1
Today…
• Turn in:
–Nothing
• Our Plan:
–Notes
–Start Homework
• Homework (Write in Planner):
–Work on homework
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Nine
Unit 6 Test Results (6% curve)
A
B
C
D
F
2
4
4
4
2
4
Average 75%
High Score 94%
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General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Nine
3
Chapter Nine
Chemical Bonds
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Chapter Nine
4
Bonding Review
• Lewis dot structures: the symbol represents
the nucleus and core electrons and dots
represent the valence electrons
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General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Nine
5
Example 9.1
Give Lewis symbols for magnesium,
silicon, and phosphorus.
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Chapter Nine
6
Bonding Review
• Octet Rule – Eight is great,
except for hydrogen and helium
2 will do!
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General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Nine
7
Bonding Review
• Forces called chemical bonds hold atoms together
in molecules and keep ions in place in solid ionic
compounds.
• Chemical bonds are electrostatic forces; they
reflect a balance in the forces of attraction and
repulsion between electrically charged particles.
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Chapter Nine
8
In this unit…
• We will look at intramolecular forces
and intermolecular forces:
– Intra- means “within” a molecule
• Examples are ionic, covalent, and metallic
bonding
– Inter- means “between” molecules
• Examples are hydrogen bonding, dipole
interactions, and dispersion forces
• First let’s look at intramolecular forces.
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Chapter Nine
9
Bonding Review
• Ionic bonding – a bond between a metal and
a nonmetal where one atom gives an
electron and one takes an electron.
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Chapter Nine
10
Bonding Review
• When atoms lose or gain electrons, they may
acquire a noble gas configuration, but do not
become noble gases.
• Because the two ions formed in a reaction between
a metal and a nonmetal have opposite charges,
they are strongly attracted to one another and form
an ion pair.
• The net attractive electrostatic forces that hold the
cations and anions together are ionic bonds.
• The highly ordered solid collection of ions is
called an ionic crystal.
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Chapter Nine
11
Bonding Review
Na donates an
electron to Cl …
… and
opposites
attract.
Sodium
reacts
violently in
chlorine gas.
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Chapter Nine
12
This is funny…
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General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Nine
13
This is funny too…
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General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Nine
14
I can’t get enough…
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General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Nine
15
Using Lewis Symbols
to Represent Ionic Bonding
• Lewis symbols can be used to represent ionic
bonding between nonmetals and: the s-block
metals, some p-block metals, and a few d-block
metals.
• Instead of using complete electron configurations
to represent the loss and gain of electrons, Lewis
symbols can be used.
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Chapter Nine
16
Example 9.2
Use Lewis symbols to show the formation of
ionic bonds between magnesium and nitrogen.
What are the name and formula of the
compound that results?
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Chapter Nine
17
Try it Out!
• Use Lewis symbols to show the
formation of ionic bonds between
sodium and phosphorus. What are the
name and formula of the compound that
results?
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General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Nine
18
Lattice Energy
• Energy released when positive and
negative ions form crystal lattice
due to their attraction for each
other.
• Stability of ionic compounds (high
melting point, brittle) because of
large lattice energy
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Chapter Nine
19
Coulomb’s Law
• Describes the electrostatic interaction
between charged particles.
• Says that the force of attraction or
repulsion between two point charges is
directly proportional to the product of
magnitude of each charge and
indirectly proportional to the square of
distance between them
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Chapter Nine
20
Coulomb’s Law
• Simply put:
– The higher the charges on ions, the
greater the lattice energy.
– The higher the distance between
ions, the smaller the lattice energy.
– If the charges are opposite in sign the
forces are attractive, if they are the
same sign the forces are repulsive.
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Chapter Nine
21
Examples
• Which has more lattice energy: LiBr or
CaO?
– CaO because they both have a charge of 2
while Li and Br have a charge of 1.
• Which has more lattice energy: NaCl or
CsCl?
– NaCl because all ions have the same charge,
but Cs is much larger. Therefore the distance
between atoms in CsCl is greater and the lattice
energy is smaller.
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Nine
22
Sample AP Exam Question
• The melting point of MgO is higher than that of NaF.
Explanations for this observation include which of the
following?
I. Mg+2 is more positively charged than Na+1
II. O-2 is more negatively charged than F-1
III. The O-2 is smaller than the F-1 ion.
A.
B.
C.
D.
E.
