Drawing Lewis Structures
 Find
total # of valence e-.
 Arrange
atoms - singular atom is usually
in the middle.
 Form
bonds between atoms (2 e-).
remaining e- to give each atom
an octet (recall exceptions).
 Distribute
 If
there aren’t enough e- to go around,
form double or triple bonds.
Polyatomic Ions
 To
find total # of valence e-:
• Add 1e- for each negative charge
• Subtract 1e- for each positive
charge
 Place
brackets around the ion and
label the charge
VSPER Model

1. Draw the Lewis structure
 2. Count total number of electron pairs around
the CENTRAL ATOM. Arrange them to minimize
the electron shell repulsion
 Note: a double or triple bond is counted as
one bonding pair when predicting geometry
 3. Describe the molecular geometry in terms of
the angular arrangement of the bonding pairs
Polarity of Molecules
The "charge distribution" of a molecule is chosen by:
• The shape of the molecule
• The polarity of its bonds


A Polar Molecule:
The center of the overall negative charge on the
molecule does not coincide with the center of overall
positive charge on the molecule
 The molecule can be oriented such that one end has a
net negative charge and the other a net positive charge,
i.e. the molecule is a dipole (di-pole means two poles
i.e. positive and negative)
Polarity of Molecules
A
Nonpolar molecule
 Has no charges on the opposite ends of the
molecule
 Or, has charges of the same sign on the
opposite ends of the molecule
Determining Polarity

Each polar bond in a polyatomic molecule will have a dipole

The overall dipole of the molecule will be the sum of the
individual dipoles Vectors
In carbon dioxide the oxygens have a partial negative
charge and the carbon a partial positive charge, the
molecule has no dipole, due to the center of positive and
negative charges being the same, canceling out
Water has a dipole and will orient in an electrical field


Ch. 12 - Liquids & Solids
I. Intermolecular Forces
(pages 411- 414)
Intramolecular and
Intermolecular Forces
 Covalent
& Ionic bonds - the force that
holds atoms together making molecules.
• These are intramolecular forces.
 There are also forces that cause
molecules to attract each other. These
are called intermolecular forces.
Definition of IMF
 Attractive
forces between molecules.
 Much
weaker than
chemical bonds
within molecules.
 a.k.a.
van der Waals forces
Types of IMF
London Dispersion Forces
The distribution of electrons around an atom, at a given
instant in time, may not be symmetrical
• Both electrons may be on 1 side of the nucleus
• Atom would have an apparent dipole moment at that
instant in time (i.e. an instantaneous dipole moment)
View animation online.




London Dispersion Forces
Due to electron repulsion, a temporary dipole on
one atom can induce a similar dipole on a
neighboring atom
This will cause the neighboring atoms to be
attracted to one another
This is called the London dispersion force (or just
dispersion force)
It is significant only when the atoms are close
together
Dipole-Dipole Forces
+
molecules
attract one another
when the partial
positive charge on
one molecule is near
the partial negative
charge on the other
molecule
View animation online.
 Polar

Hydrogen Bonding
A hydrogen atom in a polar bond (e.g. HN, H-O or H-F) can experience an
attractive force with a neighboring highly
electronegative molecule or ion which has
an unshared pair of electrons
 Hydrogen bonds are considered to be
dipole-dipole type interactions
 A bond between hydrogen and an
electronegative atom such as N, O or F is
quite polar:
Determining IMF
 NCl3
• polar = london dispersion, dipole-dipole
 CH4
• nonpolar = london dispersion
 HF
• H-F bond = london dispersion, dipoledipole, hydrogen bonding
Ionic Bonds: Ion-Dipole

Involves an interaction between a charged ion and
a polar molecule (i.e. a molecule with a dipole)
 Cations are attracted to the negative end of a
dipole
 Anions are attracted to the positive end of a dipole
 Ion-dipole forces are important in solutions of ionic
substances in polar solvents (e.g. a salt in aqueous
solvent)
Ch. 12 - Liquids & Solids
II. Physical Properties
(p. 415 - 424)
Physical Properites

