Bonding:
General Concepts
Chapter 8
1
Overview
 Types of chemical bonds, Electronegativity, Bond
polarity and Dipole Moments.
 The Ions: electron configurations, size, formula,
lattice energy calculations.
 Covalent bonds: model, bond energies, chemical
reactions.
 Lewis structure, exceptions to the octet rule,
resonance.
 Molecular structure models from Valence Shell
Electron Pair Model “VSEPR” for single and
multiple bonds.
2
Bonds
Forces that hold groups of
atoms together and make
them function as a unit.
NaCl – attraction is electrostatic since Na+ and Cl- are the
Stable forms for these elements. This is an example of
“Ionic Bonding”
3
Bond Energy
 It
is the energy required to break a
bond.
 It
gives us information about the
strength of a bonding interaction, as
well as radius.
4
Bond Length
The distance where the system
energy is a minimum.
5
Figure 8.1: (a) The interaction of two hydrogen atoms. (b) Energy
profile as a function of the distance between the nuclei of the
hydrogen atoms.
6
Change in electron
density as two hydrogen
atoms approach each
other.
7
Covalent Bond
 No electron transfer
Electrons are shared between two atoms,
positioned between the two nuclei
Example: H2, O2, H2O, CO2, etc.
8
Ionic Bonds
 Formed
from electrostatic attractions of
closely packed, oppositely charged ions.
 Formed when an atom that easily loses
electrons reacts with one that has a high
electron affinity.
9
The Ionic Bond
Li + F
1 22s22p5
1s22s1s
e- +
Li+ +
Li+ F [He]
1s
1s2[2Ne]
2s22p6
Li
Li+ + e-
F
F -
F -
Li+ F -
10
Ionic Compound
Cation is always smaller than atom from
which it is formed.
Anion is always larger than atom from
which it is formed.
11
Ionic Bonds
and
Coulomb’s Law
E = 2.31  10
19
J nm (Q1Q2 / r )
Q1 and Q2 = numerical ion charges
r = distance between ion centers (in nm)
Attractive
forces are (-), repulsive are (+).
12
Covalent or Ionic?
Covalent and ionic are simply extreme
cases.
Most molecules share electrons but
“Unequally” due to the difference in
electronegativity and electron affinity.
This will give rise to a dipole moment and
the molecule becomes polar.
13
Electronegativity


The ability of an atom in a molecule to attract
shared electrons to itself.
Linus Pauling simple model
Δ= (H  X)actual  (H  X)expected
(H-H) + (X-X)
(H-X)experimental # (H-X)expected =
2
If  = 0 => no polarity
14
15
Classification of bonds by difference in electronegativity
Difference
Bond Type
0 to 0.1
Covalent
2
0 < and <2
Ionic
Polar Covalent
Increasing difference in electronegativity
Covalent
Polar Covalent
share e-
partial transfer of e-
Ionic
transfer e16
17
Classify the following bonds as ionic, polar covalent,
or covalent: The bond in CsCl; the bond in H2S; and
the NN bond in H2NNH2.
Cs – 0.7
Cl – 3.0
3.0 – 0.7 = 2.3
Ionic
H – 2.1
S – 2.5
2.5 – 2.1 = 0.4
Polar Covalent
N – 3.0
N – 3.0
3.0 – 3.0 = 0
Covalent
18
9.5
Polar covalent bond or polar bond is a covalent
bond with greater electron density around one of the
two atoms. The molecule is called “Dipolar”.
electron poor
region
H
electron rich
region
F
e- poor
H
d+
e- rich
F
d-
Dipole Moment
19
Figure 8.2: The effect of an electric field on hydrogen fluoride molecules.
(a) When no electric field is
present, the molecules are
randomly oriented.
(b) When the field is turned on, the
molecules tend to line up with their
negative ends toward the positive
pole and their
positive ends toward the negative
pole.
20
21
Comparison of Ionic and Covalent Compounds
22
Polyatomic Molecules
May exhibit dipole moment depending on
their structure i.e. arrangement in space
23
Figure 8.4: (a) The charge distribution in the water
molecule. (b) The water molecule in an electric field.
V-Shape
24
Figure 8.5: (a) The structure and charge distribution of the ammonia molecule.
The polarity of the N—H bonds occurs because nitrogen has a greater
electronegativity than hydrogen. (b) The dipole moment of the ammonia
molecule oriented in an electric field.