II only
I and II only
I and III only
II and III only
I, II, and III
Prentice Hall © 2005
B
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Nine
23
Bonding Review
• Covalent bond – a bond between two
nonmetals where electrons are shared.
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Chapter Nine
24
Bonding Review
• Single covalent bond – a pair of electrons is
shared between two atoms (one dash/line)
Nonbonding electrons/lone pairs
Bonding electrons
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Chapter Nine
25
Bonding Review
• Some molecules require more than single
bonds to provide each atom with the required
octet (formed primarily by C, N, and O).
• Double bond – atoms share 2 pair of electrons
• Triple bond – atoms share 3 pair of electrons
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Chapter Nine
26
The Lewis Theory of
Chemical Bonding: An Overview
• Valence electrons play a fundamental
role in chemical bonding.
• In losing, gaining, or sharing electrons
to form chemical bonds, atoms tend to
acquire the electron configurations of
noble gases.
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Chapter Nine
27
Lewis Structures of
Simple Molecules
• A Lewis structure is a combination of Lewis
symbols that represents the formation of covalent
bonds between atoms.
• In most cases, a Lewis structure shows the bonded
atoms with the electron configuration of a noble
gas; that is, the atoms obey the octet rule. (H
obeys the duet rule.)
• The shared electrons can be counted for each atom
that shares them, so each atom may have a noble
gas configuration.
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Chapter Nine
28
Some Illustrative Compounds
• Note that the two-dimensional Lewis structures do not necessarily show the correct
shapes of the three-dimensional molecules. Nor are they intended to do so.
• The Lewis structure for water may be drawn with all three atoms in a line: H–O–H.
• We will learn how to predict shapes of molecules in Chapter 10.
Prentice Hall © 2005
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Chapter Nine
29
Electronegativity
• Electronegativity (EN) is a measure of the ability of an
atom to attract its bonding electrons to itself.
• EN is related to ionization energy and electron affinity.
• The greater the EN of an atom in a molecule, the more
strongly the atom attracts the electrons in a covalent bond.
Electronegativity generally
increases from left to right
within a period, and it generally
increases from the bottom to the
top within a group.
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General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Nine
30
Pauling’s Electronegativities
Electronegativity has no unit because
the values are comparative only.
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It would be a good idea to remember
the four elements of highest
electronegativity: N, O, F, Cl.
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Nine
31
Example 9.4
Referring only to the periodic table inside the front
cover, arrange the following sets of atoms in the
expected order of increasing electronegativity.
(a) Cl, Mg, Si
(b) As, N, Sb
(c) As, Se, Sb
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Chapter Nine
32
Writing Lewis Structures:
Skeletal Structures
• The skeletal structure shows the arrangement of
atoms.
• Hydrogen atoms are terminal atoms (bonded to
only one other atom).
• The central atom of a structure usually has the
lowest electronegativity.
• In oxoacids (HClO4, HNO3, etc.) hydrogen atoms
are usually bonded to oxygen atoms.
• Molecules and polyatomic ions usually have
compact, symmetrical structures.
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Chapter Nine
33
Writing Lewis Structures: A Method
1. Determine the total number of valence electrons.
2. Write a plausible skeletal structure and connect the
atoms by single dashes (covalent bonds).
3. Place pairs of electrons as lone pairs around the
terminal atoms to give each terminal atom (except
H) an octet.
4. Assign any remaining electrons as lone pairs around
the central atom.
5. If necessary (if there are not enough electrons), move
one or more lone pairs of electrons from a terminal
atom to form a multiple bond to the central atom.
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Chapter Nine
34
Some Handy Rules to Remember
–Hydrogen and the halogens bond
once.
–The family oxygen is in can bond
twice.
–The family nitrogen is in can bond
three times. So can boron.
–The family carbon is in can bond
four times.
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Chapter Nine
35
Example
Example 9.6
Write the Lewis structure of nitrogen trifluoride,
NF3.
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Chapter Nine
36
Example
Example 9.7
Write a plausible Lewis structure for phosgene,
COCl2.
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Chapter Nine
37
Example
• Example 9.6 A – Write the Lewis structure
of hydrazine N2H4.
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Chapter Nine
38
Try It Out
• Example 9.7 A – Write a plausible Lewis
structure for carbonyl sulfide, COS.
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Chapter Nine
39
Try It Out
• Example 9.7 B – Write a plausible Lewis
structure for nitrosyl chloride, NOCl.
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General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Nine
40
What if an element has charge?
