1. business and industrial
2. energy
3. oil

How To Succeed in Chem 11
• Go to every class--sit up front.
• Read the assignment before class
• Ask questions--be involved
• Do the assigned problems!!
• Work together on problems
• Make friends and study in groups.
• Review old exams and quizzes.
Announcements
– Problem solving final schedule
– Exam 1 = December 16 (evening)
– Chapter 12: Cover p. 425-451. Skip p. 452-479.
– Chapter 13: Skip all of section 13.2 and Section
13.3. Cover Section 13.4 to end of Chapter 13.
– Please don’t fall behind!!
Intermolecular Forces and
Liquids and Solids -12
Intermolecular forces are attractive forces between
molecules that explain many important physical
properties of substances.
Viscosity
Boiling and Freezing Point
Wetting or Not Wetting
Surface Tension
capillary action
Intermolecular forces play a critical role in life as we
understand it.
Polar heads in
aqueous exterior
Embedded
Proteins
Nonpolar tail
Aquoues Interior
Intermolecular forces play a critical role in life as we
understand it.
Intermolecular forces hold together the double
helix of DNA.
When Mom washes the clothes oily portion of soap
interacts with grease and dirt and removes it from our
hands and clothes.
CH3(CH2)16COO- Na+
Solids, liquids (called condensed phases) are held
together by two attractive forces intermolecular
(between molecules) and intramolecular forces
(bonds within a molecule).
• Intramolecular
forces are always
stronger than
intermolecular
forces
gaseous HCl molecules
• Intermolecular
forces control the
physical
properties of the
substance
Intramolecular forces are attractive “bonding forces”
that exist within a molecule or ionic compound holding
it together (internal chemical bonds).
Ionic Bonded
substances have
intramolecular bonds
resulting from electrons
transfer from one atom
to another.
Covalent molecules have
intramolecular bond that
result from sharing one or
more electrons pairs
between atoms usually nonmetals.
A phase is a state of matter that is homogeneous,
chemically uniform and has physically distinct
properties (density, crystal structure, index of
refraction).
GAS
Generally we recognize
4-states of matter:
1. Gas
2. Liquids
3. Solids
4. Plasma (hot ionized
gas)
LIQUID
SOLID
-very high density
-not compressibile
-low Kinetic energy
-small distance between
molecules
-IMF large
-high density
-not compressibile
-mid Kinetic energy
-molecules close but
fluid
-IMF large
-low density
-high compressibility
-high Kinetic energy
-large distance
between molecules
-no IMF
One can think of the phases of matter as “classes” of
the type of molecular motion that can be found at
different temperatures (kinetic energy of collections
of molecules).
At low temperature = low kinetic energy, the motion of molecules is
limited by intermolecular forces giving a phase of matter that is rigid and
dense. When the temperature is high, the motion of the molecules is
dominated by their translational energy, so intermolecular forces are
overcome can almost be ignored. At intermediate temperatures, molecules
translate but still stick together.
Kinetic molecular theory states that the temperature
of collection of particles (system) is proportional to
the average kinetic energy of the collection.
3
KE = Ek = 2 RT
Kinetic Energy
Temperature
KE
Temperature
Vs
IMF
Properties
The phase of a substance depends on the interplay between
KE(T), which keeps molecules far apart and moving, and IMF
which keep molecules close together and condensed.
Property
Molecular
Mobility
Density
Compressibility
Gases
Liquid
Expand to occupy Assumes shape of
container
container
Fixed Shape
molecules fixed
Very low density
High
High
High
Very Very low
Not-compressible
high molecular
speeds
Atoms and
molecules are
fixed
Low
Many
Kinetic Energy
High
Molecule slide
past one another
--flow
Medium
Intermolecular
Forces
Few and small
Many
Diffusion
Solids
N
O
IO
IM
AT
I
SIT
SU
BL
D
EP
O
ΔH < 0
ΔH > 0
Gas
N
O
TI
SA
N
O
EN
TI
D
N
RA
O
CO
AP
EV
N
You can think of the different phases as classes of
possible molecular motion due to different kinetic
energies (caused by temperature differences) and
varying degrees of intermolecular forces.
ΔH < 0
ΔH > 0
MELTING ΔH > 0
FREEZING ΔH < 0
Solid
Liquid
An Exothermic reaction gives off (releases) heat to
the surroundings.
