John C. Kotz
Paul M. Treichel
John Townsend
http://academic.cengage.com/kotz
Chapter 8
Bonding and Molecular Structure:
Fundamental Concepts
John C. Kotz • State University of New York, College at Oneonta
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CHEMICAL
BONDING
Cocaine
PLAY MOVIE
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3
Chemical Bonding
Problems and questions —
How is a molecule or polyatomic ion
held together?
Why are atoms distributed at
strange angles?
Why are molecules not flat?
Can we predict the structure?
How is structure related to chemical
and physical properties?
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4
Structure & Bonding
NN triple bond. Molecule is unreactive
Phosphorus is a
tetrahedron of P
atoms. Very
reactive!
Red
phosphorus, a
polymer. Used
in matches.
© 2009 Brooks/Cole - Cengage
PLAY MOVIE
5
Forms of Chemical Bonds
• There are 2 extreme forms of
connecting or bonding atoms:
• Ionic—complete transfer of 1 or
more electrons from one atom to
another
• Covalent—some valence
electrons shared between atoms
• Most bonds are
somewhere in between.
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6
Ionic Compounds
Metal of low IE
Nonmetal
of high EA
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2 Na(s) + Cl2(g) f
2 Na+ + 2 Cl-
7
Covalent Bonding
The bond arises from the mutual attraction
of 2 nuclei for the same electrons.
Electron sharing results.
HA + HB
HA
HB
Bond is a balance of attractive and repulsive
forces.
PLAY MOVIE
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8
Bond Formation
A bond can result from a “head-to-head”
overlap of atomic orbitals on
neighboring atoms.
••
H
+
Cl
••
••
•
•
H
Cl
•
•
••
Overlap of H (1s) and Cl (2p)
Note that each atom has a single,
unpaired electron.
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9
Chemical Bonding:
Objectives
Objectives are to understand:
1. valence e- distribution in
molecules and ions.
2. molecular structures
3. bond properties and their effect
on molecular properties.
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10
11
Electron
Distribution in
Molecules
• Electron distribution is
depicted with Lewis
electron dot
structures
• Valence electrons are
distributed as shared or
BOND PAIRS and
unshared or LONE
PAIRS.
G. N. Lewis
1875 - 1946
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Bond and Lone Pairs
• Valence electrons are distributed as
shared or BOND PAIRS and
unshared or LONE PAIRS.
••
H
Cl
•
•
••
shared or
bond pair
lone pair (LP)
This is called a LEWIS
ELECTRON DOT structure.
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Valence Electrons
Electrons are divided between core and valence
electrons
B 1s2 2s2 2p1
Core = [He] , valence = 2s2 2p1
Br [Ar] 3d10 4s2 4p5
Core = [Ar] 3d10 , valence = 4s2 4p5
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Rules of the Game
No. of valence electrons of a main group atom
= Group number
•For Groups 1A-4A, no. of bond pairs = group
number.
• For Groups 5A -7A, BP’s = 8 - Grp. No.
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Rules of the Game
•No. of valence electrons of an atom = Group
number
•For Groups 1A-4A, no. of bond pairs = group
number
• For Groups 5A -7A, BP’s = 8 - Grp. No.
•Except for H (and sometimes atoms of 3rd
and higher periods),
BP’s + LP’s = 4
This observation is called the
OCTET RULE
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Building a Dot Structure
Ammonia, NH3
1. Decide on the central atom; never H.
Central atom is atom of lowest affinity
for electrons.
Therefore, N is central
2. Count valence electrons
H = 1 and N = 5
Total = (3 x 1) + 5
= 8 electrons / 4 pairs
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Building a Dot Structure
3.
Form a single bond
between the central atom and
each surrounding atom
4.
Remaining electrons form
LONE PAIRS to complete octet as
needed.
3 BOND PAIRS and 1 LONE PAIR.
H N H
H
••
H N H
Note that N has a share in 4 pairs (8
electrons), while H shares 1 pair.