• Negative = add
electrons
• Positive = subtract
electrons
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Chapter Nine
41
Example
• Example 9.8 – Write a plausible Lewis
structure for the chlorate ion, ClO3-1.
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Chapter Nine
42
Try it Out!
• Example 9.8 B – Write a plausible Lewis
structure for the nitronium ion, NO2+1.
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Chapter Nine
43
Resonance
• Can draw more than one way because of
multiple bonds.
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Chapter Nine
44
Resonance: Delocalized Bonding
• When a molecule or ion can be represented by two or more
plausible Lewis structures that differ only in the distribution of
electrons, the true structure is a composite, or hybrid, of them.
• The different plausible structures are called resonance
structures.
• The actual molecule or ion that is a hybrid of the resonance
structures is called a resonance hybrid.
• Electrons that are part of the resonance hybrid are spread out
over several atoms and are referred to as being delocalized.
Three pairs of
electrons are
distributed among
two bonds.
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Chapter Nine
45
Example
• Example 9.10 – Write three equivalent
Lewis structures for the SO3 molecule that
conform to the octet rule.
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Chapter Nine
46
Try it Out!
• Example 9.10 A – Write three Lewis
structures for the nitrate ion, NO3-1.
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Chapter Nine
47
Molecules that Don’t Follow
the Octet Rule
• Molecules with an odd number of valence electrons have
at least one of them unpaired and are called free radicals.
• Some molecules have incomplete octets. These are usually
compounds of Be, B, or Al; they generally have some
unusual bonding characteristics, and are often quite
reactive.
• Some compounds have expanded valence shells, which
means that the central atom has more than eight electrons
around it.
• A central atom can have expanded valence if it is in the
third period or lower (i.e., S, Cl, P).
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Chapter Nine
48
Example 9.11
Write the Lewis structure for bromine pentafluoride, BrF5.
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Chapter Nine
49
Example
• Example 9.11 A – Write the Lewis structure
of phosphorus trichloride.
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Chapter Nine
50
Try It Out!
• Example 9.11 B – Write the Lewis structure
of chlorine trifluoride.
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Chapter Nine
51
Try it Out!
• Example 9.11 B – Write the Lewis structure
of sulfur tetrafluoride.
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Chapter Nine
52
Structural Formulas
• Formulas for organic
compounds that tell you how to
draw the structure.
• Example: CH3CH2OH (ethanol)
instead of C2H6O
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Chapter Nine
53
AP Exam Question
• What is the correct structural formula for 2butyne?
A. CH≡CCH2CH3
B. CH3C≡CCH3
C. CH3CH2C≡CH
D. CH3CH=CHCH3
E. CH3CH2CH2CH3
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B
Chapter Nine
54
STOP!
• Homework time….you can do all for Ch. 9
except 42, 68, and 74.
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Chapter Nine
Today…
55
• Turn in:
– Nothing
• Our Plan:
– Scavenger Hunt Review
– Notes
– Work on homework
• Homework (Write in Planner):
– Ch. 9 Homework Due Monday
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Chapter Nine
56
Remember Electronegativities?
• Arrange these elements in order
from least to most
electronegative:
–S, Ca, Cu, Cl, Al
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Chapter Nine
57
Electronegativity Difference
and Bond Type
• Identical atoms have the same electronegativity and
share a bonding electron pair equally. The bond is a
nonpolar covalent bond.
• When electronegativities differ significantly, electron
pairs are shared unequally.
• The electrons are drawn closer to the atom of higher
electronegativity; the bond is a polar covalent bond.
• With still larger differences in electronegativity,
electrons may be completely transferred from metal
to nonmetal atoms to form ionic bonds.
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Chapter Nine
58
Electronegativity Difference and
Bond Type
No sharp cutoff between
ionic and covalent bonds.
Cs—F bonds are “so polar”
that we call the bonds ____.
C—H bonds are
virtually nonpolar.
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Chapter Nine
Electronegativity Difference
•
•
•
•
59
(HE – LE/HE) X100 = % Ionic Character
< 5% = Nonpolar
5- 50% = Polar
> 50% = Ionic
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Chapter Nine
60
Depicting Polar Covalent Bonds
In nonpolar bonds,
electrons are shared
equally.
Polar bonds are also depicted
by partial positive and partial
negative symbols …
Unequal sharing
in polar covalent
bonds.
Polar bonds are often depicted
using colors to show
electrostatic potential (blue =
positive, red = negative).