2H2 (g) + O2 (g)
H2O (g)
2H2O (l) + heat
H2O (l) + heat
heat written as
a product
ΔH = <0
An Endothermic reaction absorbs heat from the
surroundings into the system.
heat + 2HgO (s)
heat + H2O (s)
2Hg (l) + O2 (g)
H2O (l)
heat written as a
reactant
ΔH = >0
Chemists measure the enthalpies (heat measured at
constant pressure) of nearly all chemical reactions
and give names to many.
The heat of fusion and the heat of vaporization are
intensive properties that tell us something about the IMF
between these molecules (tabulated in Handbook of
Physics and Chemistry).
•
Molar Heat of Fusion (∆Hfus):
The energy required to melt one
mole of solid (in kJ).
•
Molar Heat of Vaporization
(∆Hvap):
The energy (in kJ) required to
vaporize one mole of liquid.
•
Molar Heat of Sublimation
(∆Hsub):
The energy (in kJ) required to
vaporize one mole of solid to
gas.
ΔH˚vap
ΔH˚fus
The differences in energy required to
vaporize or melt a pure substance tells
us something about the IMF’s that try to
keep these molecules together in the
condensed phase.
A Small Table of Heats of Fusion and Heats of Vaporization
Notice that ΔH˚vap is greater than ΔH˚fus suggesting
that IMF are more stronger in the liquid phase.
Temperature ˚C
We can graphically show the qualitative and quantitative
aspects of a phase change by plotting the temperature
of a substance vs heat (q) added to a substance.
Heat Added (Joules)
We can graphically show the quantitative aspects of a
phase change in a “heating curve” or a plot of
temperature vs heat added to a substance.
Heating steam
past 100˚C
Boiling and
vaporizing all water
to steam at 100˚C
Heating water
to boiling
100˚C
Heating
solid ice
to 0˚C
Melting solid ice to 0˚C water
The specific heat (s) of a substance is the amount of
heat (q) required to raise the temperature of one gram
of the substance by one degree Celsius.
The molar heat capacity of a substance is the amount
of heat (q) required to raise the temperature of one
mole of the substance by one degree Celsius.
Heat
absorbed
(Joules)
q = s m ΔT
Change in Temp: ˚C or K
specific heat capacity
J/g °C or J/mol K
# of moles
or grams
The specific heat (s) of a substance
is the amount of heat (q) required
to raise the temperature of one
gram of the substance by one
degree Celsius. (How thermally
sensitive a substance is to the
addition of energy!)
Heat (q) absorbed or released:
q = s m ΔT
Here’s the same curve now applying the conservation of
energy (sum of the heats)
q = mice ∆Hfus
Phase
transition
Temperature ˚C
220
0
-100
Phase
transition
Temperature does
not change during
a phase transition.
Ice +
Water
mix
Heating
solid ice
to 0˚C
Ice
A
q = sstm mstm ∆T
q = sH2 O mH2 O ∆T
q = sice mice ∆T
100
q = mH2 O ∆Hvap
B
Temperature does
not change during
a phase transition.
Water
D
Water +
steam mix
Steam
E
C
Heating
water to
boiling
100˚C
Melting
solid ice to
0˚C water
4.12
F
38
Boiling all
water to
steam
100˚C
79.3
Heat Added (kJ/mole)
Heating
steam past
100˚C
305
309
Calculate the amount of heat required to convert 500
grams of ice at -20.0˚C to steam at 120.˚C. The specific
heat capacities of water, ice and water vapor are 4.18,
2.06 and 1.84 J/g ˚C respectively, and the latent heat of
fusion and vaporization, ΔHf and ΔHv, are 6.02 and 40.7
kJ/mol respectively.
n
!
qi = 0
sum the q’s baby
i=1
q = si mi ΔT
for non-phase transitions
qsolid=>liquid = # moles ΔH˚fusion
qliquid=>gas = # moles ΔH˚vaporization
for phase transitions
Calculate the amount of heat required to convert 500
grams of ice at -20˚C to steam at 120˚C. The specific
heat capacities of water, ice and water vapor are 4.18,
2.06 and 1.84 J/g ˚C respectively, and the latent heat of
fusion and vaporization, ΔHf and ΔHv, are 6.02 and 40.7
kJ/mol respectively.
