Chapter Eight
BONDING: GENERAL
CONCEPTS
講義
Assignment
21, 29, 43, 57, 73, 81, 91, 94, 99
Chapter 8 | Slide 2
Chemical bonds
˙Hold groups of atoms together
˙Occur when a group of atoms can lower its total energy by aggregating
˙Types of chemical bonds
˙Ionic: electrons are transferred to form ions
˙Covalent: equal sharing of electrons
˙Polar covalent: unequal electron sharing
˙Percent ionic character of a bond X─Y:
Measured dipole moment of X─Y
100%
Calculated dipole moment for X+ Y
˙Electronegativity: the relative ability of an atom to attract shared electrons
˙The polarity of a bond depends on the electronegativity difference of the bonded atoms
˙The spacial arrangement of polar bonds in a molecule determines whether the molecule has a dipole moment
Ionic bonding
˙An ion has a different size than its parent atom
˙An anion is larger than its parent ion
˙A cation is smaller htan its parent atom
˙Lattice energy: the change in energy when ions are packed together to form an ionic solid
Bond energy
˙ The energy necessary to break a covalent bond
˙Increases as the number of shared pairs increases
˙Can be used to estimate the enthalpy change for a chemical reaction
Lewis structures
˙Show how the valence electron pairs are arranged among the atoms in a molecule or polyatomic ion
˙Stable molecules usually contain atoms that have their valence orbitals filled
˙Leads to a duet rule for hydrogen
˙Leads to an octet rule for second-row elements
˙The atoms of elements in the third row and beyond can exceed the octet rule
˙Several equivalent Lewis structures can be drawn for some molecules, a concept called resonance
˙When several nonequivalent Lewis structures can be drawn for a molecule, formal charge is often used to choose the most appropriate structure(s)
VSEPR model
˙Based on the idea that electron pairs will be arranged around a central atom in a way that minimizes the electron repulsions
˙Can be used to predict the geometric structure of most molecules
Chapter 8 | Slide 3
Chapter 8 | Slide 4
How atoms are linked together?
A+BAB
EA+EB>EAB
Generally,
An+Bm+Ck+…AnBmCk…
Ionic bonds
Covalence bonds
Chapter 8 | Slide 5
Questions to Consider
•
•
•
What is meant by the term “chemical
bond”?
Why do atoms bond with each other to
form compounds?
How do atoms bond with each other to
form compounds?
Chapter 8 | Slide 6
A Chemical Bond
•
•
•
No simple way to define this.
Forces that hold groups of atoms
together and make them function as a
unit.
A bond will form if the energy of the
aggregate is lower than that of the
separated atoms.
8.1
Chapter 8 | Slide 7
The Interaction
of Two
Hydrogen
Atoms
8.1
Chapter 8 | Slide 8
The Interaction of Two Hydrogen Atoms
8.1
Chapter 8 | Slide 9
Key Ideas in Bonding
•
•
•
Ionic Bonding – electrons are
transferred
Covalent Bonding – electrons are
shared equally
What about intermediate cases?
8.1
Chapter 8 | Slide 10
Polar Covalent Bond
•
•
Unequal sharing of electrons between
atoms in a molecule.
Results in a charge separation in the
bond (partial positive and partial
negative charge).
8.1
Chapter 8 | Slide 11
The Effect of an Electric Field on
Hydrogen Fluoride Molecules
Chapter 8 | Slide 12
Polar Molecules
8.1
Chapter 8 | Slide 13
Concept Check
• What is meant by the term “chemical bond?”
• Why do atoms bond with each other to form
molecules?
• How do atoms bond with each other to form
molecules?
8.1
Chapter 8 | Slide 14
Electronegativity
The ability of an atom in a molecule to attract
shared electrons to itself.
On the periodic table, electronegativity generally
increases across a period and decreases down
a group.
(H-X)expected =
(H-H bond energy)+(X-X bond energy)
2
Δ=(H-X)actual -(H-X)expected
(Linus Pauling)
8.2
Chapter 8 | Slide 15
The Pauling Electronegativity Values
8.2
Chapter 8 | Slide 16
Concept Check
• If lithium and fluorine react, which has more
attraction for an electron? Why?
• In a bond between fluorine and iodine, which has
more attraction for an electron? Why?
8.2
Chapter 8 | Slide 17
Concept Check
• What is the general trend for electronegativity
across rows and down columns on the periodic
table?
• Explain the trend.
8.2
Chapter 8 | Slide 18
Bigger difference in electronegativity
between bonding partners means larger polarity
8.2
Chapter 8 | Slide 19
Bigger difference in electronegativity
between bonding partners means larger polarity
Bond Polarity:
Electronegativity:
Electronegativity
Difference:
H-H < S-H < Cl-H < O-H < F-H
(2.1)(2.1) (2.5)(2.1) (3.0)(2.1) (3.5) (2.1) (4.0) (2.1)
0
0.5
1.1
1.4
Polarity increases
Covalent bond  polar covalent bond
Chapter 8 | Slide 20
1.9
Exercise
Arrange the following bonds from most to least
polar:
a)
N-F
O-F
C-F
b)
C-F
N-O
Si-F
c)
Cl-Cl
B-Cl
S-Cl
8.2
Chapter 8 | Slide 21
Concept Check
Which of the following bonds would be the least
polar yet still be considered polar covalent?
Mg-O C-O O-O Si-O N-O
8.2
Chapter 8 | Slide 22
Concept Check
Which of the following bonds would be the most
polar without being considered ionic?
Mg-O C-O O-O Si-O N-O
8.2
Chapter 8 | Slide 23
Dipole Moment
•
•
Property of a molecule whose charge
distribution can be represented by a
center of positive charge and a center
of negative charge.
Use an arrow to represent a dipole
moment.
– Point to the negative charge center with
the tail of the arrow indicating the positive
center of charge.
8.3
Chapter 8 | Slide 24
Figure 8.4 An Electrostatic Potential
Map of HF
Chapter 8 | Slide 25
1 Debye
r=100 pm (=1 °A)
+e
-e
μ=1 D
Chapter 8 | Slide 26
Polar Bonds Forming Polar Molecules
Chapter 8 | Slide 27
Polar Bonds Forming Nonpolar Molecules
Total dipole moment = 0
Partial negative charge
Chapter 8 | Slide 28
Partial negative charge
Partial positive charge
Charge distribution of CO2
Chapter 8 | Slide 29
The shape of a molecule governs whether
it is polar or not.
Chapter 8 | Slide 30
Chapter 8 | Slide 31
Dipole Moment
8.3
Chapter 8 | Slide 32
Figure 8.6 a-c The Structure and
Charge Distribution of the Ammonia
Molecule
Chapter 8 | Slide 33
No Net Dipole Moment
(Dipoles Cancel)
8.3
Chapter 8 | Slide 34
e.p. Diagram HCL
Chapter 8 | Slide 35
e.p.Diagram SO3
Chapter 8 | Slide 36
e.p. Diagram CH4
Chapter 8 | Slide 37
e.p. Diagram H2S
Chapter 8 | Slide 38
Table 8.2 Types of Molecules with Polar
Bonds but No Resulting Dipole Moment
Chapter 8 | Slide 39
Linear Molecules with Two Identical
Bonds
Chapter 8 | Slide 40
Planar Molecules with Three Identical