Prentice Hall © 2005
… or with a cross-based
arrow pointing to the more
electronegative element.
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Nine
61
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Chapter Nine
62
Example 9.5
Use electronegativity values to arrange the following
bonds in order of increasing polarity:
Br—Cl, Cl—Cl, Cl—F, H—Cl, I—Cl
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Chapter Nine
63
Example & Try it out…
• Indicate the type of bond expected to form between
the two indicated atoms. If it is polar covalent, place a
“δ-” to indicate the negative end of any dipole formed.
a) H – H
b) H – O
c) H – N
d) H – C
e) H – F
f) Na – Cl
g) Cl - F
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Chapter Nine
64
Bond Order and Bond Length
• Bond order is the number of
shared electron pairs in a bond.
• A single bond has BO = 1, a
double bond has BO = 2, etc.
• Bond length is the distance
between the nuclei of two atoms
joined by a covalent bond.
• Bond length depends on the
particular atoms in the bond and
on the bond order.
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Chapter Nine
65
Prentice Hall © 2005
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Chapter Nine
Example 9.13
66
Estimate the length of (a) the nitrogen-tonitrogen bond in N2H4 and (b) the bond in
BrCl.
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Chapter Nine
67
Try it Out!
• Example 9.13 A – Estimate the oxygen to
fluorine bond length in OF2.
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Chapter Nine
68
Bond Energy
• Bond-dissociation energy (D) is the energy required to
break one mole of a particular type of covalent bond in a
gas-phase compound.
• Energies of some bonds can differ from compound to
compound, so we use an average bond energy.
The H—H bond
energy is precisely
known …
… while the O—H bond
energies for the two bonds
in H2O are different.
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Chapter Nine
69
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Chapter Nine
70
Trends in Bond Lengths and Energies
• The higher the order (for a particular type of
bond), the shorter and the stronger (higher
energy) the bond.
• A N=N double bond is shorter and stronger than a
N–N single bond.
• There are four electrons between the two positive
nuclei in N=N. This produces more electrostatic
attraction than the two electrons between the
nuclei in N–N.
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Chapter Nine
71
Calculations involving Bond Energy
• It REQUIRES energy to break a bond.
– This means that breaking a bond is an
ENDOTHERMIC process (+∆H).
• Energy is RELEASED when a bond is
formed.
– This means that forming a bond is an
EXOTHERMIC process (-∆H).
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Chapter Nine
72
Calculations Involving Bond Energy
• A reaction is endothermic if it requires
more energy to break bonds than is
given off when they are formed.
• A reaction is exothermic if it requires
less energy to break bonds than is
given off when they are formed.
– Remember the calorimetry lab?
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Nine
73
Calculations Involving Bond Energies
For the reaction N2(g) + 2 H2(g) -->N2H4(g)
When the bonds of the
product form, 163 kJ plus
4(389 kJ) of energy is
liberated.
… we must supply
946 kJ …
… plus 2(436 kJ),
to break bonds of
the reactants.
Prentice Hall © 2005
to occur …
ΔH = (+946 kJ) + 2(+436 kJ) + (–163 kJ) + 4(–389 kJ)
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Nine
74
Example 9.14
Use bond energies from Table 9.1 to estimate
the enthalpy of formation of gaseous hydrazine.
Compare the result with the value of ΔH°f
[N2H4(g)] from Appendix C.
The reaction is: N2 + H2 → N2H4
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Chapter Nine
75
Try It Out!
• Example 9.14 A – Estimate ∆H for the
reaction:
C2H6 + Cl2 → C2H5Cl + HCl
Prentice Hall © 2005
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Chapter Nine
76
Let’s Do An Activity..
• Practice what you learned today by
completing the Quick Check
handout.
• When you finish, work on your
Unit 7 Homework. It is due next
class.
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Chapter Nine
77
STOP…
Work on your Ch. 9 Homework! It is due on
Monday!
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General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Nine
78
Today…
• Turn in:
– Mark Ch. 9 HW Questions on the board
• Our Plan:
– Ch. 9 Homework Questions?
– Notes – VSEPR & Bonding Theory
– Begin Ch. 10 Homework
• Homework (Write in Planner):
– Ch. 10 HW is due next Monday!
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General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Nine
Chapter Ten
79
Bonding Theory
and Molecular Structure
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Chapter Nine
80
Molecular Geometry
• Molecular geometry is simply
A carbon
the shape of a molecule.
dioxide
• Molecular geometry is
molecule is
linear.
described by the geometric
figure formed when the atomic
nuclei are joined by (imaginary)
straight lines.
