Chemistry and Chemical Reactivity
6th Edition
1
John C. Kotz
Paul M. Treichel
Gabriela C. Weaver
CHAPTER 9
Bonding and Molecular Structure:
Fundamental Concepts
Lectures written by John Kotz
©2006
2006
Brooks/Cole
Thomson
©
Brooks/Cole
- Thomson
CHEMICAL
BONDING
Cocaine
© 2006 Brooks/Cole - Thomson
2
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?
© 2006 Brooks/Cole - Thomson
3
Structure & Bonding
NN triple bond. Molecule is unreactive
Phosphorus is a
tetrahedron of P
atoms. Very
reactive!
Red
phosphorus, a
polymer. Used
in matches.
© 2006 Brooks/Cole - Thomson
4
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.
© 2006 Brooks/Cole - Thomson
5
Ionic Compounds
Metal of low IE
Nonmetal
of high EA
© 2006 Brooks/Cole - Thomson
2 Na(s) + Cl2(g) ---> 2 Na+ + 2 Cl-
6
Covalent Bonding
The bond arises from the mutual attraction
of 2 nuclei for the same electrons.
Electron sharing results. (Screen 9.6)
HA + HB
HA
HB
Bond is a balance of attractive and repulsive
forces.
© 2006 Brooks/Cole - Thomson
7
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.
© 2006 Brooks/Cole - Thomson
8
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.
© 2006 Brooks/Cole - Thomson
9
10
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
© 2006 Brooks/Cole - Thomson
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.
© 2006 Brooks/Cole - Thomson
11
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
© 2006 Brooks/Cole - Thomson
12
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.
© 2006 Brooks/Cole - Thomson
13
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
© 2006 Brooks/Cole - Thomson
14
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
© 2006 Brooks/Cole - Thomson
15
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.
© 2006 Brooks/Cole - Thomson
H
16
17
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.
© 2006 Brooks/Cole - Thomson
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.
© 2006 Brooks/Cole - Thomson
18
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.
© 2006 Brooks/Cole - Thomson
19
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
© 2006 Brooks/Cole - Thomson
(π) bond.
20
Double and
even triple
bonds are
commonly
observed for C,
N, P, O, and S
21
H2CO
SO3
C2F4
© 2006 Brooks/Cole - Thomson
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
••
© 2006 Brooks/Cole - Thomson
••
S
OR bring in
right pair
••
O
••
•
•
22
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.
© 2006 Brooks/Cole - Thomson
23
24
Urea, (NH2)2CO
© 2006 Brooks/Cole - Thomson
25
Urea, (NH2)2CO
1. Number of valence electrons = 24 e2. Draw sigma bonds.
O
H N
H
© 2006 Brooks/Cole - Thomson
C
N
H
H
26
Urea, (NH2)2CO
3. Place remaining electron pairs in the
molecule.
••
•
•
O
••
H N
H
© 2006 Brooks/Cole - Thomson
•
•
••
C
N
H
H
27
Urea, (NH2)2CO
4. Complete C atom octet with double
bond.
••
O
••
H N
H
© 2006 Brooks/Cole - Thomson
•
•
••
C
N
H
H
Formal Atom 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)
© 2006 Brooks/Cole - Thomson
28
Carbon Dioxide, CO2
+6 - ( 1 / 2 ) ( 4 ) - 4
••
•
•
O
••
C
+4 - ( 1 / 2 ) ( 8 ) - 0
© 2006 Brooks/Cole - Thomson
O
=
•
•
0
=
0
29
30
Calculated Partial Charges in CO2
Yellow = negative & red = positive
Relative size = relative charge
© 2006 Brooks/Cole - Thomson
Thiocyanate Ion,
6 - (1/2)(2) - 6 = -1
5 - (1/2)(6) - 2 = 0
••
•
•
S
C
N
•
•
••
4 - (1/2)(8) - 0 = 0
© 2006 Brooks/Cole - Thomson
SCN-
31
32
Thiocyanate Ion, SCN-
••
••
•
•
S
C
N
•
•
•
•
••
S
C
N
••
••
•
•
S
C
N
•
•
••
Which is the most important resonance form?