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H
17
18
Sulfite ion, SO32Step 1. Central atom = S
Step 2. Count valence electrons
S= 6
3 x O = 3 x 6 = 18
Negative charge = 2
TOTAL = 26 e- or 13 pairs
Step 3. Form bonds
O
10 pairs of electrons are
now left.
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O
S
O
Sulfite ion, SO32Remaining pairs become lone pairs, first
on outside atoms and then on central
atom.
••
•
•
O
••
•
•
O
••
•
•
••
S
••
O
••
•
•
Each atom is surrounded by an
octet of electrons.
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Carbon Dioxide, CO2
1. Central atom = _______
2. Valence electrons = __ or __ pairs
3. Form bonds.
This leaves 6 pairs.
4. Place lone pairs on outer atoms.
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Carbon Dioxide, CO2
4. Place lone pairs on outer atoms.
5. So that C has an octet, we shall form
DOUBLE BONDS between C and O.
The second bonding pair forms a pi
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(π) bond.
21
Double and
even triple
bonds are
commonly
observed for C,
N, P, O, and S
22
H2CO
SO3
C2F4
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Sulfur Dioxide, SO2
1. Central atom = S
2. Valence electrons = 18 or 9 pairs
3. Form double bond so that S has an octet
— but note that there are two ways of doing
this.
bring in
left pair
••
•
•
O
••
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••
S
OR bring in
right pair
••
O
••
•
•
23
Sulfur Dioxide, SO2
This leads to the following structures.
These equivalent structures are called
RESONANCE STRUCTURES. The true
electronic structure is a HYBRID of the two.
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25
Urea, (NH2)2CO
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Urea, (NH2)2CO
1. Number of valence electrons = 24 e2. Draw sigma bonds.
O
H N
H
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C
N
H
H
27
Urea, (NH2)2CO
3. Place remaining electron pairs in the
molecule.
••
•
•
O
••
H N
H
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•
•
••
C
N
H
H
28
Urea, (NH2)2CO
4. Complete C atom octet with double
bond.
••
O
••
H N
H
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•
•
••
C
N
H
H
Atom Formal Charges
• Atoms in molecules often bear a charge (+ or -).
• The predominant resonance structure of a
molecule is the one with charges as close to 0 as
possible.
• Formal charge
= Group number
– 1/2 (no. of bonding electrons)
- (no. of LP electrons)
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Carbon Dioxide, CO2
+6 - ( 1 / 2 ) ( 4 ) - 4
••
•
•
O
••
C
+4 - ( 1 / 2 ) ( 8 ) - 0
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O
=
•
•
0
=
0
30
31
Calculated Partial Charges in CO2
Yellow = negative & red = positive
Relative size = relative charge
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Thiocyanate Ion,
6 - (1/2)(2) - 6 = -1
5 - (1/2)(6) - 2 = 0
••
•
•
S
C
N
•
•
••
4 - (1/2)(8) - 0 = 0
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SCN-
32
33
Thiocyanate Ion, SCN-
••
••
•
•
S
C
N
•
•
•
•
••
S
C
N
••
••
•
•
S
C
N
•
•
••
Which is the most important resonance form?
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•
•
Calculated Partial Charges
in SCN-
All atoms negative, but
most on the S
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••
•
•
S
••
C
N
•
•
Violations of the Octet Rule
Usually occurs with B and elements of
higher periods.
BF3
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SF4
35
Boron Trifluoride
• Central atom = _____________
• Valence electrons = __________ or
electron pairs = __________
• Assemble dot structure
The B atom has a
share in only 6 pairs of
electrons (or 3 pairs).
B atom in many
molecules is electron
deficient.
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37
Boron Trifluoride, BF3
•
•
F
•
•
+1
••
•
•
F
••
-1
B
•
•
F
•
•
••
What if we form a B—F double
bond to satisfy the B atom octet?
