Chapter 11 - Intermolecular forces of Solids and Liquids
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For gases, we used the Kinetic Molecular Theory to explain observed behaviour of gases.
This is adequate to explain the behaviour of collections of gaseous atoms or molecules since
they are characterised by large intermolecular distances.
Hence, at ordinary P and T, there are negligible attractive forces between gas molecules; gas
molecules’ behaviour can be described independently of the actual chemical species note
PV = nRT (n = # of gas particles only)
Gas molecules  low density, very compressible, whereas liquids and solids are not very
compressible.
Liquids  the molecules move freely past each other liquids take the shape of their
container, high density
Solids  the atoms or molecules are very rigidly held in place, with little freedom to move
(long range order). Density of solids is usually higher than liquids (H2O  big exception)
Intermolecular Forces
 Intermolecular forces - are defined as the attractive forces between molecules (e.g. Hbonding, dipole-dipole etc). Note  think of international  between nations.

Intramolecular forces - forces that hold atoms together in a molecule (e.g. chemical
bonding).
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In general  strength of intramolecular forces >>> strength of intermolecular
forces.
e.g., boil 1 mol of water  41 kJ of energy (overcoming intermolecular forces).

H2O (l)  H2 (g) + ½ O2 (g) rxnH = +286 kJ (overcoming intramolecular forces).
Types of Intermolecular forces
1. dipole - dipole - act between polar molecules
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e.g. NF3 or HCl
Note molecule must have dipole moments.
electrostatic forces, obey Coulomb’s Law
F  q1q2/r2
2. Ion-dipole forces - occur between an ion (cation or anion) and a dipole. Hydration of an
ionic salt
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
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NOTE: magnitude depends on
charge and size of ion
magnitude of dipole (c.f. Coulomb’s Law
above).
3. Dipole-Induced Dipole forces - Let’s place a polar molecule (possesses a dipole) near a
non-polar molecule.
2
-
+
dipole

nonpolar species (symmetrical charge distribution)
As we place the polar species near the nonpolar molecule, electron distribution of the
nonpolar species becomes distorted. Hence, we have induced a dipole
-
+
-------
++++
++++
----
++++
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The ease of inducing a dipole is related to the “polarisability” of the non-polar system.
Polarisability, the ease with which an electron distribution for a neutral molecule or
atom can be distorted, depends primarily on the magnitude of the electron-nuclear
attractive forces. These forces also depend, of course, on the magnitude (strength) of
the dipole.

Note  polarisability increases with an increase in the molar mass of the atom or
molecule.
4. Dispersion Forces - a bit more difficult to understand. These forces of attraction are also
called London forces after Fritz London: Quantum Mechanical Treatment (1930).

In a nonpolar molecule, the charge distribution is, on average, spherically symmetric.
However, we realise that electrons are constantly in motion. Therefore, at any given
time there may be an excess of electrons existing in one part of the atom or molecule.
Hence, an instantaneous dipole is produced which will then proceed to distort the
electron distribution of the neighbouring molecule, causing the atoms or molecules to
be attracted towards one another. This force is significant only when the distances
between the atoms or molecules is very small.

The strength of dispersion forces also depends on ease of polarising the atom or
molecule. Since the polarisability depends on size (mass) of atom or molecule, we
should see an increase in the magnitude of the dispersion forces with increasing
molecular mass.
i.e., large atoms 
small atoms 
e’s far from nucleus less attractive forces  easily distorted.
e’s close to nucleus  greater attractive forces  hard to distort.
 Dispersion forces are present in all atoms and molecules. They are the only
forces of attraction in nonpolar species.
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NOTE non-polar molecules
Cl2
Br2


gas
liquid
I2

solid
Why?
I2 and Br2 are more polarisable (i.e., the electron distributions are more easily distorted).
Another example, look at the melting points of the following substances.
CCl4 (-23.0C) < CBr4 (90.0C) < CI4 (171.0C)
again due to an increase in the polarisability from Cl < Br < I
Example
 What forces of attraction are present in these atoms and molecules?
a) HCl
b) Cl2
c) CCl4
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HCl  dipole-dipole and dispersion
Cl2  dispersion
CI4 dispersion (non polar molecule)
Hydrogen Bonding
 Hydrogen bonding (H-bonding) is a very important, special type of dipole-dipole interaction
between the H atom in a polar bond and an electronegative atom (N, O, F).