1. Heat ice from -20˚C to ice at 0˚C
= 500. g x 2.06 J/g ˚C x 20˚C
2. Melt ice at 0˚C to water at 0˚C
= 500. g/(18 g/mol) x 6.02 kJ/mol
3. Heat water from 0˚C to water at 100˚C= 500. g x 4.18 J/g ˚C x 100˚C
4. Evap water at 100˚C to vap at 100˚C = 500. g/(18 g/mol) x 40.7 kJ/mol
5. Heat vap from 100˚C to vap at 120˚C = 500. g x 1.84 J/g ˚C x 20˚C
1. = 20.6 kJ
2. = 167.2 kJ
3. = 209.0 kJ
4. = 1130.5 kJ
5. = 18.4 kJ
Total = 1545.6 kJ
In open containers molecules that have enough kinetic
energy can overcome IMF’s at the surface and
“evaporate” into the atmosphere.
In closed containers, molecules vaporize and condense
until there is no further change in concentration. This
forms an equilibrium “vapor pressure” over the liquid.
Evaporation
Liquid
Vapor Pressure
Liquid
The equilibrium vapor
pressure is the
pressure exerted by a
vapor over its liquid
phase (measured
under vacuum) when a
dynamic equilibrium
exists between
condensation and
evaporation.
Dynamic chemical equilibrium is reached when the rate
of evaporation and rate of condensation are equal.
Equilibrium
Time
Molecules in liquid
begin to vaporize
Molecules vaporizing and
condensing at such a rate that
no net change in numbers
occure
Dynamic equilibria is also reached in melting and
sublimation and also in most chemical reactions.
At the melting point a solid begins to
change into a liquid as heat is added. As
long no heat is added or removed melting
(red arrows) and freezing (black arrows)
occur at the same rate an the number of
particles in the solid remains constant.
aA + bB
cC + dD
Reactants
Products
Reaction Rate of the forward reaction =
= Rate of Reverse reaction
Because kinetic energy (of molecules) depends on
temperature, so does vapor pressure of a liquid.
3
RT
KE = Ek =
2
Kinetic Energy
At a higher
temperature
a larger fraction
of molecules
have sufficient KE
to escape the
condensed
phase.
Temperature
More molecules
escape at high temp
The vapor pressure of a pure liquid depends on the
intermolecular forces between molecules. The
stronger the attractive forces in the liquid phase the
lower the vapor pressure--and the less volatile it is.
2 atm
.66 atm
Which of the
following has the
highest vapor
pressure at 1 atm?
Which is the least
volatile at 1 atm?
The boiling point of a pure liquid is the temperature at
which the vapor pressure over its liquid phase is
equal to the external pressure on the liquid.
The normal boiling point is the temperature at which a
liquid boils when the external pressure is 1 atm.
Evaporation
Boiling
Vapor Pressure (torr) of Some Liquids
Vapor Pressure of Some Liquids
Heat of Vaporization, Boiling Pts of Liquids
If we plot ln (vapor pressure) vs 1/temp we observe a
linear relationship.
Vapor pressure (torr)
Vapor pressure plotted as
a function of temperature
ln (vapor pressure) plotted
as a function of 1/Temp
ln P
Temperature ˚C
1/T
The following diagram shows a close-up view of part of the
vapor-pressure curves for a solvent (red curve) and a
solution of the solvent with a second liquid (green curve).
Which solvent is more volatile?
•
The Clausius-Claperyron equation relates the vapor
pressure (P) of a pure liquid to the liquid’s temperature
(T) and the liquids molar heat of vaporization (∆Hvap).
-ΔHvap 1 
ln P =
+C
R
T
 
y =
m
x+ b
(note R = 8.31 J/K mol)
•
slope = ∆Hvap/R
ln P
By taking measurements
at two temps, we get:
P2
-ΔHvap 1
1
ln
=
− 

R
P1
 T2 T1
1/T
Vapor pressure of pure etOH is 115 torr at 34.9˚C. If ΔHvap =
40.5 kJ/mol calculate the temperature when the vapor
pressure of etOH is 760 torr. R is the gas constant given at
8.314 J/mol K
ΔHvap =40.5 kJ/mol
P1=115 torr P2=760 torr
T1= 34.9 + 273.15 = 308.0K
P2
-ΔHvap 1
1
ln
=
− 

P1
R
T
 2 T1
760 torr
ln
115 torr
=
-40.5 x103 J/mol
8.314 J/mol*K
T2 = 350K = 77˚C
1
1
T2
308K
A phase diagram summarizes the conditions at which a
substance exists as a solid, liquid, or gas phase of
matter.
A phase diagram summarizes the conditions at which a
substance exists as a solid, liquid, or gas phase of matter.