Look for a “NET DIPOLE”
25
Look for a “NET DIPOLE”
N
H
H
H
Trigonal Pyramidal Structure
26
27
The carbon dioxide molecule CO2: The opposed bond
polarities cancel out, and the carbon dioxide has no
dipole moment:
Non-polar molecule
Note: The C-O bond is polar, but the net dipole is Zero
Example: SO3, CCl4, etc.
28
29
Which of the following molecules have a dipole moment?
H2O, CO2, SO2, and CH4
O
S
dipole moment
polar molecule
dipole moment
polar molecule
H
O
C
O
no dipole moment
nonpolar molecule
H
C
H
H
no dipole moment
nonpolar molecule
30
Dipoles (polar molecules) and Microwaves
31
Compounds
Two nonmetals react: They share electrons to
achieve NGEC (Noble Gas Electron Configurations)
and form covalent bonds.
A nonmetal and a representative group metal
react (ionic compound): The valence orbitals
of the metal are emptied (cation) to achieve
NGEC. The valence electron configuration
of the nonmetal (anion) achieves NGEC.
32
Ions
Ionic compounds are always electrically
neutral e.g. they have the same amount of
+ve and –ve charges.
Common ions have noble gas configurations.
33
Electron Configurations of Cations and Anions
Of Representative Elements
Na [Ne]3s1
Na+ [Ne]
Ca [Ar]4s2
Ca2+ [Ar]
Al [Ne]3s23p1
Al3+ [Ne]
Atoms gain electrons
so that anion has a
noble-gas outer
electron configuration.
Atoms lose electrons so that
cation has a noble-gas outer
electron configuration.
H 1s1
H- 1s2 or [He]
F 1s22s22p5
F- 1s22s22p6 or [Ne]
O 1s22s22p4
O2- 1s22s22p6 or [Ne]
N 1s22s22p3
N3- 1s22s22p6 or [Ne]
34
4f
5f
35
ns2np6
ns2np5
ns2np4
ns2np3
ns2np2
ns2np1
d10
d5
d1
ns2
ns1
Ground State Electron Configurations of the Elements
-1
-2
-3
+3
+2
+1
Cations and Anions Of Representative Elements
36
Notes on Ions
 Hydrogen may form H+ or HTin forms Sn2+ and Sn4+.
Transition metals exhibit a more
complicated behavior.
37
Example of Ionic Compounds
MgO magnesium oxide is formed of Mg2+
and O2-.
CaO formed from Ca2+ and O2-.
Al2O3 is formed of 2Al3+ and 3O2-.
38
Figure 8.7: Sizes of ions related to positions of
the elements on the periodic table.
39
40
Isoelectronic Ions
Contain the the same number of electrons
8O
1s22s22p4
9F
1s22s22p5
11Na 1s22s22p63s1
12Mg 1s22s22p63s2
13Al 1s22s22p63s23p1
O2- 1s22s22p6
F- 1s22s22p6
Na+ 1s22s22p6
Mg2+1s22s22p6
Al3+ 1s22s22p6
Which one you expect to have the smallest radius?
And why?
41
Radii of Isoelectronic Ions
O2> F > Na+ > Mg2+ > Al3+
largest
smallest
13 protons vs. 10 electrons
42
Example
Choose the largest ion in each of the
following groups:
1. Li+, Na+, K+, Rb+, Cs+
2. Ba2+ , Cs+ , I- , Te2-
43
To Reiterate
Cation is smaller than parent molecule.
Anion is larger than parent molecule.
Size increase down in a group (+ve or –ve).
The larger the mass number the smaller the
size for isoelectronic cations and anions.
44
Electron Configurations of Cations of Transition Metals
When a cation is formed from an atom of a transition metal,
electrons are always removed first from the ns orbital and
then from the (n – 1)d orbitals.
Fe:
[Ar]4s23d6
Fe2+: [Ar]4s03d6 or [Ar]3d6
Mn:
[Ar]4s23d5
Mn2+: [Ar]4s03d5 or [Ar]3d5
Fe3+: [Ar]4s03d5 or [Ar]3d5
45
Why Compounds Exist?
The driving force behind the naturally occurring
compounds (such as NaCl, H2O, etc.) is to yield a
stable lower energy form.
A stable form is a an arrangement of atoms, held
together by bonding that prevent decomposition.
This bonding energy when ions condense from gas
phase into ionic solid is called Lattice Energy.
46
Lattice Energy
The change in energy when separated
gaseous ions are packed together to form an
ionic solid.
