Download Version 1.6 - Clark Science Center

Chm 118
Fall 2016
Problem Solving in Chemistry
Mr. Linck
Version: 1.6.
August 15, 2016
Written for and dedicated to MC
August, 2016
i
Contents
1 Introduction
1
2 A Simple Structural Model
2
2.1
2.2
2.3
2.4
2.5
2.6
Basis Principles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2
2.1.1
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2
Quantum Issues . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3
2.2.1
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5
Basis of Simple Structural Model . . . . . . . . . . . . . . . . . . . . . . . .
6
2.3.1
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6
The Simplest Model to Count Electrons for VSEPR Structures . . . . . . .
7
2.4.1
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8
A Slight Expansion of the Simple Model . . . . . . . . . . . . . . . . . . . .
10
2.5.1
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
10
Patching the Lewis Structure Model when Needed . . . . . . . . . . . . . .
11
2.6.1
12
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3 Using Lewis Structures to Understand Structure and Reactivity
3.1
3.2
14
Bond Lengths . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
14
3.1.1
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
14
Bond Energies and Enthalpy . . . . . . . . . . . . . . . . . . . . . . . . . .
15
3.2.1
17
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
ii
0.0
iii
3.3
Polarization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
18
3.3.1
19
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4 Acidity
4.1
4.2
4.3
4.4
4.5
4.6
4.7
20
Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
20
4.1.1
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
21
Estimating Acidity, Part I: Nature of Element . . . . . . . . . . . . . . . . .
21
4.2.1
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
23
Estimating Acidity, Part II: Charge . . . . . . . . . . . . . . . . . . . . . . .
23
4.3.1
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
23
Estimating Acidity, Part III: Resonance . . . . . . . . . . . . . . . . . . . .
24
4.4.1
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
24
Estimating Acidity, Part IV: Inductive Effects . . . . . . . . . . . . . . . . .
25
4.5.1
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
25
Hydration Energy and the Acidity of Metal Ions . . . . . . . . . . . . . . .
26
4.6.1
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
26
Distribution Diagrams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
27
4.7.1
29
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5 Symmetry
5.1
5.2
5.3
5.4
30
Definitions and Proper Symmetry Operations . . . . . . . . . . . . . . . . .
30
5.1.1
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
30
Improper Symmetry Operations: Planes of Symmetry . . . . . . . . . . . .
32
5.2.1
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
32
Improper Symmetry Operations: Center of Inversion . . . . . . . . . . . . .
33
5.3.1
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
33
Improper Symmetry Operations: Rotation-Reflection Axis . . . . . . . . . .
33
5.4.1
34
Chm 118
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Problem Solving in Chemistry
0.0
iv
5.5
5.6
5.7
5.8
The Group of Symmetry Operations . . . . . . . . . . . . . . . . . . . . . .
34
5.5.1
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
35
The Behavior of Single Objects Under Symmetry Operations . . . . . . . .
35
5.6.1
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
37
The Behavior of Several Objects Under Symmetry Operations . . . . . . . .
39
5.7.1
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
39
Classifying the Symmetry for Multiple Objects; the Combinations . . . . .
40
5.8.1
42
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6 Quantum Mechanics
6.1
6.2
6.3
6.4
6.5
6.6
6.7
6.8
6.9
45
Probability and Quantum Measurement . . . . . . . . . . . . . . . . . . . .
45
6.1.1
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
46
Waves and Interference . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
47
6.2.1
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
48
Simple Version of the Rules of Quantum Mechanics . . . . . . . . . . . . . .
49
6.3.1
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
49
A Simple but Useful Quantum Problem: The Parve on a Pole . . . . . . . .
50
6.4.1
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
50
Quantum Mechanics of the Hydrogen Atom I. Energy Levels . . . . . . . .
52
6.5.1
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
52
Quantum Mechanics of the Hydrogen Atom II. Radial Wave Function . . .
53
6.6.1
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
55
Quantum Mechanics of the Hydrogen Atom III. Angular Wave Function . .
59
6.7.1
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
60
Multi-electronic Atoms, Configurations . . . . . . . . . . . . . . . . . . . . .
63
6.8.1
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
64
Multi-electronic Atoms, Ionization Energies . . . . . . . . . . . . . . . . . .
65
Chm 118
Problem Solving in Chemistry
0.0
v
6.9.1
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
67
6.10 Multi-electronic Atoms, Valence Orbital Ionization Energies . . . . . . . . .
68
6.10.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
68
6.11 Multi-electronic Atoms, Transition Metal Ions . . . . . . . . . . . . . . . . .
69
6.11.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
69
7 Transition Metal Compounds and Color
7.1
7.2
7.3
70
Energy of Orbitals in Electrostatic Fields . . . . . . . . . . . . . . . . . . .
70
7.1.1
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
71
Metal Ions in an Octahedral and Tetrahedral Environments . . . . . . . . .
72
7.2.1
73
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Metal Ions in Other Environments: Lowering of Symmetry
. . . . . . . . .
74
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
74
The Configuration of Metal Ion Compounds . . . . . . . . . . . . . . . . . .
75
7.4.1
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
76
Configuration of Metal Ion Compounds, Spin, Field, and Ligand Strength .
77
7.5.1
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
78
Color of Metal Ion Compounds. I. Octahedral Compounds. . . . . . . . . .
79
7.6.1
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
80
Color of Metal Ion Compounds. II. Low Symmetry Compounds. . . . . . .
81
7.7.1
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
84
Color of Metal Ion Compounds. III. Intensities of Color. . . . . . . . . . . .
84
7.8.1
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
86
Magnetism in Metal Ion Compounds. . . . . . . . . . . . . . . . . . . . . . .
87
7.9.1
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
87
7.10 Crystal Field Stabilization Energies . . . . . . . . . . . . . . . . . . . . . . .
88
7.10.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
89
7.3.1
7.4
7.5
7.6
7.7
7.8
7.9
Chm 118
Problem Solving in Chemistry
0.0
vi
8 Absorbance and Kinetics
8.1
8.2
8.3
8.4
92
Using Light to See . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
92
8.1.1
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
93
Basic Kinetic Expressions and Definitions . . . . . . . . . . . . . . . . . . .
94
8.2.1
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
95
First Order Reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
96
8.3.1
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
96
Dealing with More Complex Rate Laws . . . . . . . . . . . . . . . . . . . .
97
8.4.1
98
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
9 Bonding
9.1
9.2
9.3
9.4
9.5
9.6
9.7
9.8
100
Storage of Energy in Molecules . . . . . . . . . . . . . . . . . . . . . . . . .
100
9.1.1
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
101
Using Atomic Orbitals to Make Molecular Orbitals . . . . . . . . . . . . . .
103
9.2.1
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
108
A Nomenclature for Molecular Orbitals . . . . . . . . . . . . . . . . . . . .
110
9.3.1
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
112
Molecular Orbital Energy Diagrams . . . . . . . . . . . . . . . . . . . . . .
113
9.4.1
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
113
Heteronuclear Diatomics and the First Row Diatomics with s/p Mixing . .
114
9.5.1
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
116
Hybridization as a Simplifying Tool. Part I: sp Hybridization . . . . . . . .
118
9.6.1
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
118
Hybridization as a Tool. Part II: sp2 and sp3 Hybridization . . . . . . . . .
120
9.7.1
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
120
Using Symmetry to Determine the Orbitals that can Bind. Part I. Simple
Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
121
Chm 118
Problem Solving in Chemistry
0.0
vii
9.8.1
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
122
Using Symmetry to Determine the Orbitals that can Bind. Part II. Degenerate Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
123
9.9.1
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
124
9.10 Exploring a Model for Solids . . . . . . . . . . . . . . . . . . . . . . . . . .
125
9.10.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
125
9.11 Molecular Orbital Theory of Hypervalent Compounds . . . . . . . . . . . .
126
9.11.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
126
9.12 Bonding in Metal Compounds. Part I. σ Donors . . . . . . . . . . . . . . .
128
9.12.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
129
9.13 Bonding in Metal Compounds. Part II. σ and π Donors . . . . . . . . . . .
130
9.13.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
131
9.14 Bonding in Metal Compounds. Part III. σ Donors/π Acceptors . . . . . . .
132
9.14.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
132
9.15 The Eighteen Electron Rule . . . . . . . . . . . . . . . . . . . . . . . . . . .
134
9.15.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
134
9.9
10 Entropy
136
10.1 Definitions, and the Number of Microstates in a Configuration . . . . . . .
136
10.1.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
138
10.2 A Way to Count Microstates in a Configuration and Predominant Configurations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
140
10.2.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
141
10.3 A Detour to Define Some Thermodynamic Quantities: The First Law . . .
142
10.3.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
143
10.4 Adding Heat to the Predominant Configuration. Part I. Qualitative . . . .
143
10.4.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
144
10.5 What Does a Predominant Configuration Look Like? The Boltzmann Equation145
Chm 118
Problem Solving in Chemistry
0.0
viii
10.5.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
146
10.6 Adding Heat to the Predominant Configuration. Part II. Quantitative. . . .
147
10.6.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
149
10.7 Disguising our Finding in a Word: Entropy . . . . . . . . . . . . . . . . . .
149
10.7.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
152
10.8 Entropy Defined in Terms of W; The Third Law . . . . . . . . . . . . . . .
152
10.8.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
153
10.9 Disorder is a Poor Word to Describe Entropy . . . . . . . . . . . . . . . . .
153
10.9.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
154
10.10Entropy Change in Reactions . . . . . . . . . . . . . . . . . . . . . . . . . .
154
10.10.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
155
11 The Free Energy: Entropy in Another Guise
157
11.1 Using the First Law of Thermodynamics at Constant P and T: Enthalpy . .
157
11.1.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
158
11.2 Using Enthalpy of the System to Understand the Surrounding’s Entropy
Change . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
159
11.2.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
159
11.3 Getting Rid of the Universe Altogether: Free Energy . . . . . . . . . . . . .
161
11.3.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
161
11.4 The Pressure (or Concentration) Dependence of G . . . . . . . . . . . . . .
162
11.4.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
164
11.5 Free Energy and Equilibrium . . . . . . . . . . . . . . . . . . . . . . . . . .
166
11.5.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
167
12 Equilibrium and Other Uses of the Free Energy
169
12.1 Equilibrium in Acid Solutions . . . . . . . . . . . . . . . . . . . . . . . . . .
169
12.1.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
171
Chm 118
Problem Solving in Chemistry
0.0
ix
12.2 Mixtures of Acids and Salts: Buffer Solutions . . . . . . . . . . . . . . . . .
172
12.2.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
173
12.3 Solubility Equilibrium, including Common Ion Effect . . . . . . . . . . . . .
173
12.3.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
174
12.4 Electrochemical Cells . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
174
12.4.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
176
12.5 Free Energy, Other Work, and the Nernst Equation . . . . . . . . . . . . . .
177
12.5.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
178
12.6 Using Electrochemical Cells to Solve Problems . . . . . . . . . . . . . . . .
178
12.6.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
179
Chm 118
Problem Solving in Chemistry
Chapter 1
Introduction
There are three large subject areas in chemistry that this course deals with. These are
the structure, color, and reactivity of molecules. Structure means we want to understand
the arrangement in space of the nuclei and learn what we can about where the electrons
are to be found between those nuclei. Also, how those structures influence the chemistry
of the materials. Color is of interest to us because it teaches us what the structures are,
and how easily electrons are moved around. The latter leads directly to asking questions
about reactivity: Which molecules will react with each other, what energy changes take
place during that reaction, and how fast do those reactions take place?
These are the topics in this course. We shall work our way to knowledge in each of these
areas in many small steps, coming back to a more sophisticated vision with each round of
steps. Repitition is the heart of learning.
Note to the reader: A skill that is often overlooked in scientific education is the ability to
sense when data appear weird; when they are wrong by some human error, or when they are
telling the observer to modify her understanding because it does not work. To teach this
skill there are numbers in this document that have been made weird intentionally. Look at
what you see critically and ask ”Can that be?”
1
Chapter 2
A Simple Structural Model
2.1
Basis Principles
The simplest question to ask about the structure of a molecule is ”Where are the nuclei
located in space?” If all we consider is the nuclei, then the answer is clear: as far apart as
they can get, since all nuclei are positively charged, and hence repel each other according
to Coulomb’s law, which is at the heart of all chemical understanding:
U=
1 Z1 Z2
4π0 r1,2
(2.1)
where U is the potential energy, Z1 and Z2 are the charges on the two species of concern,
r1,2 is the distance separating them, and the first term is just there to convert from units of
charge, Coulombs, and distance, to energy units. Note when both the charges are the same,
either positive or negative, then the potential energy is more positive, less stable, when the
distance is small.
So why do molecules have nuclei relatively close together? The answer is, of course, that
electrons are also present, and these negatively charged species attract nuclei, holding them
closer together in a favorable arrangement. Our question is what is that arrangement? Since
the cause is the electrons, it is not surprising that we should look to where the electrons
are in order to understand the location of the nuclei. We will see our model requires the
examination of electron pairs.
2.1.1
Exercises
2.1.1.1
Will two +1 charges have a greater potential energy at a distance of 1.0Å or 2.0Å.
2
2.2
3
Figure 2.1: Random ping-pong balls descending to two slits.
2.1.1.2
1
For this problem, assume 4π
is equal to 1. Plot the value of U as a function of ri,j for a
0
pair of positive charges of unit value.
2.1.1.3
1
For this problem, assume 4π
is equal to 1. Plot the value of U as a function of ri,j for a
0
negative charge and a positive charge, each of unit value.
2.2
Quantum Issues
To learn where electrons are, and what they do, requires a knowledge of quantum mechanics,
which is the collection of the rules that dictate the behavior of absolutely small things1 . We
will talk extensively, but mostly non-mathematically, about quantum issues in this course;
there are two aspects that we need here to proceed. The first is to acknowledge that the
behavior of electrons is ”weird,” often thought to be ”other worldly.” One aspect of this is
1
Dirac defines absolutely small things as those that cannot be observed without a change in their properties, usually position or momentum.
Chm 118
Problem Solving in Chemistry
2.2
4
Figure 2.2: Probability distribution for ping-pong balls.
that quantum mechanics tells us often that the best we can do is to find the probability of
the result of a measurement, not a determined value.
Consider dropping a bunch of randomly positioned ping-pong balls toward a floor with two
slits (holes) in it, as illustrated in Figure 2.1. Those that fall through the holes will hit
detectors below and will be counted as doing so. The distribution of ping-pong ball hits
at the detectors will be as shown in Figure 2.2, where the vertical axis is the number of
hits and the horizontal axis is the position of the hit. This is classical behavior and offers
nothing your experience does not anticipate.
If we carry out the same experiment with electrons or any other small object going through
two slits, the distribution of hits is different. That which is observed is shown in Figure 2.3.
Notice that there are regions of high probablilty of an electron hitting and regions of low
probability. The probablity of hits now looks nothing like the physical arrangement of the
slits. Rather, it appears like a diffraction pattern that would be exhibited by something like
water waves. A consequence of this is that quantum mechanics will use a wave function to
describe the behavior of electrons. This wave function is a function of the position in space
and, in general, of time. It is often written as ψ(x,y,z,t). A typical example would be
√
ψ(x, y, z, t) = 2 2Sin[πx]Sin[πy]Sin[πz]
(2.2)
which is a time-independent wave function. The probability issue we talked about above
comes about by squaring of this descriptive wave function,
P = [ψ(x, y, z, t)]2
(2.3)
where P is the probability of finding the electron at the point x, y, z at time t. We will
describe electrons in atoms and molecules by describing the corresponding wave functions.
Note there is nothing here about those descriptions, just an establishment of where we will
ultimately go. We will begin this with very few details in the next section.
The second feature that we need to deal with in a simple structural model is the fact that
electrons exhibit a feature usually called spin. This is a pure quantum mechanical feature
Chm 118
Problem Solving in Chemistry
2.3
5
Figure 2.3: Probability distribution for electrons through two slits
and has no classical analogue. The spin of electrons can take only one of two values, spin
“up”, spin of 1/2, or spin “down”, spin of -1/2. Electrons of each of these two kinds of
spin can be separated by appropriate magnetic devices. What is important to us here is the
Pauli Principle, which can be expressed in a number of equivalent ways, one of which is:
“No two electrons of the same spin can come close to each other.” Electrons are held apart
from each other because of their charge via the operation of Coulomb’s Law, equation 2.1.
Electrons of the same spin are held apart from each other by the additional operation of
the Pauli Principle.
2.2.1
Exercises
2.2.1.1
A wave function is just a recipe for finding a value at some point in space. Find the value
of ψ of equation 2.2 at the point x = 0.1, y = 0.2, z = 0.6.
2.2.1.2
Find the probability that a particle with the ψ given in equation 2.2 can be found in a small
volume centered on the point x = 0.2, y = 0.1, z = 0.6.
2.2.1.3
√
For an electron described by the wave function ψ = 2 2 Sin[2πx] Sin[πy] Sin[πz], find the
probability that the electron will be found in a small volume centered on the point x = 0.5,
y= 0.2, z = 0.2. Note: Regions of space where the probability of finding the electron is
zero2 are called “nodes.”
2
A favorite student question is ”If there is a region where there is no probability of finding the electron,
how does it get from one side to the other?” Don’t ask. Such a question is applying classical behavior to
things that do not behave that way.
Chm 118
Problem Solving in Chemistry
2.4
2.3
6
Basis of Simple Structural Model
Imagine a nucleus with four electrons around it, two of which are of up spin and two of
which are of down spin. Each set of electrons will, according to the Pauli Principle, stay
as far apart from each other as possible. If we now bring another atom up to the first,
we need electrons between the nuclei to keep them from repelling each other according to
equation 2.1. So an up spin and down spin of each set will come reasonably close (to hold
the nuclei together), but the second of up spin will stay as far as it can from the first electron
of up spin, and similarly for those of down spin. This simple model is called the VSEPR
model (for valence shell electron pair repulsion).
2.3.1
Exercises
2.3.1.1
What will be the location of two electrons of up spin near a nucleus if you require them to
be as far apart from each other as possible?
2.3.1.2
In BeCl2 there are four electrons around the Be, two of up spin and two of down spin.
Where will you find the two chlorine atoms?
2.3.1.3
What will be the arrangement for six electrons, three of each spin, in order to stay as far
apart from each other as possible?
2.3.1.4
In BF3 there are six electrons around the B, three of up spin and three of down spin. Where
will you find the three fluorine atoms?
2.3.1.5
The arrangements talked about in the earlier problems are for two dimensional structures.
In three dimensions, it is a little harder to use intuition. For eight electrons, four of each spin,
the optimium arrangement of each set of four is the tetrahedron. Look up a tetrahedron
on the web and be able to draw this arrangement.
2.3.1.6
In CH4 there are eight electrons around the C, four of up spin and four of down spin. Where
will you find the four hydrogen atoms?
2.3.1.7
Make of drawing of CH4 which shows the real geometry in space.
Chm 118
Problem Solving in Chemistry
2.4
7
Figure 2.4: Schematic representation of nitrogen atom bonding
2.4
The Simplest Model to Count Electrons for VSEPR Structures
The arguments just given require that we know how many electrons are around a given
atom to determine how the atoms attached to it will be arranged. The simplest model
for where the electrons are in a molecule is called a Lewis structure. This is well known
to students in this course, but here is a concise summary. Lewis structures of compounds
made from elements in groups IV (14) through VII (17) in the periodic table have their
electrons arranged in pairs (for reasons discussed above) and there are (usually–we pay
a lot of attention to the exceptions below) eight electrons around any given atom. (The
ubiquitous hydrogen atom has only a pair of electrons.) There is a simple consequence of
this. Consider a nitrogen atom with five electrons in the valence shell, the layer important
in bonding. One set of up spin and down spin electrons form a lone pair on the nitrogen.
Since the nitrogen atom needs three further electrons to reach the set of eight, nitrogen must
interact with three one electron donors to give, for example, NH3 ; or with a one electron
donor and an atom that donates two electrons to give, for example, HNO; or with a single
atom that donates three electrons to give, for example, N2 –see the schematic representation
in Figure 2.4. In all cases three bonds are formed; in the first case three single bonds, in
the second a single and a double bond, and in the third, a triple bond. So a nitrogen atom
always has three bonds and a lone pair. Similar arguments pertain to the other atoms
leading to the conclusion that the number of bonds around the elements in groups IV (14)
through VII (17) are four, three, two, and one, respectively.
These rules are violated if an atom carries charge, either as a result of being in a molecule
with net charge, or having formal charge. The former is illustrated by NH+
4 where the
nitrogen atom has four bonds and OH– where the oxygen atom has only one bond. Formal
charge is the charge on an atom calculated by adding to the valence shell nuclear charge (a
positive number) the charge caused by the number of lone pair electrons and one-half the
number of bonding electrons.
There is still another issue to consider to make this simple model applicable. In compounds
where there is an ambiguity, which atom is the “central” atom in the structure? That
is, should one write for CO2 the structure OOC or the structure OCO? Formal charge
answers this question. Usually one of the choice of structures will have a formal charge that
is unfavorable. A comment on what constitutes unfavorable is needed here (and will be
developed later): Elements to the right and top of the periodic table should be negatively
charged and those to the left and bottom should be positively charged. If formal charge in a
molecule is the other direction, the structure drawn is not favorable. For those of you liking
a rule, there is one that usually, but not always, works: the element of lower ionization
Chm 118
Problem Solving in Chemistry
2.4
8
energy–the energy to remove an electron from a gaseous atom–or electronegativity is in the
central position.
What to do in the VSEPR model when there are both bonding pairs of electrons and lone
pairs of electrons present in the Lewis structure? To a first approximation, one treats the
two kinds of electrons equivalently. So if the structure of CH4 is a tetrahedron, that of NH3
will also have electrons pairs arranged at the corners of a tetrahedron, but the hydrogen
atoms only occupy three of the vertices of the tetrahedron, so the shape of NH3 is trigonal
pyramidal. Likewise in H2 O, the arrangement of electron pairs is tetrahedral, but the
hydrogen atoms only occupy two of the vertices, so the shape of the molecule is “bent”.
There is an extension of the VSEPR model to predict bond angles in compounds containing
lone pairs that works reasonably, but not always; it is discussed in some of the exercises.
Finally, there is the issue of how VSEPR handles double and triple bonds. As we shall
see when we discuss bonding in a complete fashion, double and triple bonds should be
viewed as a normal bond accompanied by one or two less efficient bonds. These latter are
“not active” in a VSEPR sense. For instance, in the compound CH2 O, there is a double
bond between the carbon atom and the oxygen atom. Therefore the number of “pairs” of
electrons around the carbon atom from a VSEPR point of view is three; the geometry is
planar and triangular.
2.4.1
Exercises
2.4.1.1
Give the Lewis structure of H2 O, SCl2 , PH3 , ClF, and CH3 OH; convince yourself that all
formal charges are zero in these compounds. Use line structures and not “dot” structures.
Include lone pairs.
2.4.1.2
Show that the rule about the number of bonds holds for the compounds in the last exercise.
2.4.1.3
Draw the Lewis structure of urea, (NH2 )2 CO.
2.4.1.4
Draw the VSEPR structure of the compound in the last exercise. HINT: Consider each
non-hydrogen atom separately.
2.4.1.5
Draw the Lewis structure of AlCl3 , SiH4 , CF4 , H2 S, H3 O+ , and NH–2 .
2.4.1.6
Draw the VSEPR structure of the compounds in the last exercise.
2.4.1.7
Write Lewis structures for GeH4 , AsCl3 , and SeF2 . Use VSEPR to determine structure.
Chm 118
Problem Solving in Chemistry
2.5
9
2.4.1.8
Write Lewis structures for NCS– where the atoms are attached as indicated and the left
hand nitrogen atom is double bonded to the carbon. Note formal charges.
2.4.1.9
Write Lewis structures for NCS– where the atoms are attached as indicated and the left
hand nitrogen atom is triple bonded to the carbon. Note formal charges.
2.4.1.10
Use the results of the last two exercises and that of exercise 5 to formulate a rule for the
number of bonds to an element that is formally charged positively. For one the is formally
charged negatively.
2.4.1.11
Is OF2 likely to have an oxygen atom or a fluorine atom in the center? Why?
2.4.1.12
The Lewis structure of CO2 is not C=O=O, where some electrons are missing: you fill them
in. What’s wrong with this Lewis structure? What is the correct Lewis structure? Why is
the one we call “correct” correct?
2.4.1.13
Is the structure of NOF as written, or is it ONF, or perhaps OFN? Prove it.
2.4.1.14
Give the VSEPR structure of your answer to the last exercise.
2.4.1.15
Consider the lone pair electrons near a nitrogen atom. Do you anticipate they will be closer
to or further from the nucleus than a bonding pair of electrons between a nitrogen atom
and a fluorine atom? Which then will be more dominant in pushing other electron pairs
away in a VSEPR sense?
2.4.1.16
Consider the bonding pair electrons between a nitrogen atom and a chlorine atom. Do
you anticipate they will be closer to or further from the nucleus than a bonding pair of
electrons between a nitrogen atom and a fluorine atom? Which then will be more dominant
in pushing other electron pairs away in a VSEPR sense?
2.4.1.17
Use your results from the last two exercises to predict which bond angle will be larger,
the F-N-F in NF3 or the Cl-N-Cl in NCl3 . The observed values are 102.2o and 107.1o ,
respectively.
2.4.1.18
Predict which bond angle will be larger, the F-C-F in CHF3 or the Cl-C-Cl in CHCl3 . The
observed values are 108.8o and 110.4o , respectively.
Chm 118
Problem Solving in Chemistry
2.5
10
2.4.1.19
Predict which bond angle will be larger, the Cl-C-Cl in CHCl3 or the Cl-C-Cl in CH2 Cl2 .
The observed values are 110.4o and 111.8o , respectively.
2.5
A Slight Expansion of the Simple Model
The observant student will already have noted that we introduced BeCl2 and BF3 as examples with two and three pairs of electrons, yet this is completely impossible if one insists
upon the rule of eight, the octet, of the normal Lewis structure. In this section we address
the deviations from the octet structure that occur on both the low side (less than eight)
and the high side (more than eight).
Atoms to the left of the carbon column in the periodic table often have less than eight
electrons around the central atom in compounds. Generally the number of electrons is
twice the group number, so boron has six, magnesium four, etc. Also note that Lewis
structures are not particularly valid for compounds in which there is substantial charge
separation between the atoms in the bond: Sodium chloride is not well represented by a
simple Lewis structure. The reason for the deviation from the rule of eight is that the
nuclear charge is small for boron (compared to C, N, O, F) and hence fewer electrons can
be attracted to that nucleus. The octet rule is “out the window” for compounds of B, Be,
etc.
A more serious issue concerns compounds that violate the octet rule on the high side,
have more than eight electrons in the valence shell–see the exercises. Table 2.1 gives some
example compounds. The VSEPR model for these compounds involves determining the
most stable arrangement for five or six pairs of electrons around a central atom. The issue
of six is easy: it is the octahedron. For five pairs of electrons, the arrangement is the
trigonal bipyramid. If this is not familiar to you, look it up on the web. The tbp is more
difficult in some cases because there are two geometrically different kinds of positions, two
environments. Generally the more electronegative element goes on the axial position. Our
bonding model for these compounds, developed later, will account for this rule.
2.5.1
Exercises
2.5.1.1
Give the Lewis structure and the VSEPR of BF3 .
2.5.1.2
2–
Give the Lewis structure and the VSEPR of SO2–
4 (with some cation), SF5 , and SiF6 (with
some cation).
2.5.1.3
Give the Lewis and VSEPR structures for Cl3 PO, SF4 , and SF4 O. HINT: In the case of
SF4 you should think about the rule for the position of the most electronegative element.
Chm 118
Problem Solving in Chemistry
2.6
11
Cmpd
PF5
PBr5
AsCl5
SF6
SF5 Cl
SCl4
SO2
SeF6
SeCl4
TeF4
TeI4
Color
colorless
red-yellow
colorless
colorless
colorless
colorless
colorless
colorless
colorless
colorless
black
Properties
gas, mp −84.5o C
dec, 84o C
dec. −50o C
gas, mp −50.8o C
gas, mp −64o C
s, dec. −30o C
gas bp −10o C
gas, mp −46.6o C
solid, sublimes with dec 196o C
solid, mp 130o C
solid, dec with boiling at 283o C
Cmpd
PCl5
AsF5
SbF5
SF4
SF5 Br
SO3 ,
SO3
SeF4
SeBr4
TeBr4
Color
colorless
colorless
colorless
colorless
colorless
colorless
colorless
colorless
orange
yellow
Properties
l, bp 150o C
gas, bp −52.8o C
l, bp (with d) 140o C
gas, bp −40o C
gas, mp −79o C
mp 16.8o C
s, bp 44.4o C
gas, bp 102o C
solid and liquid only, mp 123o C
dec with boiling at 414o C
Table 2.1: Properties of Compounds that Violate the Octet.
2.5.1.4
Give the Lewis and VSEPR structures for SiF2–
6 , XeF2 , and XeOF4 .
2.5.1.5
In Table 2.1 are several compounds that violate the octet. See if you can find some consistent
“rule” that gives you some ability to predict what species will violate the octet rule and
what species will not.
2.5.1.6
Some more compounds that exist (i.e., can be put into a bottle) but violate the Lewis “octet
2–
rule” on the high side are SO2–
4 (with some anion), IF5 , ClF3 , SiF6 (with some cation).
Compounds that violate the Lewis “octet rule” and (therefore?) do not exist are: SI4 ,
–
2–
SiH2–
6 , S5 (tetrahedral form), PH5 , F3 , NF5 (all negative ions with some cation). See if you
formulation from the last exercise works. If your rule needs modification, do so.
2.6
Patching the Lewis Structure Model when Needed
A situation exists where the Lewis model breaks down completely. A good example is ozone,
O3 . The structure of ozone is a bent molecule; it is not cyclic. If you try to do a Lewis
structure of ozone (as you should) you will find that in order to preserve the sacred octet,
you must have one of the external oxygen atoms double bonded to the central oxygen atom,
and the other single bonded. Should you choose the left one as the double bonded one? or
the right one? As we shall see more extensively in the next section of this course, Lewis
structures do a reasonable job of understanding relative bond lengths. The facts are that
the single bond in H2 O2 (draw Lewis structure, always) is 1.47Å whereas the double bond
in O2 is 1.21Å; but the two bond lengths in O3 are the same, 1.278Å. Note this last value
is between the value for a single bond and that for a double bond. There is no method by
which a single Lewis structure can deal with these facts.
Chm 118
Problem Solving in Chemistry
2.6
12
Figure 2.5: Resonance in Lewis structures
The solution to this problem is to introduce a concept called resonance. The contention is
that the ozone structure with the left hand oxygen double bonded and that with the right
hand oxygen double bonded should be “averaged” to get the true structure of ozone. This
“averaging” takes place in your head.3 In the case of ozone, we end up saying that the
bonding of an external oxygen with the central atom is by “one + two bonds divided by
two or a bond and a half,” as illustrated with the average signs in Figure 2.5. Likewise,
look at the two pictures and average the charge on the external oxygen atom. What do you
get?
You must use this simplest application of resonance whenever there are two different but
equivalent ways to write a structure. Try it on NO–2 . Or on cyclic C6 H6 , a hexagon of
carbon atoms, each of which has one hydrogen atom attached to it. And resonance has an
important aspect other than accounting for the equivalency of the bond lengths: resonance
is stabilizing; compounds with resonance are more stable than they would be if there were
no resonance. We explore the reasons for this later in the course, but it is important to
keep this in mind as we progress. There are some situations in which the resonance occurs
between structures that are not equivalent. And example is in the ion NCS– as also shown in
Figure 2.5. These require us to make judgments about which structure is more important,
a process that is not always easy.
2.6.1
Exercises
2.6.1.1
Give the Lewis and VSEPR structures for NO–3 , CH3 C(O)O– (where the (O) means off the
chain of elements, in this case the chain is C-C-O ), and CO2–
3 . HINT: If you end up with a
negative formal charge on a carbon atom in the last structure, you probably are not correct.
2.6.1.2
Draw the Lewis and VSEPR structures of NO2 Cl.
3
In some simple cases like the ozone case, there are “pictures” that do the averaging for you, but these
are open to mis-interpretation and confusion. Best to do it in your head.
Chm 118
Problem Solving in Chemistry
2.6
13
Figure 2.6: Structures for exercise.
2.6.1.3
Are structures 5 and 6 in Figure 2.6 resonance structures of each other? If so, which do
you think is “more important” and why?
2.6.1.4
Write Lewis structures for HOClO3 , HOClO2 , HOClO, and ClOH, where the chlorine atom
is the central atom in the first three. HINT: The hydrogen atom is bonded to an oxygen
atom in all structures.
2.6.1.5
Write Lewis structures for ClO–4 , ClO–3 , ClO–2 , and ClO– .
Chm 118
Problem Solving in Chemistry
Chapter 3
Using Lewis Structures to
Understand Structure and
Reactivity
3.1
Bond Lengths
The distance between two atoms in a molecule is referred to as the “bond length.” These
vary according to the Lewis structure and the position in the periodic table; since atoms
lower in the table have more shells of electrons; they are bigger. The Lewis structure
influences the bond length depending on the nature of the bond between the elements,
single, double or triple. Of course, as discussed above, resonance can modify these values.
For instance, Table 3.1 gives the bond length of the compounds with nitrogen-nitrogen
bonds.
3.1.1
Exercises
3.1.1.1
The structures of species ClOn2 for n = -1, 0, and 1 are given in Table 3.2. Do these values
make sense? Why or why not? HINT: Use chemical intuition; we have no model for one of
these.
Molecule
N2 H4
N2 H2
N2
Bond Length, Å
1.45
1.21
1.098
Table 3.1: Bond Length Data for Nitrogen Compounds.
14
3.2
15
Molecule
ClO+
2
ClO2
ClO–2
Bond Length, Å
1.408
1.475
1.57
Bond Angle
119o
165o
110o
Table 3.2: Structural Data for Various Chlorine Dioxides.
Bond
H-H
F-H
I-H
C-F
N-N
Bond Energy
436
565
497
489
163
Bond
C-H
S-H
C-C
F-F
O-O
Bond Energy
416
362
331
158
146
Bond
N-H
Cl-H
C-N
Si-C
Cl-Cl
Bond Energy
391
429
305
306
242
Bond
O-H
Br-H
C-O
C-I
Si-Si
Bond Energy
463
365
358
214
226
Table 3.3: Bond Energies in kJ/mole for Various Single Bonds.
3.1.1.2
Predict the approximate nitrogen to nitrogen bond lengths in tetrazene, NH2 NNNH2 .
3.1.1.3
The P-P bond length in PH2 PH2 is about 2.2Å and that in elemental white phosphorous,
which exists as tetrameric units of formula P4 , is the same. Deduce a structure for P4 and
comment on the observation concerning the bond lengths. HINT: Be inventive and make
use of the facts: the bond lengths in the two materials are the same, therefore the bonds
are of the same kind.
3.1.1.4
The bond length found in P2 is 1.893Å. Explain the difference between that and the bond
length in P4 , which is about 2.2Å.
3.1.1.5
The S-N single bond is about 1.74Å and the S-N double bond is about 1.54Å. There are four
equivalent S-N bonds in S2 N2 with a bond length of about 1.6Å. Formulate a reasonable
Lewis structure for S2 N2 . HINT: One of the sulfur atoms in any Lewis structure you can
draw has five electron pairs around it.
3.2
Bond Energies and Enthalpy
The bond energy is the energy required to break homolytically (evenly) a chemical bond
and separate the two fragments from each other. This might be represented in the general
case as:
X2 = 2X
(3.1)
Chm 118
Problem Solving in Chemistry
3.2
16
Double Bonds
C-C
Triple Bonds
C-C
615
C-N
616
C-O
729
O-O
498
811
C-N
892
C-O
1077
N-N
945
Table 3.4: Bond Energies in kJ/mole for Various Double and Triple Bonds.
This, of necessity, produced atoms or fragments with an odd number of electrons. Such
compounds, at least when they contain atoms in the right side of the periodic table, are
called radicals. Generally radicals are unstable species, although there are some exceptions.
Two other terms useful in this discussion are diamagnetic, which for our purposes means
compounds “with only pairs of electrons,” and paramagnetic, which means “with more
electrons of one spin than the other.” Radicals are paramagnetic; most Lewis structures,
by the very nature of having eight electrons in four pairs, are diamagnetic.
The bond energy required to break the H-H bond in dihydrogen is 436 kJ/mole. Data for
some other compounds with single bonds in the Lewis structure are given in Table 3.3.
Data for compounds with double and triple bonds are in Table 3.4.
For careful work, we need to be more precise about our language concerning rupture of a
chemical bond. (The following is our first of many examinations of the beautiful construction of the human mind called thermodynamics.) If we separate two hydrogen atoms in
dihydrogen from each other, we are pulling apart a stable compound; this requires that
we put energy into the compound. Heat is one way of changing the energy of a system.
The heat added to a system at constant pressure is so useful to chemists that it has a specific name: Enthalpy is the heat added at constant pressure. Enthalpy has units of energy
per mole, kJ/mole usually. A useful measure of enthalpy is called the enthalpy of formation. This is the enthalpy change when one mole of a compound is made (formed) from
its constituent elements in their standard states. This latter is a form that has generally
be agreed upon; for instance, H2 (g) at one atmosphere pressure is the standard state of
dihydrogen, and for C(s) the standard state is graphite. Any element in its standard state
has an enthalpy of formation of zero, by definition.
There are many cases where an element exists in more than one form. For instance, solid
carbon can exist as diamond. We then ask how much heat is required at constant pressure
(usually of one atmosphere) to convert graphite to diamond. This quantity would be the
heat of formation of diamond:
C(s, graphite) = C(s, diamond)
∆Hfo = 1.9kJ/mole
(3.2)
A table of enthalpies of formation is very useful in determining the enthalpy change of
chemical reactions. Because the enthalpy of a compound is independent of how we got to
that compound, one can imagine two different paths for mixing solid MgO and gaseous CO2
to form solid MgCO3 . This first is direct reaction:
MgO(s) + CO2 (g, 1atm) = MgCO3 (s)
Chm 118
(3.3)
Problem Solving in Chemistry
3.2
17
and the second is a convoluted path:
1
MgO(s) = Mg(s) + O2 (g, 1atm)
2
CO2 (g, 1atm) = C(s, graphite) + O2 (g, 1atm)
3
Mg(s) + C(s, graphite) + O2 (g, 1atm) = MgCO3 (s)
2
(3.4)
(3.5)
(3.6)
Imagine we desire the enthalpy change for the reaction in equation 3.3. Hess’s law states
that the total enthalpy change for a series of reactions is just the sum of the enthalpy
changes for the individual reactions–enthalpy does not depend on path. Therefore, since
the sum of reactions in equations 3.4 to 3.6 is the reaction in equation 3.3, we have:
o
o
o
o
∆H3.3
= ∆H3.4
+ ∆H3.5
+ ∆H3.6
(3.7)
But we can write this in terms of enthalpies of formation as follows:
o
∆H3.3
= −∆Hfo (MgO) − ∆Hfo (CO2 ) + ∆Hfo (MgCO3 )
(3.8)
This is the familiar statement that the enthalpy change of a reaction is the enthalpy of
formation of products minus the enthalpy of formation of reactants.
3.2.1
Exercises
3.2.1.1
Why is it harder to break the N-N bond in dinitrogen than it is to break the H-H bond in
dihydrogen?
3.2.1.2
What patterns do you see in the data in Table 3.3? Do you believe enough in your patterns
to doubt any data?
3.2.1.3
Use the bond energies in the Tables 3.3 and 3.4 to compute the energy of the process
H2 CCH2 + H2 → H3 CCH3
Is the process energetically downhill? Why? HINT: Lewis structures.