3
D
218
Melting
Pressure (Atm)
Freezing
Point
Freezing
Liquid
Critical Point
Supercritical
Fluid
1.0
Solid
.43
Normal
Boiling Point
Evaporation
Condensation
0.006
Sublimation
Deposition
Gas
Triple Point
A
374
AB is the vapor pressure curve for ice; BD the vapor pressure curve
for liquid water; BC the melting point line; point B the triple point point
D labels the critical point.
Terminology in a Phase Diagram
The triple point is the single P,T point at which all three
phases are in equilibrium with one another.
The critical temperature (Tc) is the temperature above
which the gas cannot be made to liquefy, no matter how
great the applied pressure.
The critical pressure (Pc) is the minimum pressure that
must be applied to bring about liquefaction at the critical
temperature.
The normal boiling point is the temperature at which the
vapor pressure of a liquid equals 1 atm.
The normal freezing point is the temperature at which the
vapor pressure of a liquid equals 1 atm.
Water is one of the few substances on earth for which
the liquid phase is more dense than its solid phase. We
see this in a phase diagram in the slope of the solidliquid line.
more pressure at a fixed
temperature leads to a liquid
not a solid
more pressure at a fixed
temperature leads to a solid
not a liquid
Pressure (torr)
1000
K
760
A
B
D
C
Temp ˚C
From Line E to G what is occurring?
From Line J to E what is occurring?
Between what two points is the vapor pressure curve.
Which two points if the melting point curve
What is the normal boiling point of this substance?
What is the triple point of this substance?
Water Is A Unique Substance
Ice is less dense than water
Density of Water
Maximum Density
4 0C
•great solvent due to high polarity
and hydrogen bonding ability
•exceptional high specific heat
capacity
•high surface tension and capillarity
•large density differences of liquid
and solid states
A phase diagram summarizes the conditions at which a
substance exists as a solid, liquid, or gas phase of
matter.
Line AB is the vapor-pressure curve for solid-vapor (sublimation curve); BD
the vapor pressure curve for liquid-vapor water; BC the melting point line
(notice how it has a negative slope = solid is lower density); point B the
triple point: only point when three phases are in equilibrium; point D is the
critical point--the temperature at which no pressure can liquefy a gas (new
phase of matter).
Generally speaking, which phase of a substance should
have the highest density (mass/volume)? Name a
compound that does not follow this trend?
GAS
LIQUID
SOLID
Intermolecular forces are classified into four major types.
1. Ion-dipole: occur between neighboring an
ion solution and a polar molecule (dipole)
also in solution.
Na+
2. Dipole-dipole: occur between a neutral
polar molecules (same or not the same
molecules).
3. Induced-dipoles: occur when a ion or a Ion or Dipole Induced Dipole
dipole induces a spontaneous dipole in a
+
+
neutral polarizable molecule.
4. London Dispersion Forces are attractive
IMF’s that occur when spontaneous dipoles are
formed randomly or induced by other charged
species in neutral polarizable molecules.
Ion-dipole intermolecular forces are attractive
forces between an ion in solution and a neighboring
polar molecule.
Molecules
with a dipole
moment in solution
- end of dipole
attracted to + ions
+ end of dipole
attracted to -ions
-
An anion
in solution
+
A cation
in solution
Dipole-induced dipole are attractive intermolecular
forces that occur when a polar molecule (dipole)
induces a dipole by distorting an electron cloud in a
polarizable molecule.
non-polar molecule
polar molecule
induced dipole
Dipole-Dipole intermolecular forces are attractive
forces that occur between polar molecules (same or
not the same).
VS
Boiling point differences are explained by dipole-dipole
interactions
Summary of Intramolecular Types of Forces and Energies
Flowchart for classifying intermolecular forces
Intermolecular forces arise from electrostatic forces
that depend on distance between molecules.
Coulomb’s Law: The electric force acting
on a point charge q1 as a result of the presence
of a second point charge q2 some r meters
away is given by:
+
k q1 q2
F=
r2
Energy of Attraction is:
k q1 q2
E=
r
-
r
units of meters
k depends on medium (kair = 9.0 x 109 N•m2/C2)
Because Energy = Force x Distance
units of coulombs
If we can predict charge
separation or polarity in
molecules we can identify
IMF’s in play for that
molecule.
-
+
Connecting Dots: large electronegativity differences =>
polar molecules => large dipole moments => gives rise to
large IMF => decreased vapor pressure => increased
boiling points, freezing points, viscosity, surface tension.
r
a
l
o
d
n
Bo
P
Po
lar
-
Bo
nd
+
Dipole Forces +
London Dispersion
Forces
Net Dipole Moment
Polar Bond
Polar Bond
No Net Dipole Moment
No dipole forces
London
Dispersion
Forces Only
Electronegativity is an element’s inherent property to
draw electrons to itself when chemically bonded to
another atom in a molecule.