M+(g) + X(g)  MX(s)
Lattice energy is negative (exothermic)
from the point of view of the system.
47
Formation of an Ionic Solid
1.
2.
3.
4.
5.
Sublimation of the solid metal
M(s)  M(g)
Ionization of the metal atoms
M(g)  M+(g) + e
Dissociation of the nonmetal
1/2X (g)  X(g)
2
Formation of X ions in the gas phase:
X(g) + e  X(g)
Formation of the solid (LATTICE) MX
M+(g) + X(g)  MX(s)
[endothermic]
[endothermic]
[endothermic]
[exothermic]
[quite exothermic]
Lattice Energy
48
The energy changes involved in the formation of solid
lithium fluoride from its elements.
Lattice
Energy
From Gas
to Solid
49
The structure of lithium
fluoride. Called also the
NaCl structure where
each ion is surrounded
by 6 of the other ions
Applicable for all alkalimetals/halogen except
the Cesium salts.
50
Lattice Energy = k(Q1Q2 / r )
Q1, Q2 = charges on the ions
r = shortest distance between centers of the cations
and anions
The magnitude of Q’s will determine how strong is
the lattice.
51
Comparison of
the energy
changes involved
in the formation
of solid sodium
fluoride and solid
magnesium
oxide. Note all
ions are
isoelectronic
52
Electrostatic (Lattice) Energy
Lattice energy (E) increases
as Q increases and/or
as r decreases.
Cmpd
MgF2
lattice energy
2957 Q= +2,-1
MgO
3938
LiF
1036
LiCl
853
Q= +2,-2
r F < r Cl
53
Partial Ionic Character of the Covalent
Bond
Introduce the concept of percent ionic character in
a polar covalent bond
Measured dipole of X-Y
x 100
% ionic character =
Calculated dipole of X+Y-
54
Ion Pairing
Experiments showed that none of the studied systems have
100% ionic character despite large differences in electronegativity.
The reason is “Ion Pairing”
during motion, ions approach closely and for a short period
of time their charges will be neutralized before
moving away from each other.
55
Molten NaCl conducts an electric
current, indicating the presence
of mobile Na+ and Cl- ions.
The more mobile the ions the
Stronger the current and the
brighter the lamp!
56
The relationship between the ionic character of a
covalent bond and the electronegativity difference of the
bonded atoms.
Note: These are “Molten” salts
and not “Aqueous” salts!
<50%
Non-Ionic
>50%
Ionic
57
A Model for Covalent Bond
Models are attempts to explain
how nature operates on the
microscopic level based on
experiences in the macroscopic
world.
58
Fundamental Properties of Models

A model does not equal reality.

Models are oversimplifications, and are
therefore often wrong.

Models become more complicated as they age.

We must understand the underlying
assumptions in a model so that we don’t
misuse it.
59
Generally, bonds occur when collections of atoms
are more stable (lower in energy) than the separate
atoms
e.g. CH4 is 1652 KJ lower in energy than 1 mole of
C and 4 mole of H.
Stability can be determined in terms if model called
“Chemical Bond”
1652 KJ/mol
C-H bond energy =
= 413 KJ/mol
4
60
The enthalpy change required to break a particular bond in
one mole of gaseous molecules is the bond energy.
Bond Energy
H0 = 436.4 kJ
H2 (g)
H (g) + H (g)
Cl2 (g)
Cl (g) + Cl (g) H0 = 242.7 kJ
HCl (g)
H (g) + Cl (g) H0 = 431.9 kJ
O2 (g)
O (g) + O (g) H0 = 498.7 kJ
O
O
N2 (g)
N (g) + N (g) H0 = 941.4 kJ
N
N
Bond Energies
Single bond < Double bond < Triple bond
61
Using this model, one can determine other bond energies.
CH3Cl is composed of 3 C-H bond and 1 C-Cl bond
C-H bond is known 413 KJ/mol
C-Cl bond energy = 1572 – 3(413) = 339 KJ/mol
62
Average bond energy in polyatomic molecules
H2O (g)
OH (g)
H (g) + OH (g) H0 = 502 kJ
H (g) + O (g)
H0 = 427 kJ
502 + 427
= 464 kJ
Average OH bond energy =
2
63
64
Bond Energy and Bond Length
Note : The shorter the bond the higher the bond energy
65
Bond Energies and Enthalpy of
Reaction
Bond breaking requires energy (endothermic).
Bond formation releases energy (exothermic).