© 2006 Brooks/Cole - Thomson
•
•
Calculated Partial Charges
in SCN-
All atoms negative, but
most on the S
© 2006 Brooks/Cole - Thomson
33
••
•
•
S
••
C
N
•
•
Violations of the Octet Rule
Usually occurs with B and elements of
higher periods.
BF3
© 2006 Brooks/Cole - Thomson
SF4
34
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.
© 2006 Brooks/Cole - Thomson
35
36
Boron Trifluoride, BF3
•
•
F
•
•
+1
••
•
•
F
••
-1
B
•
•
F
•
•
••
What if we form a B—F double
bond to satisfy the B atom octet?
© 2006 Brooks/Cole - Thomson
37
Is There a B=F Double Bond in BF3
Calc’d partial charges in BF3
F is negative
and B is
positive
© 2006 Brooks/Cole - Thomson
38
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.
© 2006 Brooks/Cole - Thomson
MOLECULAR
GEOMETRY
MOLECULAR GEOMETRY
VSEPR
• Valence Shell Electron
Pair Repulsion theory.
• Most important factor in
determining geometry is
relative repulsion between
electron pairs.
© 2006 Brooks/Cole - Thomson
Molecule adopts
the shape that
minimizes the
electron pair
repulsions.
40
41
Electron Pair Geometries
Active Figure 9.8
© 2006 Brooks/Cole - Thomson
No. of e- Pairs
Around Central
Atom
2
42
Example
F—Be—F
Geometry
linear
180Þ
F
3
planar
trigonal
B
F
F
120Þ
H
109Þ
4
C
H
tetrahedral
H
H
© 2006 Brooks/Cole - Thomson
43
No. of e- Pairs
Around Central
Atom
2
Example
F—Be—F
Geometry
linear
180Þ
F
3
planar
trigonal
B
F
F
120Þ
H
109Þ
4
C
H
tetrahedral
H
H
© 2006 Brooks/Cole - Thomson
No. of e- Pairs
Around Central
Atom
2
44
Example
F—Be—F
Geometry
linear
180Þ
F
3
planar
trigonal
B
F
F
120Þ
H
109Þ
4
C
H
tetrahedral
H
H
© 2006 Brooks/Cole - Thomson
No. of e- Pairs
Around Central
Atom
2
45
Example
F—Be—F
Geometry
linear
180Þ
F
3
planar
trigonal
B
F
F
120Þ
H
109Þ
4
C
H
tetrahedral
H
H
© 2006 Brooks/Cole - Thomson
46
Electron Pair Geometries
Active Figure 9.8
© 2006 Brooks/Cole - Thomson
47
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
© 2006 Brooks/Cole - Thomson
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.
© 2006 Brooks/Cole - Thomson
48
Structure Determination by
VSEPR
Ammonia, NH3
The electron pair geometry is tetrahedral.
lone pair of electrons
in tetrahedral position
N
H
H
H
The MOLECULAR GEOMETRY — the
positions of the atoms — is PYRAMIDAL.
© 2006 Brooks/Cole - Thomson
49
50
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.
© 2006 Brooks/Cole - Thomson
51
Structure Determination by
VSEPR
Water, H2O
The electron pair
geometry is
TETRAHEDRAL
The molecular
geometry is
BENT.
© 2006 Brooks/Cole - Thomson
Geometries for
Four Electron Pairs
Figure 9.9
© 2006 Brooks/Cole - Thomson
52
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
© 2006 Brooks/Cole - Thomson
H
53
H
54
Structure Determination by
VSEPR
Formaldehyde, CH2O
•
•
O
•
•
The electron pair
geometry is PLANAR
TRIGONAL
C
H
H
The molecular geometry
is also planar trigonal.