Bonds 120 Degrees Apart
Chapter 8 | Slide 41
Tetrahedral Molecules with Four
Identical Bonds 109.5 Degrees Apart
Chapter 8 | Slide 42
Nonpolar
Chapter 8 | Slide 43
Polar
Chapter 8 | Slide 44
Molecular shape and polarity
Chapter 8 | Slide 45
Chapter 8 | Slide 46
BF3
Chapter 8 | Slide 47
SF4
Chapter 8 | Slide 48
SF6
..
:F:
|
..
..
:F ..
..
S
:F
..
|
..
:F
..
Chapter 8 | Slide 49
:F
..
..
-F:
..
A Bauxite Mine
Chapter 8 | Slide 50
Lithium Fluoride
Chapter 8 | Slide 51
Stable Compounds
•
Atoms in stable compounds usually
have a noble gas electron configuration.
8.4
Chapter 8 | Slide 52
Isoelectronic Series
•
A series of ions/atoms containing the
same number of electrons.
O2-, F-, Ne, Na+, Mg2+, and Al3+
8.4
Chapter 8 | Slide 53
Concept Check
Choose an alkali metal, an alkaline metal, a noble
gas, and a halogen so that they constitute an
isoelectronic series when the metals and halogen
are written as their most stable ions.
–
–
–
–
–
What is the electron configuration for each species?
Determine the number of electrons for each species.
Determine the number of protons for each species.
Rank the species according to increasing radius.
Rank the species according to increasing ionization energy.
8.4
Chapter 8 | Slide 54
Ionic Radii
8.4
Chapter 8 | Slide 55
We can “read” from the periodic table:
• Trends for:
– Atomic size, ion radius, ionization energy,
electronegativity
• Electron configurations
• Formula prediction for ionic compounds
• Covalent bond polarity ranking
8.4
Chapter 8 | Slide 56
Table 8.3 Common Ions with Noble
Gas Configurations in Ionic Compounds
Chapter 8 | Slide 57
Figure 8.8 Sizes of Ions Related to
Positions of the Elements on the
Periodic Table
Chapter 8 | Slide 58
A Chemical Reaction May Involve Many Steps:
+
sublimation (atomization)
ionization
+
dissociation of gas (atomization)
-
formation of anion
(affinity)
formation of solid
Chapter 8 | Slide 59
-
+
+
+
- + -
Figure 8.9 The Energy Changes Involved in the Formation
of Lithium Fluoride from Its Elements
dissociation of gas (atomization)
-
formation of anion
(affinity)
ionization
+
sublimation (atomization)
formation of solid
+
- +
+ +
+
The overall reaction
Chapter 8 | Slide 60
Li(s) + ½F2 (g)  LiF (s)
Figure 8.10
a & b The
Structure of
Lithium
Fluoride
Chapter 8 | Slide 61
Lattice Energy
• The change in energy that takes place
when separated gaseous ions are packed
together to form an ionic solid.
Lattice energy = k(Q1Q2/r)
8.5
Chapter 8 | Slide 62
A two-dimensional slice of an ionic crystal. The greater the distance between two ions, the
weaker their attractions. Therefore, the strongest attractions to an ion are those of the adjacent
ions of opposite charge. The blue circles denote the distances over which the two closest
repulsions occur, and the yellow circles denote the two closest attractions.
Vij 
qi q j
Chapter 8 | Slide 63
rij
N N
N N
N N
j 1 i 1
j 1 i 1
j 1 i 1
E  12 [ | Vij |   | Vij |   | Vij |]
Born-Haber Cycle for NaCl
8.5
Chapter 8 | Slide 64
Considerable energy is needed to produce cations and anions from neutral gas-phase
atoms: the ionization energy of the metal atoms must be supplied, and it is only partly
recovered from the electron affinity of the nonmetal atoms. The overall lowering of energy
that drags the ionic solid into existence is due to the strong attraction between cations
and anions in the solid. It takes 145 kJ/mol to produce the ions from the elements, and the
solid compound is 787 kJ/mol lower in energy than the separated ions.
Chapter 8 | Slide 65
Formation of an Ionic Solid
1.Sublimation of the solid metal
• M(s)  M(g)
[endothermic]
2.Ionization of the metal atoms
• M(g)  M+(g) + e
[endothermic]
3.Dissociation of the nonmetal
•
1/2X (g)
2
 X(g)
[endothermic]
4. Formation of X ions in the gas phase:
•
X(g) + e  X(g)
[exothermic]
5. Formation of the solid MX
• M+(g) + X(g)  MX(s) [quite exothermic]
Chapter 8 | Slide 66
8.5
Figure 8.11
Comparison of
the Energy
Changes
Involved in the
Formation of
Solid Sodium
Fluoride and
Solid Magnesium
Oxide
-
+
+
- ++
+
+
The overall reaction
Chapter 8 | Slide 67
This sequence of images illustrates why ionic solids are brittle. (a) The original solid
consists of an orderly array of cations and anions. (b) A hammer blow can push the ions
into positions where cations are next to cations and anions are next to anions; there are
now strong repulsive forces acting (as depicted by the double-headed arrows). (c) As a
result of these repulsive forces, the solid springs apart in fragments. (d) This chunk of
calcite consists of several large crystals joined together. (e) The blow of a hammer has
shattered the crystal, leaving flat, regular surfaces consisting of planes of ions.
Chapter 8 | Slide 68
These micrographs show the porous structure of bone. The calcium in bone is extracted
by the body if the level of calcium in the diet is low. (a) Healthy bone tissue. (b) Bone that
has suffered calcium loss through osteoporosis. The overlay shows the regular
arrangement of the calcium and phosphate ions in healthy bone.
(a)
Chapter 8 | Slide 69
(b)
Partial Ionic Character of
Covalent Bonds
• None of the bonds reaches 100% ionic character even
with compounds that have a large electronegativity
difference (when tested in the gas phase).
Chemical bonds
˙Ionic: electrons are transferred to form ions
˙Covalent: equal sharing of electrons
˙Polar covalent: unequal electron sharing
Percent ionic character of a bond X─Y:
Measured dipole moment of X─Y
100%
+ 
Calculated dipole moment for X Y
8.6
Chapter 8 | Slide 70
Figure 8.12 a-c The
Three Possible Types of
Bonds: (a) covalent
bond, (b) polar covalent
bond with both ionic and
covalent components,
(c) ionic bond, no
electron sharing.
Chapter 8 | Slide 71
Percent ionic character of a bond X─Y:
Measured dipole moment of X─Y
100%
+ 
Calculated dipole moment for X Y
8.6
Chapter 8 | Slide 72
The relationship between the ionic character of a covalent bond
and the electronegativity difference of the bonded atoms
Where are authentic ionic compounds?
Not exactly straight line:
Inexact correlation between
electronegative difference and
ionic character
This diagram is worth your meditation!
Chapter 8 | Slide 73
8.6
Ionic Compound
(GENERAL DEFINITON)
• Any compound that conducts an electric
current when melted will be classified as
ionic.
This definition would avoid the ambiguity arising from problems