3.2.1.4
The energy (actually enthalpy—heat at constant pressure) required to take P4 (g) to 2 moles
of P2 (g) is 289 kJ/mole of P4 . That required to take a mole of P2 (g) to 2 moles of P(g) is
485.3 kJ/mole of P2 . What is the energy required to take one mole of P4 (g) to four moles
of P(g)? What property of energy (enthalpy) did you use?
Chm 118
Problem Solving in Chemistry
3.3
18
3.2.1.5
Compounds of the formula CaP and SrP, which are diamagnetic, have been isolated. Describe the bonding within these compounds. HINTS: A diamagnetic compound is one in
which there are no unpaired electrons. Also, there is most likely a dominant ionic bond
between the element on the left of the periodic table (charged positively) and that on the
right (charged negatively). Finally, you might want to work with emphpolyatomic anions.
3.2.1.6
Which of the following molecules are unlikely to be “found in a bottle?” SFn where n runs
from 2 to 6. Give your reasoning.
3.2.1.7
Which of the species in the last exercise are unlikely to be diamagnetic?
3.2.1.8
If you had a compound of empirical formula SF that was diamagnetic, how would you
formulate the molecular structure? That is, give a valid Lewis structure.
3.2.1.9
Use enthalpy of formation data (https://en.wikipedia.org/wiki/Standard_enthalpy_
of_formation) to find the enthalpy change for the process
BaSO4 (s) = BaO(s) + SO3 (g, 1atm)
(3.9)
which is the thermal decomposition of barium sulfate.
3.2.1.10
Find the enthalpy change for the reaction
N2 H4 (g, 1atm) + H2 (g, 1atm) = 2NH3 (g, 1atm)
(3.10)
HINT: The enthalpy of formation of hydrazine gas is 50.6 kJ/mole.
3.2.1.11
If you did the last exercise correctly, you should be able to articulate the last sentence of
3.2 more completely and carefully. Do so.
3.3
Polarization
Looking at Lewis structures is one way to get an handle on the reactivity of molecules. For
instance, the Lewis structure of ozone, O3 , has a positive charge on the central oxygen.
Having a positive charge on an atom with high ionization energy (or, if you prefer, high
electronegativity) is not a energy stabilizing situation. High ionization potential atoms want
electrons, not a deficiency of them. In a similar fashion, reactions take place more readily,
release more energy, when they occur between an atom of high ionization energy and one
of low. This is because in the bond between those atoms, say a C-F bond, the electrons
are not distributed equally, but are pulled toward the fluorine atom (see the last exercises
Chm 118
Problem Solving in Chemistry
3.3
19
Reaction
F2 (g) + Cl2 (g) = 2FCl(g)
2F2 (g) + C(s, graphite) = CF4 (g)
F2 (g) + Be(s) = BeF2 (s)
1
2 F2 (g) + Li(s) = LiF(s)
∆H per mole of F atoms, kJ/mole of F
-25.3
-244
-513
-612
Table 3.5: Enthalpy changes for some reactions of F2 .
in section 2.4.1); this polarization is the unequal sharing of electrons in a bond. It makes
the fluorine atom negatively charged, negatively polarized, and the carbon atom positively
polarized. This arrangement is stabilizing because of the difference in nuclear charges. Note
the Lewis structure does not convey this information. Periodic position is used to determine
polarization. Upper right elements are stronger electron acceptors than lower left ones are.
3.3.1
Exercises
3.3.1.1
What is the direction of the polarization in the following bonds: C-O, B-O, F-I, C-F, O-S?
3.3.1.2
The reaction of half a mole of O2 with Be to form BeO has an enthalpy change of -610
kJ/mole, whereas that of one-third a mole of O3 with Be to form BeO has ∆H of -658
kJ/mole. Comment.
3.3.1.3
Table 3.5 contains enthalpy data for reactions of gaseous F2 . Comment on the data.
Chm 118
Problem Solving in Chemistry
Chapter 4
Acidity
4.1
Definitions
A substance is a strong acid if the reaction of the substance to produce a hydrogen ion
and some anion (there are more general definitions of acidity, which we neglect here) occurs
readily:
HA(aq) = H+ (aq) + A− (aq)
(4.1)
where the reaction is, in this case and most we consider in this document, in aqueous
solution, in water. Expressed in terms of the concentrations of materials at equilibrium, a
strong acid is one in which the concentration of protons, [H+ ], and those of anions, [A– ],
are large compared to the concentration of the original acid, [HA]. This can be made
quantitative by consideration of the so-called equilibrium constant, which is defined as:
Ka =
[H+ ][A− ]
[HA]
(4.2)
where the subscript “a” stands for “acidity” constant. A substance is a strong acid if Ka is
large and weak if Ka is small (both, of course, vague, relative words). Usually the feature
that contributes most to the value of Ka is the enthalpy change of the reaction.1
Values of Ka vary from about 1×108 to 1×10−50 . This large range can be more easily talked
about by using the concept of a pKa . The pKa is the negative logarithm (to the base 10)
of the Ka . Therefore, a Ka of 1.0×10−5 has a pKa = 5 and a Ka of 5.6×10−7 has a pKa =
6.25.
Many reactions are processes in which an acid reacts with a substance that accepts the
hydrogen ion, a base, to produce an new acid base pair:
HA(aq) + B− = HB(aq) + A− (aq)
(4.3)
In such reaction HA is clearly the acid and B– the base. The other useful language is that
HB is called the conjugate acid of the base B and A– is called the conjugate base of the acid
HA.
1
The other factor, entropy, will be introduced later in this document.
20
4.2
4.1.1
21
Exercises
4.1.1.1
What is an acid? What is a base? HINT: There are several different definitions of an acid
(base). Let’s use the easiest one for now.
4.1.1.2
2–
Write Lewis structures for SO2–
2 and SO3 .
4.1.1.3
Would you expect that the materials in the last exercise to be called “acids” or “bases”?
4.1.1.4
How can H2 O be an acid and a base?
4.1.1.5
One tenth of a mole of an acid HX is dissolved in a liter of water and at equilibrium the
[H+ ] = 1.33×10−3 M. What is the Ka of the acid?
4.1.1.6
What is the conjugate base of the acid NH+
4?
4.1.1.7
What is the pKa of an acid with a Ka of 4.4×10−3 ?
4.1.1.8
What is the Ka of an acid with a pKa of 5.9?
4.1.1.9
The general form of any equilibrium constant is the concentrations (or pressures) of the
products, raised to their stoichiometric power, divided by the concentrations (or pressures)
of reactants, similarly powered. Write the equilibrium constant for the reaction of Cu2+
with NH3 to form Cu(NH3 )2+ , all species in water.
4.1.1.10
Write the equilibrium constant for the reaction of Cu2+ with NH3 to form Cu(NH3 )2+
2 , all
species in water.
4.1.1.11
How is the equilibrium constant for the reaction of Cu(NH3 )2+ to form Cu(NH3 )2+
2 and
2+
Cu (all in water) related to the equilibrium constants of the last two exercises?
4.2
Estimating Acidity, Part I: Nature of Element
To see what factors influence the reaction that defines acidity, let’s break it down into several
steps for the acid HA, taking advantage of Hess’s law so that the sum of the equations given
Chm 118
Problem Solving in Chemistry
4.2
22
Group IV
CH4
55
SiH4
35
GeH4
25
Group V
NH3
35
PH3
27
AsH3
23
Group VI
H2 O
15.7
H2 S
7.1
H2 Se
3.8
H2 Te
2.6
Group VII
HF
3.2
HCl
-7
HBr
-8
HI
-9
Table 4.1: pKa Values for Binary Hydrogen-Element Compounds.
below equals the equation for acidity of HA:
HA(aq) = HA(g)
(4.4)
HA(g) = H(g) + A(g)
+
−
(4.5)
H(g) = H (g) + e (g)
(4.6)
−
−
(4.7)
+
+
(4.8)
−
−
(4.9)
A(g) + e (g) = A (g)
H (g) = H (aq)
A (g) = A (aq)
We are interested in the change in the enthalpy of these processes as “A” changes, so
equations 4.6 and 4.8 do not matter. Further, equation 4.4 is likely to be small, so the
important terms are those in equations 4.5, 4.7, and 4.9. If we examine a series of compounds
of approximately the same size and shape, say Hn A, then as A is changed from left to right
in the periodic table (from CH4 to NH3 , ...) bond energies (and hydration (equation 4.9)
should be approximately constant and those species that can tolerate negative charge more
easily, those with the more negative values for the enthalpy change of equation 4.7, should
be the stronger acid. Atoms that can tolerate negative charge are those to the right in the
periodic table, with large valence shell nuclear charges, or, high electronegativity. We would
therefore expect that HF would be a stronger acid than H2 O, which would be stronger than
NH3 , etc.
If we ask what would happen to the enthalpies of equations 4.5, 4.7, and 4.9 if we move
vertically in the periodic table, the situation is more complicated. Bond energies (equation 4.5) decrease, become less positive, as we move down the periodic table as does the
affinity for an electron (equation 4.7), factors that increase the acidity. Hydration energies
(equation 4.9) become less negative, which decreases the acidity. One way to remember the
facts is to say there are more terms increasing the acidity than decreasing it. Hence, HCl is
a stronger acid that HF and H2 S is a stronger acid than H2 O. Acid strength when a proton
is lost from an element in the periodic table increases as you move to the right and down
in the periodic table. Same data are given in Table 4.1.
Chm 118
Problem Solving in Chemistry
4.3
4.2.1
23
Exercises
4.2.1.1
Is HCl or H2 O the strongest acid?
4.2.1.2
–
The compound GeH4 reacts with NH3 in liquid ammonia to form NH+
4 and GeH3 ; methane
does not undergo a similar reaction. Which is the stronger acid, germane or methane?
4.2.1.3
If a substance HA had a particularly strong bond between the H and the A, how would the
acidity of HA be affected?
4.2.1.4
If a substance HA had an A– species that was strongly solvated, how would the acidity of
HA be affected?
4.2.1.5
Which would you expect to be the stronger acid, CH3 CH2 OH or CH3 CH2 NH2 ? HINT:
First determine which proton is likely to be removed in each species.
4.2.1.6
Which would you expect to be the stronger acid, CH3 CH2 OH or CH3 CH2 SH?
4.3
Estimating Acidity, Part II: Charge
In the reaction for an acid giving up a proton, equation 4.1, the positively charged proton
is being pulled away from a negatively charged A– . Imagine the original acid was H2 O.
Then the proton is being removed from OH– in the process defining the acidity of water.
Now consider what would happen if we thought about taking a proton away from the OH– :
What charge would be left on the species left behind if a proton was removed from OH– ?
The answer to that question tells you the effect of charge on acidity.
4.3.1
Exercises
4.3.1.1
Which material would you expect to be the strongest acid, H3 O+ (going to what?), H2 O
(going to what?), or OH– (and you know the extra question)?
4.3.1.2
Which material would you expect to be the strongest base, H2 O (going to what?), OH– , or
O2– ?
Chm 118
Problem Solving in Chemistry
4.4
24
Acid
H3 PO4
HClO4
HClO2
H3 BO3
HIO3
pKa
2.23
-7.0
2.95
9.14
0.77
Acid
H3 AsO4
HClO3
HOCl
HNO2
H2 SO4
pKa
2.22
-1.30
7.53
3.37
-2.0
Table 4.2: Some Acid Dissociation Constants for Oxyacids.
4.3.1.3
Is NH+
4 or NH3 the strongest acid?
4.3.1.4
What factors should you consider to determine if H2 O or NH+
4 is the stronger acid?
4.3.1.5
Which would you expect to be the stronger acid, HOCH2 CH2 O– or OH– ? HINT: Draw
Lewis structures.
4.3.1.6
Which would you expect to be the stronger acid, HOCH2 CH2 O– or HOCH2 CH2 CH2 O– ?
4.4
Estimating Acidity, Part III: Resonance
We learned earlier (see 2.6) that resonance has a stabilizing effect on the energy of a compound. Since the arguments we have been discussing about acidity are based on enthalpy
considerations, something that stabilizes compounds will place an important role. Consider
the equation defining the acidity, equation 4.1. If the material HA has resonance stability
and substance A– does not, then reactants are stabilized and HA is a weaker acid than
you would otherwise expect. If the material A– has resonance stability and substance HA
does not, then products are stabilized and HA is a stronger acid than you would otherwise
expect. If both substances, HA and A– are stabilized, then you must make a judgment
about which is stabilized the most.
4.4.1
Exercises
4.4.1.1
Which would you expect is the strongest acid, CH3 CH2 OH or CH3 C(O)OH? Why?
4.4.1.2
Which material is the stronger acid, H2 SO2 , H2 SO3 , or H2 SO4 ? Why? HINT: The S atom
is central in all Lewis structures.
Chm 118
Problem Solving in Chemistry
4.5
25
4.4.1.3
Which material is the stronger acid, HSO–2 or HSO–3 ? Why?
4.4.1.4
What is a pKa ?
4.4.1.5
Table 4.2 shows the pKa for a number of “oxy acids”. Can you find a pattern that allows
a rough prediction of the acidity?
4.4.1.6
Predict the pKa of H4 SiO4 .
4.4.1.7
Given the information in Table 4.2, if you saw data that said that arsenous acid, H3 AsO3
has a pKa of 9.23 whereas phosphorous acid, H3 PO3 has a pKa of 2.00, what might you
conclude?
4.4.1.8
The pK of the two acids in the last exercise really are what is listed there. They are true,
not intentionally wrong. Which of these materials is “normal”?
4.4.1.9
Phosphorous acid (see exercise 4.4.1.7) forms only monoanions and dianions when reacted
with bases whereas arsenous acid form both of these and trianions. Can you account for
this and the acid strength of these species (for data see exercise 4.4.1.7)? HINTS: You need
a different topology for one of them. To review, which one is odd?
4.5
Estimating Acidity, Part IV: Inductive Effects
The least important factor influencing acidity is the inductive effect. Consider the molecules
CH3 CH2 OH and CH2 FCH2 OH. The hydrogen ion that is easy to remove is on the oxygen
atom in both cases. The molecules differ in that one hydrogen in the latter has been replaced
by a fluorine atom. That fluorine atom attracts electrons toward itself in the bond to the
carbon atom, polarization, and that makes the carbon more positive. It therefore pulls
electrons from the carbon to which it is attached, making it a little more positive. That
second carbon pulls electrons better from the oxygen than normal, making the oxygen a
little more positive than normal. Hence the hydrogen ion will come off more readily.
4.5.1
Exercises
4.5.1.1
Which is the stronger acid, CF3 CH2 OH or CH3 CH2 OH? Why?
Chm 118
Problem Solving in Chemistry
4.6
26
Ion
Li+
Na+
K+
Rb+
Cs+
Fe2+
Ni2+
∆H
-123
-97
-77
-71
-63
-459
-503
Ion
Be2+
Mg2+
Ca2+
Mn2+
Zn2+
Co2+
Cu2+
∆H
-594
-459
-381
-441
-489
-491
-502
Table 4.3: Enthalpies of Hydration (kcal/mole) for Various Cations.
4.5.1.2
Which is the stronger acid, CF3 CH2 C(O)OH or CCl3 CH2 C(O)OH?
4.5.1.3
Which is the stronger acid, CF3 CH2 CH2 OH or CF3 CH2 OH?
4.6
Hydration Energy and the Acidity of Metal Ions
We saw above–section 4.2–that hydration energy plays a role in the determining of acidity.
Generally the interaction of a solvent, often water, with species, especially charged ones,
has a dramatic effect on chemical processes. In this section we examine the behavior of
metal ions when they are hydrated:
Mn+ (g) = Mn+ (aq)
(4.10)
Some data for the enthalpy change for this process are given in Table 4.3.
4.6.1
Exercises
4.6.1.1
What trends do you see in the data in Table 4.3?
4.6.1.2
Metal ions in aqueous solution typically exist as aquated species, M(H2 O)n+
6 . From where
would a proton be taken if the aquated metal ion were to act as an acid?
4.6.1.3
If the metal ion, M(H2 O)n+
6 , donates a proton to some base, what are the ligands that
remain on the resulting metal ion “complex”?
4.6.1.4
Write a chemical reaction that illustrates how a metal ion in aqueous solution, M(H2 O)n+
6 ,
acts as an acid.
Chm 118
Problem Solving in Chemistry
4.7
27
4.6.1.5
Write the expression for the equilibrium constant for the process in the last exercise.
4.6.1.6
+
Which would you expect to be the stronger acid, Cu(H2 O)+
6 or Ag(H2 O)6 ? Why?
4.6.1.7
3+
Which would you expect to be the stronger acid, K(H2 O)+
6 or Fe(H2 O)6 ? Why?
4.6.1.8
How do you account for the fact that the dominant form of V(II) in dilute acidic aqueous
2+
solution is V(H2 O)2+
6 whereas that of V(IV) is VO(H2 O)5 ?
4.6.1.9
The changes in free energy, the energy that can be converted into work other than that of
the pressure-volume kind, for the process
H+ (g) + X− (g) = H+ (aq) + X− (aq)
where X is a halogen, are -1537, -1409, -1382, and -1347 kJ/mole for X = F, Cl, Br, and I,
respectively. Plot these data versus the inverse of the ionic radius of the halide ion. Can
you guess why the plot is approximately a straight line? HINT: You can treat these data
as if they were an enthalpy for now.
4.6.1.10
Think about the extrapolation to a zero value of 1/r for the plot in the last exercise. What
physical property is that?. Give a meaning to that intercept. HINT: Think about what you
are plotting.
4.7
Distribution Diagrams
A useful tool for visualizing what is occuring in acidity reactions is a distribution diagram.
To deal with this, we need to remind ourselves about the equilibrium constant for an acid:
Ka =
[H+ ][A− ]
[HA]
(4.2 revisited)
If we take the negative of the log (to base 10) of each side, we get
− [A ]
pKa = pH − log
[HA]
(4.11)
−
[A ]
clearly changes
where pH is the common measure of [H+ ], -log([H+ ]). The ratio of [HA]
as the pH changes. A distribution diagram shows this variation in an explicit manner by
plotting the fraction of the total amount of A (sum of [A– ] + [HA]) as a function of pH. A
typical plot is shown for a metal ion with a pKa of 5.0 in Figure 4.1.
Chm 118
Problem Solving in Chemistry
4.7
28
Figure 4.1: Distribution curve for a metal ion with a pKa of 5. The left axis is the fraction of
2+
material in the form of M(H2 O)2+
6 . At pH values to the left of the line, M(H2 O)6 becomes
the dominant species.
Figure 4.2: Distribution curves for a M(H2 O)2+
6 with a pKa of, from left to right, 2 (blue),
3 (magenta), 4 (tan), and 6 (green).
Chm 118
Problem Solving in Chemistry
4.7
4.7.1
29
Exercises
4.7.1.1
Looking at Figure 4.1, what fraction of the metal ion is in the form M(H2 O)2+
6 at a pH of
4?
4.7.1.2
Looking at Figure 4.2, see if you can determine the pH at which there are equal amounts of
+
M(H2 O)2+
6 and M(H2 O)5 (OH) as a function of the pK. HINT: You could use the defining
equation as well.
4.7.1.3
Articulate why as the concentration of hydrogen ion increases, the concentration of M(H2 O)5 (OH)+
decreases. HINT: You need the equation, not a distribution diagram for this.
Chm 118
Problem Solving in Chemistry
Chapter 5
Symmetry
5.1
Definitions and Proper Symmetry Operations
Symmetry is an important tool for understanding chemistry; the application of symmetry
simplifies complicated problems. At the heart of using symmetry is the symmetry operation.
A practical definition of a symmetry operation is as follows. You look at an object and then
close your eyes. Then I do something to that object while your eyes are closed. You open
your eyes. If you cannot tell anything was done, what I did was a symmetry operation.
There are two kinds of symmetry operations, proper and improper. A proper symmetry
operation is one that you can do with your fingers on a model, a rotation, for instance. An
improper symmetry operation is one you can imagine, but not physically carry out, such as
a internal reflection plane.
Imagine a water molecule, a bent species with one of the hydrogens to the lower left and the
other to the lower right. There is a rotation by 180o about the vertical axis going through
the oxygen atom; this rotation puts the hydrogen atom that was lower left where the lower
right one was and reciprocally. You cannot tell anything was done. This rotation is called
a C2 rotation; the “C” indicating the rotation and the “2” telling you it is one of (360/2)o .
There is another proper operation that seems a little silly, but for the theory of symmetry
(called group theory) it is very important. It is the do nothing, or identity rotation. The
symbol for this rotation is “E.”
5.1.1
Exercises
5.1.1.1
A proper symmetry operation of rotation by 180o about the z axis changes the point { x y
z } to the point { -x -y z }. What does a rotation about the z axis of 90o do to the point {
x y z }? See Figure 5.1 for an illustration.
30
5.2
31
Figure 5.1: Rotation of 90o about the z axis (vertical axis) in S2+
4 . The sulfur atoms are
labeled with numbers so that the motion can be seen.
Figure 5.2: Improper symmetry operation, σ, on S2+
4 . The sulfur atoms are labeled with
numbers so that the motion can be seen.
5.1.1.2
What would you name the rotation by 90o from the last exercise?
5.1.1.3
I contend if there is a rotation of 90o there must also be a rotation of −90o . Comment.
5.1.1.4
Find a proper symmetry operation in NO–2 other than E.
5.1.1.5
Find a proper symmetry operation in NH3 other than E. Is there a second proper operation?
5.1.1.6
Demonstrate that ethylene, C2 H4 , has a C2 symmetry operation; does it have more than
one? HINT: You must know where the nuclei are in space to answer this type of question.
5.1.1.7
What are the proper symmetry operations found in the square planar compound PtCl2–
4 ?
Chm 118
Problem Solving in Chemistry
5.3
32
Figure 5.3: Another improper symmetry operation, σ, on S2+
4 . The sulfur atoms are labeled
with numbers so that the motion can be seen.
5.2
Improper Symmetry Operations: Planes of Symmetry
There are three kinds of improper symmetry operations, which are operations that cannot
be done with your fingers on a model, but must be imagined. The first is a plane of
symmetry. You must specify the plane (or use the your flattened hand to show where it
is). If the plane is the yz plane, then a reflection plane, usually labeled σ changes the point
{x y z} to {-x y z}. Consider NO–2 . This has a plane of symmetry in the molecular plane.
Also, it has a plane of symmetry perpendicular to the molecular plane and containing the
C2 axis.
5.2.1
Exercises
5.2.1.1
What is the plane of symmetry as illustrated in Figure 5.2? The vertical axis could be called
the z axis. It takes one other axis to define the plane. HINT: You could use the labels on
the sulfurs to help define your plane.
5.2.1.2
What is the plane of symmetry as illustrated in Figure 5.3? The vertical axis could be called
the z axis. It takes one other axis to define the plane. HINT: You could use the labels on
the sulfurs to help define your plane.
5.2.1.3
Is there a plane of symmetry in NH3 ? Specify where it is relative to a drawing (or, if you
prefer, in words).
5.2.1.4
Is there a plane of symmetry in C2 H4 ? Label it so a reader will understand, either in a
picture or by words.
Chm 118
Problem Solving in Chemistry
5.4
33
Figure 5.4: Improper symmetry operation, i, on the trans form of H2 O2 . The atoms are
labeled with numbers so that the motion can be seen.
5.3
Improper Symmetry Operations: Center of Inversion
The second kind of improper symmetry operation is called a center of inversion, usually
given the symbol i. This operation takes a point with coordinates of {x,y,z} and moves it
to the point {-x,-y,-z}. Or, in more prosaic terms, takes a point that is up, right, and out
and converts it to a point that is down, left, and back. In Figure 5.4 is an illustration of an
i on the trans form of hydrogen peroxide, H2 O2 .
5.3.1
Exercises
5.3.1.1
Is there a center of inversion in S2+
4 , see Figure 5.2?
5.3.1.2
Is there a center of inversion in PtF2–
4 , a square planar compound?
5.3.1.3
Is there an i in NH3 ?
5.3.1.4
Is there an i in staggered ethane, C2 H6 ? HINT: Staggered ethane appears, looking down
the C-C bond, to have the hydrogen atoms on front carbon at, say, 0, 120o , and 240o ; and
hydrogen atoms on the back carbon at 60o , 180o , and 300o .
5.3.1.5
Is there an i in eclipsed ethane, C2 H6 ? HINT: Eclipsed ethane appears, looking down the
C-C bond, to have the hydrogen atoms on front carbon at the same angles as those on the
back carbon.
5.4
Improper Symmetry Operations: Rotation-Reflection Axis
The last kind of improper symmetry operation is called a rotation-reflection axis, usually
given the symbol Sn . In this process you first rotate by an angle of 360/n, then reflect in
Chm 118
Problem Solving in Chemistry
5.5
34
Figure 5.5: Improper symmetry operation, S4 , on the B2 Cl4 . The chlorine atoms are labeled
with numbers so that the motion can be seen.
a plane perpendicular to that axis of rotation. This procedure is illustrated in Figure 5.5.
Consider a S4 about the z axis: The rotation leaves z alone and converts x to y and y to -x.
Then the reflection leaves the xy coordinates of the points alone, but inverts the z. Thus
S4 converts the point {x,y,z} to {y,-x,-z}.
5.4.1
Exercises
5.4.1.1
Use the logic concerning the point {x,y,z} as expressed above to show that S2 is the same
as i.
5.4.1.2
Is there a S4 axis in PtF2–
4 , a square planar compound?
5.4.1.3
Is there a S3 axis in NH3 ?
5.4.1.4
Is there a S6 in staggered ethane, C2 H6 ?
5.4.1.5
Is there a S6 in eclipsed ethane, C2 H6 ?
5.5
The Group of Symmetry Operations
Molecules have a certain collection, set, of symmetry operations. That set is mathematically
called a group. For instance, a water molecule has E, C2 down an axis bisecting the H-O-H
bond and containing the oxygen atom, and two planes of symmetry, one in the plane of the
molecule and the other a plane perpendicular to the molecular plane and containing the C2
axis. This set of four symmetry operations is labeled the C2v group, and is a set found for
many molecules–we will encounter it often. Note in this set there are two proper operations
and two improper ones. The general rule is that if there are any improper operations,
there will be as many improper operations as there are proper operations. Identification of
Chm 118
Problem Solving in Chemistry
5.6
35
the set of symmetry operations in a molecule is critical to applying symmetry arguments
to the molecule’s properties. For further aid in symmetry operations, see the web site
http/symmetry.otterbein.edu/galley/index.html.
5.5.1
Exercises
5.5.1.1
Does B2 H6 , called diborane, whose structure is indicated in Figure 5.6, have a plane of
symmetry? Does it have a C2 axis of rotation? Is anything wrong with the Lewis structure
of this molecule?
5.5.1.2
Find all the symmetry operations in diborane. HINT: There are eight.
5.5.1.3
Find the symmetry operations in H3 PO. HINT: You may need a Lewis structure and a
VSEPR analysis on this to get the structure; if so, do so, as you always should.
5.5.1.4
Comment on the sentence taken from The Little Book of Symmetry, published in 1879 by
Boniface Beebe: “The following molecules all contain a C3 axis of rotation: CH4 , SF6 , NH3 ,
CH2 Cl2 .”
5.5.1.5
What are the symmetry operations present in non-cyclic O3 ?
5.5.1.6
What are the symmetry operations present in cyclic O3 ? This group of symmetry operations
is called D3h .
5.5.1.7
Find the proper symmetry operations in benzene, a planar C6 H6 molecule. HINT: Don’t
forget resonance in this case, which is very important for assignment of symmetry. Why?
5.5.1.8
Find the improper symmetry operations in benzene, a planar C6 H6 molecule.
5.6
The Behavior of Single Objects Under Symmetry Operations
To use symmetry to learn about molecules, we have to understand how “objects” in the
molecule behave when the symmetry operations are applied to the molecule. Reflect upon
this for a moment: A symmetry operation leaves the molecule unchanged; an object in the
molecule, whether it is a vector, a function, a motion, are not necessarily left unchanged.
Chm 118
Problem Solving in Chemistry
5.6
36
Figure 5.6: The structure of diborane, B2 H6 . The small spheres are hydrogen atoms.
Figure 5.7: Objects within the water molecule.
Chm 118
Problem Solving in Chemistry
5.6
37
C2v
Object 1
E
1
C2z
-1
σplane
1
σperp
-1
Table 5.1: Table of characteristic numbers for motion of water molecule to the left.
To get correct answers about the behavior of the objects requires, foremost, that you keep
track of how many objects you have.
As a first example, consider the water molecule and its C2v point group. In Figure 5.7,
structure 1, are three vectors that I want to consider as one object. These vectors describe
the motion of the water molecule as a whole to the left; so the “object” is motion of the
molecule to the left. What we want to know is what happens to that object under the
symmetry operations of the water molecule? Under the identity operation, the object is
left alone; it “goes into itself.” We want to associate a number that corresponds to what
happens to the object. Under the identity operation, that number would be a one. We
write that as a 1 under the identity, see Table 5.1. Consider now the C2 rotation. This
turns all three vectors of the object into minus themselves, or the object goes into minus
itself. The appropriate number is “-1,” see Table 5.1. You can finish the arguments for the
other two symmetry operations and get the values in the table.
5.6.1
Exercises
5.6.1.1
Find the set of characteristic numbers for the single object show in Figure 5.7 given as 2.
Fill out a table as done above.
5.6.1.2
Find the set of characteristic numbers for the single object show in Figure 5.7 given as 3.
Fill out a table as done above.
5.6.1.3
Find the set of characteristic numbers for the single object show in Figure 5.7 given as 4.
Fill out a table as done above.
5.6.1.4
A molecule’s shape determines the symmetry operations it has. Objects in the molecule,
the vectors on the hydrogen atoms, the spheres, of the last few exercises, determine the
characteristic numbers we have been talking about. A table of the characteristic numbers
for C2v symmetry, that is, the symmetry of a water molecule, is in Table 5.2. The names
of the sets of characteristic numbers, A1 , . . . B2 , are the normal ones used; however, the
important point is that they are just names and for our purposes in this course it does not
matter if you use lower case or capital letters. Use a1 as a name, nothing more (although
that particular name is always given to the set of numbers that are all “ones”). Experss
your answers in the last several problems like a professional: “b2 symmetry”, for instance.
Chm 118
Problem Solving in Chemistry
5.7
38
C2v
A1
A2
B1
B2
E
1
1
1
1
C2z
1
1
-1
-1
σplane
1
-1
1
-1
σperp
1
-1
-1
1
Table 5.2: Table of characteristic numbers for C2v symmetry.
C3v
A1
A2
E
E
1
1
2
2C3
1
1
-1
3σv
1
-1
0
Table 5.3: Table of characteristic numbers for C3v symmetry.
5.6.1.5
Consider the NH3 molecule. Find the symmetry operations. Do you agree with those given
in the C3v table, Table 5.3? HINT: In that table the nomenclature 2C3 implies that there
are two different rotations of 120o , both of which have the same “number” characterizing
them within any given row.
5.6.1.6
For the symmetric stretch in the NH3 molecule–where all three hydrogen atoms move away
from the nitrogen atom along the bond axis–treated as one object, what is the symmetry?
5.6.1.7
Sketch the structure of CH2 Cl2 and find the name of the set of symmetry operations that
this molecule has.
5.6.1.8
Find the C-H stretch in CH2 Cl2 , last problem, that has b1 symmetry. HINTS: Clearly the
answer must be one of the sets of characteristic numbers that you have in the table. Let
the plane defined by the carbon and the two hydrogens be the xz plane.
5.6.1.9
Find the bend of the two C-H bonds in the plane defined by HCH in CH2 Cl2 that has a1
symmetry. HINTS: Clearly the answer must be one of the sets of characteristic numbers
that you have in front of you. A bend changes an angle but not a bond length. Hence the
vectors must be perpendicular to bonds and, in this case, centered on the hydrogen atoms.
5.6.1.10
Find the bend (or some call it a wag) of the two C-H bonds in the plane defined by HCH
in CH2 Cl2 that has b1 symmetry. HINTS: Clearly the answer must be one of the sets of
characteristic numbers that you have in front of you. A bend changes an angle but not a
bond length. Hence the vectors must be perpendicular to bonds and, in this case, centered
on the hydrogen atoms. Also, as a “bend” it must change an angle.
Chm 118
Problem Solving in Chemistry
5.7
39
C2v
Object 5
E
2
C2z
0
σplane
σperp
Table 5.4: Beginning of table of characteristic numbers for two spheres on the hydrogen
atoms of a water molecule.
D4h
A1g
A2g
B1g
B2g
Eg
A1u
A2u
B1u
B2u
Eu
E
1
1
1
1
2
1
1
1
1
2
2C4z
1
1
-1
-1
0
1
1
-1
-1
0
C2z
1
1
1
1
-2
1
1
1
1
-2
2C‘2
1
-1
1
-1
0
1
-1
1
-1
0
2C“2
1
-1
-1
1
0
1
-1
-1
1
0
i
1
1
1
1
2
-1
-1
-1
-1
-2
2S4z
1
1
-1
-1
0
-1
-1
1
1
0
σh
1
1
1
1
-2
-1
-1
-1
-1
2
2σv
1
-1
1
-1
0
-1
1
-1
1
0
2σd
1
-1
-1
1
0
-1
1
1
-1
0
Table 5.5: Characteristic numbers for D4h symmetry.
5.7
The Behavior of Several Objects Under Symmetry Operations
If we have more than one object in a molecule, things are a little different than they were in
the last section. To use symmetry in these cases we have to ask how many of the objects go
into themselves, or minus themselves, and add up those numbers; and then ask how many
objects get turned into another, in which case the object does not go into plus or minus
itself, so contributes a zero to the set. Consider the two spheres, one on each hydrogen
atom, show in Figure 5.7 given as 5. Clearly each of the two spheres go into themselves
under the identity operation, so the number under that operation is a “2.” On the other
hand, under a C2 rotation, the left hand sphere goes to the right, a zero, and the right hand
sphere goes to the left, another zero; total, zero. These data are given in Table 5.4.
5.7.1
Exercises
5.7.1.1
Find the set of characteristic numbers for the pair of objects, two spheres, one on each
hydrogen atom, show in Figure 5.7 given as 5. Fill out the rest of Table 5.4. Note that this
set of numbers is not one of the characteristic sets for C2v symmetry.
5.7.1.2
Find the set of characteristic numbers for the pair of object show in Figure 5.7 given as 2.
This is the same picture as exercise 5.6.1.1 except we are assuming we have two objects,
Chm 118
Problem Solving in Chemistry
5.8
40
Figure 5.8: Objects for vibration of atoms in a square planar molecule. See exercise 5.7.1.4
for a description.
not one. Fill out a table as done above. Note that this set of numbers is not one of the
characteristic sets for C2v symmetry
5.7.1.3
I contend that the number under the “E” symmetry operation is the number of “pictures”
that you are dealing with. See if you can offer a rationalization for this contention.
5.7.1.4
Figure 5.8 illustrates two motions of atoms in a square planar molecule such as PtF2–
4 .
Structure 1 contains two arrows that are one object; both the indicated atoms must move
as indicated at the same time. Structure 2 also contains two arrows that is a single object.
You are to consider 1 and 2 together as two objects. The symmetry of this molecule is called
D4h and has the symmetry operations given in Table 5.5. Find the set of characteristic
numbers generated by the two objects.
5.8
Classifying the Symmetry for Multiple Objects; the Combinations
We learned in section 5.6 that we could label the “symmetry” of an object in a molecule
if all symmetry operations turned that object into either itself (+1) or minus itself (-1) by
finding the set of numbers that those operations produced. In the last section, we found
that multiple objects, which always produce a integer greater than one under the identity
symmetry operation, sometimes produce a set of numbers that are not to be found in the
list of all possible characteristic numbers–exercise 5.7.1.1; and sometimes produce a set that
is found in the list of all possible characteristic numbers–exercise 5.7.1.4. In the latter case,
there is no issue; since the set of characteristic numbers from exercise 5.7.1.4 is that of the
“Eu ” row of the table, the vibration of the molecule shown in Figure 5.8 is of Eu symmetry.
The contention is that in the former case, exercise 5.7.1.1, that the answer you obtained is
Chm 118
Problem Solving in Chemistry
5.8
41
Figure 5.9: Spherical objects for determination of combinations in a square planar molecule.
See section 5.8 for a description.
the addition of two (or more, if the number under E is greater than 2) rows of the table.
We say that the two spheres on the hydrogen atoms can be combined (in a way we shall
specify) to form an A1 combination and a B1 combination.
In group theory there is a more direct way to find the characteristic
numbers within a set of arbitrary numbers rather than by just the
trial and error procedure of adding various rows. Let χC,R be the
number generated by the multiple objects, such as those in exercise 5.7.1.1 by the symmetry operation labeled C; and let χC,i be
the characteristic number for the same symmetry operation for the
ith row in the table of characteristic numbers. Then the number
of times that the ith entry is in the set generated by the multiple
objects, ni , is given by:
ni =
1X
nC χC,R χC,i
h
(5.1)
i
where h is the total number of symmetry operations in the group
and nC is the number in front of the equivalent symmetry operations
in the characteristic table (i.e., the “2” in front of C3 in the table
for C3v group).
The idea behind the combinations are most easily seen from what we have already done.
Consider the two spheres on the hydrogen atoms of the water molecule. They generated
a set of numbers that you determined in exercise 5.7.1.1. The number under the identity
symmetry operation was a 2. Whenever this is true, the combinations of the two objects
is simply the sum and the difference. In this case the sum would give a sphere on each
hydrogen atom with the same sign, say the implied positive sign that you have been dealing
with naturally. On the other hand, the difference is two spheres of equal size, but one of
which is positive and the other negative in sign. (We will see how important this is when
we deal with wave functions, where the function has a sign at a given point in space.) The
combination with both spheres positive is A1 , that with one positive and the other negative
is B1 .
Chm 118
Problem Solving in Chemistry
5.8
42
D3h
0
A1
0
A2
0
E
00
A1
00
A2
00
E
E
1
1
2
1
1
2
2C3z
1
1
-1
1
1
-1
3C2
1
-1
0
1
-1
0
σh
1
1
2
-1
-1
-2
2S3
1
1
-1
-1
-1
1
3σv
1
-1
0
-1
1
0
(x, y)
z
Table 5.6: Characteristic numbers for D3h symmetry.
There is a formula in group theory to find the combinations. It
requires that you take one of the multiple objects, any one, and
determine what happens to that object under all the symmetry
operations. For instance, imagine a square with four spheres on it,
such as show in Figure 5.9. What happens to sphere 1 under the
identity? It goes to sphere 1, itself; write that down. What happens
to sphere 1 under a clockwise C4 ? It goes to sphere 2; write that
down. Under C−4 ? To sphere 4. Etc. You now have this list of
spheres that we will label as Oi (s1) (which stands for symmetry
operation on sphere 1). You now use that list in the following sum,
where Y is the desired combination of symmetry Γ:
X
YΓ =
χOi ,Γ Oi (s1)
(5.2)
i
and χOi ,Γ is the characteristic number for the symmetry of interest
for the symmetry operation of interest. This looks complicated, but
all the quantities you need are entries in the table you prepared as
directed above or in the characteristic table for the symmetry of
interest.
5.8.1
Exercises
5.8.1.1
Show that the statement that the numbers generated in exercise 5.7.1.1 are produced by
adding the A1 and B1 characteristic numbers of the C2v point group.
5.8.1.2
Place a sphere where each of the fluorine atoms is in the square planar compound PtF2–
4 .
Find the set of numbers generated by these four objects.
5.8.1.3
Find the symmetries of the combinations of the four spheres from the last exercise. Use
either the trial and error method or the formula in equation 5.1. HINT: You do not need
to know what the combinations look like to do this problem; but see below.