F is the most
electronegative
element
Cs the
least
We identify polar bonds and use the molecular
geometry to gauge if there is a net dipole moment
(dipole) in a molecule.
δ-
δδ-
δ+
Polar with
net dipole
moment
δ+
δ+
δ+
δ-
δδδnon-polar with
no net dipole
moment
non-polar with
no net dipole moment
Polar with
net dipole
moment
Polar with
net dipole
moment
Recall your molecular and electron geometry from Chem 7.
3 EG
4 EG
2 EG
5 EG
6 EG
Summary of Intermolecular Forces (van der waals forces)
Type of Interaction
Ion
Dipole
-
+
-
Dipole
+
Energy (kJ/mol)
Highest
--an ion interacts with a
polar molecule (dipole).
+
-
+
Induced Dipole
+
-
Dipole
+
+
50-600
+
-
--two polar molecules (dipoles)
interact electrostatically.
5-25
--ion or a dipole induces a dipole
in a polarizable non-polar
molecule.
2-15
--occurs between non-polar
molecules via polarizability
--also called London Dispersion
forces
Induced Dipole Induced Dipole
-
Interaction
Dipole
Ion
-
Strength
+
Lowest
0.05-40
Summary of Intermolecular Types of Forces and Energies
Ion-dipole
H-Bond
Dipole-Dipole
Ion-induced
dipole
Dipole-induced
dipole
Dispersion
(London)
+
-
+
Hydrogen bonding is a special case of dipole-dipole
intermolecular force that occurs between a hydrogen
atom and an unshared pair of electrons in a polar N-H,
O-H, or F-H bond.
CH3CH2OH
CH3OCH3
ethanol
dimethyl ether
-24.8°C
78.3°C
The boiling points of covalent binary hydrides increase with
increasing molecular mass down a Group but the hydrides
of NH3, H2O and HF have abnormally high BP because of
hydrogen bonding.
100
?
Temperature ˚C
Increased London Forces
0
-100
Period
Which of the following substances exhibits H
bonding? For those that do, draw two molecules of
the substance with the H bonds between them.
O
(a)
C2 H6
PLAN:
(c)
CH3C NH2
Find molecules in which H is bonded to N, O or F. Draw H
bonds in the format -B:
H-A-.
SOLUTION:
(b)
(b) CH3OH
(a) C2H6 has no H bonding sites.
H
H C O H
H
H
H O C H
H
H
H
(c)
O
H N CH3C
CH3C N H
H
O
CH3C
O
H N
N H
O
H
CH3C
Which substances experience dipole-dipole intermolecular
forces? SiF4,CHBr3, CO2, SO2
SiF4, geometry tetrahedral, Si-F bonds are polar, but no molecular
dipole; bond dipoles cancel. No dipole-dipole interactions.
CO2, linear geometry, C-O bonds are polar but symmetry cancels
net diple, but no molecular dipole; bond dipoles cancel and no
dipole IMF’s
SO2, bent geometry, S-O bonds are polar and do not cancel. Sulfur
lone pair dipole only partially offsets net bond dipole. Has dipoledipole forces
CHBr3, tetrahedral geometry, C-H and C-Cl bonds are polar and do
not cancel
SO2 and CHCl3 experience dipole-dipole intermolecular forces.
Electronegativity values from a table can be used to
judge the extent of ionic, polar or covalent bond
character in a chemical bond.
χ
3.0
Mostly Ionic
2.0
χ = |ENB - ENA|
absolute value of
electronegativity
difference between
2-bond atoms
Polar Covalent
Mostly Covalent
0.0
Quiz 1
Vapor pressure of pure etOH is 115 torr at 34.9˚C. If ΔHvap = 40.5 kJ/
mol calculate the temperature when the vapor pressure of etOH is
760 torr. R is the gas constant given at 8.314 J/mol K
P1
∆Hvap
ln
=
P2
R
!
1
1
−
T2
T1
"
ΔHvap =40.5 kJ/mol
P1=115 torr P2=760 torr
T1= 34.9 + 273.15 = 308.0K
760
40.5 J/mol
ln
=
115
8.31 J/mol K!
1.888 = (−4.87 × 103 )
T2 = 350.0 K
!