• Molecular geometry is found
using the Lewis structure, but
the Lewis structure itself does
A water
NOT necessarily represent the molecule is
angular or
molecule’s shape.
bent.
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Chapter Nine
81
VSEPR
• Valence-Shell Electron-Pair Repulsion (VSEPR) is a
simple method for determining geometry.
• Basis: pairs of valence electrons in bonded atoms repel one
another.
• These mutual repulsions push electron pairs as far from
one another as possible.
When the electron
pairs (bonds) are as
B
B
B
A
B
Prentice Hall © 2005
B
A
far apart as they can
get, what will be the
B-A-B angle?
B
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Nine
82
Electron-Group Geometries
• An electron group is a collection of
valence electrons, localized in a
region around a central atom.
• One electron group:
– an unshared pair of valence electrons or
– a bond (single, double, or triple)
• The repulsions among electron
groups lead to an orientation of the
groups that is called the electrongroup geometry.
• These geometries are based on the
number of electron groups:
Prentice Hall © 2005
Electron Electron-group
groups
geometry
2
Linear
3
Trigonal planar
4
Tetrahedral
5
Trigonal
bipyramidal
6
Octahedral
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Nine
83
A Balloon Analogy
• Electron groups repel
one another in the
same way that
balloons push one
another apart.
• When four balloons,
tied at the middle,
push themselves
apart as much as
possible, they make a
tetrahedral shape.
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Chapter Nine
84
VSEPR Notation
• In the VSEPR notation used
to describe molecular
geometries, the central atom
in a structure is denoted as A,
terminal atoms as X, and the
lone pairs of electrons as E.
• The H2O molecule would
therefore carry the designation
AX2E2.
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Chapter Nine
85
VSEPR Notation
• For structures with no lone pairs on the
central atom (AXn), the molecular
geometry is the same as the electron-group
geometry.
• When there are lone pairs, the molecular
geometry is derived from the electrongroup geometry.
• In either case, the electron-group geometry
is the tool we use to obtain the molecular
geometry.
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Chapter Nine
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Chapter Nine
87
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Nine
88
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Chapter Nine
89
Prentice Hall © 2005
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Chapter Nine
90
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Chapter Nine
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Some Examples:
• Give the electron-group geometry and VSEPR
(molecular) geometry for each structure below:
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Chapter Nine
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Example 10.1
Use the VSEPR method to predict the shape of the
nitrate ion.
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Chapter Nine
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Example 10.2
Use the VSEPR method to predict the molecular
geometry of XeF2.
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Chapter Nine
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Try it out…
Example 10.3
Use the VSEPR method to describe, as best
you can, the molecular geometry of the
nitric acid molecule, HNO3.
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Chapter Nine
96
Polar Molecules and Dipole Moments
• A polar bond (Chapter 9) has separate centers of positive
and negative charge.
• A molecule with separate centers of positive and negative
charge is a polar molecule.
• The dipole moment (m) of a molecule is the product of the
magnitude of the charge (d) and the distance (d) that
separates the centers of positive and negative charge.
m = dd
• A unit of dipole moment is the debye (D).
• One debye (D) is equal to 3.34 x 10–30 C m.
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Chapter Nine
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Polar Molecules in an Electric Field
An electric field causes
polar molecules to align
with the field.
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Chapter Nine
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Bond Dipoles and Molecular Dipoles
• A polar covalent bond has a bond dipole; a
separation of positive and negative charge centers
in an individual bond.
• Bond dipoles have both a magnitude and a
direction (they are vector quantities).
• Ordinarily, a polar molecule must have polar
bonds, BUT … polar bonds are not sufficient.
• A molecule may have polar bonds and be a
nonpolar molecule – IF the bond dipoles cancel.
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Chapter Nine
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Bond Dipoles and Molecular Dipoles
• CO2 has polar bonds, but is a
linear molecule; the bond
dipoles cancel and it has no
net dipole moment (m = 0 D).
• The water molecule has
polar bonds also, but is an
angular molecule.
• The bond dipoles do not
cancel (m = 1.84 D), so
water is a polar molecule.