e.g. water H2O
O
H
H
O
H
H
O
H
H
O
H
H
Hbonds
Covalent bonds
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
Tetrahedral arrangement of H atoms around each O.
These attractive forces are strong for
1. NH3
2. H2O
3. HF
4. Alcohols (e.g., methanol, CH3OH)
5. Amines (e.g., methyl amine, CH3NH2)
6. Carboxylic acids (e.g., CH3OOH)
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Structure and properties of water
Properties (1) excellent solvent (due to dipolar)
high heat capacity (due to intermolecular H-bonding)
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Water (liquid) vs. water solid (ice)
 Liquid water exists as a “tetrahedrally” H-bonded system (alluded to earlier); solid water is
also approximately tetrahedrally bonded (i.e., 2 covalent bonds and two hydrogen bonds).
 However, the density of water >>> density of ice. This is due to the rather open structure
of water in the solid state (i.e., there is a hexagonal arrangement of water molecules in the
solid state), compared to the more random structure of liquid water.
 The density of water actually increases from 0  4C. Note: when ice melts, there are a
few molecules with enough kinetic energy to break from the H-bonded structure and they
become trapped in the “cage-like” structures.
 Hence, we must consider two processes.
1. trapping of H2O molecules
2. thermal expansion
 From 0  4C trapping dominates, yielding an increase in the number of water
molecules/unit volume (density increases slightly). After 4C thermal expansion
dominates, and the more normal pattern of decreasing density with increasing temperature
is observed.
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NOTE: SO FAR WE HAVE DEALT WITH ATTRACTIVE FORCES

We also have repulsive forces (e- -e- and nuclear-nuclear); these forces became great when
the distance between molecules is decreased. This is why we can’t compress easily
liquids and solids.
Surface Tension and Viscosity
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Surface Tension  water beads on cars, glass etc., oil (grease) can form spots on a water
surface (just look at the kitchen sink after you have washed away all the spaghetti sauce from
the dishes).
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WHY?

reason is surface tension
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definition the amount of energy needed to stretch or increase the surface by a unit of area
IMPORTANT ENERGY IS NEEDED

let’s look at the freshly waxed car.

Water molecules have a great attraction for themselves but only have very weak attractive
interactions with nonpolar substances like a wax (or air for that matter).
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Water droplet
Waxed car surface

At the surface of the droplet, the water molecules are pulled inwards (downwards) and
sideways by the H-bonding interactions (i.e., towards the centre of the droplet). At the
surface, there is only a very weak attractive interaction between the water and the air.
Note that the attractive interactions between the water and the waxed surface are also very
weak.

What shape provides the least surface contact? The sphere.