Chm 118
Problem Solving in Chemistry
5.8
43
Figure 5.10: A motion of SO3 . All arrows representing motion of oxygen atoms are equal
in length
5.8.1.4
Use equation 5.2 to find the combinations of two spheres (exercise 5.7.1.1) and see if it
agrees with the answer we stated in section 5.8.
5.8.1.5
Use equation 5.2 to find the combinations of four spheres (exercise 5.8.1.3).
5.8.1.6
Consider the SO3 molecule. Find the symmetry operations. Do you agree with those given
in Table 5.6 for D3h symmetry? HINTS: To find symmetry operations you must know the
structure of the molecule in space. In the D3h table the nomenclature 2C3z implies that there
are two different rotations of 120o , both of which have the same “number” characterizing
them within any given row.
5.8.1.7
In Figure 5.10 is a motion of the SO3 molecule. This is one object. What is the symmetry
of this (single) motion? How would you describe what is happening to the SO3 molecule?
5.8.1.8
Label the three (identical) oxygens of SO3 a, b, and c. Now for each symmetry operation
for SO3 determine how many of these oxygen atoms retain their label after the symmetry
operation has been performed. HINT: For instance, for a C2 (180o ) rotation about the SO(a) bond, O atoms b and c are not where they were, but O(a) is. So the number associated
with that rotation is “1”.
Chm 118
Problem Solving in Chemistry
5.8
44
5.8.1.9
I contend that your answer for the last problem is made up of the sum of two of the rows of
the D3h table. That is, under each symmetry operation, if you add the numbers associated
with that operation for row X and row Y of the table, you will get your answer from the last
exercise. See if you can determine which those rows are, either by trial and error or by the
use of equation 5.1. Speaking professionally, you would say “The characteristic numbers
0
00
generated by the three oxygen atoms are a2 and e where you should substitute the correct
answers.
Chm 118
Problem Solving in Chemistry
Chapter 6
Quantum Mechanics
6.1
Probability and Quantum Measurement
Quantum mechanics is the method to determine the properties of small objects. This science
is weird because the actions of small objects do not fit our notions of proper behavior, which
were hewn from what we can readily observe. Two of these actions are the probabilistic
nature of quantum measurements and the effect of the experiment on the result.
To illustrate these we use a Stern-Gerlach apparatus. For our purposes this is a box–see
Figure 6.1–with an inlet hole on the left and two exit ports on the right of the box, one on
top and one on the bottom. (We will want to turn the box on its side for some experiments,
in which case the exit holes are in front or back.) Inside the box is a magnet (and some
plumbing) that generates an inhomogeneous magnetic field. We are not going to talk about
this field except to say that a classical (big) magnet (with a property called a magnetic
moment) passing along the magnetic field of the apparatus would be deflected either up or
down by some small or large amount: a bunch of classical (big) magnets would exit in a fan
shape, including some that pass straight through. When silver atoms are passed into the
Stern-Gerlach apparatus via the left hand hole, they exit either from the top hole or the
bottom hole. In particular, none of these silver atoms try to pass straight through and run
into the wall of the box. So here is a difference between a big and small particle right away;
big ones can take any value of the magnetic moment whereas small ones seem to choose
only one of two values. This is, of course, the quantization for which quantum mechanics
is famous.
Figure 6.1: The Stern-Gerlach apparatus
45
6.1
46
Figure 6.2: Three Stern-Gerlach magnets located in series; see section 6.1.
Now set up the following experiment. Run some silver atoms through a Stern-Gerlach (SG)
device and send those that exit the top hole of the device (lets call those “spin up”) and
put them into the entrance hole of a second SG device. All of them, 100%, exit the top
hole of the second SG device. That makes sense. If we use the first device to choose spin
up silver atoms, and then measure their spin with a second up/down SG device, we find it
is spin up. The reproducibility of science; duh!
Consider a similar experiment. Take the output from the top hole of a first SG device and
input that beam into a second SG device turned on its side (so that the holes are front and
back). We observe that 50% of the silver atoms come out the front hole and 50% out the
back, in a random fashion. For instance, if we let one silver atom at a time go through
the device, we would observe hits on the front and back detectors that might be like this:
f,f,b,f,f,b,b,f,b,b,b,b,f,f. Over a long enough time, equal hits; random probability for either
a front hit or back hit for any given silver atom. All of this could be explained classically.
We simply would say that the first silver atom (in the list just above) has spin up (hence
through the upper hole of the first SG device) and spin front (hence through the front hole
of the second SG device).
If we do the experiment shown in Figure 6.2 life gets a little more messy. Here we select up
spin silver atoms from an initial SG magnet, pass these into a second SG magnet aligned
front/back, and select back spin silver atoms and pass those through a third SG magnet
oriented up/down. If our conclusion at the end of the last paragraph was correct, we would
expect the silver atoms exiting the second SG magnet to have spin up/back and to come out
of the upper hole of the third SG magnet. What happens is that there is a 50% probability
of getting a silver atom out of the upper hole, but also a 50% chance of getting one out of
the bottom hole of the third magnet. In the language of quantum philosophy, making the
front/back measurement in the second magnet removes all traces of the results of the up
spin resulting from the first magnet.
6.1.1
Exercises
6.1.1.1
A silver atom is directed to a Stern-Gerlach apparatus arranged in the up/down direction.
What is the probability that the Ag atom will come out of the upper tube?
6.1.1.2
A silver atom that came out of the upper tube of the apparatus in the last exercise is
directed to another Stern-Gerlach apparatus arranged in the up/down direction. What is
the probability that the Ag atom will come out of the upper tube?
Chm 118
Problem Solving in Chemistry
6.2
47
Figure 6.3: Two cosine waves
6.1.1.3
A silver atom that came out of the upper tube of a SG apparatus is directed to a SG
apparatus arranged in the front/back direction. What is the probability that the Ag atom
will appear from the “front” tube?
6.1.1.4
Quantum mechanics is weird, but not illogical. Try this: A silver atom from the upper tube
a SG apparatus in the up/down orientation is directed into a SG apparatus that is tilted
0.001 degrees toward the front/back position; that is, it is almost an up/down apparatus.
What is the probability a Ag atom will come out of the upper tube of the second apparatus?
HINT: I am not looking for a numerical answer, but a statement showing understanding.
6.1.1.5
Quantum mechanics is weird, but not illogical. Try this: A silver atom from the upper
tube of an up/down SG apparatus is directed into a SG apparatus that is tilted 45 degrees
(toward “front/back” with the “upper” hole becoming the “front” hole) relative to the first
SG magnet; that is, it is between a situation where the probability of the silver atom coming
out the upper tube is 1.0 and the situation where the apparatus is in a front/back position
and the probability is 0.5 from either hole. What would you guess that the probability a Ag
atom will come out of the “upper” tube (or “front” tube, depending on your perspective)
of the second apparatus? HINT: See the last hint.
6.2
Waves and Interference
The mathematics of quantum mechanics use a wave equation to describe a particle. A wave
is something that extends throughout a region of space and has values at various points in
space. For instance, the wave function Sin[π x], has a value of zero at x = 0; a value of 1 at
x = 0.5, and a value of zero at x = 1. In fact, we could simply plot Sin[π x] as a function
of x and see what value it has at any point in space. Note a function is just a recipe for
Chm 118
Problem Solving in Chemistry
6.3
48
Figure 6.4: The wave obtained by adding Cos[x] + Cos[2x]
finding a value at a point in space. The wave is not anything “real” for this description of a
quantum particle, but the square of that wave gives the probability of finding the particle
at the respective point in space.
Where the distinction between wave and probability becomes critical is when there are two
sources of the “particle,” say from one or the other of the two slits in a two slit experiment.
The quantum mechanical rules for dealing with this situation (which are supported as
correct because they always give the correct answer) is to add the waves and then square
to get the probability. This procedure gives the possibility of interference, where the two
waves add either to make a larger wave or to partially or completely cancel. Figure 6.3 is
a plot of the wave function, the wave amplitude, versus x for two cosine waves; the solid
line is that for Cos[x] and the dashed is for Cos[2x]. Look at x = 0.5; at this point there
is constructive interference as both waves are positive. On the other hand, at x = 3 there
is dominantly destructive interference. We can show this directly by actually plotting the
sum of the two waves as is done in Figure 6.4.
6.2.1
Exercises
6.2.1.1
What is the principle way in which waves that you have seen in life (those are for most
people waves in water) differ from a particle?
6.2.1.2
Do you understand the most fundamental problem with quantum mechanics when one says
that particles are described by waves?
6.2.1.3
Consider the wave functions Sin[π x] and Sin[π x + π]. What is the nature of the interference
of these waves, constructive or destructive?
Chm 118
Problem Solving in Chemistry
6.4
49
6.3
Simple Version of the Rules of Quantum Mechanics
The rules for the application of quantum mechanics require that the wave function discussed
above be manipulated by an operator. Operators, O, in the mathematical sense used here
are objects that change functions. For instance, a trivial operator could be the number “3,”
which triples a function,
O(2x + 1) = 3(2x + 1)
(6.1)
while a more subtle operator could be the differential operator,
O(2x + 1) =
∂
(2x + 1) = 2
∂x
(6.2)
Often in quantum mechanics we are looking for an operator that changes a function into a
multiple of itself, for instance,
Oψ = cψ
(6.3)
where ψ is the wave function and c is a number. When we have an operator and a function
that fulfill this relationship, it is called an eigenfunction/eigenvalue problem. In particular,
a satisfactory wave function for a one dimensional problem where the energy is constant
satisfies the eigenfunction/eigenvalue relationship
−
~2 ∂ 2
ψ + V (x)ψ = Eψ
2m ∂x2
(6.4)
where ~ is Planck’s constant divided by 2π, m is the mass of the object, the quantum
object, a parve,1 V(x) is the potential energy the parve experiences, which may change as
the parameter x is changed, and E is the stationary state energy. This equation is called
Schrödinger’s equation in one dimension.
6.3.1
Exercises
6.3.1.1
∂
Show that the operator ∂x
on the function e−kx , where k is a constant, is an eigenfunction/eigenvalue problem. What is the eigenvalue? HINT: If you don’t know calculus, skip
this exercise.
6.3.1.2
∂
Show that the operator ∂x
on the function Sin(k x), where k is a constant, is not an
eigenfunction/eigenvalue problem. HINT: If you don’t know calculus, skip this exercise.
6.3.1.3
2
∂
Show that the operator ∂x
2 on the function Sin(k x), where k is a constant, is an eigenfunction/eigenvalue problem. HINT: If you don’t know calculus, skip this exercise.
1
A parve is a name unique to this document
Chm 118
Problem Solving in Chemistry
6.4
6.4
50
A Simple but Useful Quantum Problem: The Parve on
a Pole
In this section we solve a quantum problem. It is a particularly easy problem to solve, yet
a number of the features of quantum behavior are illustrated by it. Further, the results of
this problem will be useful to us throughout the rest of this course, reaching all the way
into thermodynamics. The problem at hand is a particle, a parve, moving in one dimension
along a pole of length L with no potential energy present. Schrödinger’s equation for this
situation is given by equation 6.4 with V(x) = 0. In addition, I want to demand that the
parve stay on the pole, so I am going to insist that the potential energy outside the ends
of the pole be infinite, so there is no possibility that the parve be there. We need to find
functions that satisfy this equation. I suggest that you learned in exercise 6.3.1.3 that Sin(k
x) would satisfy Schrödinger’s equation in this case, as will the functions Cos(k x) and e−kx ,
where k is a constant. Which of these three solutions is proper for the quantum problem?
That the wave function must satisfy Schrödinger’s equation is one consideration. The
second, and an important one, is that the boundary conditions must be met. Here is
how that plays out in the parve on a pole, POP, problem. A second demand of quantum
mechanics is that the wave functions must vary smoothly in space, from one point to another.
We know from our demand that there is no probability for the parve to be outside the pole
that Px<0 = 0. The probability is given by the square of the wave function for the parve,
ψ, so Px<0 = ψ 2 = 0; which requires the ψx<0 be zero. Since the wave function must vary
smoothly with a change in x, this demands that ψx=0 = 0. That is the boundary condition.
As you will show in the exercises, of the three functions that satisfy Schrödinger’s equation,
only one satisfies this boundary condition.
There is a second boundary to our pole, the end where x = L. Outside this end of the pole,
the probability must also be zero, as must, therefore, the wave function. This demands a
second requirement on our function, that Sin(k x) = 0 at x = L; or (k L) = nπ, where
n is an integer. Look at that! The boundary condition introduces a quantum number,
n. We conclude that a legal (in that it obeys quantum rules) wave function for a POP is
ψ = Sin(n π x/L). A quantum problem solved!
6.4.1
Exercises
6.4.1.1
Show that the wave function Cos(k x) and e−kx do not satisfy the boundary condition that
ψ = 0 at x = 0.
6.4.1.2
Take equation 6.4, set V = 0, use the ψ from the end of the last section, and show that the
2 ~2 π 2
energy for a POP is given by n2mL
2 . HINT: If you don’t know calculus, skip this exercise.
Chm 118
Problem Solving in Chemistry
6.5
51
6.4.1.3
The lowest energy wave function for a parve on a pole of length L is a sine wave running
between x = 0 and x = L and spanning half a wavelength in this interval. Use the de
Broglie relationship, p = λh , where p is the momentum, p = m v, and solve the equations
for the energy of this parve. HINTS: All educated people (that means you!) should know
deBroglie’s equation. Also, all the energy of this parve is kinetic energy (E = 1/2 m v2 ).
6.4.1.4
A parve is trapped on a pole with zero potential energy. What happens to the lowest
allowed energy of the parve if the length of the pole is doubled?
6.4.1.5
Sketch the n = 5 wave function for a parve on a pole.
6.4.1.6
Given the de Broglie postulate in problem 6.4.1.3, what can you say about the kinetic energy
of a parve on a pole in the n = 5 level compared to the n = 1 level? Explain without a
calculation by thinking about the de Broglie postulate.
6.4.1.7
Compute the energy an electron would have if it was trapped on a pole of length 4.2 Å
(about the length of a four carbon fragment) in the n = 1 level using the POP model.
HINTS: You need to look up some constants and expand units like the Joule in terms of
kg, m, and sec.
6.4.1.8
What energy would it require to excite the electron from the n = 1 to the n = 2 level in
the system of the last problem?
6.4.1.9
What wavelength of light must be absorbed to provide the energy to excite the electron
from the n = 1 level to the n = 2 level in the system of the last two problems? HINT:
You should know (or learn and then know) the relevant equation, one of Einstein’s several
equations.
6.4.1.10
Given that you know what the wave function for the ground state of a POP looks like,
make a guess about what the wave function for the ground state of a parve on the surface
of a square table looks like. If you are a reasonable artist, you might be able to draw the
shape of the wave function in the third dimension given the surface is defined by x and y;
you certainly should be able to move your hands over a table to outline roughly the shape
of the wave function for a parve on a surface.
6.4.1.11
Make a guess for the wave function for a parve in a room, a three dimensional problem.
You will not be able to draw it (since that requires four dimensions), but should be able to
describe it. How many quantum numbers will be necessary for this wave function?
Chm 118
Problem Solving in Chemistry
6.6
6.5
52
Quantum Mechanics of the Hydrogen Atom I. Energy
Levels
The wave functions and the energy levels for a hydrogen atom can be solved exactly using
quantum mechanics. This solution is mathematically messy, so we won’t do it here, but
the results are worth discussing for their use in understanding all the other atoms of the
periodic table. Two features of importance to us come out of any quantum problem, the
energy of the allowed quantum levels and the nature of the wave functions. In this section,
we discuss allowed energy levels.
The allowed energy levels of the hydrogen atom are given by a particularly easy equation,
h m i e 2 2 1
En = −
2
4π0 ~ n2
(6.5)
where e is the charge on the electron and m is the mass of the electron. Notice that all the
constants in the two square brackets are known, so this equation can be written numerically
as:
1
En = −13.601 2
(6.6)
n
in units of electron volts, eV. The energy of the allowed levels in the hydrogen atom depend
inversely on the square of the quantum number n, which is an integer varying from one
upward. Thus the most stable level is -13.601 eV (n = 1), the next most stable -3.4 eV (n
= 2), etc. These energies differences agree perfectly with the positions of the wavelengths
of absorbance and emission of energy of the hydrogen atom (which move an electron to or
from higher energy levels to lower ones), verifying the results of the quantum calculation.
6.5.1
Exercises
6.5.1.1
How many quantum numbers are there in the hydrogen atom? Why? HINT: Think about
what introduces quantum numbers.
6.5.1.2
Light from the sun is partially absorbed at 656 nm. Show that this is consistent with
hydrogen atoms on the surface of the sun being excited from the n = 2 level to the n = 3
level.
6.5.1.3
There are a whole “series” of wavelengths that fail to reach earth from the sun because of
excitations from the n = 2 level of hydrogen atoms to more excited levels. Calculate a few
of these. NOTE: This series is called the Balmer series.
6.5.1.4
The ionization energy is the energy necessary to remove an electron from an atom in the
gas phase. What is the normal ionization energy of a hydrogen atom?
Chm 118
Problem Solving in Chemistry
6.6
6.6
53
Quantum Mechanics of the Hydrogen Atom II. Radial
Wave Function
The discussion of wave functions for a hydrogen atom is difficult because of our inability
to handle multidimensional space. We can deal with one or two dimensional spaces easily,
and with three dimensional space if we are careful. However, to visualize a hydrogen wave
function we would need to plot the value of the function, ψ, as a function of x, y, and z
positions in space: that takes four dimensions, with which few of us can easily accommodate.
The method of choice to avoid this problem is to plot the wave functions under some sort
of restrictive condition and then to let your imagination run free to accommodate the true
function. In this section we give one view of the hydrogen atom wave functions. It is from
these pictures that we approximate more complicated atoms. We note here that the three
quantum numbers for hydrogen atoms are usually given by a combination of numbers and
letters. The principle quantum number, n, is an integer starting at 1. The second quantum
number, usually called `, has values running from 0 to n-1, but is usually symbolized, for
historical reasons, by a letter code: The code is, ` = 0, s; ` = 1, p; ` = 2, d; ` = 3, f.
The third quantum number, m, has values from -` to `, but is often discussed in a manner
beyond what we cover here, in terms of Cartesian coordinate symbols. For instance, the
three m values for ` = 1, which rigorously are -1, 0, and 1, are often labeled x, y, and z; and
the five m values for ` = 2, which are -2, -1, 0, 1, and 2, are called xy, xz, yz, x2 -y2 and z2 .
We will not deal with any of this with mathematical rigor, but for
your enlightenment we discuss one wave function in detail. A typical
wave function is that with quantum numbers (n, l, m) of (3, 1, 0),
or what is usually called the 3pz wave function. It has the following
form:
r r −r
ψ3pz = C(z/r)(1/a)3/2 (1 − ) e 3a
(6.7)
6a a
where C and a are constants, r is the distance from the nucleus
to the electron, and z is the z component of that vector length.
Notice that the z direction is favored by this function (when z =
0, the function disappears) and that the function gets smaller as
the distance from the nucleus increases because of the exponential
decrease in the last term.
Radial wave functions give how the value of the wave function changes as you move in a
straight line away from the nucleus in some arbitrary direction (although as you will see in
the exercises, certain choices of direction lead to odd results). The results given here are
simply what comes out of solving Schrödinger’s equation; They are by no means intuitive.
Radial wave functions depend on the quantum numbers n and ` but not on the quantum
number m. There are distinct patterns that evolve as n and ` are changed. The wave
functions are shown in Figures 6.5 to 6.8, and are best viewed in color.
Another important feature of radial functions involves the square of the function, which, as
we have discussed, gives the probability. Want we will need, especially for our discussion of
multi-electronic atoms is the radial distribution curves which give the probability of finding
the electron at a given distance from the nucleus. Since the number of possible volumes
Chm 118
Problem Solving in Chemistry
6.6
54
Figure 6.5: The radial wave function for the n = 1 electron of hydrogen atom. Disregard
the vertical scale.
Figure 6.6: The radial wave functions for the n = 2 electron of hydrogen atom. Disregard
the vertical scale
Figure 6.7: The radial wave functions for the n = 3 electron of hydrogen atom. Disregard
the vertical scale
Chm 118
Problem Solving in Chemistry
6.6
55
Figure 6.8: The radial wave functions for the n = 4 electron of hydrogen atom. Disregard
the vertical scale.
Figure 6.9: The radial distribution curve for the n = 1 electron of hydrogen atom. Disregard
the vertical scale.
far from the nucleus is greater than those close, the square of the function needs to be
multiplied by 4πr2 to yield the correct function. These plots are shown in Figures 6.9-6.12
6.6.1
Exercises
6.6.1.1
Look at Figures 6.5-6.8 and find a pattern that exists among the n s wave functions. How
do they change from n = 1 to n = 4?
6.6.1.2
Look at Figures 6.5-6.8 and find a pattern that exists among the n p wave functions. How
do they change from n = 2 to n = 4?
6.6.1.3
Look at Figures 6.5-6.8 and find a pattern that exists among the n d wave functions. How
do they change from n = 3 to n = 4?
Chm 118
Problem Solving in Chemistry
6.6
56
Figure 6.10: The radial distribution curves for the n = 2 electron of hydrogen atom. Disregard the vertical scale
Figure 6.11: The radial distribution curves for the n = 3 electron of hydrogen atom. Disregard the vertical scale
Figure 6.12: The radial distribution curves functions for the n = 4 electron of hydrogen
atom. Disregard the vertical scale.
Chm 118
Problem Solving in Chemistry
6.7
57
Figure 6.13: The 3d radial wave function with three values of the wave function chosen.
See text. Disregard the values on the vertical scale.
6.6.1.4
Find the relationship between the number of nodes in an n s wave function and the n values.
6.6.1.5
How does the approximate distance of maximum in the wave function from the nucleus
change as n value changes?
6.6.1.6
See if you can find a relationship between the number of nodes in a radial wave function
and the n and ` values. HINT: You will need to use the numerical values for `, not the
letter codes.
6.6.1.7
Predict what a 5s wave function would roughly look like.
6.6.1.8
Predict what a 5g wave function would roughly look like. HINT: g is code for ` = 4.
6.6.1.9
Describe how the radial distribution curves for the n s elections differ as n changes.
6.6.1.10
Describe how the radial distribution curves for the n =4 election changes as ` changes.
Chm 118
Problem Solving in Chemistry
6.7
58
Figure 6.14: The surface with all values equal to a small number, both positive (yellow)
and negative (blue), for a 3dx2 −y2 angular wave function.
Figure 6.15: The surface with all values equal to a moderate number, both positive (yellow)
and negative (blue), for a 3dx2 −y2 angular wave function.
Chm 118
Problem Solving in Chemistry
6.7
59
Figure 6.16: The surface with all values equal to a large number, both positive (yellow) and
negative (blue), for a 3dx2 −y2 angular wave function.
6.7
Quantum Mechanics of the Hydrogen Atom III. Angular
Wave Function
The radial wave function shows how the wave function changes magnitude and sign as one
moves out from the nucleus. The angular wave function gives a sense of how the wave
function changes magnitude as one moves around the nucleus in an angular sense. There
are several ways of illustrating this. For our purposes, we are going to look at all places in
space where the wave function has the same value and connect those points with a surface;
we will do this for both positive and negative values of the same magnitude. Again, these
results come from solution to Schrödinger’s equation; you simply must learn them as they
are the key to bonding in chemical compounds. The angular functions depend on the
quantum numbers ` and m, and not on the quantum number n.
The concept is illustrated by looking at the function named dx2 −y2 . In Figure 6.13 we give
the radial function for a 3d electron. On that diagram are three lines indicating various
values of the function. Note that each line crosses the function twice. Consider now the line
with the lowest value on the absissa; we take that value, call it η, as the one of interest. If
we now change the angle and find the coordinates where the value of the function is η, and
then connect all the points of that value, we get a picture like shown in Figure 6.14. Note
that the surface is “large” because the values of r where the function has value of η are far
apart. Now consider the middle line in Figure 6.13 and the surface corresponding to its
value, shown in Figure 6.15. The surface now is somewhat smaller. Finally, the upper line
in Figure 6.13 gives the surface shown in Figure 6.16. It is possible to plot these isosurfaces
because we ask where in three dimensional space the value of the function has a certain
fixed value. This is the nature of the angular functions that we will use in this course.
So what do the angular functions look like? The s function is spherical, as shown in
Chm 118
Problem Solving in Chemistry
6.7
60
Figure 6.17: The surface with all values equal to a number for an s wave function.
Figure 6.17. Sometime people say this function is like an onion in that there are the layers
we discussed above–Figures 6.14-6.16–one after another as you move toward the nucleus.
There are three p functions, usually called px , py , and pz . Figure 6.18 gives the shape of
a px function. It doesn’t take much imagination to determine what py and pz look like; in
fact, the subscript on the value of the letter corresponding to ` exactly gives the direction
in which the function is big.
This last relationship holds for most of the d orbitals as well. There are five of these, dxy ,
dxz , dyz , dx2 −y2 , and dz 2 , For instance, the dxz orbital is big where the function xz is big,
along the diagonals of the in the xz plane. Figure 6.19 illustrates this function. The other
functions that are simple product functions of two variables are similar, and dx2 −y2 has been
illustrated above in Figure 6.16. The function that appears unique (it is not really, but to
show why not requires more math than we want to do here, although later in the course
we shall attack it) is dz 2 , which is illustrated in Figure 6.20. All of the angular functions
then are defined by the “letters” that are used to describe the m quantum number. Even
the correct signs flow from these values.
6.7.1
Exercises
6.7.1.1
Make a sketch of the 2py angular wave function. Make the sketch when you are standing
on the positive z axis.
6.7.1.2
Make a sketch of the 2py angular wave function. Make the sketch when you are standing
on the positive y axis. HINT: The ability to do such drawings will be useful when we take
up bonding.
Chm 118
Problem Solving in Chemistry
6.7
61
Figure 6.18: The surface with all values equal to a number, both positive (yellow) and
negative (blue), for a px angular wave function.
Figure 6.19: The surface with all values equal to a number, both positive (yellow) and
negative (blue), for a dxz angular wave function.
Chm 118
Problem Solving in Chemistry
6.8
62
Figure 6.20: The surface with all values equal to a number, both positive (yellow) and
negative (blue), for a dz 2 angular wave function.
6.7.1.3
Make a sketch of the 3dxy angular wave function. Make the sketch when you are standing
on the positive z axis.
6.7.1.4
Make a sketch of the 3dxy angular wave function. Make the sketch when you are standing
on the positive y axis.
6.7.1.5
How many angular nodes does a 3dxy wave function have? A 2pz ? Describe the surfaces
that defines these nodes. Do the same for a 3dz 2 wave function. What is the relationship
between the number of angular nodes and the ` quantum number?
6.7.1.6
Is there any relationship between the total number of nodes in a wave function and some
quantum number or combination thereof?
6.7.1.7
What is the difference between a 2px and a 2py wave function?
6.7.1.8
Describe the “shape” of a fxyz orbital. HINT: My intent is that you use the “name” to figure
out the shape; looking up the shape on the web is ok, but then you need to memorize it.
Chm 118
Problem Solving in Chemistry
6.8
63
6.7.1.9
Describe the difference between the dxy orbital and the dx 2−y 2 orbital.
6.8
Multi-electronic Atoms, Configurations
We use the quantum mechanical results for the hydrogen atom to generate a model to
understand multi-electronic atoms. There are two factors that go into this model: the
wave functions that the last couple of sections have described, and the Pauli principle. The
feature of the wave functions that we accented previously is that the quantum number n
is a rough measure of how far an electron is from the nucleus; this will remain important.
However, we will use a statement of the Pauli principle that differs from our previous one
(section 2.2): “No two electrons can have all four quantum numbers the same.”
We have seen the three quantum numbers that characterize a hydrogen atom electron, n, `,
and m. The fourth quantum number that we need for multi-electronic atoms is the spin of
the electron, usually called mS . Our application then goes as follows. The element with two
electrons, He, can put both electrons close to the nucleus, n = 1, into the 1s wave function
(called by chemists an “orbital”) if one of those electrons has spin up and one has spin
down. We cannot put a third electron this close to the nucleus because we have “run out of
quantum numbers,” so that next electron must go into the n = 2 level and we get a lithium
atom with the configuration of 1s2 2s1 . (We postpone putting forward the argument why
the configuration favors 2s1 rather than 2p1 for a while.) As we continue to add another
nuclear charge and another electron we move horizontally across the periodic table and get
the configurations of the elements from Be 1s2 2s2 to B 1s2 2s2 2p1 to Ne 1s2 2s2 2p6 where in
the last case we have populated all three of the 2p orbitals, 2px , 2py , and 2pz , each with two
electrons, one of spin up and one of spin down. The Pauli principle now forbids any more
electrons with n = 2, so we are forced to go to the n = 3 level and get Na, 1s2 2s2 2p6 3s1 ,
and add electrons and nuclear charges as before until we arrive at Ar with the configuration
1s2 2s2 2p6 3s2 3p6 . Going further requires that we understand the filling of 2s before 2p, so
we turn to that next.
The concept we need to apply to understand why the 2s level is filled before 2p in the
electron configuration is called penetration. It is based on Gauss’ law: To a test charge
outside the surface of a sphere, the charge on that surface acts as if it was located at the
center of the sphere. Now consider Figure 6.21 for a lithium atom and imagine a 2s electron
is three units away from a nucleus that has two 1s electrons. That pair of electrons are
between the 2s electron and the nucleus, so they act as if they were at the nucleus; thus the
net charge on the nucleus is +3 -2 = 1. The same is true of a 2p electron at this distance.
Now consider a 2s electron at a distance of 0.5 units from the nucleus. Some of the charge
probability for the pair of 1s electrons is “outside” that distance, and so has no effect on
that 2s electron. It therefore “sees” a much greater positive charge. Looking at the 2p
curve, we observe that it does not get very close to the nucleus, and is effectively screened
by the pair of 1s electrons. State this whatever way you wish: 2s penetrates better and
hence experiences a greater positive charge and a more stable orbital than 2p; or, the 2p
electron is better screened from the nuclear charge by the pair of 1s electrons and hence
Chm 118
Problem Solving in Chemistry
6.8
64
Figure 6.21: The radial probability distributions for 1s, 2s, and 2p wave functions showing
how a 2s electron has a larger probability of getting into the volume occupied by a 1s
electron than does a 2p electron. The shaded area under the 1s curve is present just to
help illustrate the penetration. Note that the 1s function has been plotted for an effective
nuclear charge of 3.
experiences a smaller positive charge and is held less tightly. The conclusion is that a 2s
electron is more stable than a 2p electron in a multi-electronic atom. This is the reason
that the configuration of a lithium atom is 1s2 2s1 rather than 1s2 2p1 ; or, as is often said,
2s fills before 2p.
A similar argument holds for the 3s electron in sodium filling before the 3p. Here the penetration is beneath both the pair of electrons in the n = 1 shell and the eight electrons in the
n = 2 shell; and from Figure 6.11 the 3s electron penetrates better than 3p with penetrates
better than 3d. So these arguments account for the configurations of the neutral atoms up
to argon. The next element in the periodic table is potassium, which has the configuration
[Ar]4s1 rather than [Ar]3d1 . Here a reasonable explanation is that 4s penetrates so much
better than 3d does that it overcomes the inherent stability of 3d over 4s because of closeness to the nucleus (measured just by n value). We could not predict this. However, it
is reasonably easy to understand. The net result is that we have in the periodic table an
electron configuration map, see Figure 6.22.
6.8.1
Exercises
6.8.1.1
When we say that a 2s electron penetrates better than a 2p, what do we mean?
6.8.1.2
Which penetrates the most, 2s or 2p in B?
6.8.1.3
Which penetrates the most, 2p or 3p in H? HINT: Careful!
Chm 118
Problem Solving in Chemistry
6.9
65
Figure 6.22: A rough sketch of the periodic table showing where electrons of various ` values
are filling in the atoms.
6.8.1.4
What is the electronic configuration of B? of N? of S?
6.8.1.5
Why is the electron configuration of Li 1s2 2s1 rather than 1s2 2p1 ?
6.8.1.6
According to the Pauli principle, which would be most stable? C (1s2 2s2 2p1x (↑)2p1y (↓))
or C(1s2 2s2 2p1x (↑)2p1y (↑))? How would the energies of these two compare to the energy
of C(1s2 2s2 2p2x )? HINT: Strictly speaking, these do not describe legitimate multielectron
wave functions, but they will suffice to illustrate the point at this stage in your education.
6.8.1.7
What is the energy ordering for the 3s, 3p, and 3d orbitals in K? What causes that order?
6.9
Multi-electronic Atoms, Ionization Energies
There is no single property more important in understanding chemistry and being able
to solve chemical problems than ionization energy, IE. Electron motion is at the heart of
chemistry and it is the ease of the movement of electrons, which is partially determined
by ionization energy, that dictates electron motion. Ionization energies of the atoms are
largely determined by the four concepts that we explore in this section.
Since Coulomb’s law (equation 2.1) is inverse in distance, the stability of an electron with
small n is much greater than that with large n. In the hydrogen atom this is very obvious,
as show by equation 6.6: the IE of hydrogen atoms from various excited states drops very
fast as n increases. For other essentially one electron species, such as lithium ion or sodium
ion, the change in IE with a change in n is damped somewhat by penetration, see below.
Thus the IE of sodium is less than that of lithium, but not by much. The second factor
Chm 118
Problem Solving in Chemistry
6.9
66
Figure 6.23: Ionization energies for the first and second row elements Li through Ne and
Na through Ar.
comes into play when we compare two atoms with the same n value. that factor is the
valence shell nuclear charge , VSNC. This is the charge that a non-penetrating electron in
the valence shell would experience if no other electrons are present in the valence shell. We
can get a sense of this factor by looking at a beryllium atom compared to a lithium atom.
In lithium the n = 2 electron has the nuclear charge of +3 partially shielded by the inner
shell (n = 1) of two electrons, so the VSNC is +1. In a beryllium atom, the nuclear charge
of +4 is also shielded by the inner shell of electrons; classically, the two electrons in the n =
2 shell would be opposite each other, which means that neither shields the other very much
from the nuclear charge. Hence the charge “seen” by one of the n = 2 electrons is about +2,
double that of the value “seen” in a lithium atom. The experimental IE of Li and Be are
5.39 eV and 9.32 eV, nearly the doubling that this crude model predicts. Similar reasoning
leads us to conclude that as we move across a row in the periodic table the VSNC increases,
leading to the general trend of higher IE to the right in the periodic table. Figure 6.23 plots
the ionization energies for the first and second row elements from Li to Ne and Na to Ar to
illustrate the general increasing IE as one move to the right in the periodic table.
Although the general trend in Figure 6.23 is a increasing IE as we move from left to right
in the periodic table, there is clearly some other factor(s) playing a role in determining
the IE because of the “glitches” in the data. The first one occurs as a drop of the IE
from beryllium to boron even though the VSNC increases; a similar feature occurs between
magnesium and aluminum in the second row. Inspection of the electronic configurations
of these elements suggest the cause: beryllium is 1s2 2s2 and boron is 1s2 2s2 2p1 . In the
first case the ionization process removes a 2s electron, which penetrates well. In the second
case, the process removes a 2p electron that does not penetrate as well. We conclude that
although the VSNC increases from Be to B, the change in penetration is sufficient to cause
the IE to drop. A similar effect occurs at Mg/Al involving 3s and 3p electrons.
The second “glitch” in Figure 6.23 occurs between nitrogen and oxygen in the first row, and
phosphorous and sulfur in the second. In all cases it is a p electron that is being removed
for both elements, so penetration differences cannot be the cause. What is changing the
normal increase with VSNC in these cases? Again, examination of the electronic configurations suggests the cause, although this time we look closely at the configuration. In the
nitrogen atom, the configuration is 1s2 2s2 2p3 , which, because of the electron-electron re-
Chm 118
Problem Solving in Chemistry
6.10
67
pulsion should probably be written as 1s2 2s2 2p1x 2p1y 2p1z . This configuration keeps electrons
far apart because of their spatial position along the x, y, and z axis, but also because of the
Pauli principle: all three 2p electrons have the same spin. At oxygen, the detailed configuration is 1s2 2s2 2p2x 2p1y 2p1z (or with the paired electrons in either of the other two equivalent
orbitals) and that last electron is easier to remove because it is paired with another in the
orbital. Hence the IE of oxygen is less than expected by the change in the VSNC.
6.9.1
Exercises
6.9.1.1
What are the four factors than influence ionization energy?
6.9.1.2
Explain why the ionization energy (generally) decreases down a column in the periodic
table. HINT: There are two opposing factors.
6.9.1.3
Explain why the ionization energy (generally) increases across a row in the periodic table.
6.9.1.4
How can you rationalize the relative IE’s of Mg and Al?
6.9.1.5
The first IE of sodium is less than that of magnesium. The second IE, which is the energy
necessary to remove an electron from a gaseous ion of charge +1, of sodium is greater than
that of magnesium. Comment.
6.9.1.6
Which has the lower second IE, Ca or Mg?
6.9.1.7
The IE of Rb[Kr 5s] is 4.176 eV whereas that of the excited state of H, H[5s] is 0.544 eV.
Account for the difference. HINT: The second case is the ionization of an excited state of
the hydrogen atom.
6.9.1.8
The IE of the excited state of Rb[Kr 5f] is 0.547 eV whereas that of the excited state of H,
H[5f] is 0.544 eV. Account for the lack of a substantial difference.
6.9.1.9
Successive ionization energies for Si are 8.15, 16.34, 33.46, 45.13, and 166.73, all in electron
volts. Take each value and divide it by the charge on the species after the electron is removed
(for example, divide 8.15 by 1 since the charge on the ion after the first ionization energy
is +1). Explain the resulting data, both similarities and differences. HINT: Don’t expect
exact agreement but be startled at how nicely the numbers arrange during this procedure;
even though I have seen it before, I always am.
Chm 118
Problem Solving in Chemistry
6.11
68
Element
H
He
Li
Be
B
C
N
O
F
Ne
s orbital
13.6
24.5
5.45
9.30
14.0
19.5
25.5
32.3
46.4
48.5
p orbital
—
—
—
—
8.30
10.7
13.1
15.9
18.7
21.5
Element
—
—
Na
Mg
Al
Si
P
S
Cl
Ar
s orbital
—
—
5.21
7.68
11.3
15.0
18.7
20.7
25.3
29.2
p orbital
—
—
—
—
5.95
7.81
10.2
11.7
13.8
15.9
Table 6.1: Valence orbital ionization energy for low atomic number elements in eV.
6.10
Multi-electronic Atoms, Valence Orbital Ionization Energies
In discussion of bonding there will be the need to know the average energy of the bonding
of an electron to an element. This is because in bonding, many of the special features that
occur in an atomic species are muted. For instance, in a bonding situation, there are eight
electrons around an oxygen atom, including six in the 2p shell. The peculiar feature of the
electron-electron repulsion that we discussed in the last section is thus absent. A method
to asses how the other factors only influence the IE is needed. This method exists and is
called the Valence Orbital Ionization Energy, VOIE. In the VOIE method the energy of a
2p electron in an atom is computed as the average value over all possible arrangements of
the electrons. Data are given in Table 6.1.
6.10.1
Exercises
6.10.1.1
From the data in Table 6.1 is the 2s VOIE or the 2p VOIE most sensitive to VSNC?
6.10.1.2
Give a reasonable explanation for your answer to the last exercise.
6.10.1.3
The VOIE of the 4p orbital of Ga is 5.93 eV and that of Kr is 14.22. Make a guess for the
value of the 4p orbital of As.
