Chemistry-140
Lecture 33
Chapter 13:
Intermolecular Forces: Liquids & Solids
 Chapter Highlights
 phases of matter & kinetic molecular theory
 intermolecular forces

metallic & ionic solids

molecular & network solids
Will not include Sections 13.3, 13.6 & 13.7 on
physcial properties of liquids and solids
Chemistry-140
Lecture 33
Kinetic Molecular Description of Liquids & Solids
 Intermolecular forces are attractive, electrostatic
interactions that occur between molecules, atoms, or ions.
Liquids:
 intermolecular forces are strong enough to hold the
molecules together yet weak enough to allow the molecules
to move.
 generally denser than gases, fairly incompressible, have
definite volumes & can flow
Chemistry-140
Lecture 33
Kinetic Molecular Description of Liquids & Solids
Solids:
 intermolecular forces are strong enough to prevent the
molecules from moving.
 generally more dense than liquids, incompressible and rigid,
do not take the shape of their containers, do not flow.
 Crystalline solids: molecules or ions are arranged in
repeating patterns possessing high symmetry.
 Amorphous solids: molecules are arranged in a random
fashion.
Chemistry-140
Lecture 33
Intermolecular Forces
 Intermolecular forces: those between molecules, atoms,
or ions that hold the particles together.
 At a given temperature these forces are
solids > liquids > gases
 In general, the stronger the intermolecular forces, the
higher the melting point and the boiling point
Chemistry-140
Lecture 33
Intermolecular Forces
 Ion-dipole forces: electrostatic attractions that form
between an ion and an oppositely charged pole of a polar
molecule.
Ion
Dipole
d-
40 - 600 kJ/mol
d+
Polar water molecule
attracted to a cation
Chemistry-140
Lecture 33
Intermolecular Forces
 Hydrated cobalt(II) chloride, [CoCl2 . 6H2O], is a solid
salt. The structure is actually [CoCl2(H2O)4] . 2H2O in
which 4 of the water molecules are attached to the Co2+ ion
via ion dipole forces.
Cl
H
O
Co
Chemistry-140
Lecture 33
Intermolecular Forces
 Dipole-dipole forces: between the d+ pole of one polar molecule
and the d- pole of another polar molecule. If, equal in mass and
size, the attraction increases with increasing polarity.
Dipole
Dipole
5 - 25 kJ/mol
including H-bonding
Chemistry-140
Lecture 33
Intermolecular Forces
 Dipole-dipole attractions between two molecules of ClBr. Cl is
more electronegative than Br which creates the permanent
dipole (d+ and d-).
Chemistry-140
Lecture 33
Intermolecular Forces
 Hydrogen bonding: a special type of dipole-dipole interaction
which exists between the hydrogen atom in a polar bond and
the unbonded electron pair of an electronegative atom or ion.
The strength (4 - 25 kJ/mol) increases in the order
N-H….Y < O-H….Y < F-H….Y
X-H….F < X-H….O < X-H….N
These trends are explained in terms of increasing
electronegativities of N < O < F.
Chemistry-140
Lecture 33
Intermolecular Forces
 Hydrogen bonding between HF molecules. A lone-pair
on F (d-) interacts with an H atom (d+) to give a linear
H-bond. This repeats in the solid to form a zig-zag
chain of HF molecules
Chemistry-140
Lecture 33
Intermolecular Forces
Water is a very unique substance!!
High specific heat
4.18 J/ g K
High heat of fusion
333 J/g
High heat of vapourization
2250 J/g
High dielectric constant
80 @ 20 oC
High surface tension
7.2 x 109 N/m
density of water = 1.00 g/mL
density of ice = 0.917 g/mL.
HOW? WHY?....
Chemistry-140
Lecture 33
Intermolecular Forces
O
H
Hydrogen
bond
Hydrogen Bonding !!!
Chemistry-140
Lecture 33
Intermolecular Forces
Hydrogen
bond
H
O
Chemistry-140
Lecture 33
Intermolecular Forces
 Dispersion forces: can exist between formally nonpolar
molecules and polar molecules due to the polarizability of the
electron clouds on the formally nonpolar molecule. Induced
dipoles, d+ and d-, are smaller than permanent dipoles in polar
molecules.
Dipole
Induced Dipole
2 - 10 kJ/mol
Chemistry-140
Lecture 33
Intermolecular Forces
 Dispersion forces: can also exist between formally nonpolar
molecules due to the polarizability of electron clouds. All liquids
and solids possess dispersion forces that tend to increase with
increasing formula weight.
Induced Dipole
Induced Dipole
0.05 - 40 kJ/mol
Chemistry-140
Lecture 33
Intermolecular Forces
Question:
List the substances BaCl2, H2, CO, HF and Ne in
order of increasing boiling point.
Chemistry-140
Lecture 33
Intermolecular Forces
Answer:
Boiling point depends, in part, on attractive forces.
BaCl2:
ionic bonding
H2:
dispersion forces (M = 2)
CO:
dipole-dipole forces (M = 28)
HF:
dipole-dipole forces (hydrogen bonding) (M = 20)
Ne:
dispersion forces (M = 20)
H2(20 K) < Ne(27 K) < CO(83 K) < HF(293 K) < BaCl2(1813 K)
Chemistry-140
Lecture 33
Final Exam
Monday December 10th, 2001
12:00 Noon
St. Denis Centre Field House
Chemistry-140
Lecture 34
Chapter 13:
Intermolecular Forces: Liquids & Solids
 Chapter Highlights