1
1
−
T2
308
"
1
1
−
T2
308
"
Quiz 1 (November 23, 2009)
Calculate the total heat in Joules needed to heat 1000
L of water at 25˚C to steam at 120˚C. Assume the
density of water is 1.0 kg/L. The specific heat
capacities of water, ice and gaseous water vapor are
4.18, 2.06 and 1.84 J/g ˚C respectively, and the latent
heat of fusion and vaporization, ΔHf and ΔHv, are 6.02
and 40.7 kJ/mol respectively.
Intermolecular forces are classified into four major types.
1. Ion-dipole: occur between neighboring an
ion solution and a polar molecule (dipole)
also in solution.
Na+
2. Dipole-dipole: occur between a neutral
polar molecules (same or not the same
molecules).
3. Induced-dipoles: occur when a ion or a Ion or Dipole Induced Dipole
dipole induces a spontaneous dipole in a
+
+
neutral polarizable molecule.
4. London Dispersion Forces are attractive
IMF’s that occur when spontaneous dipoles are
formed randomly or induced by other charged
species in neutral polarizable molecules.
London Dispersion Forces are attractive intermolecular
forces that occur when temporary dipoles are formed due
to random electron motions in all polarizable molecules.
2-neutral non-polar
polarizable molecules
spontaneous movement of electrons
forming an instantaneous
dipole moment
Induced-Dipole Moment
and IMF between molecules
Polarizability is the ease with which an electron
distribution (cloud) in the atom or molecule can be
distorted by an outside ion or dipole.
spontaneous
induced dipole that
depends on polarizability
of substance
non-polar molecule
(electron cloud)
induced dipole
dispersion
London Dispersion Force
All molecules have at least this IMF
Polarizability and London dispersion trends mirror
atomic size trends in the periodic table.
•Polarizability increases right to left
(bigger size more distortable.
•Size and
polarizability
increases
down a group
POLARIZABILITY
•Polarizability increases with molar mass (# e-) of a molecule
•Cations are less polarizable than their parent atom
because they are smaller and more compact.
•Anions are more polarizable than their parent atom
because they are larger.
Examples of increasing London dispersion forces with
increase in size and polarizability.
Increasing Size/
Polarizability
Increasing Size/
Polarizability
Larger unbranched molecules are more polarizable
than compact branched molecules.
neopentane
bp=10 oC
tetrahedral
normal pentane
bp=36 oC
extended structure
Larger polarizability in unbranched molecules
explains boiling and melting points trends in isomers.
Some Generalizations About IMF’s
1. Ion-Ion > Ion-Dipole > Dipole-Dipole > Dispersion
2. For polar molecules of approximately the same mass and
shape and volume (i.e. polarizability the same), dipole-dipole
forces dictate the difference in physical properties.
3. Hydrogen bonding occurs with polar bonds in particular HF, H-O, H-N and an unshared pair of electrons on a nearby
electrognegative atom usually F, O, or N.
4. For non-polar molecules of the same molecular mass, longer
less compact molecules are generally more polarizable and have
greater dispersion forces and show higher boiling and melting points.
5. For non-polar molecules of widely varying molecular mass,
those with more mass are typically more polarizable and
experience greater London dispersion forces and exhibit higher
boiling points and melting points.

Arrange the following substances in order
of increasing boiling points.
C2H6, NH3, Ar, NaCl, AsH3

Arrange the following substances in order of
increasing boiling points.
C2H6, NH3, Ar, NaCl
Ar < C2H6 < NH3 < NaCl
nonpolar
nonpolar
polar
ionic
London London dipole-dipole H-bonding ion-ion
Arrange the following non-polar molecules in order of
increasing melting point.
SiF4, CS2, CI4, GeCl4
Solution. None of these molecules poses a net dipole moment. Only
dispersion forces exist and these are expected to increase with
increasing molecular mass (more polarizable as a molecule gets larger).
The molar masses of these substances follow:
Substance
Molar Mass
CS2
76.131
SiF4
104.077
CI4
519.631
GeCl4 214.402
The intermolecular forces, and the melting points, should increase in the
following order: CS2 < SiF4 < GeCl4 < CI4 The experimentally
determined melting points are -110.8, -90, -49.5, and 171 oC,
respectively.
For each pair of substances, identify the dominant
intermolecular forces in each substance, and select
the substance with the higher boiling point.
(a) MgCl2 or PCl3
(b) CH3NH2 or CH3F
(c) CH3OH or CH3CH2OH
CH3
(d) Hexane
CH3CCH2CH3
(CH3CH2CH2CH2CH2CH3)
CH3
or 2,2-dimethylbutane
PLAN: •Bonding forces are stronger than nonbonding(intermolecular) forces.