such as a polyatomic ion that contains covalent bonds.
If you think a little bit hard, then you’ll find the problem
remains! What is called ‘conducts electric current’?
My offer of solution: scrap ‘ionic compound’ altogether and just
use ionic degree, but who would follow? I just use this example
to show that many problems do NOT have standard answer, the
questions are ALWAYS open. With a deepened understanding,
8.6
Progress is made…that’s all about scientific endeavor.
Chapter 8 | Slide 74
Molten, NaCl Conducts an
Electric Current, Indicating
the Presence of Mobile Na+
and Cl- Ions
Chapter 8 | Slide 75
The Covalent Chemical Bond
What is a chemical bond? That’s a big question.
Chapter 8 | Slide 76
Models
• Models are attempts to explain how
nature operates on the microscopic level
based on experiences in the macroscopic
world.
8.7
Chapter 8 | Slide 77
Fundamental Properties of Models
1. A model does not equal reality.
2. Models are oversimplifications, and are
therefore often wrong.
3. Models become more complicated and
are modified as they age.
4. We must understand the underlying
assumptions in a model so that we don’t
misuse it.
8.7
Chapter 8 | Slide 78
Table 8.4 Average Bond Energies (kJ/mol)
Chapter 8 | Slide 79
Table 8.5 Bond Lengths for Selected Bonds
Chapter 8 | Slide 80
The strengths and lengths of bonds
Chapter 8 | Slide 81
(a) The three normal vibrational modes
of H2O.
Chapter 8 | Slide 82
(b) The four normal vibrational modes
of CO2.
Infrared spectroscopy measures normal
vibrational frequencies
The infrared spectrum of tryptophan (an amino acid).
Chapter 8 | Slide 83
Bond enthalpy ΔHB
H2(g)2H(g)
ΔHo=+436 kJ/mol
ΔHB(H-H)= 436 kJ/mol
ΔHB(F-CF3)= 484 kJ/mol
ΔHB(H-CH3)= 412 kJ/mol
Chapter 8 | Slide 84
Chapter 8 | Slide 85
The bond enthalpies, in kilojoules per mole (kJ/mol), of diatomic
nitrogen, oxygen and fluorine molecules. Note how the bonds
weaken as they change from a triple bond in N2 to a single bond in F2.
Chapter 8 | Slide 86
The bond enthalpies, in kilojoules per mole (kJ/mol), of hydrogen halide
molecules. Note how the bonds weaken as the halogen atom becomes
larger.
F
Cl
Br
I
Chapter 8 | Slide 87
Bond strengths
in diatomic molecules
• The bond strength between two atoms is
measured by the bond enthalpy.
• The bond enthalpy typically increases as
the order of the bond increases, decreases
as the number of lone pairs on neighboring
atoms increases, and decreases as the
atom radius increases.
Chapter 8 | Slide 88
Bond strengths in polyatomic molecules
The bond strength between a given pair of atoms varies slightly.
The average bond enthalpy is a guide to the strength of a bond in
any molecule.
Chapter 8 | Slide 89
Multiple bond strength < bond order * single bond strength
Chapter 8 | Slide 90
Double bonds are shorter than single bonds, leading to
bond strengths larger than double of single-bond strengths.
The pairs of electrons in a multiple bond repel each other and can
weaken the bond. As a result, a double bond between carbon atoms
is not twice as strong as two single bonds would be.
The reason of
Multiple bond strength < bond order * single bond strength
Chapter 8 | Slide 91
The bond enthalpies for bonds between hydrogen and the p-block
elements. The bond strengths decrease from top to bottom of each
group as the atoms increase in size.
Chapter 8 | Slide 92
How to use average bond enthalpies
Average bond enthalpies can be used to estimate reaction
enthalpies and to predict the stability of a molecule.
Chapter 8 | Slide 93
Step 1: decide which bonds are broken
and which formed. Calculate the
change in enthalpy when the bonds
are broken in the reactants.
Step 2: calculate the bond enthalpies
for the new product bonds.
Step 3: calculate the bond formation
enthalpies for the product bonds from
the bond dissociation enthalpies
obtained in step 2 and reversing the
sign.
Step 4: add the enthalpy change required
to break the reactant bonds.
Example of using bond enthalpies
to estimate a reaction enthalpy
• Decide whether the following reaction is
exothermic or endothermic:
CH2CH3I (g) + H2O (g)CH3CH2OH (g) + HI (g)
Bonds broken: C-I (238 kJ/mol) , O-H (463 kJ/mol)
ΔHo=238 + 463 = 701 (kJ/mol)
Bond formed: H-I（299 kJ/mol）, C-O (360 kJ/mol)
ΔHo=299 + 360 = 659 (kJ/mol)
Te reaction enthalpy: ΔHr=701 - 659 = +42 (kJ/mol) endothermic
Chapter 8 | Slide 94
Classroom exercise
Which of the following reactions is
Tables 9.2,3
exothermic?
(a) CCl3CHCl2 (g) + HF(g)CCl3CHClF (g) + HCl (g)
(b) CCl3CHCl2 (g) + HF(g)CCl3CCl2F (g) + H2 (g)
(a) Bonds broken: Cl-C (338 kJ/mol), H-F(565 kJ/mol)
bonds formed: C-F (484 kJ/mol), H-Cl(431 kJ/mol)
(b) Bonds broken: H-C (412 kJ/mol), H-F(565 kJ/mol)
bonds formed: C-F (484 kJ/mol), H-H(436 kJ/mol)
Chapter 8 | Slide 95
Bond Energies
• To break bonds, energy must be added to
the system (endothermic).
• To form bonds, energy is released
(exothermic).
8.8
Chapter 8 | Slide 96
Bond Energies
ΔH = ΣD(bonds broken) – ΣD(bonds formed)
D represents the bond energy per
mole of bonds (always has a positive
sign).
8.8
Chapter 8 | Slide 97
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.
8.9
Chapter 8 | Slide 98
Localized Electron Model
• Electron pairs are assumed to be
localized on a particular atom or in the
space between two atoms:
 Lone pairs – pairs of electrons localized on
an atom
 Bonding pairs – pairs of electrons found in
the space between the atoms
8.9
Chapter 8 | Slide 99
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.
8.9
Chapter 8 | Slide 100
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.
8.10
Chapter 8 | Slide 101
Figure 8.14
G.N. Lewis
Chapter 8 | Slide 102
A Diamond Anvil Cell Used to Study
Materials at Very High Pressures
Chapter 8 | Slide 103
Covalent Bonds
• In covalent bond formation, atoms go as far as
possible toward completing their octets (duplets) by
sharing electron pairs.
Chapter 8 | Slide 104
Figure 8.13 The formation of a covalent bond between two hydrogen atoms. (a) Two
separate hydrogen atoms, each with one electron. (b) The electron cloud that forms when
the spins pair and the orbitals merge is most dense between the two nuclei. (c) The
boundary surface that we shall use to depict a covalent bond.
Chapter 8 | Slide 105
Covalent Bonds
• In covalent bond formation, atoms go as far as
possible toward completing their octets
(duplets) by sharing electron pairs.
H+H  H:H or H-H
..
:
..
..
..