The physical properties of a substance depends
upon its physical state.
 Water vapor, water and ice have the same chemical
properties, but the physical properties are different.
 Covalent bonds determine:
• 1. molecular shape
• 2. bond energies
• 3. chemical properties
 Intermolecular forces (IM) (non-covalent bonds)
influence physical properties of liquids and solids
Liquids vs. Solids
IMF Strength
Fluid
Density
Compressible
Diffusion
LIQUIDS
SOLIDS
Stronger than
in gases
Very strong
Y
N
high
high
N
N
slower than in
gases
extremely slow
Kinetic Molecular Theory

The state of a substance depends on the balance
between the kinetic energy of the individual
particles (molecules or atoms) and the
intermolecular forces

Kinetic energy keeps the molecules apart and
moving around, and is a function of the
temperature of the substance

Intermolecular forces try to draw the particles
together
Kinetic Molecular Theory
Gases have weaker IM forces than liquids
& solids due to the kinetic energy of the
molecules being greater than any attractive
forces between the molecules
 Solids and liquids have particles that are
fairly close to one another, and are thus
called "condensed phases" to distinguish
them from gases

Changing State
Temperature
 Heating and cooling can change the kinetic
energy of the particles in a substance, so we
can change the physical state of a substance
by heating or cooling it.
• Cooling a gas may change the state to a
liquid
• Cooling a liquid may change the state to a
solid
Changing State
Pressure

Increasing the pressure on a substance forces
the molecules closer together, which
increases the strength of intermolecular
forces
• Increasing the pressure on a gas may
change the state to a liquid
• Increasing the pressure on a liquid may
change the state to a solid
Liquid Properties
 Surface
Tension
• attractive force between particles in a
liquid that minimizes surface area
• energy required to increase the surface
area of a liquid by a unit amount
Liquid Properties
 Capillary
Action
• attractive force between
the surface of a liquid and
the surface of a solid due
to
• Adhesive forces bind a
substance to a surface, i.e.
IM force binding different
substances together
water
mercury
Liquid Properties
 Viscosity:The resistance of a liquid to flow is
called its viscosity
 A measure of the ease with which molecules move
past one another
 It
depends on the attractive force between the
molecules