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No net
dipole
Net dipole
Chapter Nine
100
Molecular Shapes and Dipole Moments
To predict molecular polarity:
1. Use electronegativity values to predict bond dipoles.
2. Use the VSEPR method to predict the molecular shape.
3. From the molecular shape, determine whether bond
dipoles cancel to give a nonpolar molecule, or combine to
produce a resultant dipole moment for the molecule.
Note: Lone-pair electrons can also make a
contribution to dipole moments.
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Chapter Nine
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Examples
• Would the following molecules be polar or
nonpolar?
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Chapter Nine
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Example 10.4
Explain whether you expect the following molecules
to be polar or nonpolar.
(a) CHCl3
(b) CCl4
Example 10.5
A Conceptual Example
Of the two compounds NOF and NO2F, one has m =
1.81 D and the other has m = 0.47 D. Which dipole
moment do you predict for each compound? Explain.
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Chapter Nine
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Atomic Orbital Overlap
• Valence Bond (VB) theory
states that a covalent bond is
formed when atomic orbitals
(AOs) overlap.
• In the overlap region,
electrons with opposing spins
produce a high electron
charge density.
• In general, the more extensive
the overlap between two
orbitals, the stronger is the
bond between two atoms.
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Overlap region
between nuclei has
high electron density
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Bonding in H2S
The measured bond
angle in H2S is 92°;
good agreement.
The hydrogen atoms’ s orbitals
can overlap with the two halffilled p orbitals on sulfur.
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Chapter Nine
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Important Points of VB Theory
• Most of the electrons in a molecule remain in the
same orbital locations that they occupied in the
separated atoms.
• Bonding electrons are localized in the region of AO
overlap.
• For AOs with directional lobes (such as p orbitals),
maximum overlap occurs when the AOs overlap
end to end.
• VB theory is not without its problems …
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Chapter Nine
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Hybridization
• http://www.mhhe.com/physsci/chemistry/es
sentialchemistry/flash/hybrv18.swf
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Hybridization of Atomic Orbitals
VB theory: carbon should have just two
bonds, and they should be about 90° apart.
But CH4 has four C—H bonds, 109° apart.
• We can hybridize the four orbitals holding valence
electrons; mathematically combine the wave functions
for the 2s orbital and the three 2p orbitals on carbon.
• The four AOs combine to form four new hybrid AOs.
• The four hybrid AOs are degenerate (same energy)
and each has a single electron (Hund’s rule).
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Chapter Nine
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sp3 Hybridization
• Hybridizing one s orbital with three p orbitals
gives rise to four hybrid orbitals called sp3
orbitals.
• The number of hybrid orbitals is equal to the
number of atomic orbitals combined.
• The four hybrid orbitals, being equivalent, are
about 109° apart.
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The sp3 Hybridization Scheme
Four AOs …
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… form four new hybrid AOs.
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Chapter Nine
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Methane and Ammonia
In methane,
each hybrid
orbital is a
bonding orbital
In ammonia, one of the
hybrid orbitals contains
the lone pair that is on the
nitrogen atom
Four sp3 hybrid orbitals: tetrahedral
Four electron groups: tetrahedral
Coincidence? Hardly.
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sp2 Hybridization
• Three sp2 hybrid orbitals are formed from an s orbital and
two p orbitals.
• The empty p orbital remains unhybridized. It may be used
in a multiple bond.
• The sp2 hybrid orbitals are in a plane, 120o apart.
• This distribution gives a trigonal planar molecular
geometry, as predicted by VSEPR.
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The sp2 Hybridization Scheme in Boron
A 2p orbital remains
unhybridized.
Three AOs combine to form …
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… three hybrid AOs.
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sp Hybridization
• Two sp hybrid orbitals are formed from an s orbital and a p
orbital.
• Two empty p orbitals remains unhybridized; the p orbitals
may be used in a multiple bond.
• The sp hybrid orbitals are 180o apart.
• The geometry around the hybridized atom is linear, as
predicted by VSEPR.
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sp Hybridization in Be
… with two
unused p orbitals.
Two AOs combine to form …
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… two hybrid AOs …
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Chapter Nine
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Hybrid Orbitals Involving
d Subshells
• The hybridization involving
d-orbitals is a controversial
area of chemistry. It will
not be covered in this
course. If a structure is
beyond sp3 hybridization,
we will record it’s
hybridization as N/A.
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Predicting Hybridization Schemes
In the absence of experimental evidence, probable
hybridization schemes can be predicted:
1. Write a plausible Lewis structure for the
molecule or ion.