:
. :
:
: or
F F FF
.
..
..
..
That’s why F2 is so reactive.
:
..
..
..
..
..

:
F F
Lone pairs
The Lewis structure gives not only bond positions (shared electron pairs)
But also reactivity-how reactive is the molecules and which part(s) is(are)
reactive (lone pairs).
Chapter 8 | Slide 106
The Structures of Ployatomic Species
H
H

H:C:H

|
H- C -H
|
H
H






:F O F:

:O:
H
H
|
|
|


[H- N-H]+[:O S  O:]2-[H- N-H]+
|


|
|
:O:
H
H

• Single, double, triple bonds may be
Chapter 8 | Slide 107
formed.
How to write Lewis structure of a
polyatomic species
• Count the total number of valence electrons on each atom
and divide by 2 to obtain the number of electron pairs. If the
species is polyatomic ion, add one more electron for each
negative charge or subtract one electron for each positive
charge.
• Write the chemical symbols of the atoms to show their
layout in the molecule. One can predict the most likely
arrangements of atoms by using common patterns and the
rules of thumb (largely correct but exceptions possible)
given earlier.
• Place one electron pair between each pair of bonded
atoms.
• Complete the octet (duplet for H) of each atom by placing
any remaining electron pairs as lone pairs around the
atoms. If there are not enough electrons to form octets,
form multiple bonds.
• Care for the possibility of cations (anions) and resonance.
Chapter 8 | Slide 108
How to write Lewis structure of a polyatomic species:
another summary (from Wikipedia)
•
•
•
•
•
Step one: Nitrogen is the least electronegative atom, so it is the central atom by
multiple criteria.
Step two: Count valence electrons. Nitrogen has 5 valence electrons; each oxygen has 6,
for a total of (6 × 2) + 5 = 17. The ion has a charge of −1, which indicates an extra electron,
so the total number of electrons is 18.
Step three: Place ion pairs. Each oxygen must be bonded to the nitrogen, which uses four
electrons — two in each bond. The 14 remaining electrons should initially be placed as 7 lone
pairs. Each oxygen may take a maximum of 3 lone pairs, giving each oxygen 8 electrons
including the bonding pair. The seventh lone pair must be placed on the nitrogen atom.
Step four: Satisfy the octet rule. Both oxygen atoms currently have 8 electrons assigned to
them. The nitrogen atom has only 6 electrons assigned to it. One of the lone pairs on an
oxygen atom must form a double bond, but either atom will work equally well. We therefore
must have a resonance structure.
Step five: Tie up loose ends. Two Lewis structures must be drawn: one with each oxygen
atom double-bonded to the nitrogen atom. The second oxygen atom in each structure will be
single-bonded to the nitrogen atom. Place brackets around each structure, and add the
charge (−) to the upper right outside the brackets. Draw a double-headed arrow between the
two resonance forms.
Chapter 8 | Slide 109
More Examples
NH3:
|
3+5=8 valence electrons = 4 pairs.
H- N :
|
3 pairs to form three bonds,
H
leaving one lone pair
H




H  O Br :
Hypobromous acid HBrO:
1+6+7=14 valence electrons = 7 pairs
Two pairs to form two bonds,
Leaving 5 lone pairs.
Chapter 8 | Slide 110
:O:
||
H- C
|
: O -H

Formic acid HCOOH
2+12+4=18 valence electrons
= 9 pairs
5 pairs to form five bonds,
leaving 4 lone pairs.
Classroom Exercise
• Write the Lewis structure of CH3COOH
H
:O:
|
||
H- C C
|
|
:
O
-H
H

Chapter 8 | Slide 111
Resonance: Blending of Structures

:O
|| 

[:O N  O:]


:O:
| 

[:O  N  O:]

:O:
| 

[:O N  O:]-
All valid. We cannot find two bond lengths (hypothetical N-O vs N=O)
Chapter 8 | Slide 112
Resonance: Blending of Structures
Chapter 8 | Slide 113
Resonance: Blending of Structures
Chapter 8 | Slide 114
Quiz
Write the Lewis structure of the following compounds:
(1) Formic acid (2) Nitrate (3) Carbon Monoxide
Answer:
:O:
||
H- C
|
: O-H

Chapter 8 | Slide 115

:O
|| 

[:O N  O:]


:O:
| 

[:O  N  O:]

:O:
| 

[:O N  O:]
Electrons in an Atom and in a Molecule:
from a freeman to a slave or otherwise?
• From spherically symmetrical to less or none of symmetry,
from pure to hybrid, from localized to delocalized, from
individual to collective, from neutral to charged, from simple
happy primitive tribe to adversary yet opportune land…
• From freedom to restriction or just opposite? from
independent to correlated or other way around?
Electrons have rich feeling; they try their best to keep the changes
minimized when transferred from an atom to a molecule so that an
atom in a molecule also looks the same as a free atom as possible.
Atoms seem to have some good happy memory about their free, neutral
state—after put in molecules, they tend to stay in their original state of
zero net charge by keeping their valence electrons unchanged (or if
forced
change,
minimizing it).
Chapter 8to
| Slide
116
Formal Charge
• The real number of valence electrons each atom in a molecule
“owns” may be different from the number of normal valence
electrons of that atom, rendering it positively or negatively charged.
Formal charge (FC) = number of valence electrons on the free atom (V)
- number of electrons present as lone pairs (L)
- ½ number of electrons shared in bonds (S/2)
FC  V  L  S
1
2
Formal charge can be understood as surplus or deficit of valence electrons after
deducting lone pairs and shared pairs.
Typically, the most stable Lewis structures are those in which the formal charge
of the individual nonmetal atoms are close to zero
Chapter 8 | Slide 117
Plausibility of a structure






OCO
NO N
0 0 0
-1 +2 -1




OOC





N NO
0 +2 -2
-1 +1 0

0 +2 -2
A molecule is more favored by God than others if more atoms in it
have formal charges closest to zero. An atom feels guilty if it takes
electrons from others and it feels angry if it has to give up electrons to
others.
So atoms are like you, aren’t they?
Chapter 8 | Slide 118
Classroom Exercise
• Suggest a plausible structure for the poisonous
gas phosgene, COCl2.
Resonance
states

:Cl:
0
|

O C

|
:Cl:

0
0
0
0
Chapter 8 | Slide 119
..O
:Cl:
 |
 ||
:Cl C :Cl C
||
..
|
..
:O
:Cl:


0
  ..
:Cl-O - C 0
..  |
:Cl: 0

Formal charges are all zero, too.
But why is this one not favored?
Experiment shows the oxygen has
double bond rather than single bond.
Exceptions to the Octet Rule
•
•
•
•
B, C, N, O, F observe the octet rule.
P, S, Cl can have more than 8 valence electrons.
Radicals: odd-electron species, highly reactive.
Biradicals: with two unpaired electrons.
CH3 -CH3 
 H3C  +  CH3
stress
O2 (g)  2O(g)
sunlight
O+O 2  O3  O+O 2
UV
Chapter 8 | Slide 120








O  O O: :O O  O


Chapter 8 | Slide 121
O2 is a biradical!


OO
Wrong!




O O

Chapter 8 | Slide 122
..

..
.
..
O
..
..
.
2
Classroom Exercise
Write a Lewis structure for the hydrogenperoxyl radical HOO.,
which plays an important role in atmospheric chemistry and
which in the body has been implicated in the degeneration of
neurons.
Chapter 8 | Slide 123
Case Study 8 (a) The equipment in the illustration monitors air quality from a rooftop in
Los Angeles. The concentration of NO2 in the air due to automobile traffic increases
during the day, contributing to the typical brown color of the afternoon sky.
Exhaust: N 2 (g)+O 2 (g)  2NO(g)
+Atmosphere: 2NO(g)+O2 (g)  2NO2 (g)
UV (< 400 nm)
+Sunlight:

NO

 NO 2 +  O  (smog)
Chapter 8 | Slide 124
2
Case Study 8 (b) The gas NO (left) is colorless. When it is exposed to air (right), it is
rapidly oxidized to brown NO2.
Oxygen molecules are
actually biradicals.
2NO(g)+O2 (g)  2NO2 (g)
Chapter 8 | Slide 125
Expanded valence shells
• Nonmetal atoms in Period 3 or higher can
accommodate 10, 12 or more electrons (expanded
valence shell). e.g., P, S, Cl.
P 4 (g)+6Cl2 (g)  PCl3 (l)
limited supply of chlorine
P 4 (g)+10Cl2 (g) 
 4PCl5 (s)
excessive chlorine
Chapter 8 | Slide 126
Figure 8.16 (a) A model using small spheres to represent atoms and
(b) a space-filling model of PCl5, showing how closely the chlorine
atoms must pack around the central phosphorus atom. (c) A nitrogen
atom is significantly smaller than a phosphorus atom, and five
chlorine atoms cannot pack around it.
Chapter 8 | Slide 127
Figure 8.17 Phosphorus trichloride is a colorless liquid. When it
reacts with chlorine (the pale yellow-green gas in the flask), it forms
the very pale yellow solid phosphorus pentachloride (at the bottom of
the flask).
Chapter 8 | Slide 128
Chapter 8 | Slide 129
+
Chapter 8 | Slide 130
Figure 8.18 Two circumstances in which a central atom assumes an expanded
octet. First, there are too many electrons to be accommodated in octets
because either (a) there are too many atoms attached, or (b) the central atom
must accommodate additional lone pairs. (c) Second, a resonance structure
with multiple bonds has a favorable energy.
Chapter 8 | Slide 131
S accommodates 10 electrons
Chapter 8 | Slide 132
How many electrons does Xe
accommodate?
Chapter 8 | Slide 133
Which is most plausible?
Chapter 8 | Slide 134
12 valence electrons!
Chapter 8 | Slide 135
Classroom exercise:
which is more plausible?
Chapter 8 | Slide 136
Answer
Chapter 8 | Slide 137
One more…
Chapter 8 | Slide 138
The unusual structures of group 13
halides
• Compounds of boron and aluminum may
have unusual Lewis structures in which
boron and aluminum have incomplete
octets or in which halogen atoms as
bridges.
Chapter 8 | Slide 139
Incomplete octet
Chapter 8 | Slide 140
Chapter 8 | Slide 141
+
Chapter 8 | Slide 142
F
Coordinate covalent bond
This covalent bond is
formed by two electrons
from nitrogen.
BF3 (g)+NH3 (g)  NH3BF3 (s)
Both electrons shared in the bond come from the same atom.
The coordinate bond is more extended (nonlocal) than an ordinary covalent bond.
Chapter 8 | Slide 143
Coordinate covalent bond
Chapter 8 | Slide 144
Which atom contributes the electrons that form the coordinate covalent bond?
AlCl3 ( s ) 
 Al2 Cl6 (g)
Chapter 8 | Slide 145
sublimes at 180 o C
>200 o C
Which atom contributes the electrons that form the coordinate covalent bond?
AlCl3 ( s ) 
 Al2 Cl6 (g)
Chapter 8 | Slide 146
sublimes at 180 o C
>200 o C
Writing Lewis Structures
1. Sum the valence electrons.
2. Place bonding electrons between pairs
of atoms.
3. Atoms usually have noble gas
configurations. Arrange remaining
electrons to satisfy the octet rule (or
duet rule for hydrogen).
8.10
Chapter 8 | Slide 147
Concept Check
Draw a Lewis structure for each of the following
molecules:
H2
F2
HF
8.10
Chapter 8 | Slide 148
Concept Check
Draw a Lewis structure for each of the following
molecules:
H2O
NH3
8.10
Chapter 8 | Slide 149
Exceptions
• When it is necessary to exceed the octet
rule for one of several third-row (or higher)
elements, place the extra electrons on
the central atom.
8.11
Chapter 8 | Slide 150
Resonance
• More than one valid Lewis structure can
be written for a particular molecule.
• Actual structure is an average of the
resonance structures.
8.12
Chapter 8 | Slide 151
Concept Check
Draw a Lewis structure for each of the
following molecules:
CO
CH3OH
PCl5
SF6
CO2
BF3
NO3-
8.10-8.12
Chapter 8 | Slide 152
Formal Charge
• Nonequivalent Lewis structures.
• Atoms in molecules try to achieve formal
charges as close to zero as possible.
• Any negative formal charges are
expected to reside on the most
electronegative atoms.
8.12
Chapter 8 | Slide 153
Formal Charge
• Formal charge = # valence e– on free
atom – # valence e– assigned to the atom
in the molecule.
• Assume:
– Lone pair electrons belong entirely to the
atom in question
– Shared electrons are divided equally
between the two sharing atoms
8.12
Chapter 8 | Slide 154
VSEPR Model
• VSEPR: Valence Shell Electron-Pair
Repulsion.
• The structure around a given atom is
determined principally by minimizing
electron pair repulsions.
8.13
Chapter 8 | Slide 155
Figure 8.15 The Molecular Structure of
Methane
Chapter 8 | Slide 156
Balloons Tied
Together
Naturally Form
Tetrahedral
Shape
Chapter 8 | Slide 157
Figure 8.16 a-c The Molecular Structure
of Ammonia is a Trigonal Pyramid
Chapter 8 | Slide 158
Figure 8.17 a-c The Tetrahedral
Arrangement of Oxygen In a Water
Molecule
Chapter 8 | Slide 159
Figure 8.18 The Bond Angles In the
CH4, NH3, and H2O Molecules
Chapter 8 | Slide 160
Some of the basic geometrical shapes that are used to
describe the shapes of simple molecules
Chapter 8 | Slide 161
Tetrahedral
Chapter 8 | Slide 162
Octahedral
Shape names and bond angles
(lone pairs not included)
T-shaped
Chapter 8 | Slide 163
VSEPR Model
(Valence-Shell Electron-Pair Repulsion model)
Bonding electrons and lone pairs take
up positions as far from one another as
possible, for then they repel each other
the least.
Chapter 8 | Slide 164
VSEPR Model (1):
locate the high concentrations
The positions taken up by regions of high electron concentration
(green) around a central atom.
Electron pairs or bonds are as far away from one
another as possible and so experience the minimal
repulsion from other electrons
Chapter 8 | Slide 165
VSEPR Model (2): determine
the shapes
Two ‘highs’
Five ‘highs’
Three ‘highs’
Six ‘highs’
Four ‘highs’
Seven ‘highs’
A summary of the positions taken up by regions of high electron
concentration (other atoms and lone pairs) around a central atom.
The locations of these regions are given by straight lines sticking out
of the central atom.
Use this chart to identify the arrangement of a given number of
atoms and lone pairs and then use Fig. 9.2 to identify the shape
of the molecule from the location of its atoms.
Chapter 8 | Slide 166
VSEPR Model: Example
Electron pairs or bonds are as far away from one another as
possible and so experience the minimum repulsion from other
electrons
..
..
:Cl -Be-Cl:
..
..
Shape: linear (two ‘highs’)
Chapter 8 | Slide 167
Shape: trigonal planar
(three ‘highs’)
VSEPR Model: Example
Electron pairs or bonds are as far away from one another as
possible and so experience the minimum repulsion from other
electrons
Shape: trigonal bipyramidal (five ‘highs’)
Chapter 8 | Slide 168
VSEPR Model
• In order to reduce repulsions, bonding
pairs and lone pairs take up positions
around an atom that maximize their
separations. The shape of the molecule
is determined by the location of the