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Is There a B=F Double Bond in BF3
Calc’d partial charges in BF3
F is negative
and B is
positive
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Sulfur Tetrafluoride, SF4
• Central atom =
• Valence electrons = ___ or ___ pairs.
• Form sigma bonds and distribute electron
pairs.
5 pairs around the S
atom. A common
occurrence outside the
2nd period.
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MOLECULAR
GEOMETRY
MOLECULAR GEOMETRY
VSEPR
• Valence Shell Electron
Pair Repulsion theory.
Molecule adopts
the shape that
minimizes the
electron pair
repulsions.
• Most important factor in
determining geometry is
relative repulsion between
electron pairs.
PLAY MOVIE
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42
Electron Pair Geometries
See Active Figure 8.5
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No. of e- Pairs
Around Central
Atom
2
43
Example
F—Be—F
Geometry
linear
180Þ
F
3
F
planar
trigonal
B
F
120Þ
H
4
C
H
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109Þ
tetrahedral
H
H
44
No. of e- Pairs
Around Central
Atom
2
Example
F—Be—F
Geometry
linear
180Þ
F
3
F
planar
trigonal
B
F
120Þ
H
4
C
H
© 2009 Brooks/Cole - Cengage
109Þ
PLAY MOVIE
tetrahedral
H
H
No. of e- Pairs
Around Central
Atom
2
45
Example
F—Be—F
Geometry
linear
180Þ
F
3
F
planar
trigonal
B
F
120Þ
H
4
PLAY MOVIE
© 2009 Brooks/Cole - Cengage
C
H
109Þ
tetrahedral
H
H
No. of e- Pairs
Around Central
Atom
2
46
Example
F—Be—F
Geometry
linear
180Þ
F
3
F
planar
trigonal
B
F
120Þ
H
4
PLAY MOVIE
© 2009 Brooks/Cole - Cengage
C
H
109Þ
tetrahedral
H
H
47
Electron Pair Geometries
See Active Figure 8.5
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Structure Determination by
VSEPR
••
Ammonia, NH3
H
1. Draw electron dot structure
2. Count BP’s and LP’s = 4
3. The 4 electron pairs are at the
corners of a tetrahedron.
lone pair of electrons
in tetrahedral position
N
H
H
H
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N
H
H
Structure Determination by
VSEPR
Ammonia, NH3
There are 4 electron pairs at the corners
of a tetrahedron.
lone pair of electrons
in tetrahedral position
N
H
H
H
The ELECTRON PAIR GEOMETRY is
tetrahedral.
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49
Structure Determination by
VSEPR
Ammonia, NH3
The electron pair geometry is tetrahedral.
lone pair of electrons
in tetrahedral position
N
H
H
H
PLAY MOVIE
The MOLECULAR GEOMETRY — the
positions of the atoms — is PYRAMIDAL.
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51
Structure Determination by
VSEPR
Water, H2O
1. Draw electron dot structure
2. Count BP’s and LP’s = 4
3. The 4 electron pairs are at the corners of
a tetrahedron.
The electron pair
geometry is
TETRAHEDRAL.
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Structure Determination by
VSEPR
Water, H2O
The electron pair
geometry is
TETRAHEDRAL
The molecular
geometry is
BENT.
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Geometries for
Four Electron Pairs
See Figure 8.6
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Structure Determination by
VSEPR
Formaldehyde, CH2O
1. Draw electron dot structure
•
•
H
O
•
•
C
2. Count BP’s and LP’s at C
3. There are 3 electron “lumps” around C at
the corners of a planar triangle.
•
•
O
•
•
The electron pair geometry
is PLANAR TRIGONAL with
120o bond angles.
C
H
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H
54
H
55
Structure Determination by
VSEPR
Formaldehyde, CH2O
•
•
O
•
•
The electron pair
geometry is PLANAR
TRIGONAL
C
H
H
The molecular geometry
is also planar trigonal.