H = D(bonds broken)  D(bonds formed)
Energy required
Reactants
Energy released
Products
66
Use bond energies to calculate the enthalpy change for:
H2 (g) + F2 (g)
2HF (g)
H0 = BE(reactants) – BE(products)
Type of
bonds broken
H
H
F
F
Type of
bonds formed
H
F
Number of
bonds broken
Bond energy
(kJ/mol)
Energy
change (kJ)
1
1
436.4
156.9
436.4
156.9
Number of
bonds formed
Bond energy
(kJ/mol)
Energy
change (kJ)
2
568.2
1136.4
H0 = 436.4 + 156.9 – 2 x 568.2 = -543.1 kJ
67
9.10
Localized Electron Model
A molecule is composed of atoms
that are bound together by sharing
pairs of electrons using the atomic
orbitals of the bound atoms.
68
Localized Electron Model
1.
Description of valence electron
arrangement (Lewis structure).
2.
Prediction of geometry (VSEPR model).
3.
Description of atomic orbital types used
to share electrons or hold lone pairs.
69
Lewis Structure
 Shows
how valence electrons are arranged
among atoms in a molecule.
 Reflects
central idea that stability of a
compound relates to noble gas electron
configuration.
70
Writing Lewis Structures
1. Draw skeletal structure of compound showing
what atoms are bonded to each other. Put least
electronegative element in the center.
2. Count total number of valence e-. Add 1 for
each negative charge. Subtract 1 for each
positive charge.
3. Complete an octet for all atoms except
hydrogen
4. If structure contains too many electrons, form
double and triple bonds on central atom as
needed.
71
Write the Lewis structure of nitrogen trifluoride (NF3).
Step 1 – N is less electronegative than F, put N in center
Step 2 – Count valence electrons N - 5 (2s22p3) and F - 7 (2s22p5)
5 + (3 x 7) = 26 valence electrons
Step 3 – Draw single bonds between N and F atoms and complete
octets on N and F atoms.
Step 4 - Check, are # of e- in structure equal to number of valence e- ?
3 single bonds (3x2) + 10 lone pairs (10x2) = 26 valence electrons
Actual in 3D
F
N
F
F
72
Lewis structure of F2
single covalent bond
lone pairs
F
F
lone pairs
single covalent bond
lone pairs
F F
lone pairs
73
Lewis structure of water
H
+
O +
H
single covalent bonds
H O H
or
H
O
H
2e-8e-2eDouble bond – two atoms share two pairs of electrons
O C O
or
O
O
C
double bonds
- 8e8e- 8ebonds
double
Triple bond – two atoms share three pairs of electrons
N N
triple
bond
8e-8e
or
N
N
triple bond
74
Write the Lewis structure of the carbonate ion (CO32-).
Step 1 – C is less electronegative than O, put C in center
Step 2 – Count valence electrons C - 4 (2s22p2) and O - 6 (2s22p4)
-2 charge – 2e4 + (3 x 6) + 2 = 24 valence electrons
Step 3 – Draw single bonds between C and O atoms and complete
octet on C and O atoms.
Step 4 - Check, are # of e- in structure equal to number of valence e- ?
3 single bonds (3x2) + 10 lone pairs (10x2) = 26 valence electrons
Step 5 - Too many electrons, form double bond and re-check # of e-
O
C
O
O
2 single bonds (2x2) = 4
1 double bond = 4
8 lone pairs (8x2) = 16
Total = 24
75
Carbon Dioxide CO2
C
..
O
..
C
Octet
No octet yet
..
O
..
O
BUT
Do not Break
Symmetry
C
..
O
..
76
..
..
O
..
..
..
..
O
..
or
Exceptions to the Octet Rule
The Incomplete Octet
BeH2
BF3
B – 3e3F – 3x7e24e-
Be – 2e2H – 2x1e4e-
F
B
H
F
Be
H
3 single bonds (3x2) = 6
9 lone pairs (9x2) = 18
Total = 24
F
77
Exceptions to the Octet Rule
Odd-Electron Molecules
NO
N – 5eO – 6e11e-
N
O
The Expanded Octet (central atom with principal quantum number n > 2)
SF6
S – 6e6F – 42e48e-
F
F
F
S
F
F
F
6 single bonds (6x2) = 12
18 lone pairs (18x2) = 36
Total = 48
78
Comments About the Octet Rule

2nd row elements C, N, O, F observe the octet
rule.

2nd row elements B and Be often have fewer than
8 electrons around themselves - they are very
reactive (BF3, BeH2, etc..).