phases of matter & kinetic molecular theory

intermolecular forces
 metallic & ionic solids
 molecular & network solids
Will not include Sections 13.3 & 13.6 on
physcial properties of liquids and solids
Chemistry-140
Lecture 34
Structures of Solids
 Crystalline solid: one in which the molecules, atoms, or
ions are ordered in well-defined, repeating, patterns.
ionic:
NaCl
molecular:
H 2O
metallic:
Fe
network:
C(diamond)
Amorphous solid: one in which no order exists; may have
small regions of order, but no long-range order.
glass:
borosilicate glass
polymer:
nylon
Chemistry-140
Lecture 34
Crystalline Solids
Crystalline solids can be described by unit cells.
Unit cell: the smallest
"piece" of the crystal
required to show the
repeating pattern.
Chemistry-140
Lecture 34
Crystalline Solids
 Unit cells are parallelepipeds (6-sided figures with faces
that are parallelograms) described in terms of the lengths
of the edges (a, b and c) and the angles between these
edges (a, b, and g)
a
b
c
b
a
g
Chemistry-140
Lecture 34
Crystalline Solids
 The simplest unit cell is cubic (a = b = g = 90o) and
(a = b = c). The primitive cubic unit cell consists only of
atoms at the corners of the box.
a
b
c
a
b
g
Chemistry-140
Lecture 34
Cubic Unit Cells
Primitive
Cubic
Body
Centred
Cubic
Face
Centred
Cubic
Chemistry-140
Lecture 34
Atom Sharing Between Unit Cells
 In any cubic lattice, a corner atom is shared equally
between eight unit cells.
 It therefore only contributes 1/8 of an atom
to that particular unit cell
Chemistry-140
Lecture 34
Atom Sharing Between Unit Cells
 In any cubic lattice, a face-centred atom is shared equally
between two unit cells.
 It therefore only contributes 1/2 of an atom
to that particular unit cell
Chemistry-140
Lecture 34
Atom Sharing Between Unit Cells
 The body-centred cubic unit cell has an atom at the centre
of the box in addition to the eight atoms in the corners. A
body-centred atom is not shared and therefore contributes
one full atom to the unit cell.
 Atoms may also be found on the edges of the box. An
edge-centred atom is shared equally between four unit
cells. It therefore only contributes 1/4 of an atom to that
particular unit cell.
Chemistry-140
Lecture 34
Crystal Lattice Measurements
Example 13.6:
Aluminium has a density of 2.699 g/cm3 and the
atoms are packed into a face-centred cubic unit cell.
Determine the radius of an Al atom.
Chemistry-140
Lecture 34
Crystal Lattice Measurements
Known: the density of Al
the structural make-up of the basic unit cell
We know that:
M
d
V
Mass of the unit cell = (# of Al atoms)(atomic weight of Al)
 Z  M 
M
N 
Volume of a cube = (length of the
side)3
V  l3
Chemistry-140
Lecture 34
Crystal Lattice Measurements
If we know the length of the side then we can
determine the radius of a single atom!!
Cell diagonal 
2l  4( rAl )
Cell
edge
Chemistry-140
Lecture 34
Crystal Lattice Measurements
Step 1: We know M, N and d but need to calculate Z.
Each atom at the centre of a
face contributes 1/2 of an Al
atom to the cell.
6 x (1/2) = 3 Al atoms
Each atom at the corner
contributes 1/8 of an Al atom
to the cell.
8 x (1/8) = 1 Al atom
Z = 3 + 1 = 4 Al atoms
Chemistry-140
Lecture 34
Crystal Lattice Measurements
Step 2:
Use the equation:
ZM
d
NV
Solve for V using:
ZM
V
dN
V
=
4 atoms / unit cell 26.98 g / mol
 2.699 g / cm 6.022 x 10 
3
6.640 x 10-23 cm3
23
Chemistry-140
Lecture 34
Crystal Lattice Measurements
Step 3:
Use the equation:
l3V
l  66.40 x 10
3
=
24
4.049 x 10-8 cm
Chemistry-140
Lecture 34
Crystal Lattice Measurements
Step 4:
l 2
and the equation: rAl 
4
rAl 
4.049 x 10
8
cm 2
4
= 1.432 x 10-8 cm = 1.432 Å
Chemistry-140
Lecture 34
Ionic Solids
Cl-
Cs+
CsCl Lattice
Chemistry-140
Lecture 34
Ionic Solids
Expanded NaCl Lattice
Chemistry-140
Lecture 34
Ionic Solids
The
Cl-
anions form a
face-centred unit cell
and the Na+ cations fill
the spaces, such as the
centre of the unit cell.
There must of course
be equal numbers of
cations & anions!!
(Z = 4)
ClNa+
Chemistry-140
Lecture 34
Molecular & Network Solids
 Molecular solids: those in which the particles that comprise the
crystal are molecules.
Type of
solid
Form of
unit
particles
Forces
between
particles
Properties
Molecular atoms or dispersion and fairly soft, lowmolecules dipole-dipole
moderately
forces,
high melting,
hydrogen
poor thermal
bonds
& electrical
conduction
Examples
Ar, CH4,
sucrose,
CO2
Chemistry-140
Lecture 34
Molecular & Network Solids
 Covalent network solids: those in which the crystal itself is a
large, covalently bonded, molecule.
Type of
solid
Form of
unit
particles
Forces
between
particles
Covalent- atoms
covalent
Network connected bonds
in a
covalentnetwork
Properties
Examples
very hard, very Diamond,
high melting,
quartz, SiO2
poor thermal &
electrical
conduction
Chemistry-140
Lecture 34
Textbook Questions From Chapter # 13
Intermolecular Forces:
20, 22, 26
Metallic & Ionic Solids:
43, 45, 48
Molecular & Network Solids:
52
General Questions:
60, 69
Conceptual Questions:
88
Chemistry-140
Lecture 33
Final Exam
Friday December 10th, 2001
12:00 Noon
St. Denis Centre Field House