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Contents
10-1
10-2
10-3
10-4
10-5
10-6
10-7
10-8
10-9
Lewis Theory: An Overview
Covalent Bonding: An Introduction
Polar Covalent Bonds
Writing Lewis Structures
Resonance
Exceptions to the Octet Rule
The Shapes of Molecules
Bond Order and Bond Lengths
Bond Energies
 Focus On Molecules in Space: Measuring Bond Lengths
Slide 1 of 50
General Chemistry: Chapter 10
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10-1 Lewis Theory: An Overview
 Valence e- play a
fundamental role in
chemical bonding.
 e- transfer leads to
ionic bonds.
 Sharing of e- leads to
covalent bonds.
 e- are transferred or
shared to give each atom
a noble gas configuration
 the octet.
Slide 2 of 50
General Chemistry: Chapter 10
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Lewis Symbols
A chemical symbol represents the nucleus
and the core e-.
Dots around the symbol represent valence e-.
•
• Si •
•
• Al •
•
Slide 3 of 50
• As •
•
P•
•
••
• Se
•
•
••
••
• Bi •
•
• Sb •
•
••
I
••
•
General Chemistry: Chapter 10
••
Ar
••
•
••
••
•N•
•
••
••
••
••
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EXAMPLE 10-2
Writing Lewis Structures of Ionic Compounds. Write Lewis
structures for the following compounds: (a) BaO; (b) MgCl2 ;
(c) aluminum oxide.
• O•
••
2+ ••
Ba
O
••
2-
••
Ba•
••
••
BaO
•
Note the use of the “fishhook” arrow to denote a
single electron movement. A “double headed”
arrow means that two electrons move.
Slide 4 of 50
General Chemistry: Chapter 10
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EXAMPLE 10-2
••
••
2+
Mg
••
2 Cl
••
-
••
• Cl
••
••
••
Mg •
••
MgCl2
•
• Cl
••
Slide 5 of 50
General Chemistry: Chapter 10
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10-2 Covalent Bonding: An Introduction
Slide 6 of 50
General Chemistry: Chapter 10
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Coordinate Covalent Bonds
+
H
Cl
H N H
H
••
Cl
••
-
••
••
H N
H
••
H
H
Note the “double headed” arrow
showing that two electrons move.
Slide 7 of 50
General Chemistry: Chapter 10
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Multiple Covalent Bonds
Often, more than one pair of electrons must be shared
for an atom to attain an octet.
•O
••
• C•
•
••
•
•
•
••
•
••
Slide 8 of 50
••
••
O C O
•
•
•
••
•
••
O C O
••
••
O•
•
••
••
••
••
••
•
••
•
O C O
General Chemistry: Chapter 10
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Multiple Covalent Bonds
•
•
Slide 9 of 50
••
••
N N
•
•
•
•
••
N N
N N
General Chemistry: Chapter 10
••
••
•
••
•N
•
N•
••
•
••
•
Prentice-Hall © 2007
Polar Covalent Bonds
Most chemical bonds fall between 100%
ionic and 100% covalent.
In a polar covalent bond, electrons are not
shared equally between two atom.
In such a bond, electrons are displaced
toward the more nonmetallic element.
Slide 10 of 50
General Chemistry: Chapter 10
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Electronegativity: an atom’s ability to
compete for electrons with other atoms to which it is
bonded.
Slide 11 of 50
General Chemistry: Chapter 10
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Percent Ionic Character of a bond based on ∆EN
Slide 12 of 50
General Chemistry: Chapter 10
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Electrostatic potential map
There are methods that allow us to
display the electron density distribution
within molecules.
The Electrostatic Potential map is
obtained by hypothetically probing an
electron density surface with a positive
point charge.
Slide 13 of 50
General Chemistry: Chapter 10
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10-3 Polar Covalent Bonds and
Electrostatic Potential Maps
We can map the electron density
throughout a molecule.
The electrostatic potential
is the work
done in moving the
unit of +ve charge
from one region of a
molecule to another.
Slide 14 of 50
General Chemistry: Chapter 10
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Polar Molecules
Slide 15 of 50
Learn how to apply
Valence Shell Electron Pair Repulsion,
VSEPR, Theory to predict the shapes of
molecules.
The three-dimensional shape of a
molecule is important in determining its
chemical behavior, particularly for
biologically important molecules.
Applying VSEPR Theory
 Draw a plausible Lewis structure.
 Determine the number of electron groups around
the central atom and identify them as bond or lone
pairs.
 Establish the electron group geometry.
 Determine the molecular geometry.
 Multiple bonds count as one group of electrons.
17
Linear
Trigonal
Planar
Tetrahedral
Trigonal
Bipyramidal
18
Octahedral
Dipole Moments and Polarity:
In molecules containing atoms with
different electronegativities, there is an
unequal sharing of electron density.
In other words, a polar covalent bond is
formed.
Dipole Moments and Polarity:
Dipoles are vector quantities that are
additive to produce an overall molecular
dipole moment.