3rd row and heavier elements CAN exceed the
octet rule using empty valence d orbitals (SF6,
PCl5, etc.).

When writing Lewis structures, satisfy octets first,
then place electrons around elements having
available d orbitals.
79
Note:
For molecules that has several Third row (or
higher) elements the extra electrons should be
placed on the central atom.
-
I
..
I
..
..
..
I
..
..
..
..
I3
80
Resonance
Occurs when more than one valid Lewis structure can be
written for a particular molecule.
These are resonance structures. The actual structure is an
average of the resonance structures. The value of the resonance
bond is in between a single bond and double bond.
C=C
>
Bond Energy
….
C C
>
C-C
81
Resonance Structure of Ozone: O3
O
O
+
-
-
O
O
+
O
O
What are the resonance structures of the
carbonate (CO32-) ion?
-
O
C
O
O
-
O
C
O
O
-
-
-
O
C
O
O
82
How the charges in the previous examples
were depicted?
Calculate the Formal charge on each atom.
83
Formal Charge
An atom’s formal charge is the difference between the
number of valence electrons in an isolated atom and the
number of electrons assigned to that atom in a Lewis
structure.
formal charge
on an atom in
a Lewis
structure
=
total number
total number
of valence
of nonbonding
electrons in
electrons
the free atom
-
1
2
(
total number
of bonding
electrons
)
The sum of the formal charges of the atoms in a molecule
or ion must equal the charge on the molecule or ion.
84
Two possible skeletal structures
of formaldehyde
(CH2O)
H
H
C
O
H
H
C
O
Calculate and minimize Formal Charges
85
H
-1
+1
C
O
formal charge
on an atom in
a Lewis
structure
H
=
C – 4 eO – 6 e2H – 2x1 e12 e-
2 single bonds (2x2) = 4
1 double bond = 4
2 lone pairs (2x2) = 4
Total = 12
total number
total number
of valence
of nonbonding
electrons in electrons
the free atom
-
1
2
(
total number
of bonding
electrons
)
formal charge
= 4 -2 -½ x 6 = -1
on C
formal charge
= 6 -2 -½ x 6 = +1
on O
86
H
H
0
C
formal charge
on an atom in
a Lewis
structure
0
O
=
C – 4 eO – 6 e2H – 2x1 e12 e-
2 single bonds (2x2) = 4
1 double bond = 4
2 lone pairs (2x2) = 4
Total = 12
total number
total number
of valence
of nonbonding
electrons in electrons
the free atom
-
1
2
(
total number
of bonding
electrons
)
formal charge
= 4 - 0 -½ x 8 = 0
on C
formal charge
= 6 -4 -½ x 4 = 0
on O
87
Formal Charge and Lewis Structures
1. For neutral molecules, a Lewis structure in which there
are no formal charges is preferable to one in which
formal charges are present.
2. Lewis structures with large formal charges are less
plausible than those with small formal charges.
3. Among Lewis structures having similar distributions of
formal charges, the most plausible structure is the one in
which negative formal charges are placed on the more
electronegative atoms.
Which is the most likely Lewis structure for CH2O?
H
-1
+1
C
O
H
H
H
0
C
0
O
88
Formal Charge
O C O
(-1)
(0)
(+1)
Not as good
O C O
(0)
(0)
(0)
Better
Charges are minimized
89
VSEPR Model
The structure around a given
atom is determined principally
by minimizing electron pair
repulsions.
90
Predicting a VSEPR Structure
1.
2.
3.
4.
Draw Lewis structure.
Put pairs as far apart as possible.
Determine positions of atoms from the
way electron pairs are shared.
Determine the name of molecular
structure from positions of the atoms.
91
92
Valence shell electron pair repulsion (VSEPR) model:
Predict the geometry of the molecule from the electrostatic
repulsions between the electron (bonding and nonbonding) pairs.