Chm 118
Problem Solving in Chemistry
6.11
6.11
69
Multi-electronic Atoms, Transition Metal Ions
One final topic concerning multi-electronic atoms, or rather ions. The configuration of K
and Ca are [Ar]4s1 and [Ar]4s2 , respectively. The configuration of most transition metal ions
of the first row, is [Ar]3dn 4s2 where n runs from 1 at Sc to 10 at Zn. A simple interpretation
of this result would suggest the filling of the 4s level before the 3d. It is easy to offer a
rationalization for this behavior. The conflict to be dealt with it that the 4s penetrates very
well whereas 3d is closer to the nucleus. If we assume that penetration “wins” this battle,
we can “account” for the data. This is not a predictable result, but consistent with what is
observed.
The real situation is much more complicated. One issue is that the
wave equation for the system is a multi-electronic wave equation and
a discussion of individual electrons is just not appropriate. Also, an
analysis2 suggests that there is an interplay between the difference
in the apparent energy of the 3d and 4s orbitals (for the reasons
talked about above) and electron repulsion differences between a 4s
and 3d electron with the other electrons in the molecule.
What is clear in all analyses of this situation is the response to an increase in the effective
nuclear charge that occurs when one goes from a neutral atom to a charged ion. The has the
effect of stabilizing small n values more than large ones, so the 3d orbital is stabilized more
than the 4s in the ions. (One way to look at this is that the compression of the inner shells
by the increased charge shrinks the inner orbitals a lot and hence lowers the importance
of penetration.) Thus if we have an atom with the configuration [Ar]3dn 4s2 , its dipositive
ion will have the configuration [Ar]3dn . The conclusion is this: The chemistry of transition
metal ions (and of transition metal compounds) is dominantly 3d chemistry, with the ions
formed by “taking out the 4s electrons first.”
6.11.1
Exercises
6.11.1.1
What is the electron configuration of V?
6.11.1.2
What is the electron configuration of V2+ ?
6.11.1.3
What is the electron configuration of Cr3+ ? of Mn4+ ?
2
Vanquickenborne, L. G.; Pierloot, K.; Devoghel, D.; Inorg. Chem., 1989, 28, 1805-1813.
Chm 118
Problem Solving in Chemistry
Chapter 7
Transition Metal Compounds and
Color
7.1
Energy of Orbitals in Electrostatic Fields
We have in the Lewis structure model a simple picture of bonding in compounds containing
elements in the upper right hand side of the periodic table. That model does not work
well for compounds involving transition metal atoms, those containing elements in the “dblock.” In this chapter we deal with a simple scheme for bonding in transition metal ions
(tmi). This model is usually called crystal field theory.
The essence of the model involves thinking about two electrostatic interactions, governed
by Coulomb’s law (section 2.1). The first is that the central ion in our picture, the ion we
are mostly interested in and almost always positively charged, attracts negatively charged
ions, say F– , or the negative end of a dipole (in, say NH3 ), toward itself. These ions or
dipoles attracted are called ligands. The second interaction is the response of the electrons
in various orbitals of the central ion to the field generated by those ligands. Generally
speaking, the various orbitals of the central ion that were degenerate, had the same energy
in the absence of the ligands, split into various new energy levels because of the ligands.
Preferential placement of electrons into the more stable of these split orbitals then causes
the compound to be more stable than before splitting.
As mentioned, this model is applied almost exclusively to tmi, but for purposes of introduction, we apply it to an ion with p orbitals. Imagine an ion at the origin of a coordinate
system with two negatively charged ligands surrounding it: one is located at (x, y, z) =
(0.0, 0.0, 1.0)Å and the other at (x, y, z) = (0.0, 0.0, -1.0)Å, see Figure 7.1. Let the ion
at the center have an electron in a 5pz orbital and let the ligands be represented by the
negative point charges. Since the 5pz orbital points at the negative charges, an electron in
it will be repelled rather severely, will have a high energy. In contrast, an electron in a 5px
orbital or a 5py orbital, which are oriented away from the point charges, will not be repelled
as severaly. Hence the degeneracy of the three 5p orbitals is lifted; two of them are more
stable than the third (5pz ).
70
7.2
71
Figure 7.1: Two orbitals in a linear crystal field arrangement.
7.1.1
Exercises
7.1.1.1
How many p electrons would iodine have in the molecule IF–2 assuming that the bonding is
ionic, that is, that the two fluorine nuclei are both ions, F– ?
7.1.1.2
Show how the p orbitals would split in energy in a linear compound, say IF–2 , under the
crystal field model where you treat the fluoride ions as point charges. HINT: Pay attention
to degeneracies.
7.1.1.3
What would be the occupancy of the various split orbitals for the iodine in the last two
exercises?
7.1.1.4
Show how the p orbitals would split in energy in a square planar complex IF–4 under the
crystal field model.
7.1.1.5
The molecule in the last problem has D4h symmetry. Find the name of the characteristic
numbers for the pz and the px , py wave functions on the iodine in this symmetry.
7.1.1.6
Speak like a professional and describe the splitting of exercise 7.1.1.4 in terms of the last
exercise. For example, “The iodine p orbitals are split into a high energy b1g and a more
stable eg by the ligands.”
7.1.1.7
What would be the splitting of the p orbitals in an octahedral compound such as SF6 ,
under, as usual, the crystal field approximation?
7.1.1.8
Which orbital, d2z or dx2 −y2 would be of highest energy is two negatively charged ligands
were placed the ±z axis?
7.1.1.9
We have two negative charges along the positive and negative z axis. Two of the d orbitals
are in Figure 7.2. Which are they? What are their relative energies under a crystal field
approximation.
Chm 118
Problem Solving in Chemistry
7.2
72
Figure 7.2: Two of the d orbitals for consideration in exercise 7.1.1.9.
7.2
Metal Ions in an Octahedral and Tetrahedral Environments
Transition metal ions exist in a number of different environments, the most common being
octahedral, in which the metal ion is surrounded by six ligands along the positive and negative Cartesian coordinates, ±x, ±y, ±z. Convince yourself that this octahedral environment
of point charges will cause dxy , dxz , and dyz to be repelled equally. We say that these three
orbitals remain degenerate. They are named the t2g set because of their characteristic numbers. Although not so obvious (some exercises below explore the arguments), the other two
d orbitals, d2z and dx2 −y2 are also degenerate. Since they point right at the ligands, they
are of higher energy than the t2g set. These two orbitals are labeled eg . To summarize, an
octahedral field of ligands split the five d orbitals into two sets, a more stable set of three,
t2g and a less stable set of two, eg .
A second common geometry for transition metal ion compounds is tetrahedral. The relative
energies are most easily seen by placing a tetrahedron within a cube. In Figure 7.3 we give
a cube with axes going through the center of the faces of the cube. Put the four ligands
on corners of the cube such that they are diagonally opposite each other on the faces of
the cube. Convince yourself that dxy , dxz , and dyz have the same energy. In tetrahedral
symmetry, these three orbitals are labeled t2 . The other two d orbitals, d2z and dx2 −y2 are
also degenerate. They have e symmetry. As suggested in an exercise, you should be able to
deduce that the t2 set is of high energy and the e set of low energy, more stable.
Chm 118
Problem Solving in Chemistry
7.2
73
Figure 7.3: A cube with axis drawn.
7.2.1
Exercises
7.2.1.1
Sketch the five angular d orbitals as they would look if you stood on the positive x axis.
HINT: You have to be able to draw these orbitals, accurately and quickly, to work with
crystal field problems.
7.2.1.2
Sketch the five angular d orbitals as they would look if you stood on the positive z axis.
7.2.1.3
Three of the d orbitals are dxy , dxz , and dyz . Note that these orbitals occur with one Cartesian coordinate times another. There are three more possible combinations of coordinates
taken pairwise, xx, yy, and zz. These are usually considered in three combinations, x2 -y2 ,
z2 -x2 , and z2 -y2 . Sketch each of these functions. HINT: Use the name to guide you to
ascertain (1) where the function is big, where it is zero; and (2) where it is positive and
where it is negative.
7.2.1.4
Which orbital is repelled the most by negative charges on the three (x, y, and z) plus and
minus Cartesian axes (octahedral symmetry), dx 2−y 2, dz 2−x 2, dz 2−y 2? HINT: Use your
pictures from the last exercise.
7.2.1.5
We now do something sneaky. Since quantum mechanics demands that there are only five
possible d orbitals and there are six possible combinations of a pair of Cartesian coordinates,
Chm 118
Problem Solving in Chemistry
7.3
74
we have to combine them in some fashion. We choose to add the last two (of the list in the
last exercise) and then divide by two since we don’t want to change the size. Add z2 -x2 to
z2 -y2 and then divide by two. What do you get?
7.2.1.6
The answer to the last exercise, z2 - 21 x2 - 21 y2 is the real form of the function we usually
abbreviate as dz 2. Since it is composed of those two orbitals that are equivalent to dx 2−y 2
but which would have double the interaction with ligands, but then is divided by two, we
have the same interaction. Does this argument convince you that dz 2 is equivalent in energy
to dx 2−y 2? If not, it is the best I can do; you will have to take a quantum course.
7.2.1.7
Can you make an argument for why the t2 set of orbitals is of higher energy in a tetrahedron
than is the e set? HINT: This is a problem in solid geometry, but can be approximated by
looking at just the wave function lobe closest to the point charges.
7.2.1.8
Which geometry, octahedral or tetrahedral, do you anticipate will have the largest gap
between the two sets of split orbitals? REMARK: Real solid geometry suggests the gap in
the tetrahedral environment is 4/9 that in the octahedral.
7.3
Metal Ions in Other Environments: Lowering of Symmetry
Both the octahedral and tetrahedral environments are of “high” symmetry, meaning that the
sets of symmetry operations are large (48 and 24, respectively). As symmetry is “lowered,”
as the number of symmetry operations decreases, degeneracies generally are dispatched.
Hence a particularly easy way to determine the energy ordering of the d orbitals in a
compound of low symmetry is to start with one of higher symmetry and carry out a motion
of ligands that changes that high symmetry to the lower one. As an example, consider a
tetrahedral compound that is distorted as indicated in Figure 7.4. This has the effect of
changing the tetrahedral compound to a distorted tetrahedron, and ultimately, to a square
planar compound. If we sketch how the various d orbital energy change as the distortion
occurs, we generate a correlation diagram; you are asked to do this in the exercises.
7.3.1
Exercises
7.3.1.1
What happens to the energy of the dxz orbital as the distortion in Figure 7.4 occurs?
7.3.1.2
What can you say about the energy of the dyz orbital (in Figure 7.4) compared to that of
dxz ?
Chm 118
Problem Solving in Chemistry
7.4
75
Figure 7.4: Distortion for problems ?? and ??.
7.3.1.3
What happens to the energy of the dxy orbital as the distortion in Figure 7.4 occurs?
7.3.1.4
Sketch how the d orbitals on a metal ion at the center of the cube shown in Figure 7.4
would change in energy as the distortion shown occurred.
7.3.1.5
A tetragonal distortion of an octahedral compound is one in which the ligands on the z axis
move away from the metal ion and those in the xy plane move toward the metal ion. Make
a correlation diagram for this distortion.
7.4
The Configuration of Metal Ion Compounds
To deal with transition metal compounds efficiently it is useful to devise a way to talk
about the electronic configuration. These are the same words that we use to describe the
configuration of atoms, 1s2 2s2 2p2 for a carbon atom, for instance. The difference being that
we want to describe the orbitals in which the electrons reside, and in compounds these are
no longer s, p, d, etc. Rather the names of the characteristic sets of numbers that are used
to describe the orbitals are used. Thus the configuration of diagram “A” in figure 7.5 is a21
a12 . The configuration for diagram “B” in Figure 7.5 is a22g b21u .
When dealing with octahedral environments, a similar notation is used. This would be, for
instance, for a d3 metal, t32g . For a a d7 metal complex in a tetrahedral environment, the
Chm 118
Problem Solving in Chemistry
7.4
76
Figure 7.5: Using characteristic number names for configurations; see problem ??.
configuration would be e4 t32 .
7.4.1
Exercises
7.4.1.1
What is the “d count” for each of these transition metal species in bonding situations:
Cr(III), Mn(II), V(II), Cu(II), Ag(I), Ag(III), Ni(III), Mo(0), W(VI), Ru(II), Os(III),
Mn(V)? HINT: Remember that the configuration of transition metal ions in compounds
is sans 4s electrons.
7.4.1.2
How many d electrons do these transition metal species have in bonding situations: V(III),
Ti(IV), Mo(V), Ni(0), Co(II), Co(III), Rh(III), Ir(IV), Fe(II), Mn(II)? HINT: “In bonding
situations” means sans 4s electrons.
7.4.1.3
Give the electronic configuration (in the tn2g /em
g notation) for an octahedral Cr(III) compound; for an octahedral Ni(II) compound.
7.4.1.4
Give the electronic configuration (in the tn2g /em
g notation) for an octahedral Cu(II) compound.
7.4.1.5
There is an ambiguity in electronic configuration for some octahedral metal compounds.
The issue is this for a d4 system. If the energy separation between t2g and the eg levels
is small, then rather than paying the price of putting two electrons in the same orbital, it
is energetically more favorable to place the fourth electron in the eg level (and gain some
stability from the Pauli principle). On the other hand, if the energy separation (often called
“the gap”) between the t2g and the eg levels is large, then more stability is gained by putting
a fourth electron in the t2g , even at the cost of putting two electrons in the same orbital.
So there are two configurations possible for a d4 system depending on the gap. Give the
electronic configuration (in the tn2g /em
g notation) for an octahedral Mn(II) compound if the
Chm 118
Problem Solving in Chemistry
7.5
77
splitting is small; for an octahedral Mn(II) compound if the splitting is large.
7.4.1.6
Give the electronic configuration (in the tn2g /em
g notation) for an octahedral Fe(II) compound
if the splitting is small; for an octahedral Fe(II) compound if the splitting is large.
7.4.1.7
Give the electronic configuration (in the em /tn2 notation) for small gap tetrahedral compounds of Cu(II), Fe(II), and Co(II).
7.5
Configuration of Metal Ion Compounds, Spin, Field, and
Ligand Strength
In exercise 7.4.1.5, and those that followed it, we discussed how the size of the gap between
the more stable and less stable levels in octahedral and tetrahedral compounds influenced
the configuration. There are two notations for this situation that are used interchangeably.
If the gap is small, then the electrons all tend to have their spins in the same direction (so
that Pauli avoidance can occur). If we assign a spin along the z axis of 21 to an electron
with spin up, then we can get a total spin along the z axis, called MS , by adding up these
values. For a d4 metal ion with a t32g /e1g configuration (with a small gap), the value of MS
would then be 2. On the other hand, if the gap was large for this d4 metal ion, then the
configuration would be t42g /e0g and the total value of MS would be only 1. (We shall see
shortly there is an easy way to measure “spin.”) Compounds of the former type are called
“high spin” complexes and those of the latter are “low spin” complexes. The other notation
for these two kinds of compounds reflect the gap size. We call the compound with the 32g /e1g
configuration a “low field” compound because the gap (caused by the field of the ligands) is
small. Contrariwise, the t42g /e0g configuration would be called a “high field” compound. It
is unfortunate that these two notations are opposite, but note that they are: “High field”
compounds are of “low spin.”
As implied above, what causes the difference between a high field compound and a low
field compound is the field generated by the ligands. Remember that there are two fields in
crystal field theory, that which attracts the ligands to the metal (not of concern to us now)
and the field produced by the ligands that causes the metal d orbitals to have difference
energies. The stronger the field created by the ligands, the greater the gap (in an octahedral
compound) between the t2g and eg levels. What ligands create a strong field? We would
anticipate, according to Coulomb’s law, section 2.1, that ligands close to the metal ion (small
r) and those with large charge (big Z1 ) would create the largest fields; we might expect, for
instance, that O2– >S2– >F– >Br– >NH3 . The first because of the high charges, and the
last because it has only a dipole and not a full charge. This expectation turns out to be
completely wrong. Here is a serious flaw in the crystal field theory.
Chm 118
Problem Solving in Chemistry
7.5
78
Compound
S
Cr(CH3 )(H2 O)2+
5
3
2
FeCl3 (H2 O)3
5
2
Co(NH3 )4 Cl+
2
0
MnF3–
6
2
Table 7.1: Spin data for some compounds
The order that is observed for the strength of the ligands is called the spectrochemical
series. We will figure out later in the course why this order is correct, but for now let us
just take it as a experimental observation for the strength of ligands. It is presented here:
I− < Br− < NCS– , S bonded < Cl− < NO–3 , O bonded < F− < OH− <
−
C2 O2–
4 , O bonded < H2 O < SCN , N bonded < NH3 , about the same as py
–
< en < NO2 , N bonded << CN− , C bonded ∼
= CO, C bonded.
3+
According to this series, the gap will be smaller in CrCl3–
6 than in Cr(H2 O)6 , even though
the former has a real charge and the latter only a dipole. Likewise, we would expect
that generally speaking, compounds such as MBr3–
6 would be low spin compounds, whereas
M(CN)3–
would
be
high
spin.
6
7.5.1
Exercises
7.5.1.1
In Table 7.1 are the values of the spin quantum number for some compounds. Verify these
values.
7.5.1.2
5
0
Is the compound Mn(CN)4–
6 with configuration t2g /eg high spin or low spin? high field or
low field?
7.5.1.3
What would be the MS value for a low spin octahedral Mn(II) compound? for a high spin
octahedral Mn(II) compound?
7.5.1.4
Almost all tetrahedral compounds are high spin. Comment.
7.5.1.5
For what n values in the dn configurations is there a possibility of high spin and low spin
compounds in octahedral geometry?
Chm 118
Problem Solving in Chemistry
7.6
79
7.5.1.6
3+
Which compound will have the greater splitting, MoI3–
6 or Mo(H2 O)6 ?
7.5.1.7
Which compound will have the larger splitting between the t2g orbitals and the eg orbitals
3–
−
of an octahedral complex, Cr(NH3 )3+
6 or Cr(NCS)6 when the NCS is nitrogen bonded.
7.5.1.8
A solid compound of formula MCl2 is dissolved in water, presumably to form the ion
5
M(H2 O)2+
6 . This compound has a spin of 2 . When treated with KCN, presumably to
4–
1
form M(CN)6 , the spin changes to 2 . What is M?
7.5.1.9
There are two fields in crystal field theory. The first is created by the charge on the metal.
What would you expect would happen to the metal ligand distance in a complex with a
metal with charge of +2 as opposed to one in which the metal is charged +3?
7.5.1.10
To continue from the last exercise, in which of the two compounds above would the ligand
exert the greater field? Which, therefore, would have the greater gap?
7.5.1.11
Think about the two electric fields that operate in the crystal field model and make an
4–
argument for whether the energy of splitting will be larger in FeF3–
6 or FeF6 ? Articulate
an eloquent answer.
7.6
Color of Metal Ion Compounds.
pounds.
I. Octahedral Com-
How do we know that any of the words written in this chapter are true? Certainly not
just because they are written. One of the most dramatic aspects of crystal field theory
is its ability to understand a striking feature of transition metal compounds. Whereas
materials such as NaCl, CaO, CCl4 , O2 are colorless, most transition metal ions have color.
2+
2+
For instance Cr(H2 O)3+
6 is steel blue, Co(H2 O)6 is pink, and Ni(H2 O)6 is green. What
causes these colors? Quantum mechanics says that if light has energy (given by the Einstein
equation, E = hν = hc
λ ), that exactly matches the energy gap between two quantum levels,
then there is a possibility for that light to be absorbed by the sample and excite the system
from the lower state to the upper state. If one color of visible light is removed from white
light, then it appears colored to our eyes. It turns out the the gap between the t2g and eg
levels of octahedral compounds has energy corresponding to visible light. So if we excite
a t2g electron to the eg level, we can expect the compound to be colored. This transition
of an electron from a t2g level to an eg one is called a d to d transition, since both of the
levels originated from d orbitals. Because the size of the gap for a given metal ion depends
on the nature of the ligand (through the spectrochemical series), we can often understand
what is happening in the immediate environment of a metal ion by observing its color.
Chm 118
Problem Solving in Chemistry
7.6
80
Compound
λmax , nm
Compound
λmax , nm
CrF3–
6
671
MoCl3–
6
520
729
MoBr3–
6
3–
MoI6
Mo(H2 O)3+
6
546
CrCl3–
6
Cr(H2 O)3+
6
Cr(NH3 )3+
6
575
464
725
383
Table 7.2: Spectra data for the lowest energy t2g to eg transition in some d3 compounds.
Data from A. B. P. Lever, Inorganic Electronic Spectroscopy, Elsevier, 1984.
7.6.1
Exercises
7.6.1.1
In Table 7.2 are the values of the highest wavelength of absorption for some octahedral
Cr(III) and Mo(III) compounds. Are these values consistent with the spectrochemical
series?
7.6.1.2
What would you expect to happen to the wavelength of absorption if Mn(Cl)4–
6 were con4–
verted to MnF6 ?
7.6.1.3
Which compound would absorb at smaller wavelength (that is, in the “blue”), Co(NO2 )3–
6
or Co(NCS)3–
6 , where both ligands are nitrogen bonded to the Co(III).
7.6.1.4
Which compound would you expect has the lowest energy of absorbance, Ni(H2 O)2+
6 or
2+
Ni(H2 O)4 , where the latter is presumably tetrahedral?
7.6.1.5
An octahedral compound of Cr(III) with six ligands, L1 , has an absorption peak near 550
nm. The ligand is changed to L2 (but the symmetry is still octahedral) and the peak is now
at 640 nm. What can you say about the ligand field strength of L1 and L2 ?
7.6.1.6
−1 and that in Cr(en)3+ at 22300 cm−1 .
The lowest energy peak in CrF3–
6 occurs at 14900 cm
3
Which ligand has the larger crystal field splitting parameter, F− or en? HINT: It probably
would be nice to know what en is.
7.6.1.7
The lowest energy peaks in solid VCl2 , VBr2 , and VI2 (where the coordination is approximately octahedral by sharing of anions between two or more metal ions) are 9300, 8600,
and 7870 cm−1 , respectively. Calculate the wavelength of the light associated with these
absorbances and explain the results.
Chm 118
Problem Solving in Chemistry
7.7
81
2+
Figure 7.6: Orbital energy shifts on changing ML2+
6 to trans-ML4 W2 and transitions.
7.6.1.8
VBr2 is a orange brown solid. (Brown is typically the color associated with something that
absorbs over most of the visible region.) When it is dissolved in water and then treated
with KCN, the solution turns pale yellow. What color of light is removed from white light
to make a yellow color? Is the light that is removed of high or low energy? Explain the
color changes described above.
7.6.1.9
−1
The lowest energy peak in Ni(py)2+
6 (where py is pyridine) occurs at 11700 cm . Describe
the change in the position of the electron that occurs as a result of this absorption of light.
HINTS: From what orbital to what orbital is the electron moving? How do the positions of
these orbitals differ from one another? This is an important concept to use to distinguish
from other sources of color in compounds.
7.7
Color of Metal Ion Compounds. II. Low Symmetry Compounds.
The discussion in the last section involves octahedral compounds. There are many more
compounds that, although six coordinate, are not strictly octahedral, such as Cr(H2 O)5 Cl2+ .
There are two extreme approximations to handling such compounds (and then the difficult
in-between, which we will not deal with). One is the separate orbital approach and the
other is the average field approach.
Let’s talk about the separate orbital approach first. Imagine we have an octahedral compound of the type ML2+
6 where L is some ligand. Now we take the two L ligands on the z
axis and replace them with a weaker ligand, W. Our compound is now trans-ML4 W2+
2 . We
use the simplest approximation to determine what happens to the d orbital energy levels.
Since the ligands in the xy plane are not changes, dx2 −y2 is not changed in energy; and
because the ligand W is weak and is on the z axis, d2z is lowered in energy. We also assume
that the orbitals that avoid the ligands, dxy , dxz , and dyz , are barely changed in energy.
This is shown in Figure 7.6. We see that the single transition found in the octahedral parent
compound becomes two transitions in the substituted compound, one at the same energy
as that found in the octahedral compound, and the other at lower energy. This situation
will pertain if the ligands L and W are widely separated in the spectrochemical series.
Before dealing with the average field model, it is useful to examine a property of the ab-
Chm 118
Problem Solving in Chemistry
7.7
82
Figure 7.7: All spectral lines in a tetragonally distorted compound from t2g derived levels
to eg derived levels.
sorbance of light in transition metal compounds. Recall our model that the strength of a
ligand depends on the distance it is from the metal ion. Also, note that all atoms are always
moving back and forth in vibrations that change the bond length. In a metal complex, one
such motion is for all of the ligands to move in and out in phase with each other. When
the ligands are all close to the metal ion, the gap is larger; when they are all far from the
metal ion, the gap is smaller. Since a photon of light could hit the metal complex at any
time, it might catch the ligands out (small gap) or it might catch the ligands in (big gap)
or anywhere in-between. As a result, a whole range of colors is absorbed rather than just a
single wavelength. It is usual for the “peaks” of transition metal compounds to be rather
broad, covering a range of wavelengths.
We deal with the average field model by looking carefully at the same tetragonal splitting
that we examined above for the separate orbital model. The difference will be that this
time we look at all possible transitions from levels that were originally t2g in the octahedral
parent compound to the levels that were originally eg . This is shown in Figure 7.7. There
are four possible transitions as shown there, all with very similar energies. The average
field contends that these will show up as one transition with average energy as shown on
the right side of Figure 7.7. To illustrate exactly how this works, consider Figure 7.8 where
broadened spectral lines–see last paragraph–are positioned at each of the four energies given
in the left side of Figure 7.7. Of course, these would not be seen as individual peaks, as
shown in Figure 7.8, but would add together to give a net absorbance peak. This is shown
in blue curve in Figure 7.9–the lower curve at 17000 cm−1 if you are looking in black and
white. We can compare this to the orange curve in that Figure which is the curve for the
average peak from the right hand side of Figure 7.7. This argument suggests that when
we have ligands that do not differ in their strength by too much, we expect a peak at
the average position of the strength of the various ligands. This is called the average field
approximation.
Chm 118
Problem Solving in Chemistry
7.7
83
Figure 7.8: Spectrum for a tetragonally distorted compound if each peak was seen separately.
Figure 7.9: Real spectrum for a tetragonally distorted compound and that for a hypothetical
average field system. The lower curve (at 17000 cm−1 ) is the sum of the curves in Figure 7.8
Chm 118
Problem Solving in Chemistry
7.8
84
Compound, X
SCN−
Br−
F−
NNN−
NC−
λmax , nm
620
322
595
585
560
Compound, X
I−
Cl−
NCS−
NO
NH3
λmax , nm
650
609
570
559
545
Table 7.3: Spectral data for Cr(H2 O)5 X2+ compounds.
7.7.1
Exercises
7.7.1.1
2+
When Cr(H2 O)2+
one of the products is Cr(H2 O)5 (NCS)2+ .
6 reacts with Co(NH3 )5 (NCS)
Can you tell whether or not the thiocyanate ion is bonded to the chromium ion through
the nitrogen atom or the sulfur atom by using spectroscopy? What data would you need?
HINT: Apply an average field argument.
7.7.1.2
Linkage isomerization occurs when a ligand can bind to a metal ion in more than one way.
Which of the following ligands might exhibit linkage isomerization? NO–2 , NCS– , S2 O2–
3 .
Use Lewis structures to make your point.
7.7.1.3
The ion Cr(H2 O)5 SCN2+ has an absorption peak in the visible region of the spectrum at
620 nm. The linkage isomer of this material, Cr(H2 O)5 NCS2+ has a peak at 570 nm. Give
an explanation. HINT: Use the average field concept.
7.7.1.4
When two pyridine ligands along the z axis in Ni(py)2+
6 – see exercise 7.6.1.9 – are replaced
by Br− ligands, two peaks appear, one at 8430 cm−1 and the other at 11550 cm−1 . Account
for this behavior. HINT: This is NOT an average field problem because we see two peaks,
rather it is a separate orbital problem.
7.7.1.5
Table 7.3 gives the lowest energy (spin-allowed) spectral peak for some compounds of the
form Cr(H2 O)5 X2+ . From these data construct a spectrochemical series. HINT: Use the
average field concept.
7.8
Color of Metal Ion Compounds. III. Intensities of Color.
Our discussion on the color of metal ion compounds has thus far been concerned with the
wavelength of the absorbance. There is a second parameter that is also important. How
intense is the color? To measure the intensity of the color of a sample in solution we put
the liquid in a container of fixed length, generally one cm, and irradiate it with light of
Chm 118
Problem Solving in Chemistry
7.8
85
wavelength λ and intensity I0 . On the backside of the sample, we place a detector that
measures the intensity of the beam after it has passed through the sample and find an
intensity of I. The transmittance is defined as the ratio of these numbers,
T = I/I0
(7.1)
and varies between 0 and 1. The absorbance is defined as Ā = -log[T] and varies between
0 (for T = 1) and a large number when T approaches 0. An intensely absorbing compound
thus has a small T and a large Ā. Clearly these parameters depend on the concentration
of the material, c, generally in M, and the length of the cell, `. This dependence is given
by Beer’s law:
Ā = λ c`
(7.2)
The parameter of interest to us is λ , the molar absorptivity, which is a measure of the
intensity of the color, or, in words that will serve us better, the allowedness of the transition:
how easy it is for the molecule to grab the photon of light and use its energy to promote
the electron to a higher energy state.
The question before us is what causes a compound to have a large λ . Be clear that λ can
never be large unless the λ at which you measure it corresponds to light of the right energy
to match the energy gap in the molecule. Beyond that, however, there are other factors that
influence the size of λ . We consider two of these with associated flavors. The first is that
λ is small unless the value of the spin of the compound does not change upon excitation.
This is called the “spin selection rule,” and we talk about “spin-allowed transitions,” where
∆S = 0; and “spin-forbidden transitions,” where ∆S 6= 0. For metal ion complexes this
means a transition from the configuration t32g /e1g to t22g /e2g could be spin-allowed, but that
a transition from t32g /e2g to t22g /e3g must be spin-forbidden.
The second factor that we consider to determine allowedness, the size of λ , is called parity.
For systems that contain the symmetry operation i, the sets of characteristic numbers
either have a positive number under i or a negative number. Those with a positive number
are called (and labeled) “g” functions; and those with a negative number are called “u”
functions. For reasons of simple calculus (but not proven here), the only transitions that
are possible are between a g function and a u function; both g to g and u to u transitions are
forbidden. Said in words, transitions are allowed when they proceed from an even function
to an odd one, or in reverse. Otherwise, they are forbidden. Here is a simple example. The
atomic functions have the following parity, which you should prove to yourself: s (even),
p (odd), d(even), f(odd). Thus a s to p transition is allowed, but a p to f transition is
forbidden. Likewise, and importantly, a d to d transition, such as the transitions we have
been talking about, are forbidden.
The last sentence in the last paragraph seems rather ridiculous. Why have we been talking
about something that is forbidden. The answer is that nothing is totally forbidden. When
we say something is strongly forbidden, we simply mean that λ is small. And when we
say something is allowed, we mean λ is large. To give us a handle on how to use our two
rules for allowedness, look at Table 7.4. Across the top are three categories of the whether
or not the parity rule holds–more explanation in a moment for the middle one. Down the
side are the two possibilities for ∆S. At each intersection is a rough (very) guide to the
approximate size of λ . The first column indicates the λ values when the transition is not
Chm 118
Problem Solving in Chemistry
7.9
86
g to u? No
g to u? Remembers it to be No
g to u? Yes
∆S = 0
1 to 100
100 to 1000
greater than 1000
∆S not 0
less than 1
less than 10
less than 100
Table 7.4: Approximate values of the molar absorptivity under different conditions of ∆S
and odd/even nature of the functions; the units are M−1 cm−1 . Note that “Remembers ...”
refers to an orbital in an environment that does not have the center of inversion symmetry
operation, but the starting function and final function are both even (or odd). Use of this
table without thinking will result in incorrect results; intensity is hard to predict.
officially g to u, that is, is forbidden when the molecule contains a center of inversion in
its symmetry. The middle column indicates the λ value when the molecule does not have
a center of inversion, but when the transition is, in essence, g to g or u to u. An example
is the d to d transition in a tetrahedral compound. This symmetry does not have a center
of inversion, but the orbitals “remember” that they came from an orbital, a d orbital, that
in spherical symmetry did have a center of inversion. The last column is for compounds
where the transition if from a g orbital to a u one, either in actuality, or by “remembering.”
Apply the appropriate ∆S to each by choosing the right row and you get an approximation
to the value of λ .
7.8.1
Exercises
7.8.1.1
A compound has a molar absorptivity of 15100 M−1 cm−1 . Using Table 7.4, what can you
say about the nature of the transition?
7.8.1.2
−1 cm−1 )
The complex ion Co(H2 O)2+
6 has absorbance peaks, nm, (molar absorptivity, M
at 12340 (2), 625 (shoulder), and 515 (4.6). Given the intensities of these peaks, what kind
of transition are they associated with? HINT: Remember the warning in Table 7.4; there
are a couple of possibilities and you need to use other thoughts and other data to sort out
which you want to use.
7.8.1.3
Compounds of Mn(II) that are high spin are generally almost colorless (which probably
means that λ is less than 1). Why?
7.8.1.4
–
A solution of Fe(H2 O)3+
6 is nearly colorless. When some NCS is added, the solution turns
blood red with a peak near 440 nm with an λ of about 4600 M−1 cm−1 . Comment.
Chm 118
Problem Solving in Chemistry
7.9
87
Figure 7.10: A Gouy balance. A diamagnetic substance is repelled from the magnetic field
and hence weighs less. A paramagnetic substance is pulled into the field and therefore
weighs more.
7.8.1.5
The ion VO3–
4 is colorless to our eye, but has a peak at 276 nm with an intensity of about
−1
8000 M cm−1 . Give the “d count” of this species and comment on the spectrum.
7.9
Magnetism in Metal Ion Compounds.
We have talked about the spin of transition metal compounds. To review, compounds with
net spin of 0 are said to be diamagnetic. and compounds with a net spin other than zero
are called paramagnetic. When a solid sample of a diamagnetic compound is placed in a
magnetic field, as illustrated in Figure 7.10 and weighted, it is found to weigh less than it
weighs in the absence of the field. On the other hand, a paramagnetic substance weighs
more in the presence of the magnetic field. From the difference in apparent weight, the
magnetic moment of the material, µ, can be found. Theory indicates an approximation can
be calculated from the spin via the “spin only magnetic moment” :
p
µS.O. = 2 S(S + 1)
(7.3)
where S is the net sum of the spins of the electrons and the units of the moment are Bohr
Magnetons, B.M. (This equation is only an approximation because orbital motion of the
electrons also contributes to the magnetic moment; it is harder to quantify.) For instance,
a compound with a configuration of t32g /e0g would have a magnetic moment of about 3.87
B.M.
7.9.1
Exercises
7.9.1.1
What will the magnetic moment be for Cr(NH3 )3+
6 ?
Chm 118
Problem Solving in Chemistry
7.10
88
7.9.1.2
A octahedral complex of Mn(II) is made and has a measured magnetic moment of 6.0. Is
it high spin or low spin?
7.9.1.3
A compound of Co(II) is synthesized and found to have a magnetic moment of 1.7. In a
simple model where it is assumed to be either octahedrally or tetrahedrally coordinated,
which is it?
7.9.1.4
VBr2 has a magnetic moment of about 3.8 B.M. When it is dissolved in water and then
treated with KCN, the color of the solution changes dramatically, but the magnetic moment
is largely unchanged. Explain.
7.9.1.5
When CrBr2 is treated in a manner similar to that of VBr2 in the last exercise, the magnetic
moment decreases. Explain.
7.9.1.6
The complex CoF3–
6 is paramagnetic. What is its electronic configuration?
7.9.1.7
The complex Co(NH3 )3+
6 is diamagnetic. What is its electronic configuration?
7.9.1.8
Do the results of the last two exercises make sense based on the spectrochemical series?
7.9.1.9
2+
Although Ni(NH3 )2+
6 is paramagnetic, Pt(NH3 )4 is diamagnetic. Explain. HINTS: First,
both have the same ”d count.” What is the value of that? The latter compound is four
coordinate; that means it is probably either tetrahedral or square planar; we looked at
square planar energy levels in exercise 7.3.1.5.
7.10
Crystal Field Stabilization Energies
There are energetic consequences that are driven by the splitting of the d orbitals in a
crystal field; that is the topic of this section. First let’s take the six negative changes in an
octahedron and smash them with a hammer into little fractional charges that are spread
evenly on the surface of a sphere surrounding the ion. Now imagine an electron in one
of the d orbitals of an ion at the center of the sphere. The energy of the electron in the
d orbital is increased because of the negative charges on the sphere. Since the symmetry
remains spherical, all five of the d orbitals are repelled the same, as shown in change from
“Free Ion” to “Spread Out Charge” in Figure 7.11. As we are free to choose any arbitrary
energy as our zero of energy, we take the spread out charge value to be zero. Now we let the
little fragments of charge on the sphere come back together to form charges at the corners
of the octahedron. This will increase the energy of the orbitals that point at the corners
Chm 118
Problem Solving in Chemistry
7.10
89
Figure 7.11: Energetic consequence of the splitting of the d orbitals by an octahedral crystal
field.
of the octahedron, the eg levels, as the charge increases there. And decrease the energy of
the orbitals that avoid the corners, the t2g . This situation is labeled “octahedral charge”
in Figure 7.11.
If all the orbitals were filled in the “spread out charge” state, the net energy would be, by
definition, zero. If all the orbitals remain filled in the “octahedral charge“ state, the energy
must still be zero since there has been no change in the net charge on the sphere. This
means that the stability of an electron in the t2g set, x, times the six electrons in that set
must equal the instability of an electron in the eg set, y, times the four electrons in it. This
equation, and the one that sets the total gap between the t2g and the eg levels, energy of
eg minus energy of t2g , as ∆0 are:
−6x = 4y
(7.4)
x + y = ∆0
(7.5)
These equations solve to yield the destabilization of the eg level as 3/5∆0 and the stabilization of the t2g level as -2/5∆0 , both relative to the zero of energy, which is the spherical
environment of the ligand charge. This means that a d1 metal ion compound in the t12g /e0g
configuration would be more stable by -2/5∆0 than it should be (if the charges produced by
the ligands were spherical). This is called crystal field stabilization energy, often abbreviated
CFSE. CFSE has a significant influence on the properties of many metal complexes.
7.10.1
Exercises
7.10.1.1
For Co(II) compounds, chloride ion has a value of ∆0 of 6600 cm−1 . Find the CFSE in
kcal/mole for a six coordinate CoCl2 . HINT: Chloride ions are shared by various Co(II)
ions in this material.
7.10.1.2
The lowest energy absorbance peak for Rh(NH3 )3+
6 occurs at 304.8 nm. This compound
has a magnetic moment of zero. Find the CFSE for Rh(NH3 )3+
6 .
Chm 118
Problem Solving in Chemistry
7.10
90
7.10.1.3
What is the change in CFSE (in units of ∆0 ) upon moving a Co(II) ion from an octahedral
environment to a tetrahedral one? HINT: Recall the tetrahedral splitting is just the inverse
of the octahedral one, so stability of the e set is -3/5∆t and instability of the t2 set is 2/5∆t .
Also, recall that ∆t = 4/9 ∆0 .
7.10.1.4
The heat of hydration, the heat evolved in taking a gas phase metal ion (spherical) and
putting it into a octahedral environment of water molecules, is more negative for Mn(II)
than for Ca(II) (presumably because of the smaller radius of Mn(II)) and smaller yet for
Zn(II). In fact the points for those three ions lie on roughly a straight line when plotted
against the atomic number. The value for V(II) is considerably more negative than the line
would suggest. Give an explanation.
7.10.1.5
Look up closest packing of anions or examine Figure 7.12. The spheres on top (bold or red
in color) are in depressions of the first layer. There are two kinds of holes in this packing,
tetrahedral and octahedral. Which is “A” and which is “B?”