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Structure Determination by
VSEPR
109˚
Methanol, CH3OH
Define H-C-H and C-O-H bond
angles
109˚
H-C-H = 109o
C-O-H = 109o
In both cases the atom is
surrounded by 4
electron pairs.
PLAY MOVIE
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57
Structure Determination by
VSEPR
H
Acetonitrile, CH3CN
H—C—C
109˚
H 180˚
One C is surrounded by 4 electron “lumps”
and the other by 2 “lumps”
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N
••
Define unique bond angles
H-C-H = 109o
C-C-N = 180o
58
Phenylalanine, an amino acid
1
H
C
H
C
C
C
C
H
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H
C
H
H 2 H
O
3
C
C
C
O
H
N
H
4
H
5
H
Phenylalanine
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Structures with Central Atoms
with More Than or Less Than 4
Electron Pairs
Often occurs with Group
3A elements and with those
of 3rd period and higher.
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61
Boron Compounds
••
•
•
•
•
Consider boron trifluoride, BF3
F
••
•
•
F
The B atom is surrounded by only
••
3 electron pairs.
B
•
•
•
•
F
••
Bond angles are 120o
Geometry described as
planar trigonal
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Compounds with 5 or 6 Pairs Around
the Central Atom
90Þ
F
F
P
Trigonal bipyramid
F
120Þ
F
F
5 electron pairs
PLAY MOVIE
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63
Molecular
Geometries for
Five Electron
Pairs
See Figure 8.8
All based on trigonal
bipyramid
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64
Sulfur Tetrafluoride, SF4
••
•F
•
••
••
•• F
••
• Number of valence
electrons = 34
• Central atom = S
• Dot structure
Electron pair geometry is
trigonal bipyramid
(because there are 5 pairs
around the S)
© 2009 Brooks/Cole - Cengage
••
S
••
F
••
••
•• F ••
••
90Þ
••
F
S
F
F
F
120Þ
65
Sulfur Tetrafluoride, SF4
Lone pair is in the equator because it requires
more room.
90Þ
••
F
S
F
F
F
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120Þ
Molecular
Geometries for
Six Electron
Pairs
See Figure 8.8
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66
All are based on the 8sided octahedron
67
Compounds with 5 or 6 Pairs
Around the Central Atom
90Þ
F
F
S
F
Octahedron
F
F
90Þ
F
6 electron pairs
© 2009 Brooks/Cole - Cengage
PLAY MOVIE
Bond Properties
• What is the effect of bonding and structure on
molecular properties?
Free rotation
around C–C single
bond
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No rotation around
C=C double bond
68
Bond Order
# of bonds between a pair of atoms
Double bond
Single bond
Acrylonitrile
Triple
bond
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70
Bond Order
Fractional bond orders occur in molecules with resonance
structures.
Consider NO2••
••
N
N
•• • •••
••
••
O
O• • O
O
••
••
••
••
The N—O bond order = 1.5
Total # of e - pairs used for a type of bond
Bond order =
Total # of bonds of that type
3 e - pairs in N— O bonds
Bond order =
2 N — O bonds
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Bond Order
Bond order is proportional to two important bond
properties:
(a)
(b)
bond strength
bond length
414 kJ
123 pm
110 pm
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745 kJ
Bond Length
• Bond length is the
distance between
the nuclei of two
bonded atoms.
PLAY MOVIE
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72
Bond Length
Bond length depends on
size of bonded atoms.
H—F
H—Cl
Bond distances measured in
Angstrom units where 1 A =
10-2 pm.
H—I
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73
Bond Length
Bond length depends on
bond order.
Bond distances measured in
Angstrom units where 1 A =
10-2 pm.
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74
Bond Strength
• —measured by the energy req’d to break a bond. See
Table 8.9
PLAY MOVIE
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Bond Strength
• —measured by the energy req’d to break a bond. See
Table 8.9.