Class
# of atoms
bonded to
central atom
# lone
pairs on
central atom
Arrangement of
electron pairs
Molecular
Geometry
AB2
2
0
linear
linear
B
B
93
Cl
Be
Cl
2 atoms
0 lone bonded
pairs ontocentral
centralatom
atom
94
VSEPR
Class
# of atoms
bonded to
central atom
# lone
pairs on
central atom
Arrangement of
electron pairs
Molecular
Geometry
AB2
2
0
linear
linear
0
trigonal
planar
trigonal
planar
AB3
3
95
96
VSEPR
Class
# of atoms
bonded to
central atom
# lone
pairs on
central atom
Arrangement of
electron pairs
Molecular
Geometry
AB2
2
0
linear
linear
trigonal
planar
tetrahedral
AB3
3
0
trigonal
planar
AB4
4
0
tetrahedral
97
98
99
VSEPR
Class
# of atoms
bonded to
central atom
# lone
pairs on
central atom
Arrangement of
electron pairs
Molecular
Geometry
AB2
2
0
linear
linear
trigonal
planar
AB3
3
0
trigonal
planar
AB4
4
0
tetrahedral
tetrahedral
AB5
5
0
trigonal
bipyramidal
trigonal
bipyramidal
100
101
VSEPR
Class
# of atoms
bonded to
central atom
# lone
pairs on
central atom
Arrangement of
electron pairs
Molecular
Geometry
AB2
2
0
linear
linear
trigonal
planar
AB3
3
0
trigonal
planar
AB4
4
0
tetrahedral
tetrahedral
AB5
5
0
trigonal
bipyramidal
trigonal
bipyramidal
AB6
6
0
octahedral
octahedral
102
103
Molecular structure of PCl6
-
Octahedral Geometry
104
Octahedral
electron
arrangement
for Xe
105
VSEPR
Class
# of atoms
bonded to
central atom
# lone
pairs on
central atom
AB3
3
0
AB2E
2
1
Arrangement of
electron pairs
Molecular
Geometry
trigonal
planar
trigonal
planar
trigonal
planar
bent
106
VSEPR
Class
# of atoms
bonded to
central atom
# lone
pairs on
central atom
AB4
4
0
AB3E
3
1
Arrangement of
electron pairs
Molecular
Geometry
tetrahedral
tetrahedral
tetrahedral
trigonal
pyramidal
107
VSEPR
Class
# of atoms
bonded to
central atom
# lone
pairs on
central atom
AB4
4
0
Arrangement of
electron pairs
Molecular
Geometry
tetrahedral
tetrahedral
AB3E
3
1
tetrahedral
trigonal
pyramidal
AB2E2
2
2
tetrahedral
bent
O
H
H
V-Shape
108
109
bonding-pair vs. bonding
pair repulsion
< lone-pair vs. bonding
pair repulsion
<
lone-pair vs. lone pair
repulsion
110
VSEPR
Class
AB5
AB4E
# of atoms
bonded to
central atom
5
4
# lone
pairs on
central atom
Arrangement of
electron pairs
Molecular
Geometry
0
trigonal
bipyramidal
trigonal
bipyramidal
1
trigonal
bipyramidal
distorted
tetrahedron
See-Saw
111
VSEPR
Class
AB5
# of atoms
bonded to
central atom
5
# lone
pairs on
central atom
0
AB4E
4
1
AB3E2
3
2
Arrangement of
electron pairs
Molecular
Geometry
trigonal
bipyramidal
trigonal
bipyramidal
trigonal
bipyramidal
trigonal
bipyramidal
distorted
tetrahedron
T-shaped
F
F
Cl
F
112
VSEPR
Class
AB5
# of atoms
bonded to
central atom
5
# lone
pairs on
central atom
0
AB4E
4
1
AB3E2
3
2
AB2E3
2
3
Arrangement of
electron pairs
Molecular
Geometry
trigonal
bipyramidal
trigonal
bipyramidal
trigonal
bipyramidal
trigonal
bipyramidal
distorted
tetrahedron
trigonal
bipyramidal
T-shaped
linear
I
I
113
I
Three possible arrangements of the
electron pairs in the I3 ion.
Least Lone
pair repulsions
114
VSEPR
Class
# of atoms
bonded to
central atom
# lone
pairs on
central atom
AB6
6
0
octahedral
octahedral
AB5E
5
1
octahedral
square
pyramidal
F
F
F
Arrangement of
electron pairs
Molecular
Geometry
Br
F
F
115
VSEPR
Class
# of atoms
bonded to
central atom
# lone
pairs on
central atom
AB6
6
0
octahedral
octahedral
AB5E
5
1
octahedral
AB4E2
4
2
octahedral
square
pyramidal
square
planar
Arrangement of
electron pairs
Molecular
Geometry
F
F
Xe
F
F
116
Possible electron-pair arrangements
for XeF4.
117
Predicting Molecular Geometry
1. Draw Lewis structure for molecule.
2. Count number of lone pairs on the central atom and
number of atoms bonded to the central atom.
3. Use VSEPR to predict the geometry of the molecule.
What are the molecular geometries of SO2 and SF4?