7.10.1.6
A spinel is a solid metal oxide with closest packed oxides of formula M(II)M(III)2 O4 where
the three M need not be the same. In a spinel, one-third of the metal ions are in tetrahedral
sites, and the others are in octahedral sites. A normal spinel has the M(II) in the tetrahedral
sites. Why would this be “normal”? HINT: Think about the stability of a dipositive and
tripositive cation in the presence of differing numbers of oxygen anions via Coulomb’s Law.
7.10.1.7
An “inverted” spinel has M(III) in the tetrahedral sites and the remainder of the M(III)
and all the M(II) in the octahedral sites. Fe3 O4 is an inverted spinel and Mn3 O4 is normal.
Why? HINTS: Something is over-riding the result of the last exercise, which is charge.
Could that something be CFSE?
7.10.1.8
Account for the fact that the spinel CoFe2 O4 is inverted.
Chm 118
Problem Solving in Chemistry
7.10
91
Figure 7.12: An example of closest packing. Red spheres are on top of black one.
Chm 118
Problem Solving in Chemistry
Chapter 8
Absorbance and Kinetics
8.1
Using Light to See
How do we know what we write is true? By probing chemical systems with light we can
learn a lot about what is going on at the molecular level. We take advantage of Beer’s law,
Ā = λ c`
(7.2 revisited)
to learn about the concentration of a material in solution. Note in this equation and
throughout this course, the symbol Ā is a symbol for an absorbance, while “A” itself might
stand for the name of a molecule. For instance, in Figure 8.1 is given the absorbance curves
of a solution of Cr(H2 O)5 SCN2+ as a function of time. Since the absorbance is changing
with time, some “chemistry” must be going on with the compound. Further, we can gain
some knowledge of what is happening from our understanding of the absorbance spectra of
transition metal ions. Since the peak at about 620 nm is disappearing and one is building at
570, the average strength of the ligand is going up. This would be consistent with the NCS–
being replaced by a water molecule, since water is higher on the spectrochemical series than
sulfur bonded NCS– is. More careful observation suggests even more information. Since
the peak for Cr(H2 O)3+
6 , the product we just postulated, occurs at 575 nm, and the new
peak is lower in wavelength than this, some ligand with greater field strength than water
must be replacing the sulfur bonded NCS– . Other experiments prove this to be true. The
dominant process occurring here is a linkage isomerization, as Cr(H2 O)5 SCN2+ is converted
into Cr(H2 O)5 NCS2+ . This example shows how powerful the color of a solution can be in
detecting chemical change.
Another feature in Figure 8.1 is important. The absorbance at a wavelength of about 593.5
does not change with time. Such points are called isosbestic points and are important
for understanding the chemistry behind the change. In the exercises you will derive the
conditions for isosbestic behavior. To do this derivation requires the equation for two
absorbing species in solution; since they act independent of each other, this is just a Beer’s
law sum:
Ā = λ,A [A]` + λ,B [B]`
(8.1)
For a system that is reacting, say conversion of molecule A to molecule B, there are a
92
8.1
93
Figure 8.1: (Idealized) Absorbance changes for a solution of Cr(H2 O)5 SCN2+ as a function of time. Each curve was recorded at a certain time. The curves at a wavelength of
greater than 593.5 nm were recorded at larger and larger times as you move downward. At
wavelengths less that 593.5, time increases for the curves as you move upward. Note the
isosbestic point, the point of no absorbance change with time, at about 593.5 nm.
couple of similar reactions that can be written that will be useful to you. At the beginning
of the reaction, there is no B, and the concentration of A would be called [A]0 . Likewise,
the absorbance at the beginning of the reaction would be Ā0 . At the end of the reaction,
assuming that it goes to completion, the absorbance is called the “infinite” absorbance, Ā∞
and there is no A, and [B]∞ = [A]0 . Therefore we have
8.1.1
Ā0 = λ,A [A]0 `
(8.2)
Ā∞ = λ,B [A]0 `
(8.3)
Exercises
8.1.1.1
−1 cm−1 at 508 nm. Find the
The ion Fe(phen)2+
3 has a molar absorptivity of 11,100 M
absorbance of a 0.02 mM solution in a 5.0 cm cell at 508 nm. What can you say about the
absorbance at 450 nm? Comment.
8.1.1.2
Use equation 8.1, and the facts that λ,A = λ,B at the isosbestic point, and that [B] is equal
to [A]0 - [A] to show that the absorbance does not change with time as A goes to B.
Chm 118
Problem Solving in Chemistry
8.2
94
8.1.1.3
Write the equation 8.1 for the reaction in Figure 8.1 for a wavelength of 610nm, with A
being Cr(H2 O)5 SCN2+ and B being Cr(H2 O)5 NCS2+ , where both λ,A and λ,B have value;
do this for some arbitrary time and label the time dependent parameters with a “sub t.”
Now put into your equation the stoichiometric relationship for [B], the expression [A]0 - [A];
and then subtract from both sides equation 8.3. Show that the result is:
[A]t =
Āt − Ā∞
λ,A ` − λ,B `
(8.4)
8.1.1.4
Write equation 8.1 for zero time, appropriately labeled, where [B] = 0. Write this equation
for infinite time, with appropriate labels, where [A] = 0 and [B] = [A]0 . Subtract the infinite
time value of absorbance from the zero time value and show, using equations 8.2-8.3, that
the difference is λ,A ` - λ,B `.
8.1.1.5
Use the results of the last two exercises to prove that the following ratio holds:
[A]t
Āt − Ā∞
=
[A]0
Ā0 − Ā∞
(8.5)
This is a very important and useful equation for using light to examine the time course of
reactions.
8.2
Basic Kinetic Expressions and Definitions
We looked at how color indicated that something was happening in the last section. Here we
become more precise in our language of describing change in a chemical system by discussing
the field of kinetic studies. First we get some sense of the issue of how fast reaction occurs
by considering the rate of reaction. This is a change in concentration of our reactant divided
by a change in time, or, in the calculus, d[A]
dt .
To get a sense of what happens in order for reaction to occur, imagine the space in a
beaker is divided up into a bunch of small compartments and imagine time divided into a
bunch of small increments. There are energy fluctuations in our space, so there is a finite
probability that a given compartment will have a slight excess of energy in any given time
increment. We are interested in the molecule A reacting. Let’s presume if the compartment
has that excess of energy and contains a molecule of A reaction occurs. Imagine the rate
(early in time when there are lots of reactants) is 0.00001 mole/sec. What do you think
the magnitude of the rate will be at a larger time when an energy fluctuation in a given
compartment might occur, but the concentration of A is lower? Clearly the rate will be
smaller. We learn that rate is concentration dependent, and it seems plausible that the rate
will depend on the concentration of A. We have what is called a rate law,
−d[A]
= k1 [A]
dt
Chm 118
(8.6)
Problem Solving in Chemistry
8.3
95
where the constant (for a given reaction and temperature), k1 , is called the rate constant .
In essence, the parameter k1 means how much energy is required for the reaction: large k1 ,
little energy; small k1 , lots of energy.
Let’s change our scenario a little. Imagine the same beaker with its compartments and
energy fluctuations. Now let the reaction be between a molecule of A and a molecule of
B to form some product. In this case assume that for reaction to take place, you must
have not only the energy, but also a molecule of A and a molecule of B. Assume when the
container has a reasonable concentration of A and B, the rate is 0.00001 moles/sec. Now, if
the container has a reasonable concentration of A, but is very dilute in B, what is the rate?
You would expect it to be lower since the three conditions needed, energy, an A, and a B,
is not met very often if [B] is small. Also, if very dilute in A but with a reasonable B, what
could we say about the rate? The rate law in this case must depend on both concentrations:
−d[A]
−d[B]
=
= k2 [A][B]
(8.7)
dt
dt
where k2 is the second order rate constant, and again is an (inverse) measure of the energy
required.
A final case. Molecule A reacts with itself to form products. What has to be in the
compartment for reaction to take place is two molecules of A and energy, so the rate law is:
−d[A]
= k2 [A]2
dt
(8.8)
How are these three types of rate laws distinguished? Equation 8.6 is called a first order
or overall first order rate law because the concentration of A is raised to the first power.
Equation 8.7 is a second order or mixed second order rate law because there are two concentrations, each raised to the first power. The language “first order in A and first order
in B” would apply to equation 8.7. Equation 8.8 is a pure second order rate law. The
language “second order in A” would apply to equation 8.8.
8.2.1
Exercises
8.2.1.1
What is the interpretation of a rate law of the form
this rate law?
−d[X]
dt
= k [X]2 ? What is the order of
8.2.1.2
What is the interpretation of a rate law of the form
this rate law?
−d[X]
dt
= k [X]? What is the order of
8.2.1.3
There are almost no known reactions that are zero order. That is, have a rate law of the
form −d[X]
= k? Use our interpretation of how a reaction proceeds to understand why this
dt
is so.
Chm 118
Problem Solving in Chemistry
8.3
8.3
96
First Order Reactions
Let’s focus first on how to identify a first order reaction such as −d[A]
dt = k[A]. The rate is
concentration (or time) dependent, so it is the slope of a plot of [A] versus time. Slopes are
hard to evaluate accurately, so another method is needed. We take advantage of the fact
that we can integrate a first order rate law and get an equation that can readily be plotted.
Here’s the math:
Z
[A]
[A]0
d[A]
= −k[A]
dt
d[A]
= −kdt
[A]
Z t
d[A]
=
−kdt
[A]
0
(8.10)
[A]
= −kt
[A]0
(8.12)
ln
(8.9)
(8.11)
We conclude that if we plot the logarithm of the ratio of [A] to [A]0 versus time, we will get
a straight line whose slope will be the negative of the first order rate constant, k. Note that
equation 8.5 allows a plot of an absorbance function rather than the concentration ratio
itself.
8.3.1
Exercises
8.3.1.1
Here is some kinetic data for a reaction in which the reagent being measured, A, starts out
at a concentration of 0.01 M. Show that the data are consistent with a first order rate law
and evaluate k.
t, min
1
3
5
7
9
[A], M
0.00885
0.00750
0.00600
0.00473
0.00402
t, min
2
4
6
8
10
[A], M
0.00806
0.00678
0.00539
0.00441
0.00397
8.3.1.2
Take the integrated form of the first order rate law, equation 8.12, and evaluate the time
at which the reaction is half finished, i.e., [A] = [A]0 /2.
8.3.1.3
Consider a first order reaction that starts with a concentration of A of magnitude [A]0 /2.
Take the integrated form of the first order rate law, equation 8.12, and evaluate the time
at which the reaction is half finished, i.e., [A] = [A]0 /4.
Chm 118
Problem Solving in Chemistry
8.4
97
8.3.1.4
Compare your results from the last two exercises. We say the “half-life” of a first order
reaction is independent of the concentration.
8.3.1.5
The great natural philosopher of rural Arkansas, Boniface Beebe, carried out a rate measurement. He weighed a portion of solid compound A, took some water from the distilled
water bottle, mixed the solid in the solvent and placed the sample into the temperature
equilibrated compartment of a spectrophotometer. The initial concentration of A was 0.01
M and the experiment was carried out at 50o C. Determine if the data are consistent with
a first order rate law. If they are not, suggest a cause for the deviation. HINT: Think
carefully about what Bonnie did in the experiment; and what he might have done.
t, min
2
6
10
14
18
8.4
[A], M
0.00985
0.00798
0.00600
0.00473
0.00402
t, min
4
8
12
16
20
[A], M
0.00926
0.00678
0.00539
0.00441
0.00397
Dealing with More Complex Rate Laws
Integration of more complex rate laws than first order is possible, but except for the second
order rate law, equation 8.8, which gives a linear plot of 1/[A] versus time (with a slope of
k), such integrations are difficult and sometimes impossible. Another method is needed to
deal with complex rate laws. That method is presented in this section where we use the
mixed second order rate law as an example.
Consider the rate law given in equation 8.7 for a reaction in which one mole of A reacts
with one mole of B. We run the reaction under a special set of conditions that allow the
simplification of the rate equations. Imagine that we start with a concentration of A of
0.0001 M and of B of 0.1 M. At the end of the reaction, when all of the limiting reagent,
A, is used up, the [B] = 0.0999 M, which is very close to the starting concentration of 0.1
M. Using this initial conditions we have managed to keep the concentration of B essentially
constant. The math becomes:
−d[A]
= k2 [A][B]
dt
−d[A]
≈ (k2 [B]0 )[A]
dt
−d[A]
= kpf o [A]
dt
(8.13)
(8.14)
(8.15)
which suggests that we can make a normal first order plot and get as a slope the quantity
kpro , which equals the true rate constant times the concentration of B. This is called the
pseudo-first-order approach because we have converted our more complex rate equation into
Chm 118
Problem Solving in Chemistry
8.4
98
a first order one by having all concentrations except one in large excess. To get the true
rate constant requires a second experiment at a different concentration (but still in excess)
of B.
8.4.1
Exercises
8.4.1.1
What will happen to the value of kpf o if the experiment described above is run with an
initial concentration of B of 0.2 M?
8.4.1.2
If the value of kpf o from the experiment described in the last exercise is 1.0×10−4 sec−1 ,
what is the true rate constant?
8.4.1.3
Let the rate law for a given reaction of A with B be first order in [A] and second order in
[B]; you run a pseudo first order reaction with [A]0 = 0.0001 and [B]0 = 0.1 and get a kpf o
of 0.01. What will the kpf o be if the initial conditions are [A]0 = 0.0001 and [B]0 = 0.2?
8.4.1.4
Here are some data for the reaction of A with B with a one to one stoichiometry. The initial
conditions are: [A]0 = 0.003 M, [B]0 = 0.06 M. Show the reaction is pseudo-first-order by
plotting ln[A]t versus time. The [A]t values as a function of time are given in the table.
HINT: What is the concentration of B at the end of the reaction?
t, min
2
6
10
14
18
[A], M
0.00251
0.00177
0.00119
0.000971
0.00054
t, min
4
8
12
16
20
[A], M
0.00205
0.00140
0.00185
0.00084
0.00058
8.4.1.5
What is kpf o in the last exercise?
8.4.1.6
For the same reaction as in exercise 8.4.1.4, but with [A]0 = 0.003, [B]0 = 0.096. Show the
reaction is still pseudo-first-order. The [A] values as a function of time are given for integral
steps of time from 2 to 20 in steps of 2 min: 0.00232, 0.00171, 0.00134, 0.000996, 0.000789,
0.000520, 0.000388, 0.000294, 0.000155, 0.000147.
8.4.1.7
For the same reaction as in exercise 8.4.1.4, but with [A]0 = 0.003, [B]0 = 0.12. Show the
reaction is still pseudo-first-order. The [A] values as a function of time are given for integral
steps of time from 2 to 20 in steps of 2 min: 0.00207, 0.00146, 0.00106, 0.000660, 0.000618,
0.000292, 0.000238, 0.000146, 0.0000645, 0.000190.
Chm 118
Problem Solving in Chemistry
8.4
99
8.4.1.8
Establish that the order with respect to [B] is one. Find the true rate constant for the
reaction in the last several exercises.
8.4.1.9
A reaction under pseudo-first-order conditions of [H+ ] has kpf o as a function of [H+ ] as
follows in pairs of [H+ ], kpf o : 0.1M, 1.3 x 10−3 sec−1 ; 0.2M, 5.2 x 10−3 sec−1 ; 0.5 M, 3.2 x
10−2 sec−1 . What is the order of the reaction with respect to [H+ ]?
8.4.1.10
What is the true rate constant for the reaction in the last exercise?
8.4.1.11
Use our method of “what has to collide” to understand what must be involved in a rate law
of the following form:
Rate = (k1 + k2 [A])[B]
(8.16)
Chm 118
Problem Solving in Chemistry
Chapter 9
Bonding
9.1
Storage of Energy in Molecules
Energy can be stored in a molecule. It is convenient for us to somewhat arbitrarily divide
that energy into a number of components. In this section we discuss these, from those of
the lowest energy to those of the maximum. However, since these energies are of molecules,
all of them are quantized.
The smallest energy gaps are those due to the translation of the molecule through space,
motion in the x, y, or z direction. A very reasonable model for this motion is that of a
parve on a pole, section 6.4, expanded to three dimensional space. Recall that such energy
depends inversely on the square of the length of the pole which gives a dependence of the
energy levels on the volume of the container in three dimensions. This means the energy
levels for a translating parve are much closer together in a large container than in a small
one. Also, the energy levels are inverse in the mass of the parve.
The next two kinds of motion that store energy do so exclusively in molecules, but not in
atoms. The first of these leads to a larger energy level separations than translation: it is
rotation of the molecule as a whole. The quantized energy levels for a three dimensional
rotating diatomic molecule is given by the equation:
EJ =
~2 J(J + 1)
2I
(9.1)
where J is the quantum number and can take values of 0, 1, 2, . . . , ~ is Planck’s constant
divided by 2π, and I is the moment of inertia, the rotational equivalent of mass in linear
processes, and is given by the expression
X
I=
mi ri2
(9.2)
i
where the sum is over all atoms, mi is the mass of the atom, and ri is the distance the atom
is from the center of mass of the molecule. The inverse term in I means that small molecules
have large energy gaps. The next important motion, leading to larger gaps, is vibrational
100
9.1
101
motion: stretching and bending of the bonds in a molecule. Quantum vibrational energy
levels are given by the expression
s k
1
En = ~
n+
(9.3)
µ
2
where k is the force constant, related to the stiffness of the bond and roughly related to the
m2
bond strength, and µ is the reduced mass (in a diatomic molecule, equal to mm11+m
), and n
2
is the vibrational quantum number, n = 0, 1, 2, . . . . Once again, the energy levels depend
inversely on the mass, and directly on the strength of the bond (though both through the
square root dependency). This means that low mass, reasonably bonded atoms, such as H,
dominate the high energy portion of the vibrational energy levels.
The fourth, and largest, energy that an atom of molecule can have is electronic energy. We
might estimate a typical value of this kind of energy (as you are asked to do in the exercises)
but moving an electron from the n = 1 level of a hydrogen atom, close to the nucleus, to
the n = 2 level, further from the nucleus. What it is important to grasp is that the energy
needed to move an electron around is considerably larger than the other kinds of energies
that we considered. A pictorial summary of the magnitudes of these energies, which you
will explore in the exercises, is given in Figure 9.1.
9.1.1
Exercises
9.1.1.1
Calculate the energy gap between the first and second energy levels for a parve, which we
let here be a Ne atom, on a pole of length 10 cm. HINT: Don’t forget to get the mass per
atom, not per mole.
9.1.1.2
Calculate the spacing between the first two translational energy levels for a CO2 molecule.
Use a POP approximation with a length of the pole of 10 cm. HINT: Watch your mass
units; not per mole!
9.1.1.3
Find the energy between the J = 0 and J = 1 rotational energy levels for CO which has an
I of 1.4 x 10−46 kg m2 .
9.1.1.4
Generally speaking, how would an increase in mass (and hence, usually in ri ), affect the
spacing of the rotational energy levels?
9.1.1.5
For CO, the force constant is 1.860 x 103 N/m. Find the energy gap between the n = 0 and
n = 1 vibrational energy levels.
9.1.1.6
What will happen to the spacing of vibrational energy levels as the atoms increase in mass?
Chm 118
Problem Solving in Chemistry
9.1
102
Figure 9.1: Schematic energy level diagram. The long, thick solid lines are electronic energy
levels; two are shown. Each of these electronic energy levels has associated with it several
vibrational levels, shown as (next to longest) lines, three in each of the electronic levels–the
lowest in on top of the electronic energy line. The rotational levels are shown (only for the
two lowest vibrational levels of the lowest electronic level) as the shortest lines. The gray
rectangles are the closely spaced translational levels, shown only for the lowest rotational
levels.
Chm 118
Problem Solving in Chemistry
9.2
103
9.1.1.7
What will happen to the spacing of vibrational energy levels as the bond becomes weaker?
9.1.1.8
Compute the energy required to excite a hydrogen atom from the n = 1 to the n = 2 level.
Recall that the energy of the hydrogen atom is given by En = 13.601/n2 eV. Express your
answer in Joules.
9.1.1.9
For the four motions: translation, rotation, vibration, and electronic, order them in terms
of the size of the energy level gaps.
9.1.1.10
The compound Ru(NH3 )4 bip2+ has an electronic energy absorption peak at 525 nm with an
of 3950 M−1 cm−1 . What is the energy gap between the two levels? What is the origin of
this transition? HINTS: The metal is Ru(II), a rather lowly charged metal; bip, bipyridyl,
has low lying empty orbitals composed mostly of p orbitals on the carbons and nitrogens.
9.1.1.11
Imagine a polyatomic gaseous molecule in a box at very low temperature. Now warm the
sides of the box a little and get the molecules in the box to start vibrating a little. When
one of our molecules of gas comes along and hits the wall, it may be given a push by the
vibrating wall. What will happen to the energy of the gaseous molecule after that push?
What kind of excitation will take place at low temperature?
9.1.1.12
Continuation of the last exercose. Warm the walls of the box more. Now we can give
a greater translational velocity to our gaseous molecule when it hits the wall. If it then
hits another gaseous molecule off center, it could get that second molecule to rotate more
violently. What happens to the energy of the gaseous molecules?
9.1.1.13
Continue the scenario of the last two exercises, adding more and more heat to the walls of
the vessel. What happens to the gaseous molecules? Do they translate? How? Do they
rotate? How? Do they vibrate? How?
9.2
Using Atomic Orbitals to Make Molecular Orbitals
We have a model of atomic orbitals derived from the solution to Schrödinger’s equation.
These orbitals have a shape dictated by the spherical field of the hydrogen nucleus. In a
molecule, the field is no longer spherical, but rather distorted in the shape of the molecule.
However, one might imagine that this undulating molecular field would not affect a wave
function very much near any given nucleus, where the dominant force is still the spherical
attraction to the nucleus. A mathematical approach that achieves this is called a linear
combination of atomic orbitals, abbreviated LCAO. In such an approach, we add the
atomic wave functions on various atoms to make a molecular orbital. Since the atomic
Chm 118
Problem Solving in Chemistry
9.2
104
functions decrease rapidly with distance from the nucleus, the dominate effect of the addition
occurs between nuclei, which is exactly where one would anticipate bonding electrons would
congregate. However, the electronic charge that concentrates between the nuclei must come
from somewhere. As we shall see, it comes and goes from behind the nuclei.
To see the essence of the LCAO method of bonding, we consider only a one dimensional
radial hydrogen atom 1s orbital on each of two nuclear centers, one at -1 units and the
other at +1 units. These two wave functions are show in Figure 9.2. Note that there is a
region in space, shaded pink in the Figure, where the two functions overlap in space. This
is called the overlap region and will be important to us because it is here that constructive
(or destructive) interference can occur. If we add these two functions as drawn they will
interact constructively since both are positive near zero. We must insure that the area
under the curve for the square of this new wave function generated by adding the two
atomic functions is unity (called normalization and it makes the total probability of finding
the electron somewhere equal to one). Adding and normalizing gives us the blue (solid
line) curve in Figure 9.3. The language we use to describe this blue curve is that it is the
probability resulting when the two waves interact. What should a non-interacting situation
look like? In this case we would have a 50% chance of finding the electron in the wave
function on the right nucleus and a 50% chance on the left. If we plot this non-interacting
curve we get the brown (dotted) line in Figure 9.3. Notice that the interaction leads to an
increased electron probability in the internuclear region, just as we hypothesized. Because
it is hard to see what happens behind the nuclei, that is, in the region of space from 1 to
3 (or from -1 to -3), I have plotted in Figure 9.4 an enlarged plot between 1.5 and 2.5.
Note that the brown dashed curve is above the blue curve in this region, meaning there is
less electron probability for the interacting system behind the nuclei than there is in the
non-interacting system. Electron density moves from behind the nuclei to between them in
the addition of the functions, in this constructive-interaction-wave function.
We made what is called a bonding molecular orbital by the addition of the two hydrogen
atom 1s wave functions. There is another way to combine the two functions, that is to
subtract them. This is shown in Figure 9.5 and has destructive interaction in the internuclear region. Not only is this a second way of “adding,” but it is also a necessary process.
The verbal way to describe the process is to say we must have conservation of orbitals.
We started with two hydrogen 1s wave functions, two orbitals, and when we get done we
must also have two orbitals. The addition of the two functions is one of the new ones;
the subtraction is the other. The logic behind this verbal description is that an orbital, a
wave function, is the only way of describing an electron. If we initially had two electrons to
describe, and did so with the two hydrogen 1s wave functions, φ1 and φ2 , we must also have
the ability to describe the two electrons when we finish. The addition and subtraction are
those two ways:
ψb = B(φ1 + φ2 )
(9.4)
ψa = A(φ1 − φ2 )
(9.5)
where A and B are normalizing constants. In the case presented here the addition leads to
a buildup of electron density between the nuclei and hence is label with a “sub-b” to reflect
that this is a bonding molecular orbital. In contrast, as we now develop, the subtraction is
actually an antibonding orbital, hence the subscript “a.”
Chm 118
Problem Solving in Chemistry
9.2
105
Figure 9.2: Two hydrogen 1s wave functions (in one dimension), one centered at x = 1 and
the other at x = -1. The pink shaded area is the area of overlap.
Figure 9.3: The blue (solid) curve is the square of the linear combination of the addition
of the two 1s functions, normalized so that the total probability of finding the electron
somewhere is one. The brown (dotted) curve is the non-interacting probability.
Chm 118
Problem Solving in Chemistry
9.2
106
Figure 9.4: Same as in Figure 9.3. This is just an expansion of the curve of that figure.
Figure 9.5: The linear combination of the subtraction of the two 1s functions.
Chm 118
Problem Solving in Chemistry
9.2
107
Figure 9.6: The blue (solid) curve is the square of the linear combination of the subtraction
of the two 1s functions, normalized so that the total probability of finding the electron
somewhere is one. The brown (dotted) curve is the non-interacting probability.
Figure 9.7: Same as in Figure 9.6. This is just an expansion of the curve of that figure.
Chm 118
Problem Solving in Chemistry
9.2
108
Figure 9.8: Overlap of two s orbitals using angular functions. The line is the internuclear
axis.
The destructive interaction resulting from the subtraction of the two functions actually
lowers the electron density between the two nuclei, relative to the non-interacting probability, as shown in Figure 9.6. Where does that electron density go? Figure 9.7 shows an
enlargement of the region behind the nuclei for the antibonding combination: the antibond
has greater probability there than does the non-interacting probability. In other words,
in the antibonding combination, electron density flows from the internuclear region to the
region behind the nuclei. This flow would tend to pull the nuclei away from each other:
truly antibonding.
The following exercises deal with the same issues just described, but use angular functions
instead of radial ones.
9.2.1
Exercises
9.2.1.1
Let atom A be at the origin of the coordinate system. Let atom B be at z = 1; y = 0; x =
0. Let the x axis be pointing “in and out”, the y axis “up and down”, and the z axis “left
and right.” Use an angular wave function picture to show the overlap of two s orbitals, one
on each atom. Compare your picture to Figure 9.8.
9.2.1.2
Same coordinate system as in exercise 9.2.1.1. Show, using angular functions, the overlap
of a pz orbital on A with an s orbital on B. HINT: Don’t forget that you have to “shade”
one side of a p orbital.
Chm 118
Problem Solving in Chemistry
9.3
109
Figure 9.9: Overlap of two p orbitals. The line is the internuclear axis.
9.2.1.3
Same coordinate system as in exercise 9.2.1.1. Show the overlap of a pz orbital on A with all
three of the p orbitals on B. Does one of your pictures look like that in Figure 9.9? Which
p orbital on B? Added or subtracted to that on A?
9.2.1.4
Same coordinate system as in exercise 9.2.1.1. Show the overlap of a py orbital on A with
a py orbital on B. Compare your answer with Figure 9.10.
9.2.1.5
Same coordinate system as in problem 9.2.1.1. Show the overlap of a s orbital on A with
all five of the d orbitals on B.
9.2.1.6
Same coordinate system as in problem 9.2.1.1. Show the overlap of a pz orbital on A with
all five of the d orbitals on B.
9.2.1.7
We have seen the overlap of functions on two separate atoms. When that overlap is constructive, what happens to the amplitude of the wave (function) in the region of overlap? What
happens to the square of the wave function? What happens to the electronic probability?
9.2.1.8
If the electronic probability (electron density) is greater between the two nuclei because of
the occupation of the molecular orbital, will the two nuclei be attracted or repelled?
9.2.1.9
We have seen the overlap of functions on two separate atoms. When that overlap is destructive, what happens to the amplitude of the wave (function) in the region of overlap? What
happens to the square of the wave function? What happens to the electronic probability?
9.2.1.10
If the electronic probability (electron density) is smaller (than the non-interacting function)
between the two nuclei because of the occupation of the molecular orbital, will the two
nuclei be attracted or repelled?
Chm 118
Problem Solving in Chemistry
9.3
110
Figure 9.10: Overlap of two p orbitals perpendicular to the molecular axis. The line is the
internuclear axis.
9.3
A Nomenclature for Molecular Orbitals
For the simplified molecular orbitals in this course, formed as they are from linear combination of atomic orbitals, we need a nomenclature system. As we shall see, there are two
aspects to our naming of the functions, the first is the dominant way the overlap occurs,
which is the subject of this section; the second we shall return to involves the symmetry of
the functions.
The overlap of two 1s orbitals in a hydrogen molecule is schematically pictured in Figure 9.8.
In that figure is a faint black circle drawn in a plane perpendicular to the internuclear axis.
What happens to the sign of the wave function as you travel around that circle? HINT:
Another view of this same circle is given in Figure 9.11. When the sign of a wave function
does not change as you “circle around,” the bonding is said to be a sigma bond (or a sigma
anti-bond), designated σ (or σ ∗ ). On the other hand, when the sign of a wave function
changes twice upon a complete “circle around,” that is, from “plus” to “minus” and back
to “plus,” the bonding is said to be a pi bond (or a pi anti-bond), designated π (or π ∗ )–see
the exercises. (We see it seldom except in transition metal ions, but when the sign changes
four times, the bond is a delta bond, designated δ (or δ ∗ ).)
Chm 118
Problem Solving in Chemistry
9.3
111
Figure 9.11: Overlap of two s orbitals viewed from behind one nucleus. The line is the
internuclear axis.
Figure 9.12: Overlap of two p orbitals viewed from behind one nucleus. The line is the
internuclear axis.
Chm 118
Problem Solving in Chemistry
9.4
112
Figure 9.13: Overlap of Two p Orbitals Viewed from Behind One Nucleus. The line is the
internuclear axis.
9.3.1
Exercises
9.3.1.1
A picture of two p orbitals overlapping is given in Figure 9.9; another view is given in
Figure 9.12. Is this a σ overlap?
9.3.1.2
Another picture of two p orbitals overlapping is given in Figure 9.10; another view of this
same overlap is given in Figure 9.13. Does the wave function change sign as you “circle
around?” Would you call this overlap a σ overlap? Or would you call it a π overlap?
9.3.1.3
Bring two hydrogen atoms with their 1s wave functions together and form the constructive
interaction. Also form the destructive interaction. What would you name those two different
molecular wave functions? Which one is more stable?
Chm 118
Problem Solving in Chemistry
9.4
113
Figure 9.14: MO diagram for H2 .
9.4
Molecular Orbital Energy Diagrams
We need a way to symbolize the relative energies of the bonding and antibonding orbitals
we have been discussing. Such a representation is made using what is called a MO diagram.
One for the H2 case is shown in Figure 9.14. The vertical axis is energy, with the more
stable orbitals being at the bottom. The energy of the orbitals are represented by lines, and
the lines on the left and right sides of the diagram represent the energies of the atomic levels
we started with. Just below those lines are given the atomic functions (with a small dot
indicating the position of the other nucleus for perspective). In the center of the diagram
are the two molecular orbitals, again indicated by lines; the stable bonding one, and the
unstable antibonding one. Beneath each of these is a “picture” of the orbital.
9.4.1
Exercises
9.4.1.1
The diagram in Figure 9.14 refers to H2 only because it has a pair of electrons in the σ
+
bond. The same diagram would work for for H+
2 , He2 , and He2 . Write the configurations
(in the sense of σ a σ ∗b (a and b are numbers) for each of these species.
9.4.1.2
+
The bond energies of H+
2 , H2 , He2 , and He2 are 61, 103, 55, and 0 kcal/mole, respectively.
Explain using MO theory.
9.4.1.3
If the H-H distance in H2 is 0.74 Å, and that in H+
2 is 1.06 Å, what would you expect for
He+
?
2
Chm 118
Problem Solving in Chemistry
9.5
114
9.4.1.4
Imagine an atom with only a pz orbital. Bring two of these together with the internuclear
axis the z axis and make the corresponding MO diagram. HINT: Read carefully: this is an
artificial problem, but an easy one.
9.4.1.5
Imagine an atom with only a pz orbital. Bring two of these together with the internuclear
axis the x axis and make the corresponding MO diagram. HINT: Read carefully: this is an
artificial problem, but an easy one.
9.4.1.6
Is there a difference between the strength of the bonding in the last two exercises? If so,
make sure your diagrams reflect that before you do the next exercise.
9.4.1.7
Imagine an atom with only p orbitals, three of them. Bring two of these together with the
internuclear axis the z axis and make the corresponding MO diagram.
9.4.1.8
Assume your molecule from the last exercise has 6 electrons. Give the configuration.
9.4.1.9
Build an MO energy diagram for a homonuclear diatomic molecule having 2s and 2p atomic
orbitals. Do not let any interaction occur between an s orbital on one center and a p orbital
on the other; This is called “ignore s/p mixing.”
9.4.1.10
Write the configurations and state the number of bonds in the homonuclear diatomics from
Li2 to Ne2 using your MO diagram from the last exercise.
9.5
Heteronuclear Diatomics and the First Row Diatomics
with s/p Mixing
In the last section you found the MO diagrams for homonuclear diatomics without s/p
mixing, that is, with no orbital interaction between an s orbital and a p orbital. Yet we
know from exercise 9.2.1.2 that an s orbital overlaps quite well with a p orbital. So there
is no justification for ignoring the interaction except that the s orbital is more stable than
the p. This is not a trivial issue and we pause to investigate it first.
Let’s consider the interaction of an 1s orbital on a hydrogen atom with the more stable 1s
orbital on a He atom. These atomic orbitals are shown in Figure 9.15 with the hydrogen
atom on the left and the He on the right. The two s orbitals still overlap and form the
bonding (constructive) and antibonding (destructive) MO. But think about the situation.
The helium 1s level is more stable than that of the hydrogen atom and if we added equal
amounts of the two functions together, as is implied in equation 9.5, then the electrons would
have equal probability of being on the hydrogen (less stable) atom and the helium (more
Chm 118
Problem Solving in Chemistry
9.5
115
Figure 9.15: MO diagram for HHe+ .
stable) atom. Why should electrons leave the comfortable environment of a dipositve charge
to wander over to a unipositive one? So the bonding MO should be more concentrated on
the helium atom. We write this as:
ψb = cH φH + cHe φHe
(9.6)
where cHe > cH . Using our concept of orbital conservation, if we used a lot of φHe in the
bonding orbital, we only have a little left to use in the antibonding one. Conversely, we
used only a little of φH in the bonding orbital, we have a lot left to use in the antibonding
one. Therefore the antibonding orbital is:
ψa = c∗H φH + c∗He φHe
(9.7)
where c∗H > c∗He . The drawings at the sides of the molecular orbitals in the center of our
MO diagram reflect this argument.
In a similar manner, we need to allow s orbitals (stable) and p orbitals (less stable) on
homonuclear diatomics to “mix.” The easiest way to do this is to take the answer without
mixing and make a correction for what mixing does; such a treatment, a two step treatment,
is called perturbation theory. The left hand side of Figure 9.16 gives us the results for a
homonuclear diatomic without s/p mixing. The first feature to note is an important result
we can use because of our development earlier in the course: only orbitals of the same
symmetry can interact. Application of symmetry to homonuclear diatomics is a little messy
in that the set of symmetry operations is infinite. You can see this by noting that a rotation
of 17.7 degrees about the internuclear axis is a symmetry operation, as is one of 17.8 degrees.
Chm 118
Problem Solving in Chemistry
9.5
116
Figure 9.16: MO diagram for homonuclear diatomic without and with s/p mixing.
To use symmetry we will just select certain symmetry operations and demand that if two
orbitals are going to interact, they (as “objects” in the previous discussion) must do the
same thing under those selected symmetry operations. The exercises lead you through the
process.
9.5.1
Exercises
9.5.1.1
Let the z axis of our diatomic molecule be the internuclear axis. We consider three symmetry
operations, C2z , σxz , and σxy , the plane bisecting the molecule and perpendicular to the
internuclear axis. Determine what happens to all eight orbitals on the left side of Figure 9.16
under these three symmetry operatons. Making a little table might be useful.
The choice of symmetry operations in this exercise are such that
they do not establish a critical feature of the π and the π ∗ levels. In
contrast to all the σ levels, the π levels are turned into each other
under certain symmetry operations, such as C4z . From our previous
discussion of objects in symmetric environments, this means they
must be “considered together.” A real physical consequence is that
when certain objects are “considered together” it leads to degenerate systems. That is what is happening here. The pair of π levels
belong to a set of characteristic numbers in this infinite group that
requires them to be degenerate in energy.
9.5.1.2
In the last exercise you presumably found that the level σ1 and the level σ2 had the same
symmetry. Hence they are capable of interacting. The red (dotted) lines in Figure 9.16 show
the interaction. Such interactions always stabilize the more stable level and destablize the
Chm 118
Problem Solving in Chemistry
9.5
117
least stable. It is as if the orbitals “repel” each other. Another way to look at this: The
lower orbital takes whatever “good” things the upper one has and ships back to the upper
any “bad” things it has. At this stage it is worth noting that how far the orbitals move,
whether the upper level is pushed above the π levels, as indicated in the Figure, is not
knowable at the level we are doing things.
9.5.1.3
It is interesting to see why the orbitals “repel” as indicated in the last problem. To do so,
we could draw the new orbital by adding and (of course) subtracting the starting functions.
Try it. Add in a purely graphical way σ1 to σ2 and see what you get.
9.5.1.4
Now subtract in a purely graphical way σ1 from σ2 and see what you get.
9.5.1.5
From the pictures in the last two exercises, can you guess which one new picture, the
addition or the subtraction, is the more stable orbital?
9.5.1.6
Do any other pair of orbitals on the left-hand side of Figure 9.16 have the same symmetry
and hence are capable of interacting? If so, draw in the interaction.
9.5.1.7
Add and subtract σ1∗ and σ2∗ and see what the functions look like. Can you determine which
is lower in energy?
9.5.1.8
The diagram you have now build is not so easily analyzed as the one without s/p mixing
because of the “good/bad character” trade off in the mixing. But there are interesting
consequences. What conclusion did you draw about the stability of Be2 in exercise 9.4.1.10?
What conclusion do you now draw? The fact is that Be2 is very slightly stable rather than
being like He2 .
9.5.1.9
Although this exercise concerns interesting aspects of s/p mixing, it involves very unstable
compounds. Nevertheless, the data support the model we have developed. If we accept that
σ2 is pushed above π, predict the magnetic characteristic–diamagnetic or paramagnetic–of
B2 for both the simple model of bonding and the model with s/p mixing. Do the same for
C2 . Experimentally, both answers agree with the model with s/p mixing.
9.5.1.10
Draw a Lewis structure for O2 . One of the great triumphs of MO theory is that it naturally
predicted that O2 was parmagnetic. Use the MO diagram to show that this is true.
9.5.1.11
How many net bonds are there in N2 ? in N2+
2 ? HINTS: Do this without s/p mixing. Give
your answer professionally: “There are two σ bonds and four π bonds.”
Chm 118
Problem Solving in Chemistry
9.6
118
9.5.1.12
How many bonds (and of what type) are there in N2–
2 ?