 For Diatomic molecules from atoms of
different electronegativity there is always a
dipole moment.
For polyatomic molecules the resultant dipole
moment is a vector sum of the individual bond
dipoles. As a result not all molecules with
polar covalent bonds have a net molecular
dipole.
Slide 22 of 50
General Chemistry: Chapter 10
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Slide 23 of 50
General Chemistry: Chapter 10
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Dipole Moments
Slide 25 of 50
General Chemistry: Chapter 10
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Dipole Moments
Slide 26 of 50
General Chemistry: Chapter 10
Prentice-Hall © 2007
Electrostatic potential map
There are methods that allow us to
display the electron density distribution
within molecules.
The Electrostatic Potential map is
obtained by hypothetically probing an
electron density surface with a positive
point charge.
Slide 27 of 50
General Chemistry: Chapter 10
Prentice-Hall © 2007
10-3 Polar Covalent Bonds and
Electrostatic Potential Maps
We can map the electron density
throughout a molecule.
The electrostatic potential
is the work
done in moving the
unit of +ve charge
from one region of a
molecule to another.
Slide 28 of 50
General Chemistry: Chapter 10
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Polar Molecules
Slide 29 of 50
Bond Order and Bond Length
Bond Order
 Single bond, order = 1
 Double bond, order = 2
Bond Length
 Distance between two nuclei
Higher bond order
 Shorter bond
 Stronger bond
Slide 30 of 50
General Chemistry: Chapter 10
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Slide 31 of 50
General Chemistry: Chapter 10
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Bond Energies
Slide 32 of 50
General Chemistry: Chapter 10
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Slide 33 of 50
General Chemistry: Chapter 10
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Slide 34 of 50
General Chemistry: Chapter 10
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A phase is a homogenous part of the
system in
contact with other parts of the system but
separated
From them by a well-defined boundary
Slide 35 of 50
General Chemistry: Chapter 10
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Intermolecular Forces: are attractive
forces between molecules. These are the
forces primarily responsible for the bulk
physical properties of matter, such as
boiling point.
Intramolecular Forces: are forces that
hold molecules together (bonds). These
stabilize individual molecules.
Types
of Intermolecular
Slide
36 of 50
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General Chemistry: Chapter Forces:
10
Intermolecular Forces: are attractive
forces between molecules. These are the
forces primarily responsible for the bulk
physical properties of matter, such as
boiling point.
Intramolecular Forces: are forces that
hold molecules together (bonds). These
stabilize individual molecules.
Types
of Intermolecular
Slide
37 of 50
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General Chemistry: Chapter Forces:
10
Ion-Ion interaction: The force of attraction
between ions is described by Coulomb’s
law.
Slide 38 of 50
General Chemistry: Chapter 10
Prentice-Hall © 2007
EXAMPLE 10-15
Calculating an Enthalpy of Reaction from Bond Energies.
The reaction of methane (CH4) and chlorine produces a mixture
of products called chloromethanes. One of these is
monochloromethane, CH3Cl, used in the preparation of silicones.
Calculate H for the reaction.
ΔHrxn =  ΔH(product bonds) - ΔH(reactant bonds)
=  ΔH bonds formed -  ΔH bonds broken
= -770 kJ/mol – (657 kJ/mol) = -113 kJ/mol
Slide 39 of 50
General Chemistry: Chapter 10
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The Lewis Structure is a representation
of a molecule that shows how the
valence electrons are arranged among the
atoms in the molecule.
Slide 42 of 50
1. Skeletal structure; determine central and terminal
atoms; low EN atoms central, O and H usually
terminal.
2. Determine # of valence electrons.
3.Determine # of electrons needed for each atom to
achieve octet.
4. # of electrons in step 3 minus # of electrons in
step 2 will give you # of bonding electrons.
5. Place bonding electrons among atoms in the
structure (connect atoms and use all of the bonding
electrons; you may get more than one plausible
structure).
Prentice-Hall © 2007
6. Did you use all of the valence electrons
from step 2? If yes, you are done, go to step 8.
if no, go to step 7.
7. Place the left over valence electrons
( from step 2) as non-bond pairs on the atoms
for each atom to achieve octet ( exception, H
must achieve duet).
8. Check if structure(s) are valid by
determining the formal charges on the atoms.
Slide 44 of 50
General Chemistry: Chapter 10
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Writing Lewis Structures
All the valence e- of atoms must appear.
Usually, the e- are paired.
Usually, each atom requires an octet.
 H only requires 2 e-.
Multiple bonds may be needed.
 Readily formed by C, N, O, S, and P.
Slide 45 of 50
General Chemistry: Chapter 10
Formal Charge: The formal charge on an atom in a Lewis
structure is the number of valence e- in the free atom minus
the number of e- assigned to that atom in the Lewis structure.
FC = #valence e- - #lone pair e- -
1
2
#bond pair e-
Formal charges is a measure of the extent to which an atom
has gained or lost an electron in the process of forming a
covalent bond.