Of course, gravity pulls the water droplet towards the surface.
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Attractive forces between like molecules are known as cohesive forces. Attractive fornces
between unlike molecules (or between molecules and a solid surface) are known as adhesive
forces. In the case of the water droplet on the waxed surface, the cohesive forces are
significantly larger than the adhesive forces!
n a glass capillary, however, water forms a atom 
adhesion vs. cohesion force
Water vs. Hg in a capillary
H2O in a capillary, adhesion > cohesion  meniscus  water pulled up along wall
For Hg in a capillary, cohesion > adhesion  depression occurs when a capillary tube is
dipped in Hg.
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Viscosity
 The measure of the resistance of a fluid to flow.
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gases  very low viscosity
liquids  variable e.g., benzene has a low viscosity, whereas molasses has a high viscosity.
Intermolecular forces are very important!
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e.g. ethylene glycol (antifreeze)
H
H
H
C
C
H
OH OH
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Intermolecular H - bonds attractive forces high, therefore the viscosity is high.
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benzene,
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Characterised by dispersion forces and it is a planar, flat molecule  viscosity is low. It is
relatively easy for the benzene molecules to slide past one another
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What about the following series?
hexane
octane
decane
CH3CH2CH2CH2CH2CH3
CH3(CH2)6CH3
CH3(CH2)8CH3
0.326 * 10-2 g/cms
0.542 * 10-2 g/cms
0.92 * 10-2 g/cms
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WHY?
Two reasons (1) molar mass increases  increasing dispersion forces. In
addition, as the chain lengths increase, they become entangled! The ease of entanglement
increases with increasing chain length.
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Viscosity  as T . WHY?
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H2O as example. T , H-bonds break, attractive forces decrease  viscosity decreases.
Changes of State
 evaporation  molecules have enough kinetic energy to escape form the surface.
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condensation  as vapour pressure increases, K.E. goes down slightly for some molecules
and more return to the liquid state
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When the rate of evaporation = rate of condensation, we have obtained the equilibrium
vapour pressure.
Vapour pressures and intermolecular forces
 Note that molecules in the liquid phase escape from the surface of the liquid to the vapour
phase. Only a few molecules will have a sufficient amount of kinetic energy to overcome
the intermolecular forces present in the liquid phase to escape to the vapour phase.
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vapour
liquid
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Hence, the magnitude of the vapour pressure of the liquid is an indication of the strength
of intermolecular forces. If the intermolecular forces are strong, the substance has a low
vapour pressure; if the intermolecular forces are weak, the substance will have a high
vapour pressure. Note that there is an inverse relationship between intermolecular forces
and vapour pressure.
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How do vapour pressures vary with temperature?
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Vapour pressures increase with increasing temperature. As T increases, the kinetic energy
increases, and hence, more molecules are able to overcome the intermolecular forces
present at the surface of the liquid and escape into the vapour phase.
Molar heat of vaporization
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H2O (l)  H2O(g) rxnH  vapH, molar heat of vaporisation. (energy required to vaporize
1 mole of liquid). The enthalpy of vaporisation is an excellent indication of the strength of
the intermolecular forces.
Note  magnitude of vapH is large, strong intermolecular attractive forces, whereas if the
enthalpy of vaporisation is small, the intermolecular forces are weak.
How do we determine vapH?
Use the Clausius-Claypeyron equation
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lnP = -vapH/RT + C (linear equation); P = vapour pressure, T = temp in K, and C is a
constant.
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look at two different temperatures (T1 and T2)
lnP1 = -vapH/RT1 + C
lnP2 = -vapH/RT2 + C
lnP2/P1 = -vapH/R (1/T2 - 1/T1)
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Hence, a plot of ln P vs. 1/T shuld be linear with a slope equal to
-vapH/R.
Example The vapour pressure of water at 308 K is 42.18 mm of Hg. Given that the
vapH of H2O is 40.79 kJ/mol, calculate the vapour pressure of H2O at 338 K.
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n
n
P2   vap H  1 1 
=
*  - 
P1
R
 T 2 T1 
P2  40.7 kJ/mol * 1000J/kJ  1
1 
=
*

K
P1
338K 308K 

8.314 J
mol
n
4.10 =
P2
P
= 1.41  e -1.414) = 1
P1
P2
P2
 P 2 = 42.18 mm Hg  4.10
P1
= 173 mm of Hg
ANS:
Boiling Points and Intermolecular Forces
 What about the point at which P2 = the external pressure? This is defined as the boiling point
of the solution. Hence, boiling points are dependent on the external pressure (boiling points
generally decrease with decreasing external pressure). When the external pressure is 1.00
atm, we define this temperature as the normal boiling point, which for water is exactly
100.0C or 373.15 K at 1 atm pressure. Hence, normal boiling points as well as vapH values
give an indication of the strength of intermolecular forces.
 Note  strong intermolecular forces  high enthalpies of vaporisation and high boiling
points. If the intermolecular forces are weak, the normal boiling points as well as the
enthalpies of vaporisation are small.
Examples
 Ar and CH4 and ethane have low b.p.s. (dominated by dispersion forces)