2. Use the VSEPR method to predict the electrongroup geometry of the central atom.
3. Select the hybridization scheme that corresponds
to the VSEPR prediction.
4. Describe the orbital overlap and molecular
geometry.
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Chapter Nine
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Chapter Nine
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Hybridization is Confusing…
• Try Mrs. C’s “Easy Rule”.
– Count the things stuck to the central atom.
Lone pairs count as things stuck.
•
•
•
•
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2 things = sp
3 things = sp2
4 things = sp3
5+ things stuck = N/A
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Chapter Nine
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Examples
• What is the central atom hybridization for
each of the structures below:
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Chapter Nine
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Hybrid Orbitals and
Multiple Covalent Bonds
• Covalent bonds formed by the end-to-end overlap
of orbitals are called sigma (s) bonds.
• All single bonds are sigma bonds.
• A bond formed by parallel, or side-by-side, orbital
overlap is called a pi (p) bond.
• A double bond is made up of one sigma bond and
one pi bond.
• A triple bond is made up of one sigma bond and
two pi bonds.
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Sigma and pi bonds
• Sigma bonds are stronger than pi bonds.
• When bonds are broken, pi bonds are
always broken first.
• A triple bond has 2 pi (π) bonds and 1
sigma (σ) bond. It is stronger than a
single bond because each pi bond has to
be broken and then a sigma bond has to
be broken.
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VB Theory for
Ethylene, C2H4
π-bond has two lobes
(above and below
plane), but is one bond.
Side overlap of 2p–2p.
The hybridization and
bonding scheme is
described by listing each
bond and its overlap.
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Chapter Nine
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VB Theory: Acetylene
Two π-bonds (above and
below, and front and
back) from 2p–2p
overlap …
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… form a cylinder
of π-electron
density around the
two carbon atoms.
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Chapter Nine
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Example 10.7
Formic acid, HCOOH, is the simplest carboxylic acid.
(a) Predict a plausible molecular geometry for this
molecule.
(b) Propose a hybridization scheme for the central
atoms that is consistent with that geometry.
(c) How many sigma and pi bonds are there?
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Chapter Nine
125
Stop!
• You can now complete the Ch.
10 Homework. It is due on
Monday!
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Chapter Nine
Today…
126
• Turn in:
– Nothing
• Our Plan:
– Review Activity
– Pre-Lab
– VSEPR Lab
• Homework (Write in Planner):
– Nothing
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Chapter Nine
127
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Nine
128
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Chapter Nine
129
Lab Color Key
•
•
•
•
•
•
•
•
•
H – white
O – red
P/Xe (if trigonal bipyramidal electron geometry) – gray
F – green or orange
Cl – purple
N – blue
S/Xe (if octahedral electron geometry) – yellow
C – black
P (if tetrahedral) – black
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Today…
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• Turn in:
– Nothing
• Our Plan:
– VSEPR Lab
– Work on Homework
• Homework (Write in Planner):
– VSEPR Lab due next class
– Homework due next class
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Chapter Nine
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Lab Color Key
•
•
•
•
•
•
•
•
•
H – white
O – red
P/Xe (if trigonal bipyramidal electron geometry) – gray
F – green or orange
Cl – purple
N – blue
S/Xe (if octahedral electron geometry) – yellow
C – black
P (if tetrahedral) – black
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Chapter Nine
Today…
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• Turn in:
– Lab Report in basket (rubric on top)
– Mark Homework Questions
• Our Plan:
– Go over Ch. 10 HW Questions
– Notes – Intermolecular Attraction
– Begin Homework (if time)
• Homework (Write in Planner):
– Ch. 11 Homework Due Friday!
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Chapter Nine
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So far this unit…
• We have covered INTRAMOLECULAR forces –
the forces WITHIN molecules (ionic and covalent
bonding). For example, we studied how the
oxygen and hydrogen atoms interact to form a
water molecule.
• Today we are going to cover another important
concept, but much more difficult to visualize –
INTERMOLECULAR forces.
• Intermolecular forces are forces BETWEEN
molecules. For example, we will study how two
water molecules interact with one another.
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Intermolecular Attractions
• The attractions between
molecules are much weaker
than the covalent bonds within
the molecules.
• These attractions vary
depending on the nature of the
molecules.
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Dipole-Dipole
• Many molecules have permanent dipoles.
• These polar covalent molecules form
attractions between the positive end of one
molecule and the negative end of the
adjacent molecule.