Decreases with increasing temperature - the
increasing kinetic energy overcomes the attractive
forces and molecules can more easily move past
each other
Types of Solids
 Crystalline
- repeating geometric
pattern
• covalent network
• metallic
• ionic
• covalent molecular
 Amorphous
decreasing
m.p.
- no geometric pattern
Ch. 12 - Liquids & Solids
III. Changes of State
(p. 425 – 430)
A. Phase Changes
Phase Changes
 Evaporation
• molecules at the surface gain enough
energy to overcome IMF
 Volatility
• measure of evaporation rate
• depends on temp & IMF
A. Phase Changes
 Equilibrium
• trapped molecules reach a balance
between evaporation & condensation
A. Phase Changes
p.478
 Vapor
v.p.
Pressure
• pressure of vapor above
a liquid at equilibrium
• depends on temp & IMF
• directly related to volatility
temp
v.p.
IMF
temp
v.p.
A. Phase Changes
 Boiling
Point
• temp at which v.p. of liquid
equals external pressure
• depends on Patm & IMF
• Normal B.P. - b.p. at 1 atm
Patm
b.p.
IMF
b.p.
A. Phase Changes
 Melting
Point
• equal to freezing point
IMF
 Which
m.p.
has a higher m.p.?
polar
• polar or nonpolar?
• covalent or ionic?
ionic
A. Phase Changes
 Sublimation
• solid  gas
• v.p. of solid equals
external pressure
 EX:
dry ice, mothballs,
solid air fresheners
Phase Diagrams
Phase Diagrams
•Equilibrium can exist not only between the liquid and vapor
phase of a substance but also between the solid and liquid
phases, and the solid and gas phases of a substance.
•The "triple point" is the particular condition of temperature and
pressure where all three physical states are in equilibrium
•The Critical Point: The temperature above which the gas
cannot be liquefied no matter how much pressure is applied
(the kinetic energy simply is too great for attractive forces to
overcome, regardless of the applied pressure)
B. Heating Curves
Gas - KE 
Boiling - PE 
Liquid - KE 
Melting - PE 
Solid - KE 
B. Heating Curves
 Temperature
Change
• change in KE (molecular motion)
• depends on heat capacity
 Heat
Capacity
• energy required to raise the temp of 1
gram of a substance by 1°C
• Specific heat - different values for each
substance
B. Heating Curves
 Phase
Change
• change in PE (molecular arrangement)
• temp remains constant
 Heat
of Fusion (Hfus)
• energy required to melt 1 gram of a
substance at its m.p.
• Hfus of ice = 6.009 kJ/mol
B. Heating Curves
 Heat
of Vaporization (Hvap)
• energy required to boil 1 gram of a
substance at its b.p.
• Hvap for water = 40.79 kJ/mol
• usually larger than Hfus…why?
 EX:
sweating,
steam burns
C. Properties of Water
• Hfus of ice = 6.009 kJ/mol
• Hvap for water = 40.79 kJ/mol
• To calculate the amount of heat
energy required to melt or freeze a
substance, convert grams to moles
and multiply by molar heat of fusion
or vaporization
C. Properties of Water
• How much heat energy is required
to melt 25 grams of ice
ice at 0oC to
liquid water at a temperature of
0oC?
25 g H2O 1 mol H2O
6.009 kJ
18.02 g H2O 1 mol H2O
= 8.3 kJ
C. Properties of Water
• How much heat energy is required
to change 500.0 grams of liquid
steam at 100oC?
water at 100oC to steam
500.0 g H2O 1 mol H2O
40.79 kJ
18.02 g H2O 1 mol H2O
= 1132 kJ
How to Calculate Heat
in Changes of State
ΔHvap =
40.79 kJ/mol
D
C
ΔHfus = 6.009
kJ/mol
B
q = mCΔT
CH2O(l) = 4.184 J/goC
A
q = mCΔT
CH2O(s) = 2.108 J/goC
E
q = mCΔT
CH2O(g) =
1.87 J/goC
Calculating Heat Changes
 Example
problem:
 How many joules are required to heat
10.0 grams of ice from -20.0oC to steam
at 110.oC?
A
+ B + C + D + E (make sure they
are all same units!)
A: Energy Change as ice is heated
GIVEN:
WORK:
M = 10.0 g
q = mCΔT
CH2O(s) =
2.108 J/goC
TF = 0°C
Ti = -20 °C
Q=?
(10.0 g)(2.108 J/goC)(0-(-20 °C))
q=(10.0 g)(2.108 J/goC)(20 °C)
q = 421.6 Joules
B: Phase Change
ΔHfusion : amount of heat energy
required to melt a substance
10 g H2O 1 mol H2O
6.009 kJ
1000 J
18.02 g H2O 1 mol H2O 1 kJ
ΔHfusion = 3334.5 Joules
C: Energy Change as liquid is heated
GIVEN:
WORK:
M = 10.0 g
q = mCΔT
CH2O(l) =
4.184 J/goC
TF = 100°C
Ti = 0 °C
Q=?
(10.0 g)(4.184 J/goC)(100 – 0 °C)
q=(10.0 g)(4.184 J/goC)(100 °C)
q = 4184 Joules
D: Phase Change
ΔHvaporization : energy required to boil 1
gram of a substance at its b.p.
10 g H2O 1 mol H2O
40.79 kJ
1000 J
18.02 g H2O 1 mol H2O 1 kJ
ΔHvapor = 22636 Joules
E: Energy Change as a gas is heated
GIVEN:
WORK:
M = 10.0 g
q = mCΔT
CH2O(g) =
1.87 J/goC
TF = 110°C
Ti = 100 °C
Q=?
(10.0 g)(1.87 J/goC)(110-100 °C)
q=(10.0 g)(1.87 J/goC)(10 °C)
q = 187 Joules
How to Calculate Heat in
Changes of State
A
+B+C+D+E=
 421.6+3334.6+4184+22636+187
 30763.2
J = 30800 J