atoms attached to the central atom.
Chapter 8 | Slide 169
Classroom Exercise
Predict the structure of SiCl4 and AsF5
Chapter 8 | Slide 170
Classroom Exercise
Predict the structure of SiCl4 and AsF5
Cl
|
Cl- Si -Cl
|
Si: 3s23p2
F
F
SiCl4
Chapter 8 | Slide 171
F
As
Cl
AsF5
F
F
As: 3d104s24p3
Molecules with Multiple Bonds
The VSEPR model does not distinguish single and multiple
bonds. A multiple bond is treated as just another region of
high electron concentration.
..
..
O =C= O
..
(Two ‘highs’)
Chapter 8 | Slide 172
..
(Three ‘highs’)
The VSEPR model does not distinguish single and multiple bonds. A
multiple bond is treated as just another region of
high electron concentration.
Two central atoms
Chapter 8 | Slide 173
Example
HC  CH
Two central atoms, no lone pairs.
The most possible VSEPR arrangement:
Chapter 8 | Slide 174
Formaldehyde
H
CH2O
|
..
H- C = O
..
CH2
O
Chapter 8 | Slide 175
Classroom Exercise
Predict the structure of HCN
H-C  N :
Chapter 8 | Slide 176
Different resonance structures correspond to a
single shape
Chapter 8 | Slide 177
How about if the central atoms have lone
pairs?
Lone pairs on attached atoms have little effect on molecular shape,
but the lone pairs on central atoms may have significant effect.
• A: central atom
• X: atom bonded to the central atom
• E: lone pair on the central atom
The single nonbonding electron on radicals is treated as a ‘lone pair’.
Chapter 8 | Slide 178
Lone Pair on the Central Atom
AX3E
Chapter 8 | Slide 179
Repulsion Order:
Lone pair/lone pair > lone pair/bonding pair > bonding pair/bonding pair
Chapter 8 | Slide 180
Example
H2O
..
H-O-H
..
AX2E2
Shape: Angular (not tetrahedron!)
Chapter 8 | Slide 181
Three views of water molecular shape
Chapter 8 | Slide 182
NH3
H
|
H- N -H
..
AX3E
Chapter 8 | Slide 183
Triangular pyramid
(NOT tetrahedron!)
Classroom Exercise
Predict the shape of
NO2..  -
..
AEX2
O = N  O:
..
..
..
NO2Smaller
angle
Chapter 8 | Slide 184
Angular (NOT planar triangle!)
AX4E
Axial lone pair
Chapter 8 | Slide 185
Equatorial lone pair
AX4E2
Square planar
(NOT octahedron!)
Chapter 8 | Slide 186
How to Use VSEPR Model
1. Write Lewis structure
and determine the number of
electron pairs
2. Maximize the separations.
3. Decide the positions
of lone pairs (on the central atom).
4a. Name the shape (without
considering the lone pair).
4b. Consider distortion using
repulsion order.
Lone
pair/lone pair > lone pair/bonding pair > bonding pair/bonding pair
Chapter 8 | Slide 187
Example: SF4
AX4E
Equatorial lone pair
Bent seesaw
Chapter 8 | Slide 188
T-shaped
(NOT triangular bipyramid!)
ClF3
..
..
:F:
|
..
:F- Cl - F:
..
..
AX3E
Chapter 8 | Slide 189
..
T-shaped
(NOT triangular bipyramid!)
Classroom Exercise: XeF4
..
:F:
|
..
..
..
:F - .. Xe - F:
..
..
|
:F:
..
Square planar
(NOT octahedron!)
Chapter 8 | Slide 190
AX4E2
Quiz
• Write the VSEPR formula of water and
sulfur tetrafluoride, draw and name their
structures.
Chapter 8 | Slide 191
Answer
• Write the VSEPR formula of water and
sulfur pentafluoride. Draw and name
their structures.
..
H-O-H
..
AX2E2
AX4E
Angular (not tetrahedron!)
Chapter 8 | Slide 192
T-shaped
(not triangular bipyramid!)
Chapter 8 | Slide 193
VSEPR: Two Electron Pairs
8.13
Chapter 8 | Slide 194
VSEPR: Three Electron Pairs
8.13
Chapter 8 | Slide 195
VSEPR: Four Electron Pairs
8.13
Chapter 8 | Slide 196
VSEPR: Iodine Pentafluoride
8.13
Chapter 8 | Slide 197
VSEPR
8.13
Chapter 8 | Slide 198
NH3
Chapter 8 | Slide 199
PH3
Chapter 8 | Slide 200
Predicting a VSEPR Structure
1. Draw Lewis structure.
2. Put electron pairs as far apart as
possible.
3. Determine positions of atoms from the
way electron pairs are shared.
4. Determine the name of molecular
structure from positions of the atoms.
8.13
Chapter 8 | Slide 201
Queen Bee
Chapter 8 | Slide 202
Concept Check
Determine the shape for each of the following
molecules, and include bond angles:
HCN
PH3
SF4
O3
KrF4
8.13
Chapter 8 | Slide 203
Concept Check
True or false:
A molecule that has polar bonds will always be
polar.
-If true, explain why.
-If false, provide a counter-example.
8.13
Chapter 8 | Slide 204
Concept Check
True or false:
Lone pairs make a molecule polar.
-If true, explain why.
-If false, provide a counter-example.
8.13
Chapter 8 | Slide 205
Chemical bonds
˙Hold groups of atoms together
˙Occur when a group of atoms can lower its total energy by aggregating
˙Types of chemical bonds
˙Ionic: electrons are transferred to form ions
˙Covalent: equal sharing of electrons
˙Polar covalent: unequal electron sharing
˙Percent ionic character of a bond X─Y:
Measured dipole moment of X─Y
100%
Calculated dipole moment for X+ Y
Review
˙Electronegativity: the relative ability of an atom to attract shared electrons
˙The polarity of a bond depends on the electronegativity difference of the bonded atoms
˙The spacial arrangement of polar bonds in a molecule determines whether the molecule has a dipole moment
Ionic bonding
˙An ion has a different size than its parent atom
˙An anion is larger than its parent ion
˙A cation is smaller htan its parent atom
˙Lattice energy: the change in energy when ions are packed together to form an ionic solid
Bond energy
˙ The energy necessary to break a covalent bond
˙Increases as the number of shared pairs increases
˙Can be used to estimate the enthalpy change for a chemical reaction
Lewis structures
˙Show how the valence electron pairs are arranged among the atoms in a molecule or polyatomic ion
˙Stable molecules usually contain atoms that have their valence orbitals filled
˙Leads to a duet rule for hydrogen
˙Leads to an octet rule for second-row elements
˙The atoms of elements in the third row and beyond can exceed the octet rule
˙Several equivalent Lewis structures can be drawn for some molecules, a concept called resonance
˙When several nonequivalent Lewis structures can be drawn for a molecule, formal charge is often used to choose the most appropriate structure(s)
VSEPR model
˙Based on the idea that electron pairs will be arranged around a central atom in a way that minimizes the electron repulsions
˙Can be used to predict the geometric structure of most molecules
Chapter 8 | Slide 206
Chapter Eight
Bonding: General
Concepts
案例/討論
Figure 8.1 a & b (a) The Interaction of Two
Hydrogen Atoms (b) Energy Profile as a
Function of the Distance Between the Nuclei of
the Hydrogen Atoms
Chapter 8 | Slide 208
Figure 8.2 The Effect of an Electric
Field on Hydrogen Fluoride Molecules
Chapter 8 | Slide 209
Figure 8.3 The Pauling
Electronegativity Vaules
Chapter 8 | Slide 210
Figure 8.4 An Electrostatic Potential
Map of HF
Chapter 8 | Slide 211
Figure 8.5 a-c The Charge Distribution
in the Water Molecule
Chapter 8 | Slide 212
Figure 8.6 a-c The Structure and
Charge Distribution of the Ammonia
Molecule
Chapter 8 | Slide 213
Figure 8.7 a-c The Carbon Dioxide
Molecule
Chapter 8 | Slide 214