Slide 46 of 50
 Sum of FC is the overall charge.
 FC should be as small as possible.
 Negative FC usually on most electronegative elements.
 FC of same sign on adjacent atoms is unlikely.
+
+
•• -
Slide 47 of 50
••
••
O≡N—O
••
Exceptions to the Octet Rule
Odd e- species.
••
••
•
N=O
••
H
••
•
Slide 48 of 50
O—H
••
•
H—C—H
Prentice-Hall © 2007
Exceptions to the Octet Rule
Incomplete octets.
••
••
••
F
+ F
••
••
••
B
B
Slide 49 of 50
F
F
F
F
••
••
••
+
-
F
F
-
B
F
Exceptions to the Octet Rule
Expanded octets.
P
Cl
F
S
P
Cl
Cl
Cl
F
F
F
••
Cl
••
••
Cl
F
••
Cl
••
••
Cl
••
••
••
••
••
••
F
Learn how to apply
Valence Shell Electron Pair Repulsion,
VSEPR, Theory to predict the shapes of
molecules.
The three-dimensional shape of a
molecule is important in determining its
chemical behavior, particularly for
biologically important molecules.
Valence Shell Electron Pair Repulsion
VSEPR Theory
• Electron Pairs in the Valence Shell of the central atom
Repel each other whether they are in chemical bonds (bond
pairs) or unshared (lone pairs).
• Electron pairs assume orientations about an atom to
minimize repulsions.
..
..
..
..
..
52
Balloon Analogy
Slide 53 of 50
General Chemistry: Chapter 10
Prentice-Hall © 2007
Applying VSEPR Theory
 Draw a plausible Lewis structure.
 Determine the number of electron groups around
the central atom and identify them as bond or lone
pairs.
 Establish the electron group geometry.
 Determine the molecular geometry.
 Multiple bonds count as one group of electrons.
54
Linear
Trigonal
Planar
Tetrahedral
Trigonal
Bipyramidal
55
Octahedral
Methane, Ammonia and Water
Slide 56 of 50
Table 10.1 Molecular Geometry as a
Function of Electron Group Geometry
Slide 57 of 50
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Applying VSEPR Theory
 Draw a plausible Lewis structure.
 Determine the number of e- groups and identify
them as bond or lone pairs.
 Establish the e- group geometry.
 Determine the molecular geometry.
 Multiple bonds count as one group of electrons.
 More than one central atom can be handled
individually.
Slide 63 of 50
Dipole Moments
Slide 64 of 50
General Chemistry: Chapter 10
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Dipole Moments
Slide 65 of 50
General Chemistry: Chapter 10
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The difference in electronegativity values between two atoms in a
bond (ΔEN) may be used to predict if a bond is:
a) non-polar, covalent b) polar, covalent c) ionic
For the following compounds predict whether the bond type is
a), b) or c) and justify your answer.(a bond is
considered to be ionic if ΔEN > 1.7). Find the melting point of these
compounds from the library ('The Handbook of Chemistry and Physics')
and confirm whether the melting point agrees with the predicted bond ty
NCl3 AlCl3 SO3 KI CH4
Slide 66 of 50
CF4
General Chemistry: Chapter 10
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Slide 67 of 50
General Chemistry: Chapter 10
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Of the following compounds of sulfur, which, if any,
do not possess a permanent dipole. Take the Pauling
electronegativity value for a lone pair = 3.7.
SF2
SF3+
Slide 68 of 50
SF3−
SF4
SF5+
General Chemistry: Chapter 10
SF5− SF6
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Bond Order and Bond Length
Bond Order
 Single bond, order = 1
 Double bond, order = 2
Bond Length
 Distance between two nuclei
Higher bond order
 Shorter bond
 Stronger bond
Slide 69 of 50
General Chemistry: Chapter 10
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Slide 70 of 50
General Chemistry: Chapter 10
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Bond Energies
Slide 71 of 50
General Chemistry: Chapter 10
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Slide 72 of 50
General Chemistry: Chapter 10
Prentice-Hall © 2007
EXAMPLE 10-15
Calculating an Enthalpy of Reaction from Bond Energies.
The reaction of methane (CH4) and chlorine produces a mixture
of products called chloromethanes. One of these is
monochloromethane, CH3Cl, used in the preparation of silicones.
Calculate H for the reaction.
ΔHrxn =  ΔH(product bonds) - ΔH(reactant bonds)
=  ΔH bonds formed -  ΔH bonds broken
= -770 kJ/mol – (657 kJ/mol) = -113 kJ/mol
Slide 73 of 50
General Chemistry: Chapter 10
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Focus on Molecules in Space: Measuring
Bond Lengths
Slide 74 of 50
General Chemistry: Chapter 10
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End of Chapter Questions
•Testing your decisions:
–If you get an error or a nonsense result, then
climb back to an intersection where you KNOW
you were correct, and take another route.
a
b
c
d?
e?!#
e
f
Answer
Slide 75 of 50
General Chemistry: Chapter 10
Prentice-Hall © 2007