•Hydrogen bonding is a strong type of dipole-dipole force.
•Dispersion forces are decisive when the difference is molar mass or
molecular shape.
For each pair of substances, identify the dominant
intermolecular forces in each substance, and select
the substance with the higher boiling point.
SOLUTION:
(a) Mg2+ and Cl- are held together by ionic bonds while PCl3 is covalently
bonded and the molecules are held together by dipole-dipole
interactions. Ionic bonds are stronger than dipole interactions and so
MgCl2 has the higher boiling point.
(b) CH3NH2 and CH3F are both covalent compounds and have bonds
which are polar. The dipole in CH3NH2 can H bond while that in CH3F
cannot. Therefore CH3NH2 has the stronger interactions and the higher
boiling point.
(c) Both CH3OH and CH3CH2OH can H bond but CH3CH2OH has more
CH for more dispersion force interaction. Therefore CH3CH2OH has the
higher boiling point.
(d) Hexane and 2,2-dimethylbutane are both nonpolar with only
dispersion forces to hold the molecules together. Hexane has the larger
surface area, thereby the greater dispersion forces and the higher boiling
point.
Describe
the intermolecular
that
exist between
What type
of intermolecular
forcesforces
can you
recognize
in
each ofionic
the following
molecules?
the following
and covalent
compounds.
HBr(g) HBr is a polar molecule: dipole-dipole forces. There are
also dispersion forces between HBr molecules.
CH4
CH4 is non-polar: London dispersion forces.
S
SO2
O
O
SO2 is a polar molecule: dipole-dipole forces. There are
also dispersion forces between SO2 molecules.
Namequestions
the types ofon
intermolecular
the following
Any
homeworkforces
from in
Chapter
11?
compounds.
CH3CH2CH2CH3
CH3OH
CH3CONH2
CH3COOH
CH3CH2CH2CH2OH
CH3CH2COCH3
In which substances would hydrogen bonding
forces occur between molecules?
C2H6, HCHCl3, CH3CH2OH, HNO3, PH3
Solution. Hydrogen is bonded to one of the very
electronegative atoms in CH3CH2OH and HNO3.
Hydrogen bonding should occur in both of these
substances.
Water Is Wild and Unique and Taken For Granted
--High H-bonding = high cohesive forces responsible for
the transport of water in roots and xylem of trees and
plants.
--Base-pairing in double-stranded DNA
--Heat capacity, high heat of vaporization, inverted
density of water and ice.
Three common physical properties of liquids that depend
on the magnitude of IMF’s include surface tension,
capillarity and viscosity.
surface tension
Viscosity
capillarity
Surface tension is the amount of energy required to
stretch or increase the surface area of a liquid.
molecules at
the surface feel
a net force
downward
molecules in
the interior
experience
equal force in
3-D
We say that the leave
has a “low surface
energy” as water will not
spread out (it takes
energy to spread the
water out)
Observed Physical Properties and IMF’s
Surface Tension of Some Liquids
Substance
Formula
Surface Tension
(J/m2)
at
Major IMF’s
200C
CH3CH2OCH2CH3
1.7x10-2
dipole-dipole; dispersion
ethanol
CH3CH2OH
2.3x10-2
H bonding
butanol
CH3CH2CH2CH2OH
2.5x10-2
H bonding; dispersion
water
H 2O
7.3x10-2
H bonding
mercury
Hg
48x10-2
metallic bonding
diethyl ether
The surface tension of a liquid is a function of temperature.
Why might this be? (what forces compete here?)
Hot water works better
than cold water in
cleaning your clothes or
hands as it can more
effectively “wet” dirt
(water gets into pores
and not get “stuck”).
Capillary effect occurs when the adhesive IMF’s
between a liquid and a substance are stronger than the
cohesive IMF’s inside the liquid.
Cohesion is the intermolecular attraction between like molecules.
Adhesion is an attraction between unlike molecules
Cohesion
Adhesion
– Capillary rise implies that the:
• Adhesive forces > cohesive forces
– Capillary fall implies that the:
• Cohesive forces > adhesive forces
Water will rise to different heights depending on the
diameter of the capillary. Why?
•
Viscosity is the measure of a liquid’s resistance to flow
relative to one another, and is thus related to the
intermolecular forces.

Oil for your car is bought based on
this property: 10W30 or 5W30
describes the viscosity of the oil at
high and low temperatures (W
means winter dudes higher the
number the more viscous).