• Dipole-dipole attractions are effective over
short distances between molecules and
decrease as the distance between molecules
increases.
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Chapter Nine
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Dipole-Dipole
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Dipole-Dipole Simulation
• http://intro.chem.okstate.edu/ap/HCldipole.
html
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Hydrogen Bonding
• Strong dipole-dipole attractions occur in
molecules containing hydrogen covalently
bonded to a small very electronegative
atom (nitrogen, oxygen, or fluorine).
• The hydrogen in these bonds has a partial
positive charge, since there are no inner
core electrons and the shared electrons are
strongly attracted to the small, very
electronegative atoms.
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Hydrogen Bonding
• The attractions in hydrogen bonds are
four to five times stronger than other
dipole-dipole attractions.
• Hydrogen bonding is responsible for
the unusual properties of water and is
very important in biological systems
such as proteins and DNA/RNA.
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Chapter Nine
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Hydrogen Bonding
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Chapter Nine
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Hydrogen Bonding
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Hydrogen Bonding Simulation
• http://intro.chem.okstate.edu/ap/HHbond.ht
ml
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Chapter Nine
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London Dispersion Forces
• Nonpolar molecules do not have permanent
dipoles.
• At any given instant, the electrons may be
unevenly distributed within an atom or molecule
giving it a partial negative end.
• This results in atoms being attracted to one
another for an instant (can even happen in noble
gases).
• The greater the molar mass (more electrons) the
stronger the dispersion forces.
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Chapter Nine
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London Dispersion Forces
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Chapter Nine
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Dispersion Forces Simulation
• http://intro.chem.okstate.edu/ap/LondonDis
p.html
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Chapter Nine
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Intermolecular Attraction
• Melting point and boiling point of
covalent molecules increase with the
strength of the forces holding them
together.
• Intermolecular forces depend on molar
mass, molecular shape, and other
factors.
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General Order of Strength
1.Ionic
2.Hydrogen Bonds
3.Dipole-dipole (Polar
molecules)
4.Dispersion Forces
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Intermolecular Attraction
• Specifically, the stronger the force,
the higher the melting and boiling
point.
• If the forces are the same, larger
molecules with less compact
shapes have higher melting and
boiling points.
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Molar Mass & Boiling Point
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Example
• Draw the Lewis structures for each of these
compounds and indicate which
intermolecular forces (if any) are involved.
a) NI3
b) PI3
c) BF3
d) CH3COOH
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Example
• Based on the structures and intermolecular
forces you just indicated, which of those
compounds will have the HIGHEST
melting point? Consider mass as well.
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Sample AP Exam Question
• Use principles of atomic structure, bonding and/or
intermolecular forces to respond to each of the
following. Your response must include specific
information about all substances referred to in
each question.
– At a pressure of 1 atm, the boiling point of NH3 (l)
is 240 K, whereas the boiling point of NF3 (l) is
144 K
• Identify the intermolecular force(s) in each
substance.
• Account for the differences in the boiling points of
the substances.
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Sample AP Exam Question
• Which of the following describes the changes in
forces of attraction that occur as H2O changes phase
from a liquid to a vapor?
a) H – O bonds break as H – H and O – O bonds form.
b) Hydrogen bonds between H2O molecules are broken
c) Covalent bonds between H2O molecules are broken.
d) Ionic bonds between H+ ions and OH- ions are
broken.
e) Covalent bonds between H+ ions and H2O molecules
become more effective.
B
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Today…
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• Turn in:
– Nothing
• Our Plan:
– Intermolecular Forces Review + Lab Activity
– Work on Ch. 11 Homework
• Due Next Class
– Investigation 5 Pre-Lab
• Homework (Write in Planner):
– Have Lab Report ready to go
– Ch. 11 HW Due
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Chapter Nine
Today…
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• Turn in:
– Mark HW Questions on the board
• Our Plan:
– Homework Questions
– Investigation 5
• Homework (Write in Planner):
– Lab Report due Monday
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Today…
• Turn in:
– Mark HW Questions on the board
• Our Plan:
– Unit Review – Scavenger Hunt
– Work on Final Study Guide
• Homework (Write in Planner):
– Final next class
– Breakfast Club Wednesday at 6 am
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Today…
• Turn in:
– Nothing
• Our Plan:
– Last minute questions?
– Final Exam
• Homework (Write in Planner):
– AP Exam Review Packet due the
Monday that you return from break!
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Chapter Nine