O
S
AB2E
bent
F
O
F
S
F
AB4E
F
distorted
tetrahedron
118
The molecular structure of
methanol CH3OH.
(a) The arrangement of
electron pairs and atoms
around the carbon atom.
(b) The arrangement of
bonding and lone pairs around
the oxygen atom.
(c) The molecular structure.
119
Links
http://www.molecules.org/VSEPR_table_c.html
http://www.molecules.org/VSEPR_table.html
120
QUESTION
Which of the following statements is incorrect?
1) Ionic bonding results from the transfer of electrons
from one atom to another.
2) Dipole moments result from the unequal
distribution of electrons in a molecule.
3) The electrons in a polar bond are found nearer to
the more electronegative element.
4) A molecule with very polar bonds can be nonpolar.
5) Linear molecules cannot have a net dipole
moment.
121
ANSWER
5)
Linear molecules cannot have a net dipole
moment.
Section 8.1 Types of Chemical Bonds (p. 348)
A linear molecule with two different atoms will
have a dipole moment.
122
QUESTION
Atoms having greatly differing electronegativities
are expected to form:
1) no bonds.
2) polar covalent bonds.
3) nonpolar covalent bonds.
4) ionic bonds.
5) covalent bonds.
123
ANSWER
4)
ionic bonds.
Section 8.2 Electronegativity (p. 352)
If two atoms have greatly differing
electronegativities the more electronegative
atom will pull on the bonding electrons so
strongly the electrons will transfer from one atom
to the other.
124
QUESTION
Choose the compound with the most ionic bond.
1) LiCl
2) KF
3) NaCl
4) LiF
5) KCl
125
ANSWER
2)
KF
Section 8.2 Electronegativity (p. 353)
Use the electronegativity chart (Figure 8.3) to
determine the greatest difference in
electronegativities of atoms in the binary ionic
compounds.
126
QUESTION
For the elements Rb, F, and O, the order of
increasing electronegativity is:
1) Rb < F < O.
2) Rb < O < F.
3) O < F < Rb.
4) F < Rb < O.
5) none of these.
127
ANSWER
2)
Rb < O < F.
Section 8.1 Types of Chemical Bonds (p. 352)
Electronegativities increase moving up a column
and to the right in the periodic table.
128
QUESTION
Based on electronegativities, which of the
following would you expect to be most ionic?
1) N2
2) CaF2
3) CO2
4) CH4
5) CF4
129
ANSWER
2)
CaF2
Section 8.1 Types of Chemical Bonds (p. 352)
CaF2 contains a metal and nonmetal part. This
must be an ionic compound.
130
QUESTION
The electron pair in a C-F bond could be
considered:
1) closer to C because carbon has a larger radius
and thus exerts greater control over the shared
electron pair.
2) closer to F because fluorine has a higher
electronegativity than carbon.
3) closer to C because carbon has a lower
electronegativity than fluorine.
4) an inadequate model since the bond is ionic.
5) centrally located directly between the C and F.
131
ANSWER
2)
closer to F because fluorine has a higher
electronegativity than carbon.
Section 8.1 Types of Chemical Bonds (p. 352)
The fluorine is more electronegative than carbon
and will draw the bonding electrons closer to
itself on average.
132
QUESTION
How many of the following molecules possess
dipole moments?
BH3, CH4, PCl5, H2O, HF, H2
1)
2)
3)
4)
5)
1
2
3
4
5
133
ANSWER
2)
2
Section 8.3 Bond Polarity and Dipole Moments
(p. 354)
Among the central atom molecules, only water
has lone pairs distorting its shape into a bent
configuration. HF is a linear molecule with two
atoms with different electronegativities.
134
QUESTION
Which of the following molecules has a dipole
moment?
1) BCl3
2) SiCl4
3) PCl3
4) Cl2
5) none of these
135
ANSWER
3)
PCl3
Section 8.4 Ions (p. 357)
Only PCl3 has a lone pair that forces the bonds
into a trigonal pyramidal configuration.
136
QUESTION
Which of the following has the smallest radius?
–
1) Br
2–
2) S
3) Xe
2+
4) Ca
5) Kr
137
ANSWER
4)
Ca
2+
Section 8.4 Ions (p. 360)
The calcium ion has lost its two outermost
electrons and the remaining electrons are drawn
even closer to the nucleus.
138
QUESTION
Which of these is an isoelectronic series?