e.p. Diagram HCL
Chapter 8 | Slide 215
e.p.Diagram SO3
Chapter 8 | Slide 216
e.p. Diagram CH4
Chapter 8 | Slide 217
e.p. Diagram H2S
Chapter 8 | Slide 218
Figure 8.8 Sizes of Ions Related to
Positions of the Elements on the
Periodic Table
Chapter 8 | Slide 219
Figure 8.9 The Energy Changes
Involved in the Formation of Lithium
Fluoride from Its Elements
Chapter 8 | Slide 220
Figure 8.10
a & b The
Structure of
Lithium
Fluoride
Chapter 8 | Slide 221
Figure 8.11
Comparison of
the Energy
Changes
Involved in the
Formation of
Solid Sodium
Fluoride and
Solid
Magnesium
Oxide
Chapter 8 | Slide 222
Figure 8.12
a-c The
Three
Possible
Types of
Bonds
Chapter 8 | Slide 223
Figure 8.13 The Relationship Between the
Ionic Character of a Covalent Bond and the
Electronegativity Difference of the Bounded
Atoms
Chapter 8 | Slide 224
Figure 8.15 The Molecular Structure of
Methane
Chapter 8 | Slide 225
Figure 8.16 a-c The Molecular Structure
of Ammonia is a Trigonal Pyramid
Chapter 8 | Slide 226
Figure 8.17 a-c The Tetrahedral
Arrangement of Oxygen In a Water
Molecule
Chapter 8 | Slide 227
Figure 8.18 The Bond Angles In the
CH4, NH3, and H20 Molecules
Chapter 8 | Slide 228
Figure 8.19 a &
b In a Bonding
Pair of
Electrons the
Electrons are
Shared by Two
Nuclei (b) In a
Lone Pair, Both
Electrons Must
Be Close to a
Single Nucleus
Chapter 8 | Slide 229
Figure 8.20 a & b Possible Electron
Pair Arrangements for XeF4
Chapter 8 | Slide 230
Figure 8.21 a-c Three Possible
Arrangements of the Electron Pairs in
the I3- Ion
Chapter 8 | Slide 231
Figure 8.22
a-c The
Molecular
Structure of
Methanol
Chapter 8 | Slide 232
Quartz
Chapter 8 | Slide 233
Linear Molecules with Two Identical
Bonds
Chapter 8 | Slide 234
Planar Molecules with Three Identical
Bonds 120 Degrees Apart
Chapter 8 | Slide 235
Tetrahedral Molecules with Four
Identical Bonds 109.5 Degrees Apart
Chapter 8 | Slide 236
A Bauxite Mine
Chapter 8 | Slide 237
Lithium Fluoride
Chapter 8 | Slide 238
Molten, NaCl
Conducts an
Electric
Current,
Indicating the
Presence of
Mobile Na+
and Cl- Ions
Chapter 8 | Slide 239
Figure 8.14
G.N. Lewis
Chapter 8 | Slide 240
A Diamond Anvil Cell Used to Study
Materials at Very High Pressures
Chapter 8 | Slide 241
Balloons Tied
Together
Naturally
Form
Tetrahedral
Shape
Chapter 8 | Slide 242
Queen Bee
Chapter 8 | Slide 243
NH3
Chapter 8 | Slide 244
PH3
Chapter 8 | Slide 245
Table 8.1 The Relationship Between
Electronegativity and Bond Type
Chapter 8 | Slide 246
Table 8.2 Types of Molecules with Polar
Bonds but No Resulting Dipole Moment
Chapter 8 | Slide 247
Table 8.3 Common Ions with Noble
Gas Configurations in Ionic Compounds
Chapter 8 | Slide 248
Table 8.4 Average Bond Energies
(kj/mol)
Chapter 8 | Slide 249
Table 8.5 Bond Lengths for Selected
Bonds
Chapter 8 | Slide 250
Table 8.6
Arrangements
of Electron
Pairs Around
an Atom
Yielding
Minimum
Repulsion
Chapter 8 | Slide 251
Table 8.7
Structures of
Molecules
that Have
Four Electron
Pairs Around
the Central
Atom
Chapter 8 | Slide 252
Table 8.8
Structures of
Molecules
with Five
Electron Pairs
Around the
Central Atom
Chapter 8 | Slide 253
Chapter Eight
Bonding: General
Concepts
問題
Question
• Which of the following elements forms the
most ionic bond with chlorine?
–K
– Al
–P
– Kr
– Br
Chapter 8 | Slide 255
Answer
•a)K
•Section 8.2, Electronegativity
•The most ionic bond results from the largest dipole
moment and the greatest difference in
electronegativity. With chlorine, potassium has the
greatest difference in electronegativity.
Chapter 8 | Slide 256
Question
•As a general pattern, electronegativity is
inversely related to
–
–
–
–
Chapter 8 | Slide 257
ionization energy.
atomic size.
the polarity of the atom.
the number of neutrons in the nucleus.
Answer
• b) atomic size.
• Section 8.2, Electronegativity
• Electronegativity increases from left to right and
from bottom to top in the periodic table. Atomic
size increases from top to bottom in the periodic
table. Thus electronegativity and atomic size are
inversely related.
Chapter 8 | Slide 258
Question
• Which of the following is the most polar
bond without being considered ionic?
– C—O
– Mg—O
– N—O
– O—O
– O—F
Chapter 8 | Slide 259
Answer
•a) C—O
•Section 8.3, Bond Polarity and Dipole Moments
•Given that O is common to all of the bonds, the relative
difference in electronegativity for C, Mg, F, and N will
determine the polarity. Decreasing polarity results from
lowering the difference in electronegativity. While the
greatest difference in electronegativity is between Mg and O,
this bond is considered to be ionic, not polar.
Chapter 8 | Slide 260
Question
• Which of the following is the least polar
bond, yet still considered polar covalent?
– C—O
– Mg—O
– N—O
– O—O
– O—F
Chapter 8 | Slide 261
Answer
•c) N—O
•Section 8.3, Bond Polarity and Dipole Moments
•Given that O is common to all of the bonds, the relative
difference in electronegativity for C, Mg, F, and N will
determine the polarity. Decreasing polarity results from
lowering the difference in electronegativity. While the
smallest difference in electronegativity is between O and O
(both have the same electronegativity value, so the
difference is zero), this bond is not considered to be polar,
yet is perfectly covalent.
Chapter 8 | Slide 262
Question
• What is the correct order of the following
bonds in terms of decreasing polarity?
– N—Cl, P—Cl, As—Cl
– P—Cl, N—Cl, As—Cl
– As—Cl, N—Cl, P—Cl
– P—Cl, As—Cl, N—Cl
– As—Cl, P—Cl, N—Cl
Chapter 8 | Slide 263
Answer
•e)As—Cl, P—Cl, N—Cl
•Section 8.3, Bond Polarity and Dipole Moments
•Given that Cl is common to all of the bonds, the
relative difference in electronegativity for As, P, and
N will determine the polarity. Decreasing polarity
results from lowering the difference in
electronegativity.
Chapter 8 | Slide 264
Question
• In which case is the bond polarity incorrect?
a) +H—Fb) +K—Oc) +Mg—Hd) +Cl—Ie) +Si—S-
Chapter 8 | Slide 265
Answer
•d)
+Cl—I-
•Section 8.3, Bond Polarity and Dipole Moments
•Bond polarity results from the more electronegative element
pulling the electron density toward it, resulting in a partial
negative charge. Because F, O, H, Cl, and S are more
electronegative than H, K, Mg, I, and Si, respectively, they
will carry the – charge.
Chapter 8 | Slide 266
Question
• Which of the following has the largest
radius?
– S2– Cl– Ar
– K+
– Ca2+
Chapter 8 | Slide 267
Answer
•a)
S2-
•Section 8.4, Ions: Electron Configurations and
Sizes
•All of these ions have similar sizes as atoms, so
the charge determines the largest radius. Adding
electrons increases the size.
Chapter 8 | Slide 268
Question
• Which of the following ionic compounds has the
largest lattice energy (i.e., the lattice energy most
favorable to a stable lattice)?
–
–
–
–
–
CsI
NaCl
LiF
CsF
MgO
Chapter 8 | Slide 269
Answer
•e)MgO
•Section 8.5, Formation of Binary Ionic
Compounds
•Lattice energy is directly proportional to