© 2006 Brooks/Cole - Thomson
55
Structure Determination by
VSEPR
H
Methanol, CH3OH
Define H-C-H and C-O-H bond
angles
H-C-H = 109o
C-O-H = 109o
In both cases the atom is
surrounded by 4
electron pairs.
© 2006 Brooks/Cole - Thomson
109˚
••
H—C—O—H
••
H
109˚
56
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”
© 2006 Brooks/Cole - Thomson
N
••
Define unique bond angles
H-C-H = 109o
C-C-N = 180o
57
Phenylalanine, an amino acid
1
H
C
H
C
C
C
C
H
© 2006 Brooks/Cole - Thomson
H
C
H
H 2 H
O
3
C
C
C
O
H
N
H
4
H
5
H
Phenylalanine
© 2006 Brooks/Cole - Thomson
58
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.
© 2006 Brooks/Cole - Thomson
59
60
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
© 2006 Brooks/Cole - Thomson
61
Compounds with 5 or 6 Pairs
Around the Central Atom
90Þ
F
F
P
Trigonal bipyramid
F
120Þ
F
F
5 electron pairs
© 2006 Brooks/Cole - Thomson
62
Molecular
Geometries for
Five Electron
Pairs
Figure 9.11
All based on trigonal
bipyramid
© 2006 Brooks/Cole - Thomson
63
Sulfur Tetrafluoride, SF4
••
•F
•
••
••
•• F
••
• Number of valence
electrons = 34
• Central atom = S
• Dot structure
Electron pair geometry
--> trigonal bipyramid
(because there are 5 pairs
around the S)
© 2006 Brooks/Cole - Thomson
••
S
••
F
••
••
•• F ••
••
90Þ
F
••
S
F
F
F
120Þ
64
Sulfur Tetrafluoride, SF4
Lone pair is in the equator
because it requires more
room.
90Þ
F
••
S
© 2006 Brooks/Cole - Thomson
••
S
•• F ••
••
F
F
F
••
•F
•
••
•• ••
F
••
120Þ
••
F
••
••
65
Molecular
Geometries for
Six Electron
Pairs
Figure 9.14
© 2006 Brooks/Cole - Thomson
All are based on the 8sided octahedron
66
Compounds with 5 or 6 Pairs
Around the Central Atom
90Þ
F
F
S
F
Octahedron
F
F
F
6 electron pairs
© 2006 Brooks/Cole - Thomson
90Þ
Bond Properties
• What is the effect of bonding and structure on
molecular properties?
Free rotation
around C–C single
bond
© 2006 Brooks/Cole - Thomson
No rotation around
C=C double bond
67
Bond Order
# of bonds between a pair of atoms
Double bond
Single bond
Acrylonitrile
Triple
bond
© 2006 Brooks/Cole - Thomson
68
69
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
© 2006 Brooks/Cole - Thomson
70
Bond Order
Bond order is proportional to two important bond
properties:
(a)
(b)
bond strength
bond length
414 kJ
123 pm
110 pm
© 2006 Brooks/Cole - Thomson
745 kJ
Bond Length
• Bond length is the
distance between
the nuclei of two
bonded atoms.
© 2006 Brooks/Cole - Thomson
71
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
© 2006 Brooks/Cole - Thomson
73
Bond Length
Bond length depends on
bond order.
Bond distances measured in
Angstrom units where 1 A =
10-2 pm.
© 2006 Brooks/Cole - Thomson
Bond Strength
• —measured by the energy req’d to break a bond. See
Table 9.9
© 2006 Brooks/Cole - Thomson
74
Bond Strength
• —measured by the energy req’d to break a bond. See
Table 9.10.
•
BOND
STRENGTH (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.
© 2006 Brooks/Cole - Thomson
75
76
© 2006 Brooks/Cole - Thomson
77
Bond Strength
Bond
Order Length
HO—OH
O=O
••
O
••O
••
© 2006 Brooks/Cole - Thomson
•• •
O•
••
Strength
1
142 pm
210 kJ/mol
2
121
498
1.5
128
?