9.5.1.13
–1
2–
The bond lengths in O+
2 , O2 , O2 , and O2 are 1.122, 1.21, 1.26, and 1.49 Å. Comment.
9.5.1.14
The bond lengths in C2 , N2 , O2 , and F2 are 1.243, 1.298, 1.208, and 1.412 Å. Comment.
9.5.1.15
Build the m.o. diagram for V2+
2 considering only d orbitals on the atom. HINT: Remember
that a V(I) compound exhibits chemistry of d electrons only: Count all valence electrons,
but then put all of the appropriate number into the d orbitals of the vanadium.
9.6
Hybridization as a Simplifying Tool.
bridization
Part I: sp Hy-
It is relatively easy to deal with diatomic molecules, but they are of limited interest. More
compelling are polyatomic species. As the number of atoms increases, the molecular orbital
procedure becomes harder (unless there is lots of symmetry, as we shall see), and less informative for a chemist. To deal with this situation, chemists have adopted an approximation
that works reasonably well in a number of cases. This approximation makes the answers to
MO problems closer to the intuitive Lewis structures. This method is called hybridization.
It fits in at this position in our development because for diatomics, hybridization allows us
to mix s and p orbitals on an atom before we build the molecular orbitals.
The basis of hybridization is simple. We add and (as always) subtract functions on the same
atom. This process generates orbitals that are not solutions to Schrödinger’s equation, but
as we shall see, are useful in getting the bonding of molecules. The result of hybridization is
always orbitals that point in some specific direction; contrast this with an s orbital, which
points everywhere, a p orbital that points left and right, or a d orbital that points in four
directions. Orbitals that point in a specific direction simplifies problems because only atoms
in that direction bond well to the hybrid.
9.6.1
Exercises
9.6.1.1
Draw the z axis from an atom to the right and left on the paper. Draw an atomic s orbital
on an atom at the origin and let it be positive everywhere. Do you agree that this orbital
could have a value of 0.03 at the point x = 0, y = 0, z = 1? What would the value be at
the point (0,0,-1)? Draw a pz orbital at the same atomic center. Do you agree that this
second orbital could have a value of 0.02 at the point (0,0,1)? What would the value be
at the point (0,0,-1)? Now if we add these two orbitals, what would the net value be at
Chm 118
Problem Solving in Chemistry
9.6
119
the points (0,0,1) and (0,0,-1)? Choose the correct words in the following sentence: “The
orbital formed from adding the s and the p atomic orbitals is (bigger or smaller) when z is
positive and is (bigger or smaller) when z is negative.” The orbital that you have formed is
called an sp hybrid.
9.6.1.2
Do the last exercise again, except this time subtract the p from the s. HINT: Recall, “if
you add two orbitals, you must . . . ”.
9.6.1.3
Construct an sp hybrid so that it maximizes in the positive x direction. To do this, you
must understand which orbital in the last two exercises gave the directional properties.
Show that the second sp hybrid must maximize in the negative x direction. HINT: Use the
conservation of orbitals rule: if you add two functions to make one new one, you must also
subtract them to make another new one.
9.6.1.4
When you build an sp hybrid you use one s orbital and one p orbital. What orbitals are left
on the atom that “have not been used?” You will find as you progress that these orbitals
are always used for producing π bonds.
9.6.1.5
Given your answer to the last exercise, complete the following statement: “An atom in
which we assume sp hybridization will have
π bonds.
9.6.1.6
Build the m.o. diagram for N2 . Use sp mixing via hybridization before bonding. Once you
set these up, divide them into those that point at the other nitrogen atom and those that do
not. Let each of these pairs interact. The purpose of the sp hybrid on one nitrogen atom is
to get a big orbital pointing at the other nitrogen orbital, and the other way around. These
two interact strongly. On the other hand, the other sp hybrid on each nitrogen points away
from the other orbital so it does not bond very well; give them small interactions.
9.6.1.7
Build the m.o. diagram for CO. Use sp mixing via hybridization on the carbon atom before
bonding, but leave the oxygen atom “unhybridized” and use only p orbitals on the oxygen
atom.
9.6.1.8
How does the m.o. diagram of CO differ from the isoelectronic N2 ?
9.6.1.9
In going from CO to CO+ the bond length changes only slightly, from 1.128 to 1.115 Å.
How could you account for this observation? HINT: Think about what must be true of the
electron that you remove for the bond length to change very little.
9.6.1.10
It is much easier to ionize NO than to ionize CO. Why? HINT: Build an MO diagram.
Chm 118
Problem Solving in Chemistry
9.7
120
9.6.1.11
The advantage of hybrid orbitals is that bonds are localized and therefore, most of the time,
only one overlap has to be considered for each orbital. Take advantage of this truth to build
an mo diagram for C2 H2 . HINT: You will need a structure with geometry.
9.7
Hybridization as a Tool. Part II: sp2 and sp3 Hybridization
Hydribization is used extensively in organic chemistry, the chemistry of carbon compounds.
There are two other hybridization schemes that are commonly used, although neither is as
easy to comprehend as the sp hybrid case discussed above. The first is sp2 hybridization,
which mixes an s orbital with two p orbitals. These three hybrids point at the corners of a
triangle. When you want to bond a carbon atom to three other atoms, sp2 hybrids should
be used. The other hybridization mixes all three p orbitals with the s orbital; this produces
sp3 hybrids that point at the vertexes of a tetrahedron. They are used to bond a carbon
atom to four other atoms.
9.7.1
Exercises
9.7.1.1
When you build an sp2 hybrid you use one s orbital and two p orbitals. What orbitals are
left on the atom that “have not been used?” For what can you use this left over p orbital?
9.7.1.2
Complete the following statement: “An atom in which we assume sp3 hybridization will
have
π bonds.
9.7.1.3
The advantage of hybrid orbitals is that bonds are localized and therefore, most of the time,
only one overlap has to be considered for each orbital. Take advantage of this truth to build
an mo diagram for BH3 .
9.7.1.4
The advantage of hybrid orbitals is that bonds are localized and therefore, most of the time,
only one overlap has to be considered for each orbital. Take advantage of this truth to build
an mo diagram for C2 H4 .
9.7.1.5
Build an mo diagram for C2 H6 using hybrid orbitals.
9.7.1.6
Some molecules are best treated with a combination of methods, using hybrid orbitals and
symmetry. The second we turn to next, but first let’s do a separation of a problem into two
+
parts to prepare for later. Consider the molecule C3 H+
5 in the form CH2 CHCH2 . Draw
Chm 118
Problem Solving in Chemistry
9.8
121
Figure 9.17: MO diagram for water molecule.
the Lewis structure and use VSEPR to determine structure. What hybridization do you
use for the three carbon atoms? Consider only the sp2 orbitals on each carbon and the
hydrogen atoms; that is, neglect the three pz orbitals on the three carbon atoms. Make an
mo diagram for the σ part of the bonding scheme.
9.8
Using Symmetry to Determine the Orbitals that can
Bind. Part I. Simple Systems
We have seen previously only orbitals of the same symmetry can bind to each other. In
polyatomic molecules, use of symmetry greatly simplifies the approach to MO diagram. The
trick to using symmetry is to produce symmetry adapted linear combinations of atomic
orbitals, SALCAO. Consider a water molecule as our first example. This molecule has C2v
symmetry with a table of characteristic numbers given here:
where the molecular plane is the xz plane, as indicated in molecular sketch in Figure 9.17.
Let’s classify the orbitals on the oxygen atom according to their symmetry. You should do
this. What I got is indicated in the last column of Table 9.1; for instance, the pz orbital
has a1 symmetry and the px orbital has b2 symmetry. Notice in Figure 9.17 that these are
Chm 118
Problem Solving in Chemistry
9.8
122
C2v
A1
A2
B1
B2
E
1
1
1
1
2
C2
1
1
-1
-1
0
σyz
1
-1
1
-1
0
σxz
1
-1
-1
1
2
Os , Opz , Ha + Hb
Opy
Opx , Ha + Hb
(Ha , Hb )
Table 9.1: Characteristic number table for water molecular orbitals.
labeled as such in the right hand column. These symmetry assignments are straightforward
because the oxygen atom orbitals are turned only into plus or minus themselves.
The hydrogen atom orbitals are not so simple. The orbital on Ha is turned into that on
Hb and hence the two must be considered together, as two objects. This results in the
numbers given in the last row of Table 9.1; prove this to yourself. To find the combination
of hydrogen orbitals that have characteristic numbers from the named rows of the table, we
add and subtract the two hydrogen 1s functions, generating what are called ha1 = ha + hb
and hb2 = ha - hb . Show that these combinations do have the indicated symmetry.
Bonding can occur between orbitals of the same symmetry. To start off, we note that
there is only one b1 orbital, the py on the oxygen atom; it is therefore non-bonding and
“goes straight across” in the diagram–see Figure 9.17. There are only two orbitals of
b2 symmetry, one of the oxygen p orbitals and the antisymmetric linear combination of
hydrogen orbitals we called hb2 . These therefore “repel” each other and we get one bonding
b2 and one antibonding b2 . The only remaining symmetry is a1 ; there are three orbitals of
this symmetry, and that situation is hard to deal with in a qualitative manner, as we are
doing. The easiest way out of this problem is to remember that orbitals that differ in energy
a lot don’t interact very much. So to approximate what is happening, we neglect the very
stable oxygen atom 2s orbital, and then let the oxygen pz interact with ha1 . Figure 9.17
shows all of this. Work your way through the arguments and be sure you understand what
is happening.
9.8.1
Exercises
9.8.1.1
Find the names of the characteristic numbers for the s and p orbitals on the sulfur in SF2 .
9.8.1.2
Find the names of the characteristic numbers for the F pin orbitals in SF2 . What do the
two combinations look like? HINT: A pin orbital has the shape of a p orbital and points
from the F to the S on both F’s; hence its name “in.”
9.8.1.3
Build an m.o. diagram for SF2 molecule using only pin orbitals on the F and only p orbitals
on the S.
Chm 118
Problem Solving in Chemistry
9.9
123
9.8.1.4
What is the symmetry of NO–2 ? Let the plane of a NO–2 ion be the xz plane. Then there
are three py orbitals on the three atoms. Sketch these.
9.8.1.5
Find the three symmetry adapted orbitals that result from the three orbitals you described
in the last exercise.
9.8.1.6
Combine the SALCOs of the last problem to make an MO diagram for nitrite ion.
9.8.1.7
There are four electrons in the π system of NO–2 . (Do you see where those are from a Lewis
structure?) Is there a π bond in NO–2 ? On what atoms is it located?
9.9
Using Symmetry to Determine the Orbitals that can
Bind. Part II. Degenerate Systems
As we saw previously, when two or more functions belong to a characteristic set of numbers
with a two or greater under the identity operator, manipulation of them is a little more
difficult. We work through an example here. Imagine an s orbital on one of the hydrogen
atoms in NH3 . What happens to that s orbital when you carry out a C3 symmetry operation? Clearly it gets carried into another hydrogen atom. That means we have to consider
the two together. However, a C−3 symmetry operation takes the initial hydrogen atom
into the third one. So all three must be considered together. Using the C3v table given in
Table 9.2, we can find the numbers associated with these three functions as indicated in
the last row of the table. Examining this set of numbers, it is clearly composed of a1 and e
characteristic sets. Verify this for yourself. Since we have three functions, the easy rule of
adding and subtracting doesn’t work and we need to be more innovative (or use equation 5.2
of section 5.8). It is easy, always, to get the a1 function as it has to have the same sign
everywhere, and all functions of equal magnitude. In this case that function would be:
ψa1 = h1sa + h1sb + h1sc
(9.8)
Since the pz function on the nitrogen atom of the NH3 has a1 symmetry (we will ignore the
nitrogen atom 2s function as we did previously to make matters simpler), these two orbitals
bind to each other to make a bonding and antibonding pair.
What does the e set of the hydrogens interact with? The answer, although it is somewhat
messier to show, is the N px and py orbitals, which have characteristic numbers labeled “e”.
Here’s a start to showing that: If we let one of the H atoms in NH3 have y coordinates of 0
(i.e., it is in the xz plane), then the σv operation in the xz plane takes px and py into px and
-py , respectively, hence giving a total characteristic number of 0. Also, it is clear that the
identity operator takes px and py each into themselves, generating a characteristic number
of 2. That gives us two of the characteristic numbers of the “e” set. Close enough unless
you want to be hardcore:
Chm 118
Problem Solving in Chemistry
9.10
124
C3v
A1
A2
E
E
1
1
2
3
2C3
1
1
-1
0
3σv
1
-1
0
1
Table 9.2: The C3v table for NH3 .
The operation C3 also works, but since it is a 120o rotation,√ the
orbital px goes, as does the vector x, into -1/2 of itself (and - 3/2
of py , which is why the two functions are mixed up together) as you
can show with simple plain geometry. Likewise for the vector y and
hence py . The total “functions” that go into themselves is then -1/2
-1/2 = -1, as required by the characteristic number.
9.9.1
Exercises
9.9.1.1
Build an MO energy level diagram for NH3 using the symmetry adapted functions from
above. HINT: Use only the p orbitals on the nitrogen atom.
9.9.1.2
Consider a square planar compound of a transition metal. Let the ligands have only pin
orbitals and the metal have only d orbitals. Find the symmetry of the ligand orbitals.
HINT: The D4h characteristic numbers table is given in Table 9.3.
9.9.1.3
What symmetry do the metal d orbitals have in the system defined in the last exercise?
HINTS: Some metal orbital may get turned into another under a given symmetry operation.
If so it is possible that the two together may generate a set of numbers in the characteristic
number table. We would then say, speaking professionally, that “function x and function
y have Eu symmetry. Also, pay attention to the number under the center of inversion
operator; it is often easy to distinguish between Eu and Eg on the basis of this operator.
9.9.1.4
Make an MO diagram of the square planar molecule defined in the last two exercises.
9.9.1.5
If the metal is Ni2+ and the ligands each have two electrons in their pin orbitals, put the
appropriate number of electrons into your MO diagram of the last exercise.
Chm 118
Problem Solving in Chemistry
9.10
125
D4h
A1g
A2g
B1g
B2g
Eg
A1u
A2u
B1u
B2u
Eu
E
1
1
1
1
2
1
1
1
1
2
2C4
1
1
-1
-1
0
1
1
-1
-1
0
C2
1
1
1
1
-2
1
1
1
1
-2
0
2C2
1
-1
1
-1
0
1
-1
1
-1
0
2C”2
1
-1
-1
1
0
1
-1
-1
1
0
i
1
1
1
1
2
-1
-1
-1
-1
-2
2S4
1
1
-1
-1
0
-1
-1
1
1
0
σh
1
1
1
1
-2
-1
-1
-1
-1
2
2σv
1
-1
1
-1
0
-1
1
-1
1
0
2σd
1
-1
-1
1
0
-1
1
1
-1
0
Table 9.3: The D4h table of characteristic numbers.
9.10
Exploring a Model for Solids
In this section we take what we know about MO diagrams and examine a long chain of
atoms, bonded together. This is a one-dimensional model of a solid. It is based on concepts
already known to you.
9.10.1
Exercises
9.10.1.1
This problem is fictitious to make it easier to do. Imagine a set of atoms of type A in the
plane of the paper with the x axis from left to right. These A atoms have only one p orbital
emphperpendicular to the x direction, in our case, “up and down.” We do all our lining
up of atoms along the x direction. Build a MO diagram for a dimer of A, each with one
orbital. Sketch the two molecular orbitals.
9.10.1.2
Now bring two of the dimers from the last exercise together along the x axis, each with
its two molecular orbitals. First we let orbitals close to each other (in energy) interact;
construct the m.o. scheme for the interaction of the two lowest molecular orbitals. Do so
also for the two high molecular orbitals. This molecule with the four molecular orbitals is
our tetramer. Sketch each of the four orbitals.
9.10.1.3
The tetramer of the last exercise has a plane of symmetry down the middle of the molecule
and perpendicular to the x axis. The molecular orbitals that you got there are either
symmetric (go into themselves) or asymmetric (go into minus themselves) upon reflection
in that plane. If two molecular orbitals have the same symmetry, they can interact (if they
did not previously). Do the appropriate interaction between the low levels and the high
levels from the last problem.
Chm 118
Problem Solving in Chemistry
9.11
126
Orbital
I px
I py
Iz
2 F pin , added
2 F pin , subtracted
E
σxy
σxz
σyz
Table 9.4: Selected symmetry operations for MO diagram for IF–2 .
9.10.1.4
Bring two of the simple tetramers (exercise 9.10.1.2) together to make an octamer. What
do your molecular orbitals look like? Do you see a pattern in how they wave? Do you
see why the more stable ones are stable? why the ones in the middle are there? why the
unstable ones are unstable?
9.10.1.5
What happens if you keep bringing two of what you made in this set of exercises together?
What do you get for a molecular orbital diagram? You will have to describe this rather
than draw it out. This set of molecular orbitals is called a band of bonds (and antibonds)
and is used in a theory of metal bonding called band theory.
9.11
Molecular Orbital Theory of Hypervalent Compounds
When we discussed the exceptions to the Lewis octet theory earlier (see section 2.5) we
concluded that expansion beyond the octet required electronegative elements on the outside
of an element in the second row or higher of the periodic table. Herein we establish why this
condition is needed. We use as our example molecule IF–2 , a linear molecule with the iodine
atom at the center. The exercises will lead you through a MO approach to understanding
the bonding in this species.
9.11.1
Exercises
9.11.1.1
Since the symmetry of IF–2 has an infinite number of symmetry operations (and that is a
lot to deal with), we use only selected operations. Consider only the p orbitals on the I and
let the internuclear axis be the z axis. Fill in the top part of the chart in Table 9.4.
9.11.1.2
On the fluorine atoms, let’s consider only the pin orbitals. Since there are two orbitals, we
must add them and subtract them. Describe the symmetry of the addition of the two pin
orbitals on the two fluorine atoms. Put you entry in the chart in Table 9.4.
Chm 118
Problem Solving in Chemistry
9.11
127
9.11.1.3
Describe the symmetry of the subtraction of the two pin orbitals on the two fluorine atoms.
Finish filling in the chart in Table 9.4.
9.11.1.4
Start an MO diagram by putting in the atomic orbitals on the sides. HINT: Which is more
stable, 2p on fluorine or 5p on iodine?
9.11.1.5
Can the function made by adding the two fluorine pin orbitals interact with any orbital?
Draw the appropriate line to its energy in the center of your picture.
9.11.1.6
Can px on I interact with any orbital? Draw the appropriate line to its energy in the center
of your picture.
9.11.1.7
Make an argument for why the py on I will have the same energy as the px orbital does.
9.11.1.8
In there any orbital on I that can interact with any orbital on the fluorine atoms? Indicate
your answer on your MO diagram.
9.11.1.9
Electrons are hard to count when atoms used in an MO diagram which does not use all the
orbitals of a given type. In the exercises in this section we ignored the electrons in two of
the three p orbitals of the fluorine atoms. In which of those three orbitals should we place
the “hole” present in a fluorine atom? To avoid this difficulty, it is useful to think about
the fluorine atoms as having a complete shell (six electrons in the p orbitals and hence a
-1 charge) and assign the appropriate charge to the other atom, the iodine atom. Then we
know since all orbitals on the fluorine atom are filled, that pin is filled. What charge should
we put on the iodine atom in IF–2 if we say the fluorine atoms have a charge of -1 each? How
many electrons does the iodine atom them contribute to the MO diagram? Fill in your MO
diagram with electrons.
9.11.1.10
How many “bonds” are holding the IF–2 molecule together?
9.11.1.11
How many electrons would you have to put into an m.o. diagram for IF–4 if you considered
only pin orbitals on the fluorine atoms and only p orbitals on the iodine atom?
9.11.1.12
How many electrons would you have to put into an m.o. diagram for IF–4 if you considered
only pin orbitals on the fluorine atoms and both s and p orbitals on the iodine atom?
Chm 118
Problem Solving in Chemistry
9.12
128
9.11.1.13
How many electrons would you have to put into an m.o. diagram for IO–4 if you considered
only pin orbitals on the oxygen atoms and only p orbitals on the iodine atom?
9.11.1.14
Consider the bond between the two hydrogen atoms in H2 . There are a pair of electrons
in that bond. If you were to try to assign those electrons as “belonging” to one atom or
another, how would you do it? HINT: Very easy problem.
9.11.1.15
Assume the pair of electrons in any bonding orbital (for the MO diagram in problem 9.11.1.10)
can be assigned 50% to each of the two atomic orbitals (or SALCAOs) and that electrons in
lone pairs belong to the appropriate atom. Under this assumption, what will be the charges
on the atoms in IF–2 ? Do this slowly and carefully. It is just a mechanical count of electrons.
9.11.1.16
Do the fluorine atoms in IF–2 end up with charge? Does the iodine end up with charge?
Does this result agree with the conclusion we made long ago that the external atoms should
be electronegative, should want electrons relative to the central atoms?
9.11.1.17
Build an molecular orbital diagram for linear H–3 ; place the appropriate number of electrons
in the molecular orbitals.
9.11.1.18
Is the central atom or an external atom of H–3 more negative? The MO diagram of hypervalent compounds always places negative charge on the outside atoms; that’s the source of
our rule.
9.12
Bonding in Metal Compounds. Part I. σ Donors
In this and the next three sections we put together a molecular orbital model of bonding
in transition metal compounds. This model is a more realistic model than the crystal field
theory that we dealt with earlier, section 7.1 and following sections. It will, of course,
support the splitting of the d orbitals in an octahedral field into a lower energy set of three
a higher energy set of two. In addition, it will be able to account for the spectrochemical
series, something that crystal field theory could not do. It will lead us to consider an
important model for stability of a certain class of compounds, the “eighteen electron rule.”
We use a fairly simple model to understand the pertinent features. Consider the molecule
M(NH3 )2+
6 where M is some transition metal ion. To deal with the bonding we consider
the lone pair electrons on the ammonia ligands, and treat it as a pin orbital (we would get
the same symmetry results if it was an s orbital or an sp3 hybrid). On the metal ion, we
consider only the metal d orbitals. The symmetry of the octahedral molecule is high (which
is beneficial), but that means that the manipulations are not as simple as those we have
done. I ask you believe what I am going to tell you. The symmetry of the six ligand orbitals
Chm 118
Problem Solving in Chemistry
9.12
129
Oh
A1g
A2g
Eg
T1g
T2g
A1u
A2u
Eu
T1u
T2u
E
1
1
2
3
3
1
1
2
3
3
8C3
1
1
-1
0
0
1
1
-1
0
0
6C2
1
-1
0
-1
1
1
-1
0
-1
1
6C4
1
-1
0
1
-1
1
-1
0
1
-1
0
3C2
1
1
2
-1
-1
1
1
2
-1
-1
i
1
1
2
3
3
-1
-1
-2
-3
-3
8S3
1
1
-1
0
0
-1
-1
1
0
0
6S4
1
-1
0
-1
1
-1
1
0
1
-1
6σd
1
-1
0
1
-1
-1
1
0
-1
1
3σh
1
1
2
-1
-1
-1
-1
-2
1
1
Table 9.5: The Oh table of characteristic numbers.
in octahedral symmetry–see Table 9.5–is a1g + eg + t1u . The appropriate SALCAOs are
schematically indicated in Figure 9.18. The symmetry of the metal d orbitals is, as indicated
long ago, eg and t2g . Clearly the only interaction is between the two eg orbitals.
Since the metal ion’s orbitals are less stable than the ligand orbitals, the metal eg levels
get pushed up in energy and the ligand eg levels get stabilized. That is to say, for the two
molecular orbitals of eg symmetry, the upper one, closest in energy to the original metal d
orbitals, is primarily metal in character–see section 9.5 and Figure 9.15. The lower eg orbital
is primarily ligand in character. All other orbitals “go straight across.” The ammonia ligand
is called a σ donor: Electrons move from ligand to metal because of charge, so the ligand
is a donor. The only orbital available to the ammonia to donate the electrons is cylindrical
about the metal-ammonia axis, hence the σ designation. There are relatively few pure σ
donor ligands as most ligands have more than one lone pair, which, as we will see in the
next section, changes the bonding pattern.
9.12.1
Exercises
9.12.1.1
Using the information given above, make an MO diagram for the M(NH3 )2+
6 compound.
9.12.1.2
The ligand orbitals are of σ type, symmetrical about the ligand-metal bond. So the molecular orbital that is mostly metal eg is
in character. Choices are σ, σ ∗ , π, π ∗ , nonbonding.
9.12.1.3
The metal t2g level is
in character. Choices are σ, σ ∗ , π, π ∗ , non-bonding.
9.12.1.4
All of the ammonia orbitals are filled. Assume that the metal ion brought in three electrons.
Put the electrons into your MO diagram.
Chm 118
Problem Solving in Chemistry
9.13
130
Figure 9.18: Schematic of ligand orbitals for an octahedral environment. Symbols give the
sign of the wave function at the indicated position. The “double plus” is bigger than a
“plus.”
9.12.1.5
Compare the configuration of the d3 compound of the last exercise with the answer you
would get with crystal field theory.
9.12.1.6
In crystal field theory we introduced a parameter called ∆ that was the difference in energy
between the eg levels and the t2g levels. Since that same difference exists in the molecular
orbital approach (although now the eg level is just “mostly” metal in character), it would
seem reasonable to use the same nomenclature for this gap. For the compound M(NH3 )2+
6 ,
what determines the size of ∆? Remember your answer because we shall use it later.
9.13
Bonding in Metal Compounds. Part II. σ and π Donors
We now deal with the bonding in a compound of the type MF4–
6 . This analysis follows
closely the one in the last section; in fact, half of the work is identical to that in the last
section. Consider fluoride ion as a typical σ donor, through the pin orbital, and a π donor
through the filled p⊥ orbitals, those perpendicular to the metal-ligand bond. There are two
of these orbitals per fluorine ion, for a total of 12 orbitals. As you will show below, the p⊥
orbitals can interact with dxy , dxz , and dyz , the t2g set of orbitals. The symmetry agrees,
in that the SALCAOs generated by the 12 p⊥ orbitals are t1g + t1u + t2g + t2u . Clearly
the symmetry allowed interaction between the t2g on the metal and that on the fluoride
Chm 118
Problem Solving in Chemistry
9.14
131
ligands forms a bonding/antibonding pair. The σ aspect of bonding is identical to that in
the last section.
9.13.1
Exercises
9.13.1.1
Sketch a p⊥ on one fluoride ion interacting with dxz on the metal. This is typical of the
type of interaction found between the t2g of the metal and those of the ligands.
9.13.1.2
What kind of bond (or antibond) did you draw in the last exercise, σ or π?
9.13.1.3
Using the information given above, make an MO diagram for the MF4–
6 compound. HINT:
Be careful about how far things gets pushed up or down. Remember that the overlap is
better for a σ interaction than for a π one.
9.13.1.4
The ligand orbitals that are of σ type, are symmetrical about the ligand-metal bond. So
the molecular orbital that is mostly metal eg is
in character. Choices are σ, σ ∗ , π,
∗
π , non-bonding.
9.13.1.5
The ligand orbitals that are of π type, p⊥ , make π bonds and antibonds. Therefore, the
mostly metal t2g level is
in character. Choices are σ, σ ∗ , π, π ∗ , non-bonding.
9.13.1.6
All of the fluoride orbitals are filled. Assume that the metal ion brought in three electrons.
Put the electrons into your MO diagram.
9.13.1.7
Compare the configuration of the d3 compound of the last exercise with the answer you
would get with crystal field theory.
9.13.1.8
In crystal field theory we introduced a parameter called ∆ that was the difference in energy
between the eg levels and the t2g levels. Since that same difference exists in the molecular
orbital approach (although now both the eg level and the t2g level are “mostly” metal in
character), it would seem reasonable to use the same nomenclature for this gap. For the
compound MF4–
6 , what determines the size of ∆? This is more complicated than the answer
to the similar question in the last section. Remember your answers because we shall use
them later.
Chm 118
Problem Solving in Chemistry
9.14
9.14
132
Bonding in Metal Compounds. Part III. σ Donors/π
Acceptors
In order to examine the third kind of bonding between metal and ligand we need to understand ligands with π bonds. We use as our example CN– . This is a heteronuclear diatomic
and will have an energy level diagram roughly like that seen previously, Figure 9.16; refer to
that diagram for the following arguments. In the case of CN– we have ten valence electrons
so the configuration of the molecule is σ12 σ1∗2 π 4 σ22 . Since the nitrogen atom has a greater
VOIE, the π level is distorted toward the nitrogen atom whereas π ∗ is distorted toward the
carbon atom. The former, being filled, is dominately between the nuclei and, since it is
distorted toward the nitrogen atom, does not overlap well with metal orbitals; we neglect
it in our treatment. The latter, the π ∗ , is on the outside of nuclei and localized on the
carbon atom, which is the atom which the cyanide ion uses to bind to metal. We must pay
attention this high energy empty orbital in our treatment. It is of note that this orbital
is the lowest occupied molecular orbital , called the LUMO. Since these π ∗ levels have the
same symmetry as any other π orbital on a ligand, we know that symmetry from the last
section: t1g + t1u + t2g + t2u . The only orbital of importance for bonding with the metal d
orbitals then is t2g . Since this orbital is empty on the ligand it serves the role of accepting
electron density from the metal. We call ligands like cyanide ion π acceptors.
Within the σ electrons, the most important ones are those in the highest occupied molecular
orbital, the HOMO, which is σ2 . Careful analysis shows this orbital is dominately an orbital
on carbon atom pointing away from the nitrogen atom; ideal for σ donation to the metal.
As before the symmetry of these σ donor orbitals is a1g + eg + t1u and the only component
that is important for bonding to the d orbitals of the metal is the eg one.
The situation simplifies down to this if we ignore all the orbitals on the ligands that are of
the wrong symmetry to bind to the metal: We have a stable eg level on the ligands filled
with electrons, the usual eg and t2g on the metal (with as many electrons as the metal has),
and a high energy t2g on the ligands that is empty. In the exercises you will build an MO
diagram from these orbitals.
9.14.1
Exercises
9.14.1.1
Sketch a π ∗ on one cyanide ion ligand interacting with dxz on the metal. This is typical of
the type of interaction found between the t2g of the metal and those of the ligand.
9.14.1.2
What kind of bond (or antibond) did you draw in the last exercise, σ or π?
9.14.1.3
Using the information given above, make an MO diagram for the M(CN)4–
6 compound.
HINT: Be careful about how far things gets pushed up or down. Remember that the
overlap is better for a σ interaction than for a π one.
Chm 118
Problem Solving in Chemistry
9.14
133
9.14.1.4
The ligand orbitals that are of σ type, are symmetrical about the ligand-metal bond. So
the molecular orbital that is mostly metal eg is
in character. Choices are σ, σ ∗ , π,
∗
π , non-bonding.
9.14.1.5
The ligand orbitals that are of π type, π ∗ , make π bonds and antibonds. In this case the
mostly metal t2g level is
in character. Choices are σ, σ ∗ , π, π ∗ , non-bonding.
9.14.1.6
All of the cyanide ion orbitals of type σ2 are filled and all of those of type π ∗ are empty.
Assume that the metal ion brought in three electrons. Put the electrons into your MO
diagram.
9.14.1.7
Compare the configuration of the d3 compound of the last exercise with the answer you
would get with crystal field theory.
9.14.1.8
In crystal field theory we introduced a parameter called ∆ that was the difference in energy
between the eg levels and the t2g levels. Since that same difference exists in the molecular
orbital approach (although now both the eg level and the t2g level are “mostly” metal in
character), it would seem reasonable to use the same nomenclature for this gap. For the
compound M(CN)4–
6 , what determines the size of ∆? This is more complicated than the
answer to the similar question in the last two sections.
9.14.1.9
In this and the last two sections we now have three kinds of ligand systems, all of which are
σ donors. They differ in being π neutral (NH3 ), a π donor (F– ), and a π acceptor (CN– ).
One can roughly look at the three classes of ligands as having the same effect on the eg
orbitals of the metal (though see exercise 9.14.1.13). They differ in what they do to the t2g
levels. Articulate that difference.
9.14.1.10
Look at the spectrochemical series and see if you can see how the conclusions from the last
exercise determine the size of ∆ as the ligand is changed.
9.14.1.11
What properties does a ligand have that makes it occur low in the spectrochemical series?
9.14.1.12
What properties does a ligand have that makes it occur high in the spectrochemical series?
9.14.1.13
Which would you expect would be the better σ donor, F– or CN– ? Why?
9.14.1.14
Does your result from the last exercise enhance the effect of π difference, or subdue it?
Chm 118
Problem Solving in Chemistry
9.15
134
9.14.1.15
For a metal complex with π accepting ligands, such as CN– or CO, how many electrons are
needed to fill all the stable orbitals? For purposes of counting here, we only consider those
orbitals on the ligands that are σ donor, π donor, or π acceptor with respect to the metal
atom or ion; also, a stable orbital might be said to be any orbital found in a MO diagram
to be below the original energy of the metal orbitals.
9.15
The Eighteen Electron Rule
In the last exercise of the last section you established that 12 electrons are needed to fill all
of the σ levels of the ligands, including those that are non-bonding as well as those that are
bonding. On the metal, you found that if the t2g were stabilized by a π accepting ligand,
those bonding orbitals on the metal held 6 electrons. We conclude that for the most stable
compound (with π accepting ligands, 18 electrons are needed. This analysis is valid enough
that there is a rule called the eighteen electron rule for metal containing compounds. The
exercises illustrate the extent and range of its validity.
9.15.1
Exercises
9.15.1.1
To use the eighteen electron rule you have to decide what charge to give to the ligands
around the metal. You can make this assignment arbitrarily, but you must be consistent.
Usually, it is best to make a reasonable assignment: Cl– is negative; CO and NH3 are
neutral. All of these ligands contribute two electrons to the complex. How many electrons
would H– contribute?
9.15.1.2
Does the compound Mn(CO)5 H obey the eighteen electron rule if you consider the H atom
as an H– ion?
9.15.1.3
Does the compound Mn(CO)5 H obey the eighteen electron rule if you consider the H atom
as an atom? How can the answer to both this question and the one of the last exercise be
“yes?”
9.15.1.4
Which of the following are valid 18 electron rule compounds? Cr(CO)6 , Fe(CO)6 , Mn(CO)5 Cl,
Fe2 (CO)9 , (HINT: think metal/metal bond), Ni(CO)5 , HCo(CO)4 ? If it is not valid, what
closely related compound would be valid?
9.15.1.5
Compounds having NO as a ligand are tricky, although those having NO+ are easy. Comment.
Chm 118
Problem Solving in Chemistry
9.15
135
9.15.1.6
Which of the following are valid 18 electron rule compounds? Mn(CO)5 (NO), Fe(CO)3 Cl3 ,
Ni(CO)4 , Co2 (CO)8 ? If it is not valid, what closely related compound would be valid?
9.15.1.7
Should the compound CrF3–
6 obey the eighteen electron rule? It so, we are in trouble,
because this is a perfectly stable compounds. Comment.
Chm 118
Problem Solving in Chemistry
Chapter 10
Entropy
10.1
Definitions, and the Number of Microstates in a Configuration
Entropy is a very confusing concept. Although there are strong voices at work attempting
to wean people away from the concept that it is “disorder,” that concept is still prevalent.
Even if it were valid to accept such a definition, how do you measure “disorder?” In this
Chapter we will look at a way of assessing entropy that has a strong conceptual foundation
even if it is impractical in practice; then we will tie the strong foundation to the experimental
measure, gaining the best of both worlds.
We start by considering a set of oscillators with their associated quantum levels; these
oscillators are localized in space so we can tell which one is the first, A, which the second,
B, etc. Although it is not essential to the development, it makes it easier for us if we deal
with oscillators rather than parves moving through space because the quantum levels in
oscillators are evenly spaced, as shown in Figure 10.1. One of our oscillators is in one or
another of those quantum levels, but only one level; this is not to be confused by the figures
we will use soon which portray a collection of oscillators. The first question we ask is how
many different ways can we give a total of 2 units of energy to four oscillators. One way
would be to give oscillator A 2 units of energy and all the rest zero units. This is shown
in Figure 10.2. We shall call this picture of one way to give four particles (the number of
particles is N) two units of energy a microstate of the system.
A second way for an N = 4/ET = 2 system to exist is shown in Figure 10.3 where it is
particle B that has the two units of energy. If you step back far enough the labels on the
oscillators (balls) are no longer visible and we have a picture that corresponds to both of the
microstates at once. Such a picture that shows the arrangement of the oscillators without
their labels is called a configuration, Figure 10.4. Be sure you see the difference between
a microstate and a configuration. One important question we want to ask is: How many
different microstates lead to this same configuration? In this case it is relatively easy to
determine the answer by just counting. This configuration has one oscillator, the one with
two units of energy, in a unique position. That one oscillator is either A or B or C or D. So
136
10.1
137
Figure 10.1: Energy levels for a harmonic oscillator. Note the even spacing. Also, the lowest
level has arbitrarily been assigned “zero” units of energy. Energy is in some arbitrary unit.
Figure 10.2: A microstate of the system with four particles, N = 4, and two units of energy,
ET = 2 units. Energy is in some arbitrary unit.
Chm 118
Problem Solving in Chemistry
10.1
138
Figure 10.3: Another microstate of the system with four particles, N = 4, and two units of
energy, ET = 2 units. Energy is in some arbitrary unit.
there are four microstates in this configuration.
The second important question is: Are there any other configurations with the same N and
the same ET ? A little thought will give you the answer. We could give oscillators A and
B each one unit of energy and let C and D be without any energy. Another way of saying
that, consistent with our pictures, is that we put oscillator A and B in level e1 and C and
D in level e0 . Is this configuration different? How many microstates are in this second
configuration. We can count these, as you are asked to do in the exercises.
Finally, an important fact. Each microstate is as probable as any other microstate. Here we
enter the statistical issue of probability that will play such an important part of what follows.
If we throw a unit of energy at our four oscillators, any one of them is as likely as the other
to pick it up. So each microstate is equally probable. It follows if two configurations differ
in the number of microstates that the configuration with the higher number of microstates
is more probable. Configurations differ in probability. To summarize: Each microstate is
equally probable; configurations differ in their probability.
10.1.1
Exercises
10.1.1.1
Craps is a game in which rolling a “7” as the sum of the two dice has consequences. What
is the probability that one will roll a “7” with two dice? What is the probability that one
Chm 118
Problem Solving in Chemistry
10.2
139
Figure 10.4: The configuration for four particles, N = 4, and two units of energy, ET = 2
units. Energy is in some arbitrary unit.
will roll a “2” with two dice?
10.1.1.2
What is most probable in craps, rolling a “7” or rolling a “2” (with two dice)?
10.1.1.3
What is the difference between a microstate and a configuration?
10.1.1.4
How many microstates are in the configuration (with N = 4/ET = 2) in which two oscillators
each have one unit of energy and two oscillators have zero units of energy. Brute force
counting. We will get a formula soon.
10.1.1.5
Are there any other configurations for the N = 4/ET = 2 system?
10.1.1.6
How many configurations are there for N = 4/ET = 3 system? HINT: Don’t do the number
of microstates at this point.
10.1.1.7
How many configurations are there for N = 10/ET = 3 system? HINT: Don’t do the number
of microstates at this point.
Chm 118
Problem Solving in Chemistry
10.2
10.2
140
A Way to Count Microstates in a Configuration and
Predominant Configurations
In the last section we determined the number of microstates in some configuration by brute
force counting. This method becomes increasingly harder as the number of oscillators
increases (and we will want to increase the number up to Avogadro’s number, No !). We
hence switch to a mathematical formula that evaluates the number of microstates in a
configuration. This formula uses factorials. Recall that N! is one times two times three
times ... up to N. Two features: Factorials get big very quickly. My calculator bombs
out at 70!. Imagine No ! Secondly, 0! is defined as one.1 Given these two statements, the
formula that correctly tells us the number of microstates in a configuration for our row of
oscillators is:
N!