charge, and the species with the highest
charge is MgO.
Chapter 8 | Slide 270
Question
•
•
•
•
Bond
H—H
F—F
H—F
Average Bond Energy (kJ/mol)
432
154
565
•Given the average bond energies above, estimate ∆H for
the following reaction:
•
H2 + F2  2HF
•
– -21 kJ
– 21 kJ
– 544 kJ
– -544 kJ
– 1151 kJ
Chapter 8 | Slide 271
Answer
•d)
–544 kJ
•Section 8.8, Covalent Bond Energies and
Chemical Reactions
•∆H = [432 kJ + 154 kJ] – 2(565 kJ) = –544 kJ
Chapter 8 | Slide 272
Question
• Which of the following does not contain at
least one double bond in the Lewis
structure?
– H2CO
– C2H4
– CO2
– C3H8
Chapter 8 | Slide 273
Answer
•d)
C 3H 8
•Section 8.10, Lewis Structures
•Choice (a) contains a C=O double bond,
choice (b) contains a C=C double bond, and
choice (c) contains two C=O double bonds.
Chapter 8 | Slide 274
Question
• Which of the following statements concerning resonance
structures is correct?
– The concept of resonance is used because the Lewis structure
model is incomplete when describing bonding in a molecule.
– For a species having three resonance structures, it is best to think
of the species as existing as each of these structures one-third of
the time.
– All charged molecules have resonance structures.
– The octet rule must not be violated in writing resonance structures.
Chapter 8 | Slide 275
Answer
•a) The concept of resonance is used because the Lewis
structure model is incomplete when
describing bonding
in a molecule.
•Section 8.12, Resonance; Section 8.11, Exceptions to the
Octet Rule
•Lewis structures show static, localized electrons. These
structures are incomplete because electrons are better
thought of as delocalized in the molecule.
Chapter 8 | Slide 276
Question
• Which of the following has a Lewis
structure most like that of CO32-?
– CO2
– SO32– NO3– O3
– NO2
Chapter 8 | Slide 277
Answer
•c) NO3•Section 8.10, Lewis Structures; Section 8.11,
Exceptions to the Octet Rule
•The number of atoms and the electron count are
the same for NO3- and CO32-; thus these two ions
have similar Lewis structures.
Chapter 8 | Slide 278
Question
• Which molecule or ion violates the octet
rule?
– H2O
– NO3– PF3
– I3– None of these
Chapter 8 | Slide 279
Answer
•d)
I3-
•Section 8.10, Lewis Structures; Section 8.11,
Exceptions to the Octet Rule
•Violations of the octet rule involve having
more than 8 electrons around an atom, which
is the case with I3-.
Chapter 8 | Slide 280
Question
• How many resonance structures does the
molecule SO2 have?
–0
–1
–2
–3
–4
Chapter 8 | Slide 281
Answer
•c)
2
•Section 8.10, Lewis Structures; Section 8.12,
Resonance
•The two resonance structures (without the
lone pairs) are O–S=O and O=S–O.
Chapter 8 | Slide 282
Question
•
Which of the following molecules is polar?
–
–
–
–
XeF4
XeF2
BF3
NF3
Chapter 8 | Slide 283
Answer
•d) NF3
•Section 8.3, Bond Polarity and Dipole Moments; Section
8.13, Molecular Structure: The VSEPR Model
•A polar molecule results from a species with a nonzero
dipole moment. The square planar (XeF4), linear (XeF2), and
trigonal planar (BF3) geometries have zero dipole moments.
The trigonal pyramidal geometry (NF3) has a nonzero dipole
moment.
Chapter 8 | Slide 284
Question
•Which of the following molecules does not
have a dipole moment?
–
–
–
–
–
Chapter 8 | Slide 285
H2S
H2O
H2Xe
All of the above have a dipole moment.
None of the above has a dipole moment.
Answer
•c) H2Xe
•Section 8.3, Bond Polarity and Dipole Moments;
Section 8.13, Molecular Structure: The VSEPR
Model
•Dipole moments result from differences in
electronegativity and geometry. Both H2S and H2O
are bent molecules, whereas H2Xe is linear.
Chapter 8 | Slide 286
Question
• Which of the following molecules has a
dipole moment?
– CF4
– SF4
– XeF4
– All of the above have a dipole moment.
– None of the above has a dipole moment.
Chapter 8 | Slide 287
Answer
•b) SF4
•Section 8.3, Bond Polarity and Dipole Moments; Section
8.13, Molecular Structure: The VSEPR Model
•Dipole moments result from differences in electronegativity
and geometry. Both the tetrahedral (CF4) and square planar
(XeF4) geometries result in dipole moments of zero. The
distorted tetrahedral geometry of SF4 produces a dipole
moment.
Chapter 8 | Slide 288
Question
•Which of the following molecules has a
dipole moment?
–
–
–
–
Chapter 8 | Slide 289
BCl3
SiCl4
PCl3
Cl2
Answer
•c) PCl3
•Section 8.3, Bond Polarity and Dipole Moments; Section
8.13, Molecular Structure: The VSEPR Model
•The trigonal planar (BCl3), tetrahedral (SiCl4), and linear
(Cl2) geometries result in dipole moments of zero. The
trigonal pyramidal geometry (PCl3) results in a nonzero
dipole moment.
Chapter 8 | Slide 290
Question
•Which of the following statements about the
species N2, CO, CN–, and NO+ is false?
–
–
–
–
Chapter 8 | Slide 291
All are isoelectronic.
Each contains a triple bond.
All are linear.
The bond in each species is polar.
Answer
•d) The bond in each species is polar.
•Section 8.3, Bond Polarity and Dipole Moments; Section
8.13, Molecular Structure: The VSEPR Model
•The species N2, CO, CN–, and NO+ all have 10 valence
electrons, which requires using a triple bond. Because they
all involve only two atoms, these species are linear.
Because the atoms are the same in N2, it does not have
polar bond and is not a polar molecule.
Chapter 8 | Slide 292
Question
• What is the approximate measure of the
bond angles about the carbon atom in the
formaldehyde molecule, H2CO?
– 120°
– 60°
– 109°
– 180°
– 90°
Chapter 8 | Slide 293
Answer
•a)
120°
•Section 8.13, Molecular Structure: The
VSEPR Model
•According to the VSEPR model, C in
formaldehyde has three electron pairs, which
are oriented 120° from one another.
Chapter 8 | Slide 294
Question
• What type of structure does the XeOF2
molecule have?
– Pyramidal
– Tetrahedral
– T-shaped
– Trigonal planar
– Octahedral
Chapter 8 | Slide 295
Answer
•c) T-shaped
•Section 8.13, Molecular Structure: The VSEPR
Model
•According to the VSEPR model, Xe has five
electron pairs with three bonding pairs, which
means its geometry is T-shaped.
Chapter 8 | Slide 296
Question
• Which of the following statements best describes BF3
and NF3? (Note: Geometry refers to the electron pair
arrangement, and shape refers to the atom
arrangement.)
– They have variable geometries and shapes due to
their potential resonance structures.
– They have the same geometry and different shapes.
– They have the same geometry and the same shape.
– They have different geometries and the same
shape.
– They have different geometries and different
shapes.
Chapter 8 | Slide 297
Answer
•e)They have different geometries and different
shapes.
•Section 8.13, Molecular Structure: The VSEPR
Model
•BF3 has a trigonal planar geometry and a trigonal planar
shape. NF3, owing to the lone pair on the nitrogen, has a
tetrahedral geometry and a trigonal pyramidal shape.
Chapter 8 | Slide 298