Bond Strength
• Measured as the energy req’d to break a bond. See
Table 9.10
© 2006 Brooks/Cole - Thomson
78
Bond Strength
• Measured as the energy req’d to break a bond. See
Table 9.10.
•
BOND
STRENGTH (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.
© 2006 Brooks/Cole - Thomson
79
80
Using Bond Energies
Estimate the energy of the reaction
H—H(g) + Cl—Cl(g) ----> 2 H—Cl(g)
Net energy = ∆Hrxn =
= 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
© 2006 Brooks/Cole - Thomson
81
© 2006 Brooks/Cole - Thomson
82
Using Bond Energies
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 = ∆H = +678 kJ - 864 kJ = -186 kJ
© 2006 Brooks/Cole - Thomson
83
Using Bond Energies
Estimate the energy of the reaction
2 H—O—O—H ----> 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?
© 2006 Brooks/Cole - Thomson
84
Using Bond Energies
2 H—O—O—H
----> 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
© 2006 Brooks/Cole - Thomson
85
Using Bond Energies
2 H—O—O—H
----> 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.
© 2006 Brooks/Cole - Thomson
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?
© 2006 Brooks/Cole - Thomson
86
Bond Polarity
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)
© 2006 Brooks/Cole - Thomson
87
88
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)
© 2006 Brooks/Cole - Thomson
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.
© 2006 Brooks/Cole - Thomson
89
+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, .
© 2006 Brooks/Cole - Thomson
90
91
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
© 2006 Brooks/Cole - Thomson
Linus Pauling, 1901-1994
The only person to receive two unshared Nobel prizes
(for Peace and Chemistry).
Chemistry areas: bonding, electronegativity, protein
structure
© 2006 Brooks/Cole - Thomson
92
Electronegativity
Figure 9.14
© 2006 Brooks/Cole - Thomson
93
94
Electronegativity, 
See Figure 9.14
• 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).
© 2006 Brooks/Cole - Thomson
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.”
© 2006 Brooks/Cole - Thomson
95
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.
© 2006 Brooks/Cole - Thomson
96
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.
© 2006 Brooks/Cole - Thomson
97
Dipole Moments
Why are some molecules polar but
others are not?
© 2006 Brooks/Cole - Thomson
98
Molecular Polarity
Molecules will be polar if
a)bonds are polar
AND
b)
the molecule is NOT “symmetric”
© 2006 Brooks/Cole - Thomson
All above are NOT polar
99
Polar or Nonpolar?
Compare CO2 and H2O. Which one is polar?
© 2006 Brooks/Cole - Thomson
100
Carbon Dioxide
101
• 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
© 2006 Brooks/Cole - Thomson
-0.75
+1.5
-0.75
102
Consequences of H2O
Polarity
© 2006 Brooks/Cole - Thomson
103
Polar or Nonpolar?
• Consider AB3 molecules: BF3, Cl2CO, and NH3.
© 2006 Brooks/Cole - Thomson
Molecular Polarity, BF3
F
B
F
F
B—F bonds in BF3 are polar.
But molecule is symmetrical and
NOT polar
© 2006 Brooks/Cole - Thomson
B atom is
positive and
F atoms are
negative.
104
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.
© 2006 Brooks/Cole - Thomson
B atom is
positive but H
& F atoms are
negative.
105
Is Methane, CH4, Polar?
Methane is symmetrical and is NOT polar.
© 2006 Brooks/Cole - Thomson
106
Is CH3F Polar?
C—F bond is very polar.
Molecule is not symmetrical and
so is polar.
© 2006 Brooks/Cole - Thomson
107
108
CH4 … CCl4
Polar or Not?
•
Only CH4 and CCl4 are NOT polar. These are the only two molecules that are “symmetrical.”
© 2006 Brooks/Cole - Thomson
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.
© 2006 Brooks/Cole - Thomson
109
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.
© 2006 Brooks/Cole - Thomson
110