•
BOND
Bond dissociation
enthalpy (kJ/mol)
H—H
C—C
C=C
CC
NN
436
346
602
835
945
The GREATER the number of bonds (bond order) the
HIGHER the bond strength and the SHORTER the bond.
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77
© 2009 Brooks/Cole - Cengage
Bond Strength
Bond
Order
HO—OH
O=O
••
O
••O
••
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•• •
O•
••
Length
Strength
1
142 pm
210 kJ/mol
2
121
498
1.5
128
?
78
Using Bond Dissociation
Enthalpies
Estimate the energy of the reaction
H—H(g) + Cl—Cl(g) f 2 H—Cl(g)
Net energy = ∆rH =
= energy required to break bonds
- energy evolved when bonds are made
H—H = 436 kJ/mol
Cl—Cl = 242 kJ/mol
H—Cl = 432 kJ/mol
PLAY MOVIE
© 2009 Brooks/Cole - Cengage
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80
Using Bond Dissociation Enthalpies
Estimate the energy of the reaction
H—H + Cl—Cl ----> 2 H—Cl
H—H = 436 kJ/mol
Cl—Cl = 242 kJ/mol
H—Cl = 432 kJ/mol
Sum of H-H + Cl-Cl bond energies
= 436 kJ + 242 kJ = +678 kJ
2 mol H-Cl bond energies = 864 kJ
Net = ∆rH = +678 kJ - 864 kJ = -186 kJ
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Using Bond Dissociation
Enthalpies
Estimate the energy of the reaction
2 H—O—O—H f O=O + 2 H—O—H
Is the reaction exo- or endothermic?
Which is larger:
A)
energy req’d to break bonds
B)
or energy evolved on making bonds?
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81
Using Bond Dissociation
Enthalpies
2 H—O—O—H
f O=O + 2 H—O—H
Energy required to break bonds:
break 4 mol of O—H bonds = 4 (463 kJ)
break 2 mol O—O bonds = 2 (146 kJ)
TOTAL ENERGY to break bonds = 2144 kJ
TOTAL ENERGY evolved on making O=O bonds
and 4 O-H bonds bonds = 2350 kJ
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82
Using Bond Dissociation
Enthalpies
2 H—O—O—H
83
f O=O + 2 H—O—H
Net energy = +2144 kJ - 2350 kJ = - 206 kJ
The reaction is exothermic!
More energy is evolved on
making bonds than is
expended in breaking bonds.
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Molecular Polarity
Water
Boiling point =
100 ˚C
Methane
Boiling point
= -161 ˚C
Why do water and
methane differ so
much in their
boiling points?
Why do ionic compounds
dissolve in water?
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84
Bond Polarity
85
HCl is POLAR because it
has a positive end and a
negative end.
+d
-d
••
••
H Cl
••
Cl has a greater share in
bonding electrons than
does H.
Cl has slight negative charge (-d) and H has
slight positive charge (+ d)
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86
Bond Polarity
• Three molecules with polar,
covalent bonds.
• Each bond has one atom
with a slight negative charge
(-d) and and another with a
slight positive charge (+ d)
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Bond Polarity
This model, calc’d using CAChe
software for molecular
calculations, shows that H is +
(red) and Cl is - (yellow). Calc’d
charge is + or - 0.20.
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87
+d
-d
••
••
H Cl
••
Bond Polarity
Due to the bond polarity, the H—Cl
bond energy is GREATER than
expected for a “pure” covalent
bond.
BOND
“pure” bond
real bond
ENERGY
339 kJ/mol calc’d
432 kJ/mol measured
Difference = 92 kJ. This difference is
proportional to the difference in
ELECTRONEGATIVITY, .
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89
Electronegativity, 
 is a measure of the ability of an
atom in a molecule to attract
electrons to itself.
Concept proposed by Linus
Pauling 1901-1994
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Linus Pauling, 1901-1994
PLAY MOVIE
The only person to receive two unshared Nobel prizes
(for Peace and Chemistry).
