size of a chlorine atom is so large that the approach of hydrogen atom is
made
difficult. In ethyl ethanoate, no hydrogen atom is electropositive enough to form
hydroqen
bond with the partially negative oxygen atoms
<2> However, trichloroethane and ethyl ethanoate can interact through hydrogen bonding:
<3> By considering the temperature change when two liquids are mixed, the energy given out
when the hydrogen bond formed. (i.e. the hydrogen bond strength ) can be estimated.
L.S.T. Leung Chik Wai Memorial School
F.6 Chemistry
Chapter 10: Intermolecular Forces
Chapt.10: p.17
Experiment
(a) When 17.7 cm3 ( 0.1 mol) of trichloromethane were mixed with 7.9 cm3 ( 0.1 mol) of ethyl
ethanoate in an insulated boiling tube calorimeter ( of negligible heat capacity ), there was a
temperature rise of 17 0C.
Given the following data
Liquid
R.M.M
Density / g cm-3
Specific heat capacity / kJ kg-1 K-1
Trichloromethane
119.5
1.48
0.98
Ethyl ethanoate
88
0.90
1.92
(1) Determine the strength of the bond ( in kJ mol-1 , formed between trichloromethane and ethyl
ethanoate.
(2) How does this value compare with the strength of an ordinary covalent bond?
(3) Suggest two sources of errors in the determination of the hydrogen bond strength.
L.S.T. Leung Chik Wai Memorial School
F.6 Chemistry
Chapter 10: Intermolecular Forces
Chapt.10: p.18
(F) Hydrogen bonding in ice, protein and DNA (deoxyribonucleic acid )
(1) Hydrogen bonding in ice
Each water molecule has two hydrogen atoms and two non-bonded electron pairs on the
oxygen atom. Each water molecule can form 2 hydrogen bonds.
-
In ice lattice, each water molecule is surrounded tetrahedrally by four neighbouring water
molecules. The position of the water molecules have been fixed by hydrogen bonds.
-
This results in an open lattice of hexagonal network with very inefficient packing of
molecules. Therefore, ice has a low density and floats on water.
The smaller density of ice compared to water can be explained in terms of hydrogen bonds
present in water and in the open hexagonal network of ice respectively.
-
On melting, the hydrogen bonds in the ice lattice break down. Water molecules can become
more closely packed, so that water has a higher density.
-
At 4 0C , water molecules pack most compactly, and the density of water reaches a maximum.
(1 g cm-1 )
Exercise
Both water and hydrogen fluoride form hydrogen bonds. Explain why ice floats on water whereas
solid hydrogen fluoride does not float on hydrogen fluoride liquid?
L.S.T. Leung Chik Wai Memorial School
F.6 Chemistry
Chapter 10: Intermolecular Forces
Chapt.10: p.19
Hydrogen bonding play a important role in the structure of many important biochemical substance such
as proteins and deoxyribonucleic acid (DNA)
(2) Hydrogen bonding in protein
Protein consists of long chain of formula
-
Protein chain are held in close proximity to one another by hydrogen bonds formed between
N-H and C=O groups of the neighbouring chains.
L.S.T. Leung Chik Wai Memorial School
F.6 Chemistry
Chapter 10: Intermolecular Forces
Chapt.10: p.20
(3) Hydrogen bonding in DNA
-
In DNA, the 2 helical nucleic acid chain held together by hydrogen bonds.
-
These hydrogen bonds are formed between specific base pairs on the DNA chains.
-
The double helix structure maintained by hydrogen bonds between base pairs in helical
chain is very important for life.