Change ethane to ethanol  ethanol has strong H-bonding, whereas ethane is
characterised only by dispersion forces. Ethane boils at -88.6C; ethanol boils at 78.5C.
Critical Temperatures and Critical Pressures
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Critical pressure - the minimum pressure that has to be applied to liquefy the gas at the
critical temperature (or highest temperature where the liquid substance can exist)
Critical temperature - the temperature above which a gas will not liquefy no matter how great
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the pressure.
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e.g. CO2  familiar with two forms, the solid (known as ‘dry ice’) and the gas
(atmospheric CO2).
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Liquid CO2 cannot be made at atmospheric pressure (note CO2 has a Pc = 73 atm).
Supercritical CO2 ( i.e., CO2 at a temperature above its Tc = 31.0C) is used to
decaffeinate coffee!!
Solid liquid  m
 H2O (s)  H2O (l)
fusH  molar heat of fusion  energy required to convert 1
mole of water form the solid to liquid state.

Note fusH << vapH . This has to do with the differences in intermolecular distances
(packing of molecules) in the various states. When we heat a solid, the molecules are
packed close together. As we progress to the liquid state, the molecules are still fairly
close together (i.e., strong attractive forces). However, as we progress to the gaseous
state, we find the intermolecular distances are large and the molecules are far apart
(therefore, weak interactions).
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Heating/Cooling Curves - graphically show the heating, melting, and vaporising
processes. Note that the melting points and the boiling points have invariant (constant)
temperatures.
Solid vapour  m
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

CO2 (s) ⇌ CO2 (g)
subH  molar heat of sublimation  energy required to convert
1 mole of the substance from the solid to gaseous state. The reverse process (where the
gas directly to the solid state is known as deposition).
Good examples, CO2 at atmospheric pressure, I2(s)
Note
subH = fusH + vapH from Hess’s Law
Example

Calculate the heat required to evaporate 37.3 g of H2O at its boiling point.

Process


H2O (l) ⇌ H2O (g)
g of H2O = 37.3  moles of H2O = 37.3g * 1 mol/18.02g = 2.07 moles H2O
 energy (heat) needed = 2.07 moles of H2O (l) * 40.79 kJ/mol = 84.2 kJ of heat
vapH = 40.7 kJ/mol
Phase Changes
 Transform one phase to another, e.g., ice  water is an example of a solid  liquid phase
change (needs heat!!).
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
Phase Diagrams  give the overall relationship among the solid, liquid, and vapour
phases at many temperatures and pressures.

Look at some typical phase diagrams.

external pressure (usually atm)  y axis; T(C) x-axis

water
1. Below OC generally, solid (note the pressure dependence of the melting point of
water).
2. At 100C. Intersect the liquid-vapour  m line, p = 1 atm exactly. (note the
pressure dependence of the boiling point discussed previously).
3. All three curves intersect at a specific point called the triple point. This is a
very important point as it is the only point where all three phases are in  m
with one another. For water the triple point is at 0.006 atm and 0.01C.
4. Note that the solid-vapour  m exists only below the triple point.
5. What happens if we lower the external pressure to 0.10 atm?
6. m.p. increases, b.p. decreases  very important

look at CO2 phase diagram
1. The liquid region (phase) lies well above atmospheric pressure as the triple point for
CO2 is above atmospheric pressure (5.2 atm and -57C).
2. The solid-liquid boundary line has a positive slope (i.e., melting point increases with
increasing pressure). Compare to water.
3. Note the critical point for CO2 and compare with water.
Crystal Structure
 crystalline solid - rigid and long range order
 amorphous solid - lacks a long-range 3-D structure

In a crystalline solid - atoms, molecules or ions occupy specific positions. Arranged so
that the net attractive forces are a maximum (lowest energy state).