W =
(10.1)
n0 ! × n1 ! × n2 ! × . . .
where W is the number of microstates in the configuration, N is the total number of oscillators and n0 is the number in the e0 level, n1 the number in the e1 level, etc. So to do the
calculation in the last section:
W =
24
4!
=
= 4
3! × 0! × 0! × 1!
6
(10.2)
With the equation in hand, we can calculate the number of microstates in configurations
easily (as long as the numbers don’t get too large: the bane of factorials). Let’s agree on
a shorthand to designate the configuration. We list a set of numbers with slashes between
them. The numbers are the number of oscillators in the various energy levels, starting
at the lowest energy level, the one with zero units of energy. In this nomenclature, the
configuration given in Figure 10.4 is 3/0/1/0/0.
For any listing of the N/ET , one can write the configurations that are possible. We can
never do this with large numbers of oscillators because it becomes impossible to imagine all
of the possible configurations. We are going to use a small number of oscillators to establish
features that are strictly true only for a large number of oscillators. If you make up your own
examples, beware that sometimes the general truths, valid for a large number of oscillators,
do not hold for a small number. To get started, let’s examine the configurations possible
for the system N = 10/ET = 3. The three configurations and the number of microstates
for each is given in the first row of Table 10.1. You should verify these results are correct.
In the following rows of this table are the number of microstates in the three configurations
for an increasing number of particles, up to 70, holding the total energy constant at 3
units. This example shows that a configuration becomes predominant as the number of
particles increases. In this case, it is configuration C, which with ten particles is only slightly
favored over configuration B, but becomes increasingly favored as N goes up. This is shown
graphically in Figure 10.5. Notice that a N = 10, configurations B and C are almost equally
probable, but by N = 70, configuration C occurs over 91 percent of the time. Our rule: As
the number of oscillators goes up, a configuration (or a set of configurations closely related in
the case of a large number of oscillators), becomes overwhelmingly predominant. This means
1
There are web pages that address why this has been defined this way and why that definition is useful.
Chm 118
Problem Solving in Chemistry
10.2
141
N
10
20
30
40
50
60
70
A
(N-1)/0/0/1
10
20
30
40
50
40
70
B
(N-2)/1/1/0
90
380
870
1560
2450
3540
4830
C
(N-3)/3/0/0
120
1140
4060
9880
19600
34220
54740
Table 10.1: The number of microstates in the various configurations for the N = N/ET =
3 system.
Figure 10.5: The fraction of the microstates in the various configurations for N particles
with three units of energy, ET = 3 units; see text.
that for practical purposes we are always working with the predominant configuration; the
others are so overwhelmingly disfavored that we never find the collection in them.
10.2.1
Exercises
10.2.1.1
Find the number of microstates, W, in the configuration 5/1/1/1/0.
10.2.1.2
Find the number of microstates, W, in the configuration 6/3/1/0/0.
10.2.1.3
Consider 10 particles with a total energy of 4 units. Find all the configurations and the W
of each. Identify the most probable configuration and note W for it.
Chm 118
Problem Solving in Chemistry
10.3
142
10.2.1.4
Consider 20 particles with a total energy of 4 units. Find all the configurations and the W
of each. Identify the most probable configuration and note W for it.
10.2.1.5
Consider 70 particles with a total energy of 4 units. Find all the configurations and the W
of each. Identify the most probable configuration and note W for it.
10.2.1.6
From the last three exercises, what is the predominant configuration? What do you conclude
about how W for the most probable configuration changes as N goes up?
10.3
A Detour to Define Some Thermodynamic Quantities:
The First Law
We are about to ask the question: What happens to the value of W of the predominant
configuration when we add energy to the system? Adding energy to the system occurs
either in the form of heat or work. In that sentence are three words that we must be careful
to define in a precise manner. First, the system. The system is that part of the universe
that we are studying. It could be a beaker of liquid or a rock or a room with its contents.
The rest of the universe, that part we are not studying, is called the surroundings. You
define the system; nature does not. Heat is something we all think we know about, but it
is a hard concept to define. I like the definition given by Fenn2 : “Heat is the interaction
(what happens) between a hot object and a cold object that are in contact with each
other.” To make sense of this, we need to understand what is meant by an interaction.
This is when there is relation between a observable change in the system and one in the
surroundings. The most obvious such change relative to heat is, of course, temperature.
When a hot object (high temperature) is placed next to a cold object (low temperature),
the temperature of both change in the expected direction. The interaction that causes those
changes in temperature is heat. Fenn points out that one way to test for a heat interchange
is to insulate the system and ask if the rate of the change in the property is lowered. We
will use the symbol “q” for the total heat absorbed by the system in a given process, and
δq for the amount of heat absorbed in a given small step of that process.
Work is defined as a force operating over a distance. Also, any interaction that is not heat
is work. If we restrict ourselves to systems of gases, as we shall mostly do, then work is the
external pressure (force per unit area) times the change in volume (where the distance of
our definition comes in because of the volume divided by the area of the pressure term) of
the gas. For gases then, the work done on the gas for a small change in volume is
δw = −Pext dV
(10.3)
where the minus sign arises because a positive change in volume means the gas did work,
not the other way around. Please note that the external pressure may be a function of the
2
Fenn, J. B., “Engines, Energy, and Entropy. A Thermodynamic Primer,” W. H. Freeman and Co., 1982,
p. 5.
Chm 118
Problem Solving in Chemistry
10.4
143
volume of the gas; we will find it necessary for this to be true later. Integration of this
expression gives us the total work, “w.”
This restriction to gases also allows us to examine how the interaction called work affects
the kind of systems that we are investigating in this chapter. Since for gases work involves
a change in volume, and a change in volume influences the quantum levels for translation
directly: A smaller volume (shorter pole for a POP) causes the separation of the quantum
levels to become larger, but, importantly for us, to change the spacing. On the other hand,
one can add energy by heat interaction and, if there is no volume change, there is no change
in energy level spacings. It is this last kind of process that we want to talk about in our
discussion of configurations.
The discussion above suggests a relationship that is known as the first law of thermodynamics. If the only two methods of interaction to give energy to a system are heat and work,
then we should be able to express the energy change in terms of those variables. That is
the first law in differential form:
dE = δq + δw
(10.4)
where δq is the heat absorbed by the system, and δw is the work done on the system. In
terms of finite differences
∆E = q + w
(10.5)
10.3.1
Exercises
10.3.1.1
If the external pressure is constant at 1.0 atm, calculate the work done on the system when
a gas is compressed from 2.0 L to 1.0 L. Your units are peculiar here, L atm. It turns out
1 L atm is about 101 J.
10.3.1.2
In the last exercise, comment on the sign of your answer.
10.3.1.3
Imagine a system where after a process in which an external pressure of 1.0 atm compresses
an ideal gas from 2.0 L to 1.0 L there is no change in energy. (The energy of an ideal gas is
a function of temperature only; no change in temperature, no change in energy.) Find the
heat absorbed by the system during the process.
10.4
Adding Heat to the Predominant Configuration. Part
I. Qualitative
We now return to our discussion of configurations and ask the question: What happens to
the number of microstates in the predominant configuration if heat is added to the system.
From the last section, we know that addition of energy as heat changes the population of
the levels, but does not change the energy level spacings. So we can use our equally spaced
Chm 118
Problem Solving in Chemistry
10.4
144
Energy
1
2
3
4
5
Predominant Configuration
19/1/0/0/0
18/2/0/0/0
17/3/0/0/0
16/4/0/0/0
16/3/1/0/0
W
20
190
1140
4845
19380
Wi
Wi−1
—
9.5
6.0
4.25
4.00
Table 10.2: The change in the number of microstates in the predominant configuration as
heat is added to a system. Energy units arbitrary.
oscillator levels (convenient) and work out the answer by brute force. We start with a N
= 20 system with ET = 1. The only configuration for this system is 19/1/0/0 and this
has a W = 20. Then we add one unit of energy in the form of heat (no change in energy
spacings) and get two possible configurations, 18/2/0/0/0 and 19/0/1/0/0. The first is the
predominant configuration. (Remember, we are doing this with a small number of particles
to illustrate the principle. With a large number, there would be a configuration that is
overwhelming predominant.) This has a W = 190. We continue this process as show in
Table 10.2. The W values of the predominant configuration for each case up to ET = 5 are
given in the table along with, in the last column, the ratio of the W for a given amount of
energy to that with one unit less energy. Notice that this ratio starts large and becomes
smaller as the amount of energy increases. Since we are adding heat to our system, the
energy content is indicative of the “temperature” of the system. This allows the following
important conclusion: The number of microstates in the predominant configuration goes up
when heat is added, but it goes up by a larger factor at low temperature than it does at high
temperature.
10.4.1
Exercises
10.4.1.1
Imagine twelve particles in the set of energy levels. Find the configuration that is most
probable for a system with a total of two units of energy.
10.4.1.2
Same number of particles as in exercise 10.4.1.1. Find the configuration that is most probable for a system with a total of three units of energy.
10.4.1.3
Same number of particles as in exercise 10.4.1.1. Find the configuration that is most probable for a system with a total of four units of energy.
10.4.1.4
Show that W increases in going from the most probable configuration of exercise 10.4.1.1 to
that of 10.4.1.2.
Chm 118
Problem Solving in Chemistry
10.5
145
10.4.1.5
Show that the ratio of W10.4.1.3 /W10.4.1.2 is less than W10.4.1.2 /W10.4.1.1 , where the subscripts
refer to exercise numbers, and the W’s are for the most probable configuration, hereafter
Wmpc , and then latter still, just W for reasons that will become clear.
10.4.1.6
How do you state the conclusion of the exercises in this section in general terms? HINT:
There are two issues concerning the Wmpc , how it changes with an increase in E by a constant
amount, and how it changes at a low E (temperature) versus a high E (temperature).
10.5
What Does a Predominant Configuration Look Like?
The Boltzmann Equation
We have been dealing with predominant configurations by trial and error. It would be nice
if we could find a property of predominant configurations so that we could assess how the
populations of the individual levels are related when a configuration is predominant. To
be concrete, let’s examine the N = 26/ET = 7 situation. I claim that the predominant
configuration of this system is 20/5/1/0/0. See if you can verify that this configuration has
a larger W than 19/7/0/0/0 or 21/3/2/0/0. In what follows, I want to continue to use the
symbol “n” for the number of particles in a given level; therefore, I am going to use the
letter “v” to denote the level name. Now examine the ratio nv=1 /nv=0 which is the ratio of
the number of particles in two levels separated by one unit of energy. Compare that to the
ratio nv=2 /nv=0 where the levels are separated by two units of energy. Those ratios differ,
suggesting that the ratio of the number of particles in two levels depends on the difference
in the energy between those two levels,
nv=1
= f (e1 − e0 )
nv=0
(10.6)
This result leads to an interesting conclusion: We can write the following:
nv=2
nv=2 nv=1
=
×
nv=0
nv=1 nv=0
(10.7)
which upon insertion of equation 10.6 into equation 10.7 gives
f (e2 − e0 ) = f (e2 − e1 ) × f (e1 − e0 )
(10.8)
Now look at this required functional form. It is of the type
which suggests that
ea+b = ea × eb
(10.9)
nv=1
= e−β(e1 −e0 )
nv=0
(10.10)
with β some constant. Since this holds for any pair of values (with the appropriate energy
difference), the parameters “1” and “0” can be replaced by “i” and “‘j.”
Chm 118
Problem Solving in Chemistry
10.5
146
A book by Nash3 derives equation 10.10 in an elegant fashion from
consideration of the moment of oscillators from one level to another.
Nash also does an neat calculation showing that the predominant
configuration for Avogadro’s number of oscillators is more predominant than one that differs from it by about 1 part in 1010 (in each
energy level) by a factor of about 10430 !! That is predominant.
Using several methods that we won’t discuss here, we can show that β = kb1T where kb
is Boltzmann’s constant, 1.38×10−23 J/K. So our final equation for the appearance of the
predominant configuration is what is know as (one of) Boltzmann’s equations,
−(ej −e0 )
nv=j
= e kb T
nv=0
(10.11)
This equation gives the relative populations in two levels in the predominant configuration,
so it defines what the predominant configuration looks like. Note that for our equally
spaced energy level system of oscillators, that the population goes down by the same factor
for each increase in v value. Therefore a configuration such as 50/10/2/0/0 would be a
predominant configuration as the decrease is by a factor of 5 for each increase in level,
whereas 50/4/5/0/0 would not be.
10.5.1
Exercises
10.5.1.1
You have a set of equally spaced energy levels, the lowest of which has zero units of energy
and each energy level gap is one unit. If this system is in the predominant configuration
and there are 1920 particles in the e=0 level and 480 in the e=1 level, how many are in the
e=5 level? HINT: Don’t try to calculate this by evaluating factorials!
10.5.1.2
Consider a predominant configuration. If you have 1000 particles in the n = 0 level, the
spacing between the levels is 4.14×10−21 J/K, and the temperature is 300 K, what is the
number of particles in n = 1? in n = 2?
10.5.1.3
For ease of this problem, imagine a system with translational energy spacing of two units,
comprised of 20/2/0 particles for a total energy of 4 units. Hold the energy constant, but
increase the size of the container such that the spacing between the levels becomes one unit.
What configuration is roughly consistent with 4 units of energy now?
10.5.1.4
What is W for each of the configurations of exercise 10.5.1.3?
10.5.1.5
Does W increase in going from the small container of exercise 10.5.1.3 to the large one?
3
Nash, L. K., “A Statistical Approach to Classical Chemical Thermodynamics,” Addison-Wesley Publishing Co., 1971.
Chm 118
Problem Solving in Chemistry
10.6
147
10.5.1.6
If we make the argument that systems will evolve to the predominant configuration (because
it has the greatest number of equally probable microstates), will a gas expand to occupy a
larger volume if given the opportunity? In view of the last problem, do you see why?
10.6
Adding Heat to the Predominant Configuration. Part
II. Quantitative.
We follow Nash4 to get a quantitative relationship between adding heat energy to a predominant configuration and the change in the number of microstates of that configuration.
The actual derivation is of secondary importance. What is critical to our ultimate goal of
being able to think about entropy are the two assumptions that go into the derivation. The
first of these involves the constancy of β, of temperature, while the energy is added; the
second demands a constancy of the energy level spacing, which, in turn, means that the
interaction that increases the energy is heat, not work.
To proceed with the math, we first take the logarithm of both sides of equation 10.1 and
differentiate it for changes in the number of oscillators in various levels, noting that d ln(N!)
= 0. We then use Stirling’s approximation5 to get rid of the factorials in the differential
term:
X
ln(W ) = ln(N !) −
ln(ni !)
i
dln(W ) = −
X
dln(ni !)
i
=−
X
=−
X
(10.12)
(ln(ni )dni + ni dni /ni − dni )
i
ln(ni )dni
i
At this point we introduce Boltzmann’s equation in the form of equation 10.10 to produce
X
X
dln(W ) = −
(ln(n0 )dni ) +
(βei dni )
i
= −ln(n0 )
i
X
X
dni + β
(ei dni )
i
(10.13)
i
where the transformation from the first equation of 10.13 to the second is only valid if β
is a constant; that is, temperature is a constant. This is a critical point for our future use
of this equation. The first term in the last equation of 10.13 is zero since the sum of the
changes of all oscillators in all levels must be zero since the number of oscillators is constant.
Also because we have an expression for the total energy,
X
ET =
(ei ni )
(10.14)
i
4
5
Ibid.
Stirling’s approximation is that ln (a!) = a ln (a) - a
Chm 118
Problem Solving in Chemistry
10.6
148
and differentiation of both sides of equation 10.14 this gives the sum in the second term of
the second equation in 10.13 if the energy levels are constant!. Our final equation is:
dln(W ) = βdET
(10.15)
or, if we specifically recognize that the energy change is heat, we can substitute the symbol
for a small amount of heat into the expression to give us
dln(W ) = βδq
(10.16)
The change of the logarithm of the number of microstates in the predominant configuration
is equal to the heat absorbed at constant temperature times β (or the heat absorbed at
constant temperature divided by kb T). The last two conditions because of the conditions
used in the derivation: constant temperature so that we can factor the β out and heat as
the interaction so that the spacing of the energy levels does not change.
Our knowledge to here: There are a bunch of configuration possible for a given N and a
given ET . We know one of these (or a collection of very closely related ones) is the predominant configuration, which is the one that systems will adopt because it is overwhelmingly
predominant. We also know what the predominant configuration “looks like.” Further, we
know how the value of W in the predominant configuration changes with an input of heat
energy.
The highly restrictive nature of our model thus far causes some
doubt about whether it will be useful in “real life.” Again we can
use the arguments advanced by Nash6 to lift the specific restrictions.
We note that if we imagine a system, call it C (for complex), where
the spacings of the levels do change during the process, and put it in
thermal contact with a system of the type described above, called S
(for simple), then at thermal equilibrium we know that there will be
no change in Wtotal upon moving a few particles to different levels.
Therefore
dln(Wtotal ) = dln(WC ) + dln(WS ) = 0
dln(WC ) = −dln(WS )
(10.17)
= −βδqS
However, qS = - qC , and hence the same equation holds for the
change in the number of microstates in the predominant configuration of the complex system as is true for the simple system.
Now imagine that we consider a system (that part of the universe that we are interested in)
and its surroundings, which we imagine are very large, in essence the rest of the universe.
Further these surroundings are so large that no matter what happens to the temperature
of the system, that of the surroundings is constant. We know that the condition for a
spontaneous process is that the ln Wtotal will increase. Therefore
dln(Wtotal universe ) = dln(Wsys ) + dln(Wsurr ) ≥ 0
(10.18)
Here is our basic equation for a spontaneous process.
6
Ibid.
Chm 118
Problem Solving in Chemistry
10.7
10.6.1
149
Exercises
10.6.1.1
One conclusion of this section is equation 10.15. What assumptions are made to arrive at
this critical equation?
10.6.1.2
Carbon tetrachloride, CCl4 , is a liquid at room temperature. Gaseous CCl4 surely has a
larger W that does the liquid because of the translational energy levels in the gas. Boniface
Beebe, the great natural philosopher from rural Arkansas, thought about these facts and
said ”Because W is greater for gaseous CCl4 than for the liquid, CCl4 is always a gas.”
Comment.
10.6.1.3
Show that equation 10.18 requires that a cup of hot chai will cool if left in a room at ambient
temperature.
10.7
Disguising our Finding in a Word: Entropy
Equation 10.18 is the condition for a spontaneous process. In concept it is perfectly clear
what is happening in this equation–the probability of population of energy levels in the
predominant configuration rather than others–but in practice equation 10.18 is not very
useful if we want numbers to express the process since for real systems W is too large to
calculate easily. What we do is to define a new concept that is related to W in order to
make calculations of numerical values easier. That new concept is entropy, which we define:
S = kb ln(W )
(10.19)
This definition lowers the tremendous value of W by taking the logarithm of it, and by multiplying by the very small value of Boltzmann’s constant. Note also that from equation 10.16
that the change in entropy is given by
dS = kb dln(W )
= kb βδq
δq
=
T
(10.20)
where all the conditions on equation 10.16 hold for the last equation of the set 10.20.
Using this definition of entropy allows us to recast equation 10.18 in terms of entropy
changes:
dSuniverse = dSsys + dSsurr ≥ 0
(10.21)
for a spontaneous process.
Out choice of a system and big surroundings, which maintain a constant temperature (because of their size) allows an easy calculation of the ∆Ssurr . It is merely the heat flow
Chm 118
Problem Solving in Chemistry
10.7
150
Figure 10.6: The expansion of a gas from V = 1 L, P = 3 atm. to V = 3 L against an
external pressure of 1 atm.
divided by the temperature of the surroundings. For instance, if during a process a system
absorbs 25 J of heat from the surroundings held at 298K, then the entropy change of the
surroundings is -25J/298K or -0.084 J/K; note the negative sign occurs because the heat
absorbed by the surroundings is the negative of the heat absorbed by the system.
Determining the entropy change for the system is more of a challenge. The cause of this
challenge is the requirement from the last sections that the temperature remain constant
when we use equation 10.16. Figure 10.6 shows an ideal gas at a pressure of 3 atm in
a volume of 1 L expanding against an external pressure of 1 atm to a final volume of 3
L, all in a temperature bath at 298K. This happens very fast and the time course of the
volume of the gas in shown schematicallly by the blue curve in in Figure 10.7. Since the
gas does work nearly instantaneously on the surroundings, the energy of the gas must drop.
The surroundings slowly supply heat to the gas to raise the temperature back to the initial
value of 298; the time dependence of the temperature is shown (schematically, no scale of
temperature is given!) in the purple curve in Figure 10.7. The real heat absorbed by the
gas in the piston is not at a constant temperature and hence cannot be used to calculate
the entropy change of the gas.
One important feature. Entropy is proportional to W and W depends on the state of
the system, not where it was some instant ago. Hence entropy itself is a state function,
dependent only on the state of the system and not its history. So if we could figure out a
way to expand the gas from the initial conditions of 1 L, 3 atm to the final condition of 3L,
1 atm, such that the temperature did not change, then we could use that heat to calculate
the entropy change and that entropy change would be the same no matter how we actually
carry out the expansion. There is a way to do this in principle and Sir Issac Newton showed
us how to do the calculation. The way is very slowly. If we expand the gas a very small
amount, say by dV against an external pressure, Pext which is essentially the same as the
internal pressure, P, then the temperature change will be very small, and we can calculate
the work of that small expansion, and since we have an ideal gas at constant temperature
(∆E = 0), we can calculate the heat absorbed by the system in that very small expansion
Chm 118
Problem Solving in Chemistry
10.7
151
Figure 10.7: A schematic illustration of the change in volume with time for the expansion
illustrated in Figure 10.6. On the same plot, but without units on the vertical axis, is the
temperature of the gas as a function of time; see text.
(usually called, and so labeled below, a reversible expansion):
δw = −Pext dV
δw = −Psys dV
nRT
dV
δw = −
V
Z Vf
dV
wrev = −nRT
V
Vi
Vf
wrev = −nRT ln( )
Vi
qrev = −wrev
Vf
qrev = nRT ln( )
Vi
(10.22)
This leads us to the equation for the entropy of the system,
∆S =
qrev
T
(10.23)
Adding this to the value for the surroundings gives the entropy of the universe, and from
that we can test spontaneity.
Chm 118
Problem Solving in Chemistry
10.8
10.7.1
152
Exercises
10.7.1.1
Boniface Beebe, the wisest of the Arkansarian philosophers, noted that the entropy of
CCl4 (g) was 309.6 J/(K mole) and that of CCl4 (l) was 214.4 J/(K mole). He said: ”Clearly
CCl4 will exist as a gas since the entropy is greater for a gas.” Comment.
10.7.1.2
Calculate the work done on an ideal gas (in a piston and cylinder) when 1.0 moles at a
pressure of 20.0 atm. and T = 200K is expanded against an external pressure of 1.0 atm
isothermally. Find the heat.
10.7.1.3
Do the expansion in the last exercise, but in this case let the external pressure be 10.0 atm
until the piston stops moving, then reduce the external pressure to 1.0 atm. Find the work
and heat for the overall process.
10.7.1.4
Do the expansion in the last exercise, but in this case let the external pressure be 15.0 atm
until the piston stops moving, then reduce the external pressure to 10 atm until the piston
stops moving, then reduce the pressure to 1.0 atm. Find the work and heat for the overall
process.
10.7.1.5
In view of the last several exercises, are work and heat dependent upon the path?
10.7.1.6
Expand the gas in exercise 10.7.1.2 from the same initial state to the same final state, but
do it reversibly. What is the work? What is the reversible heat? What can you say about
the reversible work compared to that found in the last several exercises?
10.7.1.7
Compute the entropy change of the system for the gas in exercise 10.7.1.2.
10.7.1.8
Compute the entropy change of the surroundings and of the universe for the gas in exercise 10.7.1.2. Does experience tell you that the process is spontaneous? Does entropy
agree?
10.8
Entropy Defined in Terms of W; The Third Law
The definition of entropy:
S = kb ln(W )
(10.24)
where kb is Boltzmann’s constant and the W is for the predominant configuration creates
a situation that is unusual. Entropy has an absolute zero. When W = 1, S = 0 J/K. We
Chm 118
Problem Solving in Chemistry
10.9
153
get a W = 1, of course, when all the particles are in their lowest possible energy level.
That occurs at low temperature, specifically 0K. The third law of thermodynamics states
this: the entropy of any pure substance is zero at absolute zero. This leads to the concept
of absolute entropies. Since entropy increases as you add heat to a system (and adding
heat to a system generally raises its temperature), entropy of a substance can be given an
absolute value dependent upon the temperature. Tables of absolute entropies exist, but
read section 10.10 before using.
10.8.1
Exercises
10.8.1.1
Determine which system has the highest entropy, Ag(s) at 298K or Ag(s) at 340K? How
did you reach your conclusion?
10.8.1.2
The entropy of CCl4 is about 210 J/(mole K) at 65o C, 214 J/(mole K) at 75o C, but is 309
J/(mole K) at 80o C. Explain these data.
10.8.1.3
Which would have the higher entropy at 298o K, NH3 (g) or Ne(g)? Why? HINT: As always,
think population of energy levels.
10.8.1.4
The entropies of NaF(s), MgO(s), and AlN(s) are, respectively, 51.5, 26.8, and 20.2 J/(mole
K) at 298o K. Give a rationalization.
10.8.1.5
Determine which system has the highest entropy Na(s) at 371K or Na(`) at 371K? How did
you reach your conclusion?
10.9
Disorder is a Poor Word to Describe Entropy
For many years entropy has been described by the word “disorder,” and pictures in text
books have shown a student’s very messy dorm room (or one could use a picture of my desk)
as an example of high entropy. This word is fraught with difficulties if used to describe
entropy and should be avoided. Look at Figure 10.8 and determine which side is most
“disordered.” I say that the right hand picture has roughly even spacing of the particles,
and looks “ordered” to me, whereas the left hand side has all the particles bunched to
the left and is thereby “disordered”. But if you use a simple calculation of W (divide the
container into two parts and determine the probability all the particles will be in the left
hand part compared to the probability they will be equally distributed) you will find that
the right hand picture has the highest entropy. Entropy is better described as the dispersion
of energy among various energy levels.
Chm 118
Problem Solving in Chemistry
10.10
154
Figure 10.8: The Folly of Using “Disorder” after Lambert, F. L.; J. Chem. Ed, 2002, 79,
187-192.
10.9.1
Exercises
10.9.1.1
If you have a bottle of Li(g) and a bottle of the same size of Cs(g), both at the same
temperature, determine which is moving more rapidly. HINT: All the energy is kinetic and
total energy is proportional to temperature.
10.9.1.2
Given your answer to the last exercise, which would you imagine is more chaotic, less
ordered, Li(g) or Cs(g). What does “disorder” predict for the highest entropy?
10.9.1.3
Now consider Li(g) and Cs(g) from the point of view of microstates, not “disorder.” If you
use a POP model, which has energy levels more closely spaced (in the same size container),
Li(g) or Cs(g)? Which do you predict has the highest W and the highest entropy?
10.9.1.4
Look up some data to determine which system has the highest entropy Cs(g) or Li(g) at
298K, both at the same pressure?
10.10
Entropy Change in Reactions
We need to talk about properties of systems that depend on the amount of material present,
an extensive property. Volume is extensive as is energy. Properties that do not depend upon
the amount of material present are call intensive, such as temperature or pressure. What
kind of property is entropy? As chemists we most often want to express the value of a
thermodynamic quantity for a reaction of some sort. Since the quantity of material that
we deal with is variable, the most useful kind of quantity is an intensive one. What would
happen if we expressed entropy per mole of material transferred from the left of our equation
to the right? What kind of property—intensive or extensive—would that quantity be? We
have a symbol for the change in entropy of a substance, dS or ∆S. Now we need one for the
quantity we just invented, one to be used with reactions. There are several suggestions for
such symbols in use, including ∆r S, which is used in advanced books, but is often regarded
g which was suggested but never adopted, and even ∆S, which is often
as rather clumsy, ∆S,
used in introductory texts (and courses) and which leaves it up to the user to understand
Chm 118
Problem Solving in Chemistry
10.10
155
the context–probably seldom the case. What would I mean if I told you the hunk of gold in
going from state a to state b has ∆S of -20 J/K? Am I giving you an intensive or extensive
property? Work you way through the following exercises to get a feel for this dual use of
the same symbol.
A second issue we need to address is that changes in chemical reactions depend upon the
conditions of the reagents and products, their pressures, temperatures, etc. Generally, it has
been agreed that in the standard state all materials are pure gases are at one atmosphere
pressure and all concentration are equal to one molar, and, usually, the temperature is
298.15K If this is so, the thermodynamic parameter is given a “super o”.
10.10.1
Exercises
10.10.1.1
What would I mean if I told you that ∆r S for
H2 O(s, T=273K) = H2 O(`, T=273K)
was 22 J/(mole K)?
10.10.1.2
Given the information in the last exercise, how would you interpret the following sentence:
“The entropy change, ∆S, for melting water at 273K is 11 J/K.”
10.10.1.3
What would you conclude with the information that “∆S for the melting of water at 273K
is 22 J/(mole K)”?
10.10.1.4
Would the entropy change of the following reaction at 298.15K be characterized by a ∆S o ?
C(s) + 2Cl2 (g, P = 0.1 atm) = CCl4 (`)
10.10.1.5
Use a table of absolute entropies to predict the value of ∆S o for
2K(s) + F2 (g) = 2KF(s)
Give a rationalization for your answer. HINT: If the text you are using does not have a
good table of thermodynamic values, try the web page at
http://chemistrytable.webs.com/enthalpyentropyandgibbs.htm
or other sites found by searching for “enthalpy of formation” or “absolute entropy”.
Chm 118
Problem Solving in Chemistry
10.10
156
10.10.1.6
Use a table of absolute entropies to predict the value of ∆S o for
NH3 (g) + HBr(g) = NH4 Br(s)
Give a rationalization for the sign of your answer.
10.10.1.7
Use a table of absolute entropies to predict the value of ∆S o for
CH3 CHCH2 (g) = cyclic-C3 H6 (g)
Give a rationalization for the sign of your answer.
10.10.1.8
Use a table of absolute entropies to predict the value of ∆S o for
3 H2 (g) + Fe2 O3 (s) = 2Fe(s) + 3H2 O(g)
Give a rationalization for the sign of your answer.
10.10.1.9
Comment on which direction the reaction
2NO(g) + Cl2 (g) = 2NOCl(g)
will proceed if all reagents are under standard conditions and you are at some “high enough”
temperature. HINT: You need to apply your thoughts about ∆Suniv since you are being
asked about spontaneity.
10.10.1.10
Comment on which direction the reaction
SrSO4 (s) = SrO(s) + SO3 (g)
will proceed at some “high enough” temperature.
10.10.1.11
Comment on which direction the reaction
SO3 (g) = SO2 (g) + 21 O2 (g)
will proceed at some “high enough” temperature.
Chm 118
Problem Solving in Chemistry
Chapter 11
The Free Energy: Entropy in
Another Guise
11.1
Using the First Law of Thermodynamics at Constant P
and T: Enthalpy
We saw in section ?? that the first law of thermodynamics says that the change in energy
of a system is equal to the heat absorbed by the system plus the work done on the system.
To remind you, energy is a state function–it does not depend on the path taken to get from
the initial state to the final state–whereas heat and work are not state functions. For our
purposes here, we consider systems in which the only work is “pressure-volume” work, that
of pushing back the atmosphere, δw = - Pext dV. Since most experiments are done on the
open bench top in which the external pressure is constant, we are going to assume that
condition. Also, we assume that the system is in a freely floating piston so that the external
pressure is the same as the internal. This essentially defines the path. Here is the algebra:
∆E = δq + δw
Z
= δqP − Pext dV
Z
= δqP − Pext dV
(11.1)
= δqP − P ∆V
∆E + P ∆V = δqP
where we labeled the δq value for the constant pressure in the second equation, took advantage of the constancy of the external pressure in the third and set the internal pressure
as equal to the external in the fourth. The final equation has δqP equal to combination of
a bunch of state functions, and hence must also be a state function. We label this state
function the change in enthalpy, ∆H, which was introduced in section 3.2: δqP is ∆H.
More generally the enthalpy, H, is defined as E + PV, which, under conditions of constant
pressure leads to our definition of the change in enthalpy. The words exothermic (giving
157
11.1
158
off heat) and endothermic (absorbing heat) are commonly used to describe the sign of the
enthalpy change.
11.1.1
Exercises
11.1.1.1
What is the change in enthalpy equal to?
11.1.1.2
The enthalpy change for a reaction is -5.4 kJ/mole. Is the process exothermic? Is the
process spontaneous? HINT: Second is a trick question.
11.1.1.3
Use tables of enthalpies of formation to determine the enthalpy change for the process
2SO2 (g) + O2 (g) = 2SO3 (g)
under standard conditions. What do those numbers tell you about the spontaneity of this
reaction?
11.1.1.4
Give a reason why the enthalpy change is what it is in the reaction in the last exercise.
HINT: Think about what is happening.
11.1.1.5
What is the sign of the enthalpy of vaporization of a liquid, any liquid? Give a molecular
explanation for your answer.
11.1.1.6
The enthalpy of vaporization of water is 40.7 kJ/mole and that of fusion of water is 6.01
kJ/mole. Assuming these values are independent of temperature, what is the enthalpy of
sublimation of ice to water vapor at -10o C?
11.1.1.7
State Hess’s Law. Be articulate. (We covered this in section 3.2.)
11.1.1.8
Describe the process of combustion in terms of what chemicals come together and what
products are formed. What would the enthalpy of combustion be?
11.1.1.9
The standard enthalpy of combustion of propane is -2220; of C(gr), -394; and of dihydrogen,
-286; all in kJ/mole of the substance named. Find the enthalpy change for the conversion
of graphite into propane by reaction with dihydrogen. HINT: Use the last two exercises.
11.1.1.10
Find the enthalpy change for the reaction (not balanced)
Chm 118
Problem Solving in Chemistry
11.2
159
B2 O3 (s) + CaF2 (s) = BF3 (g) + CaO(s)
HINT: You will need to look up some numbers.
11.1.1.11
Describe how a calorimeter works. How would you calibrate a calorimeter?
11.2
Using Enthalpy of the System to Understand the Surrounding’s Entropy Change
There are two factors that allow us to use a system parameter, the enthalpy change, to
assess a surrounding parameter, the entropy change of the surroundings. The first is that
at constant pressure the enthalpy change is the heat absorbed by the system. Since what is
absorbed by the system came from the surroundings, then the negative of the heat absorbed
by the system is the heat absorbed by the surroundings. Secondly, remember that we defined
the surroundings so that it is big. Therefore the heat that is absorbed by the surroundings is
absorbed at constant temperature and we can use equation ?? to compute the surroundings
entropy.
This allows us to use equation 10.21 to get the entropy change of the universe for some
process:
∆Suniverse = ∆Ssys + ∆Ssurr
∆Hsurr
= ∆Ssys +
T
(11.2)
−∆Hsys
= ∆Ssys +
T
∆H
= ∆S −
T
where in going from the third to the fourth equation we have understood the lack of a
subscript to mean a system parameter.
11.2.1
Exercises
11.2.1.1
Calculate the entropy change of the surroundings if we start with one mole of an ideal gas
in a cylinder with a piston at a pressure of 10 atm and a temperature of 298K held in place
with a pin. We remove the pin and let it expand against an external pressure of one atm
in a large temperature bath at 298K.
11.2.1.2
Calculate the entropy change of the system for the apparatus described in the last exercise.
11.2.1.3
Calculate the entropy change of the universe for the last two exercises.
Chm 118
Problem Solving in Chemistry
11.3
160
11.2.1.4
Calculate the entropy change of the universe when a mole of liquid water is converted to a
mole of gaseous water at a pressure of one atm against an external pressure of 1 atm and
a temperature of 298K. HINTS: Remember that the temperature does not change during
such a conversion. You will have to look up a number.
11.2.1.5
Repeat the last exercise—find the entropy change of the universe–at a temperature of 373K
assuming that the enthalpy and entropy of the conversion are not a function of temperature.
11.2.1.6
Seldom are we interested in gases expanding. In most instances of interest to a chemist, we
are focussed on a reaction. The concept of spontaneity in a reaction depends, as we shall
see, on the states of the various components. For a while we will deal with the question of
spontaneity when all chemicals are in their standard states, one atm pressure and pure. To
do this, we can proceed in one of two equivalent thinking processes: (1) We can imagine
a very large vessel of reactants and products, with all species at one atm pressure and
then allow a mole of reactants to move to products without that conversion causing any
significant change in pressure of any reagent or product; or (2) we can move a small amount
of reactants to products so that pressures do not change, compute the entropy change
for that small amount, and convert it to a “per mole” number. The description of these
systems is “all species in their standard state”. Compute the enthalpy change when CaCO3
decomposes into CaO and CO2 (g) with all species in their standard state. If this occurs at
25o C, what is the change in the entropy of the surroundings?
11.2.1.7
Find the entropy change of the system for the reaction in the last exercise.
11.2.1.8
Is the reaction of the last two exercises spontaneous at room temperature? What would
happen at a higher temperature? Show your answer to this second question mathematically
assuming that the system properties, ∆S and ∆H, are independent of temperature.
11.2.1.9
The reaction to produce formaldehyde is
H2 (g) + CO(g) = H2 CO(g)
At T = 25o C, ∆H o = 1.96 kJ/mole and ∆S o = -109.6 J/mole. Find the change in entropy
of the surroundings when one mole of formaldehyde is produced at 25o C under standard
conditions.
11.2.1.10
If the temperature for the formation of formaldehyde (see last problem) is raised to 50o C,
what will happen to the value of the change in entropy of the surroundings provided ∆H o
and ∆S o are approximately independent of temperature over this range.
Chm 118
Problem Solving in Chemistry
11.3
161
11.2.1.11
Is the production of formaldehyde (see exercise 11.2.1.9) spontaneous at 25o C?
11.3
Getting Rid of the Universe Altogether: Free Energy
We now are going to manipulate the last equation in the set 11.2 so that there is no mention
of the universe. We first multiply by a constant T and then by negative one, which change
the sign on the inequality for spontaneous processes:
∆H
≥0
T
= T ∆S − ∆H ≥ 0
∆Suniverse = ∆S −
T ∆Suniverse
(11.3)
−T ∆Suniverse = ∆H − T ∆S ≤ 0
∆G = ∆H − T ∆S ≤ 0
where in going from the third equation to the fourth we have defined ∆G as -T∆Suniv ; ∆G
is a state function since it is equal to a combination of state function values. We call ∆G
the free energy change and it is less than zero for a spontaneous process.
Another approach (somewhat more sophisticated) is as follows. Let’s
invent a new function called G that is composed of previously defined functions:
G = E + PV − TS
(11.4)
We take the total differential of this equation and then assert the
conditions that T and P are constant to get an expression for dG,
dGP,T = dE + P dV − T dST
(11.5)
Insert the definition of dH at constant P,
dH = dE + P dV
(11.6)
dGP,T = dHP − T dST
(11.7)
and we get
or, for a finite change in the standard state
∆Go = ∆H o − T ∆S o
(11.8)
There are tables of free energies of formation, the free energy change when a compound is
made from its elements in their standard state.
11.3.1
Exercises
11.3.1.1
Take equation 11.8 and change the enthalpy change from a system value to one for the
surroundings, then divide all terms by T. Show that this equates -∆Go /T with ∆S o univ .
Chm 118
Problem Solving in Chemistry
11.4
162
11.3.1.2
What is the criterion for spontaneity in terms of WP C ?