In general, viscosity decreases as temperature increases
and visa versa.
Solids are broadly classified into two general catagories:
crystalline and amorphous.
Crystalline
-have ordered internal structure
with repeating units called “unit
cells”.
-Several sub-classes of
crystalline solids are recognized.
-Often flat-faceted stones
(crystals)
-90% of all solids take this form.
Amorphous
-Void of long range order or
internal structure.
-Fluids that are solid
We categorize crystalline solids into a several classes.
Type
Unit of
structure
Atomic
Atoms
Covalent Molecules
Network
Ionic
Metallic
Atoms
Positive &
negative
ions
Atoms
Interparticle
Forces
Dispersion
Dispersion,
dipole-dipole,
H bonds
Covalent bond
Physical
Behavior
Soft, very low mp,
poor thermal &
electrical conductors
Group 8A(18)
[Ne-249 to Rn-71]
Fairly soft, low to
moderate mp, poor
thermal & electrical
conductors
Nonpolar - O2[-219],
C4H10[-138], C6H14
[-95] Polar CHCl3
[-64], HNO3[-42],
H2O[0.0]
Very hard, very high mp,
usually poor thermal and
electrical conductors
Hard & brittle, high mp,
Ion-ion
good thermal & electrical
attraction
conductors when molten
Metallic
bond
Examples (mp,0C)
Soft to hard, low to very
high mp, excellent
thermal and electrical
conductors, malleable
and ductile
diamond,
graphite, quartz,
silca, silicon
carbide
NaCl [801]
CaF2 [1423]
MgO [2852]
Na [97.8]
Zn [420]
Fe [1535]
Properties of Various Crystalline Solids
Type
Example
Structural Unit
Properties
Ionic
NaCl, K2SO4
CaCl2
Cations and
Anions
Metallic
Fe, Ag, Cu,
Na, K, etc.
Metal Atoms
Malleable, ductile,
conduct electricity
Molecular
any covalent
molecule
Molecules
Non-crystalline,
poor conductors
Network
Graphite, quartz,
diamond, mica
Atoms in an infinite
2D or 3D network
Non-crystalline,
poor conductors
Amorphous
Glass, wax
nylon, rubber
Covalent but no
long range regularity
Non-crystalline,
poor conductors
Hard, brittle
high melting,
low solid
conductivity
Every crystal lattice can be thought as being made of
one singular repeating 3-D shape that we call a “unit
cell”.
lattice point
unit
cell
portion of a 3-D lattice
Crystalline materials are highly-ordered atoms possessing
a repeating structure in space at the atomic level called a
lattice.
The highly-ordered nature of these crystals is reflected
in the real-world.
The structure of any crystalline solid is described by
identifying a “unit cell” which is a 3-D volume pattern
that when translated up/down/right/left builds the
remaining part of the solid in 3-dimensions.
unit
cell
lattice point
unit
cell
portion of a 3-D lattice
2-D analogy for unit
portion
of a 2-D lattice
cell
pattern.
Translating this
pattern gives the
pattern again.
In 1850, Bravais
found that all
crystalline solids are
one of 7-possible
basic unit-cells
(parallelpipeds).
Rhombohedral Hexagonal
Monoclinic
We will only cover these!
Tetragonal
Cubic
Orthorhombic
Triclinic
Because it describes many common crystalline
substances our focus will be the cubic system and its 3
possible “unit cells”.
Simple cubic
Body-centered cubic
Face-centered cubic
Cubic System and 3-Unit Cells
Simple cubic
Bodycentered
cubic
Facecentered
cubic
The simple cubic system has as its unit cell atoms
occupying each of the eight corners of a cube. The
distance from atom to atom along the lattice is the
same in every direction with angle of 90˚ throughout.
Simple Cubic
1/8 atom at
8 corners
Atoms/unit cell = 1/8 * 8 = 1
coordination number = 6
The body-centered cubic has a coordination number
of eight, 1/8 atoms at the eight corners and 1 atom at
the center of each unit cell.
Body-centered
1/8 atom at
8 corners
1 atom at
center
coordination number = 8
Atoms/unit cell = (1/8*8) + 1 = 2
The face-centered cubic has a coordination number
of twelve, 1/2 atom at the six faces, 1/8 atom at the
eight corners of a unit cell.
1/8 atom at
8 corners
1/2 atom at
6 faces
coordination number = 12
Atoms/unit cell = (1/8*8)+(1/2*6) = 4