the centre of the positions occupied by atoms (molecules or ions)  lattice point

unit cell - basic repeating arrangement of the atoms, molecules or ions.
There are seven types of unit cells. The crystal lattice is a 3-D array of points, each rep.
identical environment in the crystal.
1. cubic a = b = c;
 =  =  = 90
2. tetragonal
a = b  c;
 =  =  = 90
3. orthorhombic a  b  c;
 =  =  = 90
4. rhombohedral a = b = c;
 =  =   90
5. monoclinic
a  b  c;
 =  = 90
  90
6. triclinica  b  c;
      90
7. hexagonal
a = b  c;
 =  = 90  = 120
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
look at a cubic unit cell
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
simple cubic
body-centred cubic
face-centred cubic

NOTE  each atom is shared by a number of unit cells.

example  simple cubic cell vs a face-centred cube (Slide).
Examples of Atoms and Molecules with differing simple unit cells
 gold  face centred cubic cell
 tungsten  body-centred cubic cell
 NaCl  ionic salt  face centred cubic structure
Packing Spheres
 closest packing  refers to the most efficient arrangement of spheres  try to imagine the
difference packing golf balls in a large box (more eff  golf balls fill holes)



Two Types
1. ABA  hexagonal close – packing (hcp)
2. ABC  cubic close - packing (ccp).
Note  in hcp the spheres in alternate layers occupy the same vertical positions, whereas
in ccp, for every fourth layer, the spheres occupying the same vertical space. For both
structures  each sphere is surrounded by 6 other spheres in the same layer, 3 spheres in
the layers immediately above and below it. The co-ordination # = 12
Coordination #  the # of atoms (or ions) surrounding an atom (or ion) in the crystal
lattice.

Compare with a simple cubic cell (scc).

co-ordination # = 8 (8 nearest neighbours)
scc has a lower packing efficiency
i.e., the hcp and ccp “plug” the holes better

What can we learn about an atom from its crystal structure?
Very important we can calculate the atomic radius!
Example
1. Simple cubic system
Simple cube a = b = c;
Length of edge  a = 2r  if we know a we can calculate the atomic radius.
2. Body Centred Cube.
b2 = 2a2; b = 2 a; c2 = a2 + b2; c2 = a2 + 2a2 = 3a2
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 c = 3 * a; a = c/ 3; but c = 4r; a = 4r/ 3
3. Face Centred Cube  1 = 8 r
cubic unit structure  cell sides are equal
c = 4r  c2 = 16r2 = a2  a = 8 r
Types of Crystals
1. Ionic  e.g. NaCl
2. covalent  e.g. SiO2
3. molecular  e.g. I2, H2O (icc), sucrose
4. metallic  e.g. Au, Ag, Ni ...
Types of Crystals and General Properties
Type
Units at Lattice Points
Forces
General Properties
Examples
(1)
Ionic
“+” and “-“ ions
electrostatic
high m.p., hard
doesn’t conduct
heat and electricity
well
NaCl
CsCl
NaI
(2)
Covalent
Atoms
covalent bonds
hard, high m.p.,
doesn’t conduct
heat and electricity
SiO2
diamond
graphite

molecular 
molecules or atoms; dispersion, dipole-dipole H-bonds; soft, low m.p.,
poor conductor of heat and electricity, e.g., I2, H2O, sucrose

metallic 
Atoms; metallic bonds; soft  hard, low to high m.p.s, good conductor of
heat and electricity, e.g., Ni, Al, Ti, etc

Amorphous Solids  lack a regular 3-D arrangement. Best-known example  glasses

Glasses are optically transparent fusion products of inorganic materials that have cooled to a
rigid state without recrystallising. Glasses have some liquid and some solid properties
Coloured glasses  due largely to the presence of metal ions.
e.g.
green glass - iron oxide (Fe2O3)
yellow glass  UO2
blue glass  Co(II) and Cu(II) oxides