Chemistry areas: bonding, electronegativity, protein
structure
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90
Electronegativity
See Figure 8.11
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92
Electronegativity, 
See Figure 8.11
• F has maximum .
• Atom with lowest  is the center atom in
most molecules.
• Relative values of  determine BOND
POLARITY (and point of attack on a
molecule).
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Bond Polarity
Which bond is more polar (or DIPOLAR)?
O—H
O—F

3.5 - 2.1
3.5 - 4.0

1.4
0.5
OH is more polar than OF
-d
O
+d
H
+d
O
-d
F
and polarity is “reversed.”
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Molecular Polarity
Molecules—such as HI and H2O—
can be POLAR (or dipolar).
They have a DIPOLE MOMENT. The polar HCl molecule
will turn to align with an electric field.
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Molecular Polarity
The magnitude of the
dipole is given in
Debye units. Named for
Peter Debye (1884 -
1966). Rec’d 1936
Nobel prize for work on
x-ray diffraction and
dipole moments.
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Dipole Moments
Why are some molecules polar but
others are not?
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Molecular Polarity
Molecules will be polar if
a)
bonds are polar
b)
AND the molecule is NOT “symmetric”
All above are NOT polar
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Polar or Nonpolar?
Compare CO2 and H2O. Which one is polar?
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Carbon Dioxide
99
• CO2 is NOT polar
even though the CO
bonds are polar.
• CO2 is symmetrical.
Positive C atom
is reason CO2
and H2O react to
give H2CO3
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-0.75
+1.5
-0.75
100
Polar or Nonpolar?
• Consider AB3 molecules: BF3, Cl2CO, and NH3.
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Molecular Polarity, BF3
F
B
F
F
B—F bonds in BF3 are polar.
But molecule is symmetrical and
NOT polar
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B atom is
positive and
F atoms are
negative.
101
Molecular Polarity, HBF2
H
B
F
F
B—F and B—H bonds in HBF2
are polar. But molecule is NOT
symmetrical and is polar.
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B atom is
positive but H
& F atoms are
negative.
102
Is Methane, CH4, Polar?
Methane is symmetrical and is NOT polar.
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Is CH3F Polar?
C—F bond is very polar.
Molecule is not symmetrical and
so is polar.
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105
CH4 … CCl4
Polar or Not?
• Only CH4 and CCl4 are NOT polar. These are the only two
molecules that are “symmetrical.”
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Substituted Ethylene
• C—F bonds are MUCH more polar than
C—H bonds.
• Because both C—F bonds are on same
side of molecule, molecule is POLAR.
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Substituted Ethylene
• C—F bonds are MUCH more polar than C—H bonds.
• Because both C—F bonds are on opposing ends of
molecule, molecule is NOT POLAR.
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Visualizing Charges and Polarity
Electrostatic Potential Surfaces
Electrostatic potential surfaces (EPS) can be used to
visualize
a) Where the charges lie in molecules
b) The polarity of molecules
The HF molecule
See page 382
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109
Visualizing Charges and Polarity
F
H
• The boundary surface around the molecule is
made up of all the points in space where the
electron density is a given value (here 0.002 e-/A3).
• The colors indicate the potential experienced by a
H+ ion on the surface. More attraction (a negative
site) is red, and repulsion (a positive site) is blue.
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Visualizing Charges and Polarity
• As expected, the surface near O in H2O and the N
is CH3-NH2 is red because that is the more
electronegative atom.
• H2O is polar - with O more negative and H more
positive. CH3-NH2 is also polar.
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Visualizing Charges and Polarity
BF3
Cl2CO
• The EP surfaces show BF3 is not polar (but the B
is slightly positively charged), whereas Cl2CO is
polar with the O more negative than Cl.
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112
Visualizing Charges and Polarity
• The EP surfaces show cis-C2H2Cl2 (left) is polar
whereas trans-C2H2Cl2 (right) is not polar.
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