11.3.1.3
What is the criterion for spontaneity in terms of ∆S univ ?
11.3.1.4
What is the criterion for spontaneity in terms of ∆Go ?
11.3.1.5
For the reaction
2 CuCl(s) + Cl2 (g) = 2 CuCl2 (s)
compute the values of ∆H o and ∆S o by looking up the appropriate values from tables of
thermodynamic parameters. Use these and equation 11.8 to compute ∆Go . Compute ∆Go
from data in thermodynamic tables and show agreement. HINT: The value of ∆H o and
∆Go for Cl2 are zero, but the value for the So is not.
11.4
The Pressure (or Concentration) Dependence of G
Clearly entropy is a function of pressure. You can see this in our original model, the number
of microstates in the predominant configuration: To increase the pressure requires that we
add particles to a fixed volume which will increase W. To see how to deal with the pressure
dependency, we work with the free energy function. There are some tricky concepts in this
development; if you miss those, be sure to appreciate the end result, equation 11.17.
We start with equation 11.4 and take the total differential, then use
the first law for a reversible process,
dG = dE + P dV + V dP − T dS − SdT
dG = δqrev − P dV + P dV + V dP − T
δqrev
− SdT
T
(11.9)
dG = V dP − SdT
where the substitution of dS of the system for its equivalent,
was used in going from the first equation to the second.
δqrev
T
We now take the third equation in the set 11.9 and integrate it
between the limits of pressure = Po and some arbitrary pressure, P
at constant T (so the second term drops out), for an ideal gas, for
which we can substitute nRT/P for V. This gives us
P
o
G = G + RT ln
(11.10)
Po
Chm 118
Problem Solving in Chemistry
11.4
163
where Go is the G when P = Po , the pressure in some “standard”
state (usually 1 atm). This free energy is a state property of the system and hence must be dependent on those parameters that change
the state of the system. If we consider a mixture of chemicals, say
A, B, and C, then G must be a function of T and P, as always, but
also the number of moles of each of the various components. We
could write this as
G = G(T, P, nA , nB , nC )
(11.11)
If we take the total differential of this equation at constant T and
P we get
X ∂G dG =
dni
∂n i nj ,T,P
i
(11.12)
X
=
µi dni
i
where i runs over, in our specific case, the three components A, B,
and C, and nj is all of those n except ni . In the second equality I
have inserted, as is conventionally done, the chemical potential, µi ,
∂G
which is simply the name for the value of ( ∂n
)nj ,T,P , which is also
i
called a partial molar free energy (see exercise 11.4.1.1).
To make further progress we need to have a method to determine
the extent of reaction. This is usually done by defining a parameter
called the extent of reaction with the symbol ξ. For clarity, let’s
consider a specific reaction: A + B = C, and let’s define ξ through
the equation
1
dξ = dni
(11.13)
νi
where νi is the stoichiometric coefficient of reagent i and is a negative quantity for reactants. We can then express the change in any
component’s number of moles with dξ by solving equation 11.13 for
dni . At this point you should do exercises 11.4.1.2 to 11.4.1.5. We
can now solve equation 11.13 for dni and insert those values into
equation 11.12 to give us
X
dG =
µi νi dξ
i
dG X
=
µi νi
dξ
(11.14)
i
To solve the second equation we need to express the partial molar
free energies in terms of the pressures of the gases. We simply
argue that since these are free energies, they should obey the same
equation that a free energy obeys, namely, equation 11.10; so we
Chm 118
Problem Solving in Chemistry
11.4
164
can put in for the µi the value µoi + RT ln
h
Pi
Pio
i
to give us
dG X
Pi
o
µi + RT ln
νi
=
dξ
Pio
i
(11.15)
X
X
Pi
o
νi RT ln
=
µi νi +
Pio
i
i
P o
o
which, if we define
i µi νi as ∆r G and move the νi inside the
logarithm term as an exponent gives us our final equation, which is
written in terms of ∆r Go or some such symbol–see section 10.10–for
the reaction νA A + νB B = νC C + νD D:
νC νD νA νB P P P PoB
dG
o
= ∆r G = ∆r G + RT ln CνC DνD oA
(11.16)
dξ
PoC PoD PAνA PBνB
To summarize our discussion above, the pressure dependence of the free energy change for
a reaction when the standard state is taken to be one atmosphere (Po = 1) is given by
Pressure of products to stoichiometric power
o
(11.17)
∆G = ∆G + RT ln
Pressure of reactants to stoichiometric power
where we have adopted the usual practice and removed the “r” subscripts. It is the same
equation for species whose concentration can vary as it is for species whose pressure varies.
This equation is the gives the pressure (concentration) dependence of the free energy. The
argument of the logarithm in equation 11.17 is often called Q for simplicity. The equation
then reads
∆G = ∆Go + RT ln(Q)
(11.18)
11.4.1
Exercises
11.4.1.1
Look at the definition of µi in equation 11.12 and see if you can justify the following verbal
description of a partial molar quantity, stated for reasons of concrete appreciation in terms
of a partial molar volume: “The partial molar volume of substance i is the change in volume
(dV) of the mixture (of which it is a component) when a small number of moles (dni ) of
substance i is added, holding the other number of moles of other species constant.”
11.4.1.2
As an exercise in the use of ξ, let the reaction be A + B = C and the initial concentrations
of A and B be 1.0M and that of C be zero. At the point of reaction where the concentration
of A is 0.5M, evaluate ξ. Use that value to find the concentration of B and the concentration
of C at that point.
11.4.1.3
Same reaction as the last problem, but the initial concentrations are A = 1.0M, B = 0.7M
and C = 0.0M. Find ξ when the concentration of A is 0.5M and use that value to find the
concentrations of B and C at that point.
Chm 118
Problem Solving in Chemistry
11.4
165
11.4.1.4
For the reaction A + 3B = 2C, show that the extent of reaction varies from zero at the
beginning of the reaction to 1 at the end (no limiting reagent left) if the initial number of
moles of A, B, and C are respectively 1, 4, and 1M.
11.4.1.5
For the reaction A + 3B = 2C, show that the extent of reaction varies from zero at the
beginning of the reaction to 13 at the end (no limiting reagent left) if the initial number of
moles of A, B, and C are respectively 3, 1, and 2.
11.4.1.6
Find ∆Go for the reaction 2Mg(s) + O2 (g) = 2MgO(s).
11.4.1.7
Give an explanation of the spontaneity of the reaction in exercise 11.4.1.6 in terms of the
entropy of the universe and its components. Be complete.
11.4.1.8
Find the free energy change for the reaction in exercise 11.4.1.6 if the pressure of O2 (g) is
1.0 ×10−4 atm. Is the reaction spontaneous under these conditions? HINT: Since the two
solids have rather small change in volume with a change in pressure, their Go values are
nearly independent of P and do not appear in equation 11.17 etc.
11.4.1.9
Find ∆Go for the reaction CH3 OH(g) = CO(g) + 2H2 (g) at 298K.
11.4.1.10
Find ∆G for the reaction in exercise 11.4.1.9 if the pressure of CH3 OH(g) is 0.1 atm, that
of CO(g) is 1.0×10−2 atm, and that of H2 (g) is 5.0×10−3 atm.
11.4.1.11
Assuming that ∆H o and ∆S o are independent of T, find ∆Go for the reaction in exercise 11.4.1.9 at 50o C and 150o C.
11.4.1.12
A long tube is prepared containing SiCl4 (g); at one end ot this tube is placed some impure
Si(s) and that end of the tube is heated. Over time at the cool end of the tube pure Si(s)
forms. Account for this result using thermodynamic arguments. HINT: Think about the
conproportionation of Si(IV) and Si(0):
SiCl4 + Si = 2SiCl2
11.4.1.13
This and the next four exercises give a derivation that suggests that equation 11.17 is
correct. Write the chemical equation that represents benzene liquid vaporing to benzene
gas at a pressure of one atmosphere. HINT: We can express the full conditions of a gas, say
X, in a reaction via X(g, P = 0.5 atm), for instance.
Chm 118
Problem Solving in Chemistry
11.5
166
11.4.1.14
Write the chemical equation that represents benzene liquid vaporing to benzene gas at a
pressure of P atmosphere.
11.4.1.15
We can couple the two equations from the last two exercises by letting benzene liquid at
P = 1 atm. go to benzene liquid at P = P atm. What is your estimate of ∆H for this
process. HINT: Think about what energy changes take place, especially if we treat the gas
as ideal. What is your guess about ∆S for this process? HINT: Is anything happening to
the benzene liquid?
11.4.1.16
To complete the cycle we have been working on, we could let benzene gas at P = 1 atm.
go to benzene gas at P = P atm. To be concrete, if P = 0.1, and we work at constant
temperature, we are changing the pressure of benzene gas at constant temperature. What
does this do to the energy under the assumption of ideal gas behavior? HINT: Note if H =
E + PV, then ∆H o = ∆E o + ∆P V o = ∆E o + ∆nRT o ; what is the consequence if ∆T o
= 0? What does it do to the entropy?
11.4.1.17
Write the complete thermodynamic cycle from the information in the last four exercises.
Let the value of ∆G at P = 1 atm. be ∆Go and the value of ∆G at P = P be ∆G. Write
one in terms of the other and additional terms as appropriate. You have just produced
what we derived more rigorously earlier in this section.
11.4.1.18
Free energies of formation of benzene liquid and benzene gas at 298K are 124.5 and 129.7
kJ/mole, respectively. Find the value of ∆G when the pressure of benzene gas is 10 mm
Hg. HINT: 760 mm Hg is one atmosphere.
11.4.1.19
Use data from the last exercise to find the value of ∆G when the pressure of benzene vapor
is 400 mm Hg at 25o C.
11.5
Free Energy and Equilibrium
Equation 11.17 gives us the information we need to know to predict the spontaneity of
a reaction at constant T and P. When the value of ∆G is less than zero, the reaction is
spontaneous; when greater, non-spontaneous. What about the other condition, when ∆G
= 0. Under that condition, the process is at equilibrium and there is no net reaction in
either direction. The mathematics of this situation are interesting; when ∆G is zero, then
" ν
#
νD
C
νA νB
PCeq
PDeq
PoA
PoB
o
∆G = −RT ln
(11.19)
νC νD
νA
νB
PoC
PoD PAeq
PBeq
where the terms PCeq is the equilibrium pressure of substance C, etc. This is a rather
bizarre situation. The appropriate ratio of equilibrium pressures is governed by the free
Chm 118
Problem Solving in Chemistry
11.5
167
energy change when all the pressures are 1 atm. If we define an equilibrium constant as we
did in section 4.1, then we can recognize that equation 11.19 can be written as
∆Go = −RT ln(K)
(11.20)
We can combine equations 11.18 and 11.20 to give a succinct expression for the free energy
change:
Q
(11.21)
∆G = RT ln
K
11.5.1
Exercises
11.5.1.1
The reaction
I2 (g) + Cl2 (g) = 2ICl(g)
has an equilibrium constant of 7.76×108 . What is the value of ∆Go ?
11.5.1.2
Is the reaction in exercise 11.5.1.1 spontaneous when the pressure of ICl is 0.001 atm and
the other pressures are 1.0 atm? When the pressure of ICl is 10 atm and the other pressures
are 1.0 atm?
11.5.1.3
Offer a rationalization as to why the standard free energy change for the reaction in exercise 11.5.1.1 is so small in magnitude.
11.5.1.4
For the reaction
ZnF2 (s) = Zn2+ (aq) + 2F– (aq)
the value of K is 0.003. Find the value of ∆Go for this reaction. HINT: The role of
concentrations and pressures are interchangeable in our equations for free energy.
11.5.1.5
If you make 50 mL of a 2 M solution Zn(NO3 )2 and 50 mL of a 4.0M solution of CsF, and
mix them, what will happen? HINT: Ignore any reactions of nitrate ion and cesium ion,
but pay attention to the last exercise; you might consider using equation 11.21.
11.5.1.6
Assuming that ∆H o and ∆S o are independent of temperature, find the boiling point of
CCl4 (`).
Chm 118
Problem Solving in Chemistry
11.5
168
11.5.1.7
The normal boiling point of trimethylphosphine is 38.4o C. Its vapor pressure at -45.2o C is
0.017 atm. Find ∆H o , ∆S o , and the vapor pressure at 15o C.
11.5.1.8
Imagine you have a chemical reaction with a ∆Go 1 and an associated K1 . You also have
a second chemical reaction with a a ∆Go 2 and an associated K2 . Show, using the additive
property of ∆Go , that the equilibrium constant for the reaction that is the sum of chemical
reactions 1 and 2, K, is K = K1 K2 .
11.5.1.9
Use equation 11.8 and the relationship between ∆Go and lnK to show if you determine K
at various temperatures and then plot ln[K] versus 1/T that you can obtain ∆H o and ∆S o
from the corresponding line.
Chm 118
Problem Solving in Chemistry
Chapter 12
Equilibrium and Other Uses of the
Free Energy
12.1
Equilibrium in Acid Solutions
We learned in the last chapter how to use free energy changes under standard conditions to
learn about the equilibrium constant for a reaction. To express that equation in a different
way, the equilibrium constant is given by
K=e
−∆Go
RT
(12.1)
Once you have a value of ∆Go you can get an equilibrium constant, and, having that,
you can determine the concentrations of the various components of the solution. It is this
determination for weak acids that we deal with in this section.
Suppose we have an acid, HA, in an equilibrium HA = H+ + A− , with a Ka = 1.0×10−5 ,
Ka =
[H + ][A− ]
[HA]
(12.2)
where the brackets stand for the concentrations of the various ions and molecular species
in solution. Let us suppose that the solution at hand was made by dissolving 1.0 moles of
HA in a liter of water. Since water itself is an acid, H2 O = H+ + OH− , we have two acids
present, as well as two “existence” equations, one stating that all the moles of “A” that we
added must still be there, in either the form of HA or the form of A− ; this is called a “mass
balance” equation. The second “existence” equation concerns the charge of the solution;
that must be balanced, or, said another way, the count of cations (times their charge) must
equal the count of anions (times their charge); this is the “charge balance” equation. These
four equation are:
[H + ][A− ]
Ka =
[HA]
Kw = [H + ][OH − ]
(12.3)
TA = [HA] + [A− ]
[H + ] = [A− ] + [OH − ]
169
12.1
170
The four equations in equations 12.3 have four unknowns in them and hence can be solved
for those four unknowns. These equations generate a cubic equation to solve, so they are
not much fun to solve (rigorously) by hand. Also, it is of considerable interest to apply
chemical intuition to the solution as this allows one to grasp what is happening in the
solution. Let’s commence that exercise. We added an acid to water, so the concentration
of a weak base such as OH– is unlikely to be very large. Let us guess it is zero compared
to the concentration of [A− ]. That makes the fourth equation of equations 12.3 state that
[H+ ] = [A− ] which in turn makes the first equation of the set become
Ka =
[H + ]2
= 1.0 x10−5
[HA]
(12.4)
We can now use the second equation of equations 12.3 and the simplified fourth equation
again to get rid of the [HA] from the denominator of equation 12.4 to give us
Ka =
[H + ]2
= 1.0 x10−5
TA − [H + ]
(12.5)
which is has only one variable, [H+ ], and is a quadratic equation in that variable. This can
be solved more easily than the cubic equation, but is still messy. A further simplification,
using intuition, appears if we note that our acid is weak, has a small Ka . That means that
the concentration of hydrogen ion is not likely to be very large. So let us guess that [H+ ] is
small compared to the total molarity of acid, TA . Then the equation is
Ka =
[H + ]2
= 1.0 x10−5
TA
(12.6)
and the solution is [H+ ] = 3.162 x 10−3 M. (The rigorous answer is 3.157 x 10−5 M, but we
will see how to improve the approximate answer shortly.) We can (and should) now check
our two guesses. Is 0.003162 small compared to 1.00, the second guess we made. Yes. Then
we use the second equation in equations 12.3 to compute [OH− ] (3.16 x 10−12 ) and ask if
it is small compared to [A− ] = [H+ ] = 3.16 x 10−3 ? Yes. The guesses are ok.
Let’s consider a couple of changes in order to see how they influence our answers. First,
what is the acid is weaker, say Ka = 1.00 x 10−8 . Then following the same procedure gives
[H+ ] = 1.000 x 10−4 M by the guesses and rigorously the same (with round-off). So the
weaker the acid, the better the guesses. The second change worth making is to lower the
concentration of the acid, lower the value of TA . If that value is 0.01 for the acid with a
Ka of 1.0 x 10−5 , then the guesses yield an answer of 3.162 x 10−4 versus a rigorous answer
of 3.113 x 10−4 . This error is about 5 parts in 300 or slightly over one percent. Too much?
Depends on your needs. But clearly the stronger the acid and the more dilute the total
acid added, the more problematic the accuracy. Fortunately, checking your approximations
always tells you if you are in trouble or not.
There is a fast way to fix calculations of this sort when the guesses are poor. Let’s do a
case to see how this is done. Let’s consider an acid with a Ka = 2.00 x 10−3 and a TA of
0.01M. Making our two guesses and solving for [H+ ] as above gives 4.47 x 10−3 M. Is that
small compared to TA of 0.010M. No. Rather than going back and solving the quadratic, it
is faster to simply put the answer for hydrogen ion concentration into the additive portion
Chm 118
Problem Solving in Chemistry
12.1
171
Figure 12.1: A graph showing how the guessed value of the [H+ ] approaches the true value
as the number of iterations increases for the example in the text. The straight line is the
rigorous answer.
of the appropriate equation similar to equation 12.5 and solve again:
[H + ]2
[H + ]2
=
= 2.0 x10−3
+
TA − [H ]
0.01 − 0.00447
(12.7)
which leads to [H+ ] = 0.00332M; do it again to get [H+ ] = 0.00365M; and again for
0.00356M; etc. You will zig-zag toward the correct answer, which in this case is 0.00358M,
as shown in Figure 12.1. Incidentally, the method of calculation is roughly how modern
computer programs compute molecular properties using quantum mechanics.
Finally, recall that the results of a calculation of [H+ ] are often expressed in terms of the
negative logrithm of that number, or in terms of the “pH.”:
pH = −log([H + ])
12.1.1
(12.8)
Exercises
12.1.1.1
Find the pH of a solution prepared by dissolving 0.10 moles of HNO2 in a liter of water.
The Ka for HNO2 is 5.1×10−4 .
12.1.1.2
Find the pH of a solution prepared by dissolving 0.01 moles of HF in a liter of water. The
Ka for HF is 6.75×10−4 .
Chm 118
Problem Solving in Chemistry
12.2
172
12.1.1.3
Given the definition of pH, what do you imagine a pKa = 6.35 means?
12.1.1.4
The acid H2 S has two protons that can be removed. The first has a pKa = 7.00 and the
second a pKa of 12.92. This diprotic acid hence has five equations necessary to define the
system. Write those five equations.
12.1.1.5
Show from your equations of the last exercise, good chemical intuition and good guesses
that the solution to the following problem is no more difficult than the previous exercises:
Find the pH of a solution prepared by dissolving 0.010 moles of H2 S in a liter of water.
HINT: S2– is a base.
12.1.1.6
The two pKa of oxalic acid, HOC(O)C(O)OH, are 1.25 and 4.28, a difference of three
units. The two pKa of adipic acid, HOC(O)CH2 CH2 CH2 CH2 C(O)OH are 4.42 and 5.41, a
difference of one unit. Comment.
12.2
Mixtures of Acids and Salts: Buffer Solutions
Let’s apply the logic of the last section to a slightly more complex problem; that application
yields an extraordinary easy solution. We are going to prepare a solution that has 0.50 moles
of acid HA and 0.50 moles of NaA dissolved in one liter of water where HA is a weak acid.
Also, note that generally speaking, salts of Group I cations like Na+ , completely dissociate
from anions, so the only role for Na+ in these problems is as an additional mass balance
equation and in the charge balance equation.
We then have for this problem the four equations of equation 12.3 with a modification to
the charge balance equation to include sodium ions, and a new mass balance equation for
Na+ :
[H + ][A− ]
Ka =
[HA]
Kw = [H + ][OH − ]
(12.9)
TA = [HA] + [A− ]
[N a+ ] + [H + ] = [A− ] + [OH − ]
TN a = [N a+ ]
To solve this equation lets think, as we did before, about what we added to the solution.
We added an acid, so we might guess that [OH− ] might be small compared to [A− ]. But
we also added a base (NaA) to the solution, so we might think that [H+ ] would be small
compared to [Na+ ]. If we make both of these guesses in the fourth equation of the set 12.9,
then, since the fifth equation fixes the sodium ion concentration, the [A− ] is fixed. And
once that is true, [HA] is fixed by the third equation. Putting this into the first equation
Chm 118
Problem Solving in Chemistry
12.3
173
gives us:
Ka =
[H + ][N a+ ]
(TA − [N a+ ])
(12.10)
which in the case of the numbers we are using gives us that [H+ ] = Ka .
Solutions such as this one, made by mixing an acid and a salt containing its conjugate base
are called buffer solutions. As you will show in an exercise, these resist a change in [H+ ]
when either acid or base is added.
12.2.1
Exercises
12.2.1.1
Here is a variation on the theme of section 12.1. Find the pH of a solution made from
100.0 g of Na(CH3 COO) dissolved in a liter of water given that the pKa of acetic acid is
4.75. HINTS: (1) The equations are those in 12.9. (2) Since you added a base, a logical
approximation might be what? (3) Concentrated solution, so likely that [A− ] is large
compared to what? (4) Rigorous answer is 9.4216.
12.2.1.2
Find the pH when 1.0 moles of acetic acid (pKa = 4.75) and 0.5 moles of sodium acetate
are placed in a liter of water.
12.2.1.3
Determine the pH when the solution in the last exercise is treated with 0.10 moles of NaOH;
assume no change in volume.
12.2.1.4
Determine the pH of the solution in exercise 12.2.1.2 is treated with 0.10 moles of HCl
assuming no change in volume. HINT: You need to modify the charge balance equation for
the presence of Cl– and to add another mass balance equation for chloride.
12.3
Solubility Equilibrium, including Common Ion Effect
Another common type of equilibrium that commonly occurs involves the limited dissolving
of certain solids. The classic example is silver chloride, which dissolves very slightly in
aqueous solution:
AgCl(s) = Ag+ (aq) + Cl– (aq)
The equilibrium constant for this reaction (classically called a Ksp for “solubility product”)
is 1.8×10−10 . To solve problems of this sort is very easy with our logic of previous sections.
We have two unknowns, the concentration of silver ions and the concentration of chloride
ions, and we have the Ksp and the charge balance equation. (See the next example to be
Chm 118
Problem Solving in Chemistry
12.4
174
sure you know how to deal with the charge balance equation when the charges on the two
ions differ.) Solution is trivial in this case to give [Ag+ ] = 1.34×10−5 .
Of more interest is asking what the solubility of a compound is in a solvent that contains
one of the ions of the compound. This kind of problem has what is called the “common ion
effect.” For instance, what is the solubility of ScF3 is in a solution that is 0.01M in NaF?
We now have three unknowns, [Sc3+ ], [F– ], and [Na+ ]; and three equations:
Ksp = [Sc3+ ][F− ]3 = 4.2 × 10−18
3[Sc3+ ] + [Na+ ] = [F− ]
(12.11)
+
[Na ] = 0.010
Note that the second equation, the charge balance equation, needs a “3” in front of the
scandium ion because the equation is counting charge and each scandium ion carries three
charges. Chemical intuition makes the solving of this set of equations very easy. Since ScF3
is not very soluble, one guesses that the [Sc3+ ] is small compared to that of [Na+ ]; therefore,
[F– ] = [Na+ ] =0.010M and from the first equation, [Sc3+ ] = 4.2×10−12 . Does our guess
check out?
One final word. The problem just quoted is not correct, although this is the way it is often
done. The reason is that fluoride ion is a weak base; to solve this problem correctly required
that the equilibrium
HF(aq) = H+ (aq) + F– (aq)
and then, of course, the equation for water as an acid, must be included. Such problems
become difficult very quickly. In this case the rigorous answer is essentially the same as the
approximation above because fluoride is not a very strong base.
12.3.1
Exercises
12.3.1.1
The Ksp for PbSO4 is 1.6×10−8 . Find the solubility of PbSO4 in water.
12.3.1.2
Find the solubility of PbSO4 in a solution that is 0.010M in Na2 SO4 .
12.3.1.3
The Ksp for PbCl2 is 1.6×10−5 . Find the solubility of PbCl2 in water.
12.4
Electrochemical Cells
To examine the other crucial area where free energy calculations are important, we need to
establish the concept of the electrochemical cell; a schematic version of a cell is shown in
Chm 118
Problem Solving in Chemistry
12.4
175
Figure 12.2: A schmatic diagram of an electrochemical cell. Note the assumed direction of
flow of electrons is indicated by the arrow.
Figure 12.2. In this cell a piece of zinc metal is suspended in a solution of zinc ions and a
piece of copper metal is suspended in a solution of cupric ions. (Note in both sides there
is some “innocent” or “spectator” anion, unspecified, to balance charge.) The two pieces
of metal are connected by a wire that allow electrons to flow (through a volt meter, which
measures the tendency for the electrons to flow). The two beakers are connected by a “salt
bridge” that allows ions to migrate from one beaker to another in order to keep charge
balance on the two sides, but does not allow mass transport of material from one beaker to
the other. When we draw a cell this way, or give the abbreviation to be suggested soon, it
is always assumed that charge flows from the left hand cell to the right hand cell through
the external circuit. (If the potential of the cell reads “positive”–you can look at this as
the electrons pushing the meter in a clockwise fashion, then that is the real direction the
electrons would flow.)
In order to get electrons to flow in the assumed direction, definite processes have to happen
in each of the two cells. In the left hand cell, electrons must be produced to provide for
those that are leaving from left to right; therefore the chemistry must be:
Zn(s) = Zn2+ (aq) + 2e−
where the “2e− ” are electrons that flow down the wire from left to right. Likewise, electrons
must be consumed in the right hand cell, so the reaction is:
Cu2+ (aq) + 2e− = Cu(s)
Chm 118
Problem Solving in Chemistry
12.4
176
These two reactions are isolated from each other and are called “half-cell reactions.” Each
of them has a certain tendency to consume electrons (the reaction as written in the case
of the copper cell, but the reverse of the reaction given in the zinc cell) and that tendency
is measured by the half-cell reduction potential. This can be abbreviated as Cu2+ /Cu and
o = 0.340V and E o = -0.763V if all concentrations
Zn2+ /Zn. The values of these two are ECu
Zn
are 1.0 M (and are relative to an assumed value of 0.00V for the H+ (aq)/H2 (g)). Since the
zinc reaction is going in the opposite direction, the value for its potential must change sign,
and then the two can be added to give the net voltage of the cell of 1.103V.
There is a common abbreviation for electrochemical cells. For the cell above it looks like
this:
Zn(s) | Zn2+ (aq, 1.0M) || Cu2+ (aq, 1.0M) | Cu
where the single vertical line is a phase separation (in this case between solid and solution)
and the double vertical lines in the middle represent the salt bridge. Note that our convention about the direction of electron flow remains the same: from left to right in the external
circuit, which now must be imagined.
Please note that the potential represents the tendency to push electrons and hence is independent of the number of electrons involved in the reaction that creates that potential.
Hence in the cell
Zn(s) | Zn2+ (aq, 1.0M) || Ag+ (aq, 1.0M) | Ag
the cell potential is 0.763V from the Zn side (sign reversed from conventional “reduction
potential” because it is an oxidation) plus 0.800V for the silver side, for a total of 1.563V,
even though the number of electrons differ in the two reactions. When you compute a free
energy change–see the next section–the number of electrons is introduced. The number of
electrons affects the free energy, not the potential.
12.4.1
Exercises
12.4.1.1
Make a sketch (as, for example, in Figure 12.2) for the electrochemical cell:
Ag(s) | Ag+ (aq, 1.0M) || Fe2+ (aq, 1.0M),Fe3+ (aq, 1.0M) | Pt
where the Pt in the right hand cell is simply there to get the electrons to the reactive
centers, the ferrous and ferric ions.
12.4.1.2
What is the potential of the cell in the last exercise if all concentrations are 1.0M and the
o = 0.800V and E o
EAg
F e3+ is 0.771V?
Chm 118
Problem Solving in Chemistry
12.5
177
12.4.1.3
What is the potential of the cell
Ag(s) | Ag+ (aq, 1.0M) || Cl– (aq, 1.0M),Cl2 (g, 1.0 atm) | Pt
if the potential for the reaction
Cl2 (g) + 2e− = 2Cl– (aq)
is 1.358V.
12.5
Free Energy, Other Work, and the Nernst Equation
The general equation for the free energy, G, is
G = H − TS
(12.12)
which, if we take the total differential at constant T and P and insert the first law with
the work term being the total work of all kinds, and demand that our process be reversible,
we get (in a similar fashion to equation 11.9)
dG = dE + P dV − T dS
dG = δqrev + δwP V + δwother + P dV − T
δqrev
T
(12.13)
dG = −P dV + δwother + P dV
dG = δwother
The free energy change is equal to the work other than pressure volume work, that is done
on the system. The work that we are interested in here is electrical work, which is given by
the product of the potential change (equivalent to the force), E, times the charge that flows
(equivalent to distance moved), nF, where n is the number of moles of electrons and F is
the Faraday, the charge per mole; and in order to get work done on the system we need a
negative sign in front:
δwelect = −nFE
(12.14)
This leads us to the relationship between the potential and the free energy
∆G = −nFE
(12.15)
with a similar equation with super ‘o’ (see equation 12.16) to reflect the situation when the
chemicals are in their standard states.
We can calculate the free energy change for exercise 12.4.1.3 using the equation 12.16:
∆Go = −nFE o
(12.16)
where, if we write the “cell reaction” as
Chm 118
Problem Solving in Chemistry
12.6
178
2Ag(s) + Cl2 (g) = 2Ag+ (aq) + 2Cl– (aq)
the value is - (2)(96,500 kJ/V)(0.558V) = - 107.7 kJ. Note if we wrote the cell reaction as:
Ag(s) + 0.5Cl2 (g) = Ag+ (aq) + Cl– (aq)
then the number of electrons that “move down the wire” is one mole and the free energy
change is - 53.8 kJ, as you would expect.
Finally we manipulate the equation for the concentration dependence of ∆G, equation 11.18,
to give us the same information in terms of cell potential:
∆G = ∆Go + RT ln(Q)
−nFE = −nFE o + RT ln(Q)
RT
E = Eo −
ln(Q)
nF
(12.17)
where the last expression in equation 12.17 is called the Nernst equation.
12.5.1
Exercises
12.5.1.1
Write the cell reaction and find the free energy change for the cell:
Cu(s) | Cu2+ (aq, 1.0M) || Fe2+ (aq, 1.0M), Fe3+ (aq, 1.0M) | Pt
if E o for Cu2+ /Cu is 0.340 and that for Fe3+ /Fe2+ is 0.771.
12.5.1.2
Find the free energy change for the cell of the last exercise if the [Cu2+ ] = 0.01M, [Fe2+ ] =
1.0M, and [Fe3+ ] = 0.0001M.
12.6
Using Electrochemical Cells to Solve Problems
The Nernst equation, the last expression in equation 12.17, is very useful. Electrochemical
cells can be constructed to learn about process that are hard to determine in other ways.
Under these circumstances, the cell potential, E, is measured and some other quantity can
then be evaluated. The exercises give several examples.
Chm 118
Problem Solving in Chemistry
12.6
179
Experiment
A
B
C
[H+ (aq)
1.0 M
0.10 M
0.01M
E
2.193V
2.134V
2.075V
Table 12.1: The observed voltages for cells described in exercise 12.6.1.3.
12.6.1
Exercises
12.6.1.1
Consider the electrochemical cell
Ag(s) | Ag+ (aq, 1.0M) || Ag+ (aq, unknown concentration) | Ag
which has E of 0.142V at 298K. What is the concentration of silver ion in the right hand
compartment? This kind of problem is called a concentration cell and is integral to nerve
firings.
12.6.1.2
Consider the electrochemical cell
Cu(s) | Cu2+ (aq, 0.010M) || Ag+ (aq), IO–3 (aq, 0.10M) | Ag
which has E of 0.154V at 298K. What is the value of the equilibrium constant for the
dissolving of AgIO3 ? HINT: This is really a concentration cell in which the unknown is the
concentration of silver ion in equilibrium with solid AgIO3 and a solution of iodate ion of
0.10M.
12.6.1.3
Imagine that the form of Cl(V) in solution is not known. We build an electrochemical cell
as follows:
Zn(s) | Zn2+ (aq, 1.00M) || Cl(V)(aq, 0.010M), H+ (aq, C), Cl– (aq, 1.00M) | Pt
where the concentration of the hydrogen ion in the right hand compartment, C, varies.
Table 12.1 gives the potential observed as a function of the hydrogen ion concentration.
Show that this data is consistent with ClO–3 as the form of Cl(V), but not with ClO+
2 or
ClO3–
.
4
Chm 118
Problem Solving in Chemistry
Index of Important Concepts and
Terms
C2v , 37, 121
C3v , 38, 123
D3h , 42
D4h , 39
Oh , 129
One object, 37
Several objects, 39
Characteristic sets in combination,
formula, 41
Charge balance equation, 169
Chemical potential, 163
Color
Average field, 81
Separate orbital, 81
Common ion effect, 174
Concentration cell, 179
Configuration
Definition, 136
Nomenclature, 140
Predominant, 140
Predominant
Adding heat, 144
Adding heat to, 148
Appearance, 146
Symmetry notation, 75
Conjugate acid, 20
Conjugte base, 20
Correlation diagram, 74
Coulomb’s law, 2
Counting electrons, 127
Crystal field splitting
Low symmetry, 74
Octahedral, 72
Tetrahedral, 72
δ bond, 110
π acceptor ligand, 132
π donor ligand, 130
σ bond, 110
σ donor ligand, 128
pi bond, 110
Absorbance, 85
Absorbance of light, 79
Absorbance peak width, 82
Acid, 20
Acidity
Effect of charge, 23
Effect of resonance, 24
Hess’s law for, 21
Inductive effects, 25
Nature of Element, 22
Allowedness, 85
Average field, 81
Balmer series, 52
Band theory, 126
Beebe, Boniface, 35, 97, 149, 152
Beer’s law, 85, 92
Boltzmann’s equation, 146
Bond energies, 113
Bond energy, 15
Bond length, 14
Bond lengths, 113
Boundary conditions, 50
Buffer soltuion, 173
Cell reaction, 177
Characteristic combination, formula, 42
Characteristic numbers
180
12.6
Crystal field stabilization energy (CFSE),
88
Crystal field theory, 70
Crystal field theory
Color in low symmetry, 81
Color in octahedral compounds, 79
Flaw, 77
de Broglie postulate, 51
Degenerate orbitals, 70
Diamagnetic, 16, 18, 87
Diprotic acids, 172
Disorder, 136
Distribution curves, 27
Eighteen electron rule, 128, 134
Electrochemical cell, 174
Electron spin, 4
Endothermic, 158
Energy levels
Electronic, 101
Rotational, 100
Translational, 100
Vibrational, 100
Energy levels in hydrogen atom, 52
Enthalpy, 16
Enthalpy of formation, 16
Enthalpy of reaction, 17
Entropy
Absolute, 153
Defined, 149
Not disorder, 153
Reversible expansion, 151
State function, 150
Entropy change in reaction, 155
Equilibirum
Weak acid, 169
Equilibrium, 166
Equilibrium
Aqueous, 169
Guesses to solve, 170
Equilibrium constant, 20
Ethane
Eclipsed, 33
Staggered, 33
Exothermic, 158
Extensive property, 154
Extent of reaction, 163
Chm 118
181
Factorials, 140
Faraday, 177
First law of thermodynamics, 143
First order, 95
Formal charge, 7
Four equations/four unknowns, 169
Free energy, 161
Free energy
Equilibrium, 166
Of formation, 161
Partial molar, 163
Pressure dependence, 162
g functions, 85
Gap size
Ligand effect, 78
Metal effect, 79
Gauss’ law, 63
Group
C2v , 34
D3h , 35, 42
D4h , 39
Half-cell reactions, 176
Half-cell reduction potential, 176
Half-life, 97
Heat
Definition, 142
Hess’s law, 17
Heteronuclear diatomic, 132
Heteronuclear diatomics, 114
High field complex, 77
High spin complex, 77
Highest occupied molecular orbital, 132
HOMO, 132
Homonuclear diatomics, 113
Hybridization, 118
Hybridization
sp, 118
sp2 , 120
sp3 , 120
Hydration energies, 22
Hydration energy, 26
Hypervalent compounds, 10, 126
Integrated rate law
First order, 96
Second order, 97
Problem Solving in Chemistry
12.6
Intensive property, 154
Ionization energy
Definition, 52
Distance effect, 65
Effect of electron repulsion, 66
Effect of penetration, 66
Effect of VSNC, 66
Four important factors, 65
second, 67
Isomerization
Linkage, 92
Isosbestic behavior, 92
LCAO, 103
Lewis structure, 7
Ligands, 70
Linear combination of atomic orbitals,
103
Low field complex, 77
Low spin complex, 77
Lowest unoccupied molecular orbital, 132
LUMO, 132
Magnetic moment
Spin only, 87
Mass balance equation, 169
Microstate
Definition, 136
Probability, 138
MO diagram, 113
Molar absorptivity, 85
Multi-electronic atoms
Configurations, 63
Nernst equation, 178
Nomenclature of orbitals
δ, 110
π, 110
σ, 110
Normalization, 104
Operator, 49
Orbital, 63
Orbital conservation, 104
Orbital interaction
Antibonding, 104
bonding, 104
Symmetry restriction, 115
Overlap, 104
Chm 118
182
Paramagnetic, 16, 87
Parity, 85
Parve, 49
Parve on a pole, 50, 100
Pauli Principle, 5
Pauli principle, 63
Penetration, 63
Periodic table, 64
Perturbation theory, 115
pH, 171
pKa , 172
pKa , 20
Planck’s constant, 49
Polarization, 19
POP, 50
Probability, 4
Probability, quantum, 48
Quantum numbers, 53
Radial distribution curve, 53
Radicals, 16
Rate constant, 95
Rate law
First order, 94
Meaning of sum in, 99
Second order, 95
Rate of reaction, 94
Resonance, 12
Reversible expansion, 150
SALCAO, 121
Schrödiner’s equation, 49
Schrödinger’s equation, 103
Screening, 63
Second order, 95
Separate orbital approach, 81
Solids, 125
Solubility equilibrium, 173
Solubility product, 173
Spectrochemical series, 78, 128
Spin, 4
Spin allowed, 85
Spin forbidden, 85
Spin selection rule, 85
Spinel, 90
Spontaneous process, 148
Spontaneous reaction
Problem Solving in Chemistry
12.6
Entropy criterion, 149
Standard state, 155
State function, 157
Stern-Gerlach experiment, 45
Stirling’s approximation, 147
Symmetry adaped LCAO, 121
Symmetry operation
Center of inversion, 33
Definition, 30
Group, 34
Identity, 30
Improper, 30
Plane, 32
Proper, 30
Rotation, 30
Rotation-Reflection, 33
System
Definition, 142
Thermodynamics, 16
Third law of thermodynamics, 153
Transition metal compounds
MO theory, 128
Chm 118
183
Transmittance, 85
Two slit experiment, 4
u functions, 85
Valence orbital ionization energy, 68
Valence shell electron repulsion theory, 6
Valence shell nuclear charge, 66
Vibration in metal complexes, 82
VOIE, 68
VSEPR, 6
VSNC, 66
W
Calculation of, 140
Wave equation, 47
Wave function, 4
Wave function
Radial, 53
Wave function, angular, 59
Work
Definition, 142
electrical, 177
other, 177
Problem Solving in Chemistry