+
+
+
+
1) Na , K , Rb , Cs
+
2+
2–
2) K , Ca , Ar, S
+
2+
2–
–
3) Na , Mg , S , Cl
4) Li, Be, B, C
5) None of these
139
ANSWER
2)
+
2+
K , Ca , Ar, S
2–
Section 8.4 Ions (p. 360)
All of the species in an isoelectronic series must
have the same number of electrons with the
same electron configurations.
140
QUESTION
When electrons in a molecule are not found
between a pair of atoms but move throughout
the molecule, this is called:
1) ionic bonding.
2) covalent bonding.
3) polar covalent bonding.
4) delocalization of the electrons.
5) a dipole moment.
141
ANSWER
4)
delocalization of the electrons.
Section 8.1 Types of Chemical Bonds (p. 348)
Delocalization increases the strength of bonds.
Molecules with delocalization tend to be
unusually stable.
142
QUESTION
Which of the following has the smallest radius?
1) F
2) Ne
2–
3) O
2+
4) Mg
+
5) Na
143
ANSWER
4)
Mg
2+
Section 8.4 Ions (p. 360)
The loss of electrons will always decrease the
size of the ion and the addition of electrons will
always increase the size of the ion.
144
QUESTION
Which of the following atoms cannot exceed the
octet rule in a molecule?
1) N
2) S
3) P
4) I
5) All of the atoms (1–4) can exceed the octet
rule.
145
ANSWER
1)
N
Section 8.11 Exceptions to the Octet Rule (p. 380)
Nitrogen does not have d orbitals to allow extra
bonds to go beyond the octet rule.
146
QUESTION
Choose the electron dot formula that most
accurately describes the bonding in CS2. (Hint:
Consider
formal
charges.) ..
..
..
..
1) :S  C  S:
2) :C  S  S:
..
..
.. .. ..
3) :S
4) :S
..  C  S:
..
..  C  S:
..
..
5) :S
..  C  S:
147
ANSWER
1)
..
..
:S  C  S:
Section 8.12 Resonance (p. 383)
The correct structure follows the octet rule and
also has the highest symmetry of the group. It’s
atoms all have a formal charge of zero as well.
148
QUESTION
In the Lewis structure for elemental nitrogen
there is (are):
1) a single bond between the nitrogens.
2) a double bond between the nitrogens.
3) a triple bond between the nitrogens.
4) three unpaired electrons.
5) none of these.
149
ANSWER
3)
a triple bond between the nitrogens.
Section 8.10 Lewis Structures (p. 375)
Only a triple bond will allow both atoms to
complete their octet.
150
QUESTION
In the reaction between magnesium and sulfur,
the magnesium atoms:
1) become anions.
2) become cations.
3) become part of polyatomic ions.
4) share electrons with sulfur.
151
ANSWER
2)
become cations.
Section 8.5 Formation of Binary Ionic
Compounds (p. 362)
Magnesium is a metal. Metals will lose electrons
when forming ionic compounds with nonmetals.
152
QUESTION
In the Lewis structure for SF6, the central sulfur
atom shares __________ electrons.
1) 4
2) 8
3) 10
4) 12
5) none of these, because SF6 is an ionic
compound
153
ANSWER
4)
12
Section 8.11 Exceptions to the Octet Rule
(p. 380)
Fluorine is so electronegative its atoms will only
form single bonds. Therefore the sulfur atom
must have six bonds, one to each F atom.
154
QUESTION
Which of the following Lewis structures best
describes BF3?
1)
2)
155
QUESTION
(continued)
3)
4)
156
QUESTION
(continued)
5)
157
ANSWER
Section 8.11 Exceptions to the Octet Rule
(p. 380)
1)
The structure whose
atoms have the lowest
formal charges is the
most stable molecule.
158
QUESTION
The bond angle in H2Se is about:
1) 120.
2) 60.
3) 180.
4) 109.
5) 90.
159
ANSWER
4)
109.
Section 8.13 The VSEPR Model (p. 389)
The electron pair arrangement around selenium
is tetrahedral. Two of the electron pairs are
bonding pairs, so the angle of H-Se-H is 109.
160
QUESTION
According to VSEPR theory, which of the
following species has a square planar molecular
structure?
1) TeBr4
2) BrF3
3) IF5
4) XeF4
5) SCl2
161
ANSWER
4)
XeF4
Section 8.13 The VSEPR Model (p. 391)
XeF4 has four bonding pairs of electrons, with
two nonbonding pairs keeping the bonds set in a
plane.
162